Reseach Report, Keio University, Dept. Math. KSTS/RR-18/002 November 22, 2018 ( http://www.math.keio.ac.jp/academic/research_pdf/report/2018/18002.pdf) Linguistic Copenhagen interpretation of quantum mechanics: Quantum Language [Ver. 4] Shiro ISHIKAWA (ishikawa@math.keio.ac.jp) [November, 2018] Department of mathematics, Faculty of science and Technology, Keio University, 3-14-1, Hiyoshi, Kouhokuku, in Yokohama, 223-8522, Japan Abstract Recently we proposed" quantum language" (or," the linguistic Copenhagen interpretation of quantum mechanics"), which was not only characterized as the metaphysical and linguistic turn of quantum mechanics but also the linguistic turn of Descartes=Kant epistemology. Namely, quantum language is the scientific final goal of dualistic idealism. It has a great power to describe classical systems as well as quantum systems. Thus, We believe that quantum language is the language in which science is written. The purpose of this preprint is to examine and assert our belief (i.e.,"proposition in quantum language" ⇔" scientific proposition (i.e., proposition which can be tested by experiment )"). Preface; What is science? This is the lecture note for graduate students. This lecture has been continued, with gradually improvement, for about 15 years in the faculty of science and technology of Keio university 1. In this lecture, I explain "quantum language"(="measurement theory"="linguistic Copenhagen interpretation of quantum mechanics"), which was proposed as the language in which science is written by myself. Quantum language is a language that is inspired by the Copenhagen interpretation of quantum mechanics, but it has a great power to describe classical systems as well as quantum 1 This preprint is the 4th version of Refs. [53, 54, 55]: S. Ishikawa, Linguistic interpretation of quantum mechanics; Quantum Language, Research Report, Dept. Math. Keio University, (http://www.math.keio.ac. jp/en/academic/research.html) [53] : [Ver.1]; KSTS/RR-15/001 (2015); 416 p (http://www.math.keio.ac.jp/academic/research_pdf/ report/2015/15001.pdf) [54] : [Ver.2]; KSTS/RR-16/001 (2016); 426 p (http://www.math.keio.ac.jp/academic/research_pdf/ report/2016/16001.pdf) [55] : [Ver.3]; KSTS/RR-17/007 (2017); 434 p (http://www.math.keio.ac.jp/academic/research_pdf/ report/2017/17007.pdf) Roughly speaking, we say that [Ver. 2]="[Ver.1]+ Sec.11.3( Wave function collapse)", [Ver. 3]="[Ver.2]+ Sec.4.5( Bell's inequality)", [Ver. 4]="[Ver.3]+ Sec.10.8 (Brain in a Vat, Five-minute hypothesis, etc.)". Also, for my recent results, see my homepage ( http://www.math.keio.ac.jp/~ishikawa/indexe.html) 1 systems. In this lecture, I assert that quantum language, roughly speaking, has the three aspects as follows. The three aspects of quantum language 1©: the standard interpretation of quantum mechanics (i.e., the true colors of the Copenhagen interpretation) thus, in this paper we consider that "the linguistic Copenhagen interpretation"= "the linguistic interpretation" ="the Copenhagen interpretation" 2©: the final goal of the dualistic idealism (Descartes=Kant philosophy) 3©: theoretical statistics of the future And therefore, I think that " 1©:quantum information theory" ∪ " 2©:dualistic idealism" ∪ " 3©:statistics" ⊂"quantum language" Thus I conclude The main assertion of this lecture Quantum language is the language in which science is written That is, the following (i) and (ii) are equivalent: (i) proposition in quantum language (ii) scientific proposition (i.e., proposition which can be tested by experiment ) The purpose of this lecture is to examine and explain these assertions I believe that making such a language is exactly the true purpose of the philosophy of science. . Philosophy of science: What is science? Our original motivation is to answer the question "What is science?". It is well known that the famous answer "falsifiability" is due to Popper (cf. [73]). However his answer was too literature-like. And thus, most scientists did not show much interest in "falsifiability". Hence, some may, from the scientific point of view, prefer the following answer(A): (A) Science is an academic field with statistics as language For example (A1) Economics is to describe economic phenomena in statistics. (A2) Psychology is to describe psychological phenomena in statistics. 2 (A3) Biology is to describe Biological phenomena in statistics. (A4) Newton mechanics is to describe mechanical phenomena in statistics (= dynamical system theory). (An) * * * * * * Although most scientists may be interested in the above answer (A) rather than "falsifiability" (cf. [73]), I think that it is not enough (for example, the definition of statistics is not clear). In this paper, I propose that (B) Science is an academic field described by quantum language For example (B1) Economics is to describe economic phenomena by quantum language. (B2) Psychology is to describe psychological phenomena by quantum language. (B3) Biology is to describe Biological phenomena by quantum language. (B4) Newton mechanics is to describe mechanical phenomena by quantum language. (B5) Quantum mechanics is to describe quantum mechanical phenomena by quantum language. (Bn) * * * * * * The reader would be convinced that the answer (B) is better than the answer (A). Also, the following may be regarded as the supplementary reader of this text: • [49]: S. Ishikawa, History of Western Philosophy from the quantum theoretical point of view, Research Report (Department of mathematics, Keio university, Yokohama), (KSTSRR-16/005, 2016, 142 pages) (KSTS-RR-16/005, 2016, 142 pages) (http://www.math.keio.ac.jp/academic/research_pdf/report/2016/16005.pdf) • [50]: S. Ishikawa, History of Western Philosophy from the quantum theoretical point of view [Ver. 2], Research Report (Department of mathematics, Keio university, Yokohama), (KSTS-RR-17/004, 2017, 132 pages) (http://www.math.keio.ac.jp/academic/research_pdf/report/2017/17004.pdf) 3 i

Chapter 1 My answer to Feynman's question Dr. R. P. Feynman (one of the founders of quantum electrodynamics) said the following wise words:(]1) and (]2): 1 (]1) There was a time when the newspapers said that only twelve men understood the theory of relativity. I do not believe there ever was such a time. There might have been a time when only one man did, because he was the only guy who caught on, before he wrote his paper. But after people read the paper a lot of people understood the theory of relativity in some way or other, certainly more than twelve. On the other hand, I think I can safely say that nobody understands quantum mechanics. and (]2) We have always had a great deal of difficulty understanding the world view that quantum mechanics represents. * * * * * * I cannot define the real problem, therefore I suspect there's no real problem, but I'm not sure there's no real problem. In this lecture, I will answer Feynman's question (]1) and (]2) as follows. ([) I am sure there's no real problem. Therefore, since there is no problem that should be understood, it is a matter of course that nobody understands quantum mechanics. This answer may not be uniquely determined, however, I am convinced that the above ([) is one of the best answers to Feynman's question (]1) and (]2). The purpose of this lecture is to explain the answer ([). That is, I show that If we start from the answer ([), we can double the scope of quantum mechanics. And further, I assert that Metaphysics (which might not be liked by Feynman ) is located in the center of science. In this lecture, I will show the above. 1The importance of the two (]1) and (]2) was emphasized in Mermin's book [70] 1 1.1 Quantum language (= measurement theory) 1.1 Quantum language (= measurement theory) 1.1.1 Introduction In this lecture, I will explain "quantum language" (= measurement theory (=MT)=Linguistic Copenhagen interpretation ), which is located as illustrated in the following figure: Figure 1.1. [The location of quantum language in the history of world-description (cf. refs.[32, 53]) ] Parmenides Socrates 0©:Greek philosophy Plato Aristotle Schola-−−−−→ sticism 1© −−→ (monism) Newton (realism) 2© → relativity theory −−−−−−→ 3© → quantum mechanics −−−−−−→ 4© −→ (dualism) Descartes Locke,... Kant (idealism) 6©−→ (linguistic view) linguistic philosophy language−−−−−→ 8© language−−−−−−→ 7©  5©−→ (unsolved) theory of everything (quantum phys.)  10©−→ (=MT) quantum language (language) Figure 1.1: The history of the world-view statistics system theory language−−−−−→ 9© (Descartes, Locke may belong to substance dualism) the linguistic world view ( dualism, idealism ) the realistic world view (monism, realism) It should be noted that the above figure implies the following three: [ 7© ]: to clarify the Copenhagen interpretation of quantum mechanics, that is, the linguistic Copenhagen interpretation is the true figure of so-called Copenhagen interpretation [ 8© ]: to clarify the final goal of the dualistic idealism (Descartes=Kant epistemology) (cf. ref. [49, 51]) [ 9© ]: to reconstruct statistics in the dualistic idealism Therefore, Figure 1.1 is all in this lecture. 2 Ishikawa's Homepage Chap. 1 My answer to Feynman's question ♠Note 1.1. If most physicists feel something like metaphysics in quantum mechanics, the reason is due to Figure 1.1. That is, we consider that there are two "quantum mechanics", that is, "(realistic) quantum mechanics" in 5© and "(metaphysical) quantum mechanics" in 10©. Namely, • quantum mechanics  "(realistic) quantum mechanics" in 5© "(metaphysical) quantum mechanics" in 10© The former is not completed yet. The latter is "the usual quantum mechanics" studied in undergraduate course of university. In this lecture, we are not concerned with the former. ♠Note 1.2. If readers are familiar with quantum mechanics, it may be recommended to read the following short papers before reading this lecture text. (a) Ref. [31]: S. Ishikawa, A New Interpretation of Quantum Mechanics: Journal of quantum information science: Vol.1(2), pp.35-42, 2011 (b) Ref. [32]:S. Ishikawa, Quantum Mechanics and the Philosophy of Language: Reconsideration of traditional philosophies, Journal of quantum information science, Vol. 2(1), pp.2-9, 2012 (c) Ref. [48] S. Ishikawa, Linguistic interpretation of quantum mechanics; Projection Postulate, Journal of quantum information science, Vol. 5, No.4 , 150-155, 2015, DOI: 10.4236/jqis.2015.54017 (http://www.scirp.org/Journal/PaperInformation.aspx?PaperID=62464) (d) Ref. [52] Ishikawa,S., Bell's inequality should be reconsidered in quantum language , Journal of quantum information science, Vol. 7, No.4 , 140-154, 2017, DOI: 10.4236/jqis.2017.74011 (http://www.scirp.org/Journal/PaperInformation.aspx?PaperID=80813) The similarities and differences between the linguistic interpretation and so called Copenhagen interpretation have been clarified in the above (c). 1.1.2 From Heisenberg's uncertainty principle to the linguistic interpretation As explained in §4.2, (A) In 1991(cf. ref. [23])2, I found the mathematical formulation of Heisenberg's uncertainty principle (i.e., ∆x *∆p ≥ ~/2 in (4.36)), which clarified that • under what kind of condition does Heisenberg's uncertainty principle hold? 2Ref.[23]:S. Ishikawa, "Uncertainty relation in simultaneous measurements for arbitrary observables" Rep. Math. Phys. Vol.29(3), pp.257–273, 1991, 3 Ishikawa's Homepage 1.1 Quantum language (= measurement theory) I thought that this result is interesting. However, from immediately after the discovery (A), the interpretation of quantum mechanics began to worry me. There are many interpretations of quantum mechanics, for example, "the Copenhagen interpretation", "the many world interpretation", "the probabilistic interpretation", etc. In the applied field of quantum mechanics, we can expect that the same conclusion is derived from different interpretations. In this sense, the problem of "the interpretation of quantum mechanics" is not serious. However, concerning Heisenberg's uncertainty principle, this problem is important. That is because the meaning of "errors" in Heisenberg's uncertainty principle depend on the interpretation of quantum mechanics ( for example, the meaning of "errors (∆x and ∆p)" depends on the acceptance of "the collapse of wave function" or not ) . Thus, • I want to establish the "standard" interpretation of quantum mechanics. In what follows, let me mention my idea (i.e., the linguistic interpretation of quantum mechanics): Recalling that quantum mechanics was called "matrix mechanics" (when quantum mechanics was proposed (i.e., 1920s), I consider that (B1) from the mathematical point of view, quantum mechanics is the theory of "square matrix" On the other hand, (B2) from the mathematical point of view, classical mechanics is the theory of "diagonal matrix" Thus, we have the following problem: (C) What is the interpretation which is common to both quantum system (B1) and classical system (B2)? And we conclude that (D) the answer to the question (C) is uniquely determined as "quantum language", where quantum language can describe classical systems as well as quantum systems. Since quantum language is not physics but language (= metaphysics), quantum language (= the linguistic interpretation of quantum mechanics) is completely different from other quantum interpretations. In this sense, we are convinced that 4 Ishikawa's Homepage Chap. 1 My answer to Feynman's question (E) quantum language (= the linguistic interpretation of quantum mechanics ) is forever, even if some propose the "final" interpretation of quantum mechanics in the realistic view (i.e., 5© in Figure 1.1 ) 5 Ishikawa's Homepage 1.2 The outline of quantum language 1.2 The outline of quantum language 1.2.1 The classification of quantum language (=measurement theory) Quantum language (= measurement theory ) is classified as follows. (A) measurement theory (=quantum language)  pure type (A1) { classical system : Fisher statistics quantum system : usual quantum mechanics mixed type (A2) { classical system : including Bayesian statistics, Kalman filter quantum system : quantum decoherence Therefore, we have two kinds of quantum language, i.e., pure measurement theory and mixed measurement theory. The former is formulated as follows. (A1) pure measurement theory (=quantum language) := [(pure)Axiom 1] pure measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } the manual to use spells And the mixed measurement theory (or, statistical measurement theory) is formulated as follows. (A2) mixed measurement theory (=quantum language) := [(mixed)Axiom(m) 1] mixed measurement (cf. §9.1) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } the manual to use spells 1.2.2 Axiom 1 (measurement) and Axiom 2 (causality) Since the pure measurement theory is the most fundamental, we mainly devote ourselves to pure measurement theory. Although it is impossible to read Axiom 1 ( measurement: §2.7) and Axiom 2 (causality; §10.3) at the present time, we present them as follows. 6 Ishikawa's Homepage Chap. 1 My answer to Feynman's question (B):Axiom 1 (measurement) pure type (This will be able to be read in §2.7 ) With any system S, a basic structure [A ⊆ A]B(H) can be associated in which measurement theory of that system can be formulated. In [A ⊆ A]B(H), consider a W ∗-measurement MA ( O=(X,F, F ), S[ρ] ) ( or, C∗-measurementMA ( O=(X,F, F ), S[ρ] ) ) . That is, consider • a W ∗-measurement MA ( O, S[ρ] ) ( or, C∗-measurement MA ( O=(X,F, F ), S[ρ] ) ) of an observable O=(X,F, F ) for a state ρ(∈ Sp(A∗) : state space) Then, the probability that a measured value x (∈ X) obtained by the W ∗-measurement MA ( O, S[ρ] ) ( or, C∗-measurement MA ( O=(X,F, F ), S[ρ] ) ) belongs to Ξ (∈ F) is given by ρ(F (Ξ))(≡ A∗(ρ, F (Ξ))A) (1.1) (if F (Ξ) is essentially continuous at ρ, or see Definition 2.14 ). And (C): Axiom 2 (causality) (This will be able to be read in §10.3) Let T be a tree (i.e., semi-ordered tree structure). For each t(∈ T ), a basic structure [At ⊆ At]B(Ht) is associated. Then, the causal chain is represented by a W ∗sequential causal operator {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 ( or, C∗sequential causal operator {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 ) Here, note that (D1) the above two axioms are kinds of spells (i.e., incantation, magic words, metaphysical statements), and thus, it is impossible to verify them experimentally. In this sense, the above two axioms correspond to "a priori synthetic judgment" in Kant's philosophy (cf. [62]). Therefore, (D2) what we should do is not to understand the two, but to learn the spells (i.e., Axioms 1 and 2) by rote. 7 Ishikawa's Homepage 1.2 The outline of quantum language Of course, the "learning by rote" means that we have to understand the mathematical definitions of followings: • basic structure [A ⊆ A]B(H), state space Sp(A∗), observable O=(X,F, F ), etc. ♠Note 1.3. If metaphysics did something wrong in the history of science, it is because metaphysics attempted to answer the following questions seriously in ordinary language: (]1) What is the meaning of the keywords (e.g., measurement, probability, causality) ? Although the question (]1) looks attractive, it is not productive. What is important is to create a language to deal with the keywords. So we replace (]1) by (]2) How are the keywords (e.g., measurement, probability, causality) used in quantum language ? The problem (]1) will now be solved in the sense of (]2). ♠Note 1.4. Metaphysics is an academic discipline concerning propositions in which empirical validation is impossible. Lord Kelvin (1824–1907) said Mathematics is the only good metaphysics. Here we step forward: (]) Quantum language is another good metaphysics. Lord Kelvin might think that Kant philosophy (Critique of Pure Reason [62]) is not good metaphysics. However, I consider that a priori synthetic judgment (i.e., axiom which cannot be examined by experiment) corresponds to [Axiom 1 and Axiom 2]. That is, a priori synthetic judgment ( Kant philosophy ) ←→ (correspondence) Axiom 1 and Axiom 2 (quantum language) See ref. [32]:S. Ishikawa, Quantum Mechanics and the Philosophy of Language: Reconsideration of traditional philosophies, Journal of quantum information science, Vol. 2(1), pp.2-9, 2012 8 Ishikawa's Homepage Chap. 1 My answer to Feynman's question 1.2.3 The linguistic interpretation Axioms 1 and 2 are all of quantum language. Therefore, (]) after learning Axioms 1 and 2 by rote, we need to brush up our skills to use them through trial and error. Here, let us recall a wise saying • Experience is the best teacher, or custom makes all things and our experience • A manual helps us to master the rules quickly. Thus, we understand to master the linguistic interpretation of quantum mechanics = to make practice with a manual to use Axioms 1 and 2 Although the linguistic interpretation (= the linguistic Copenhagen interpretation ) is composed of many statements, the simplest and best representation may be as follows. (E):The linguistic Copenhagen interpretation ) (This will be explained in §3.1) Only one measurement is permitted. We can also choose apparently opposite viewpoints concerning the linguistic interpretation, though they look a bit too extreme. (E1) Through trial and error, we can do well without the linguistic interpretation. (E2) All that are written in this note are a part of the linguistic interpretation. They are viewpoints obtained from the opposite standpoints. In this sense, there is a reason to regard this lecture note as something like a cookbook. 9 Ishikawa's Homepage 1.2 The outline of quantum language ♠Note 1.5. Kolmogorov's probability theory (cf. [63] ) starts from the following spell: (]) Let (X,F, P ) be a probability space. Then, the probability that a event Ξ(∈ F) happens is given by P (Ξ) And, through trial and error, Kolmogorov found his extension theorem, which says that (]) Only one probability space is permitted. This surely corresponds to the linguistic interpretation "Only one measurement is permitted." That is, (the most fundamental theorem) Probability theory (Only one probability space is permitted) (correspondence)←→ (the linguistic interpretation) Quantum language (Only one measurement is permitted) In this sense, we want to assert that (]) Kolmogorov is one of the main discoverers of the linguistic interpretation. Therefore, we are optimistic to believe that the linguistic interpretation "Only one measurement is permitted" can be, after trial and error, acquired if we start from Axioms 1 and 2. That is, we consider, as mentioned in (H1), that we can theoretically do well without the linguistic interpretation. 1.2.4 Summary Summing up the above arguments, we see: 10 Ishikawa's Homepage Chap. 1 My answer to Feynman's question (F): Summary ( All of quantum language ) Quantum language (= measurement theory ) is formulated as follows. measurement theory (=quantum language) := [Axiom 1] Measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } manual to use spells (1.2) [Axioms]. Here (F1) Axioms 1 and 2 are kinds of spells, (i.e., incantation, magic words, metaphysical statements), and thus, it is impossible to verify them experimentally. In this sense, I consider that a priori synthetic judgment (Kant philosophy) −−−−−−−−−→ quantization Axioms 1 and 2 (quantum language) Therefore, what we should do is not "to understand" but "to use". After learning Axioms 1 and 2 by rote, we have to improve our skills to use them through trial and error. [The linguistic interpretation]. From a pure theoretical point of view, we do well without the interpretation. However, (F2) it is better to know the linguistic interpretation of quantum mechanics (= the manual to use Axioms 1 and 2), if we want to make quick progress in using quantum language. The most important statement in the linguistic interpretation (§3.1) is Only one measurement is permitted. After all, we think that Descartes philosophy [dualistic idealism] −→  Continental Rationalism [Axioms] British empiricism [Linguistic interpretation]  −→ Kant philosophy[quantum language] 11 Ishikawa's Homepage 1.3 Example: measurement of "Cold or Hot" 1.3 Example: measurement of "Cold or Hot" Axioms 1 and 2 (mentioned in the previous section ) are too abstract. And thus, I am afraid that the readers feel that it is too hard to use quantum language. Hence, let us add a simple example in this section. It is sufficient for the readers to consider that our purpose in the next chapters is • to bury the gap between Axiom 1 and the following simple example (i.e., "Cold" or "Hot"). Example 1.2. [The measurement of "Cold or Hot" for the water in a cup] Let testees drink water with various temperature ω ◦C (0 5 ω 5 100). And assume: you ask them "Cold or Hot ?" alternatively. Gather the data, ( for example, gc(ω) persons say "Cold", gh(ω) persons say "Hot") and normalize them, that is, get the polygonal lines such that fc(ω) = gc(ω) the numbers of testees fh(ω) = gh(ω) the numbers of testees (1.3) And fc(ω) =  1 (0 5 ω 5 10) 70−ω 60 (10 5 ω 5 70) 0 (70 5 ω 5 100) , fh(ω) = 1− fc(ω) 1 fc fh 0 10 20 30 40 50 60 70 80 90 100 Figure 1.2: Cold or hot? Therefore, for example, (A1) You choose one person from the testees, and you ask him/her whether the water (with 55 ◦C) is "cold" or "hot" ?. Then the probability that he/she says [ "cold" "hot" ] is given by [ fc(55) = 0.25 fh(55) = 0.75 ] 12 Ishikawa's Homepage Chap. 1 My answer to Feynman's question In what follows, let us describe the statement (A1) in terms of quantum language (i.e., Axiom 1). Define the state space Ω such that Ω = interval [0, 100](⊂ R(= the set of all real numbers)) and measured value space X = {c, h} ( where "c" and "h" respectively means "cold" and "hot"). Here, consider the "[C-H]-thermometer" such that (A2) for water with ω ◦C, [C-H]-thermometer presents [ c h ] with probability [ fc(ω) fh(ω) ] . This [C-H]-thermometer is denoted by O = (fc, fh) Note that this [C-H]-thermometer can be easily realized by "random number generator". Here, we have the following identification: (A3) (A1) ⇐⇒ (A2) Therefore, the statement (A1) in ordinary language can be represented in terms of measurement theory as follows. (A4) When an observer takes a measurement by [[C-H]-instrument] measuring instrumentO=(fc,fh) for [water] (System (measuring object)) with [55 ◦C] (state(= ω ∈ Ω) ) , the probability that measured value [ c h ] is obtained is given by [ fc(55) = 0.25 fh(55) = 0.75 ] This example will be again discussed in the following chapter(Example 2.31). 13 Ishikawa's Homepage

Chapter 2 Axiom 1 - measurement Quantum language (= measurement theory ) is formulated as follows. • measurement theory (=quantum language) := [Axiom 1] Measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } manual to use spells Measurement theory asserts that • Describe every phenomenon modeled on Axioms 1 and 2 (by a hint of the linguistic interpretation)! In this chapter, we introduce Axiom 1 (measurement). Axiom 2 concerning causality will be explained in Chapter 10. 2.1 The basic structure[A ⊆ A ⊆ B(H)]; General theory The Hilbert space formulation of quantum mechanics is due to von Neumann. I cannot emphasize too much the importance of his work (cf. [83]). 2.1.1 Hilbert space and operator algebra Let H be a complex Hilbert space with a inner product 〈*, *〉, where it is assumed that 〈u, αv〉 = α〈u, v〉 (∀u, v ∈ H,α ∈ C(= the set of all complex numbers)). And define the norm ‖u‖ = |〈u, u〉|1/2. Define B(H) by B(H) = {T : H → H | T is a continuous linear operator} (2.1) B(H) is regarded as the Banach space with the operator norm ‖ * ‖B(H), where ‖T‖B(H) = sup ‖x‖H=1 ‖Tx‖H (∀T ∈ B(H)) (2.2) 15 2.1 The basic structure[A ⊆ A ⊆ B(H)]; General theory Let T ∈ B(H). The dual operator T ∗ ∈ B(H) of T is defined by 〈T ∗u, v〉 = 〈u, Tv〉 (∀u, v ∈ H) The followings are clear. (T ∗)∗ = T, (T1T2) ∗ = T ∗2 T ∗ 1 Further, the following equality (called the "C∗-condition") holds: ‖T ∗T‖ = ‖TT ∗‖ = ‖T‖2 = ‖T ∗‖2 (∀T ∈ B(H)) (2.3) When T = T ∗ holds, T is called a self-adjoint operator (or, Hermitian operator). Let Tn(n ∈ N = {1, 2, * * * }), T ∈ B(H). The sequence {Tn}∞n=1 is said to converge weakly to T (that is, w − limn→∞ Tn = T ), if lim n→∞ 〈u, (Tn − T )u〉 = 0 (∀u ∈ H) (2.4) Thus, we have two convergences (i.e., norm convergence and weakly convergence) in B(H)1. Definition 2.1. [C∗-algebra and W ∗-algebra] A(⊆ B(H)) is called a C∗-algebra, if it satisfies that (A1) A(⊆ B(H)) is the closed linear space in the sense of the operator norm ‖ * ‖B(H). (A2) A is ∗-algebra, that is, A(⊆ B(H)) satisfies that F1, F2 ∈ A⇒ F1 * F2 ∈ A, F ∈ A⇒ F ∗ ∈ A Also, a C∗-algebraA(⊆ B(H)) is called a W ∗-algebra, if it is weak closed in B(H). 2.1.2 Basic structure[A ⊆ A ⊆ B(H)]; general theory Definition 2.2. Consider the basic structure [A ⊆ A ⊆ B(H)] ( or, denoted by [A ⊆ A]B(H)) . That is, • A(⊆ B(H)) is a C∗-algebra, and A(⊆ B(H)) is the weak closure of A. Note that W ∗-algebra A has the pre-dual Banach space A∗( that is, (A∗) ∗ = A ) uniquely. Therefore, the basic structure[A ⊆ A ⊆ B(H)] is represented as follows. (B): General basic structure:[A ⊆ A ⊆ B(H)] A∗xdual A ⊆−−−−−−−−−−−−−→ subalgebra*weak-closure A ⊆−−−−−−→ subalgebra B(H)ypre-dual A∗ (2.5) 1Although there are many convergences in B(H), in this paper we devote ourselves to the two. 16 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement 2.1.3 Basic structure[A ⊆ A ⊆ B(H)] and state space; General theory The concept of "state space" is fundamental in quantum language. This is formulated in the dual space A∗ of C∗-algebra A ( or, in the pre-dual space A∗ of W ∗-algebra A). Let us explain it as follows. Definition 2.3. [State space, mixed state space] Consider the basic structure: [A ⊆ A ⊆ B(H)] Let A∗ be the dual space of the C∗-algebraA. The mixed state space Sm(A∗) and the pure state space Sp(A∗) is respectively defined by (a) Sm(A∗) = {ρ ∈ A∗ | ‖ρ‖A∗ = 1, ρ ≥ 0 (i.e., ρ(T ∗T ) ≥ 0(∀T ∈ A))} (b) Sp(A∗) = {ρ ∈ Sm(A∗) | ρ is a pure state}. Here, ρ(∈ Sm(A∗)) is a pure state if and only if ρ = αρ1 + (1− α)ρ2, ρ1, ρ2 ∈ Sm(A∗), 0 < α < 1 =⇒ ρ = ρ1 = ρ2 The mixed state space Sm(A∗) and the pure state space Sp(A∗) are locally compact spaces (cf. ref.[87]). Assume that A∗ is the pre-dual space of A. Then, another mixed state space S m (A∗) is defined by (c) S m (A∗) = {ρ ∈ A∗ | ‖ρ‖A∗ = 1, ρ ≥ 0 (i.e., ρ(T ∗T ) ≥ 0(∀T ∈ A))} That is, we have two "mixed state spaces", that is, C∗-mixed state space Sm(A∗) and W ∗mixed state space S m (A∗). The above arguments are summarized in the following figure: (C): General basic structure and State spaces Sp(A∗) C∗-pure state ⊂ Sm(A∗) C∗-mixed state ⊂ A∗xdual A ⊆−−−−−−−−−−−−−→ subalgebra*weak-closure A ⊆−−−−−−→ subalgebra B(H)y pre-dual (2.6) S m (A∗) W ∗-mixed state ⊂ A∗ 17 Ishikawa's Homepage 2.1 The basic structure[A ⊆ A ⊆ B(H)]; General theory Remark 2.4. In order to avoid the confusions, three "state spaces" should be explained in what follows. (D) "state spaces"  Fisher statistics * * * pure state space:Sp(A∗): most fundamental Bayes statistics * * *  C∗-mixed state space:Sm(A∗) : easy W ∗-mixed state space:S m (A∗): natural, useful In this note, we mainly devote ourselves to the W ∗-mixed stateS m (A∗) rather than the C ∗mixed stateSm(A∗), though the two play the similar roles in quantum language. 18 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement 2.2 Quantum basic structure[C(H) ⊆ B(H) ⊆ B(H)] and State space If a conclusion is said previously, we say the following classification of (i.e., quantum state space and classical state space): (A) General basic structure[A ⊆ A]B(H) pure state space Sp(A∗) C∗-mixed state space Sm(A∗) W ∗-mixed state space S m (A∗) =⇒  (A1):Quantum basic structure[C(H) ⊆ B(H)]B(H) pure state space Sp(Tr(H)(≈H)) C∗-mixed state space Sm(Tr(H))(=Tr+1(H)) W ∗-mixed state space Sm(Tr(H))(=Tr+1(H)) (A2):Classical basic structure[C0(Ω) ⊆ L∞(Ω, ν)]B(L2(Ω,ν)) pure state space Ω C∗-mixed state space M+1(Ω) W ∗-mixed state space L1+1(Ω,ν) In what follows, we shall explain the above classification (A): 2.2.1 Quantum basic structure[C(H) ⊆ B(H) ⊆ B(H)]; In quantum system, the basic structure[A ⊆ A ⊆ B(H)] is characterized as [C(H) ⊆ B(H) ⊆ B(H)] (2.7) That is, we see: (B): Quantum basic structure:[C(H) ⊆ B(H) ⊆ B(H)] Tr(H)xdual C(H) ⊆−−−−−−−−−−−−−→ subalgebra*weak-closure B(H) ⊆−−−−−−→ subalgebra B(H)ypre-dual Tr(H) (2.8) Before we explain "compact operators class C(H)" and "trace class F(H)", we have to prepare "Dirac notation" and "CONS" as follows. 19 Ishikawa's Homepage 2.2 Quantum basic structure[C(H) ⊆ B(H) ⊆ B(H)] and State space Definition 2.5. [(i):Dirac notation] Let H be a Hilbert space. For any u, v ∈ H, define |u〉〈v| ∈ B(H) such that (|u〉〈v|)w = 〈v, w〉u (∀w ∈ H) (2.9) Here, 〈v| [ resp. |u〉 ] is called the "Bra-vector" [ resp. "Ket-vector" ] . [(ii):ONS(orthonormal system), CONS(complete orthonormal system)] The sequence {ek}∞k=1 in a Hilbert space H is called an orthonormal system (i.e., ONS), if it satisfies (]1) 〈ek, ej〉 = { 1 (k = j) 0 (k 6= j) In addition, an ONS {ek}∞k=1 is called a complete orthonormal system (i.e., CONS), if it satisfies (]2) 〈x, ek〉 = 0 (∀k = 1, 2, ...) implies that x = 0. Theorem 2.6. [The properties of compact operators class C(H)] Let C(H)(⊆ B(H)) be the compact operators class. Then, we see the following (C1)-(C4) ( particularly, "(C1)↔ (C2)" may be regarded as the definition of the compact operators class C(H)(⊆ B(H)) ) . (C1) T ∈ C(H). That is, • for any bounded sequence {un}∞n=1 in Hilbert space H, {Tun}∞n=1 has the subsequence which converges in the sense of the norm topology. (C2) There exist two ONSs {ek}∞k=1 and {fk}∞k=1 in the Hilbert space H and a positive real sequence {λk}∞k=1 (where, limk→∞ λk = 0 ) such that T = ∞∑ k=1 λk|ek〉〈fk| (in the sense of weak topology) (2.10) (C3) C(H)(⊆ B(H)) is a C∗-algebra. When T (∈ C(H)) is represented as in (C2), the following equality holds ‖T‖B(H) = max k=1,2,*** λk (2.11) (C4) The weak closure of C(H) is equal to B(H). That is, C(H) = B(H) (2.12) 20 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement Theorem 2.7. [The properties of trace class Tr(H)] Let Tr(H)(⊆ B(H)) be the trace class. Then, we see the following (3D1)-(D4)( particularly, "(D1)↔ (D2)" may be regarded as the definition of the trace class Tr(H)(⊆ B(H)) ). (D1) T ∈ Tr(H)(⊆ C(H) ⊆ B(H)). (D2) There exist two ONSs {ek}∞k=1 and {fk}∞k=1 in the Hilbert space H and a positive real sequence {λk}∞k=1 (where, ∑∞ k=1 λk <∞ ) such that T = ∞∑ k=1 λk|ek〉〈fk| (in the sense of weak topology) (D3) It holds that C(H)∗ = Tr(H) (2.13) Here, the dual norm ‖ * ‖C(H)∗ is characterized as the trace norm ‖ * ‖Tr such as ‖T‖Tr = ∞∑ k=1 λk (2.14) when T (∈ Tr(H)) is represented as in (D2), (D4) Also, it holds that Tr(H)∗ = B(H) in the same sense, Tr(H) = B(H)∗ (2.15) Remark 2.8. Assume that a Hilbert space H is finite dimensional, i.e., H = Cn, i.e., Cn = {z =  z1 z2 ... xn  | zk ∈ C, k = 1, 2, ..., n}. Put M(C, n) = The set of all (n× n)-complex matrices and thus, A = A = B(Cn) = C(H) = Tr(H) = M(C, n) (2.16) However, it should be noted that the norms are different as mentioned in (C3) and (D3). 21 Ishikawa's Homepage 2.2 Quantum basic structure[C(H) ⊆ B(H) ⊆ B(H)] and State space 2.2.2 Quantum basic structure[C(H) ⊆ B(H) ⊆ B(H)] and State space; Consider the quantum basic structure: [C(H) ⊆ B(H) ⊆ B(H)] and see the following diagram: (E): Quantum basic structure and State space Sp(Tr(H)) C∗-pure state ⊂ Sm(Tr(H)) C∗-mixed state ⊂ Tr(H)xdual C(H) ⊆−−−−−−−−−−−−−→ subalgebra*weak-closure B(H) ⊆−−−−−−→ subalgebra B(H)y pre-dual (2.17) S m (Tr(H)) W ∗-mixed state ⊂ Tr(H) In what follows, we shall explain the above diagram. Firstly, we note that C(H)∗ = Tr(H), Tr(H)∗ = B(H) (2.18) and Sm(Tr(H)) = S m (Tr(H)) ={ρ = ∞∑ n=1 λn|en〉〈en| : {en}∞n=1 is ONS , ∞∑ n=1 λn = 1, λn > 0} =:Tr+1(H) (2.19) Also, concerning the pure state space, we see: Sp(Tr(H)) ={ρ = |e〉〈e| : ‖e‖H = 1} =: Trp+1(H) (2.20) Therefore, under the following identification: Sp(Tr(H)) 3 |u〉〈u| ←→ identification u ∈ H (‖u‖ = 1) (2.21) we see, Sp(Tr(H)) = {u ∈ H : ‖u‖ = 1} (2.22) where we assume the equivalence: u ≈ eiθu (θ ∈ R). 22 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement Definition 2.9. Define the trace Tr : Tr(H)→ C such that Tr(T ) = ∞∑ n=1 〈en, T en〉 (∀T ∈ Tr(H)) (2.23) where {en}∞n=1 is a CONS in H. It is well known that the Tr(T ) does not depend on the choice of CONS {en}∞n=1. Thus, clearly we see that TrH ( |u〉〈u|, F ) B(H) = Tr(|u〉〈u| * F ) = 〈uFu〉 (∀||u||H = 1, F ∈ B(H)) (2.24) Remark 2.10. Assume that a Hilbert space H is finite dimensional, i.e., H = Cn. Then, M(C, n) = The set of all (n× n)-complex matrices That is, F =  f11 f12 * * * f1n f21 f22 * * * f2n ... ... . . . ... fn1 fn2 * * * fnn  ∈M(C, n) (2.25) As mentioned before, we see A = A = B(Cn) = C(H) = Tr(H) = M(C, n) (2.26) and further, under the following notations: TrD+1(Cn) = { diagonal matrixF =  f11 0 * * * 0 0 f22 * * * 0 ... ... . . . ... 0 0 * * * fnn  ∣∣∣ fkk ≥ 0, n∑ k=1 fkk = 1 } TrDP+1 (Cn) = { F =  f11 0 * * * 0 0 f22 * * * 0 ... ... . . . ... 0 0 * * * fnn  ∈ TrD+1(Cn) ∣∣∣ fkk = 1 (for some k = j),= 0 (k 6= j)} We see, mixed state space: Tr+1(Cn) = { UFU∗ : F ∈ TrD+1(Cn), U is a unitary matrix } (2.27) pure state space: Trp+1(Cn) = { UFU∗ : F ∈ TrDP+1 (Cn), U is a unitary matrix } (2.28) 23 Ishikawa's Homepage 2.3 Classical basic structure[C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] 2.3 Classical basic structure[C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] 2.3.1 Classical basic structure[C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] In classical systems, the basic structure[A ⊆ A ⊆ B(H)] is restricted to the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] And we get the following diagram: (A): Classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] M(Ω)xdual C0(Ω) ⊆−−−−−−−−−−−−−→ subalgebra*weak-closure L∞(Ω, ν) ⊆−−−−−−→ subalgebra B(L2(Ω, ν))ypre-dual L1(Ω, ν) (2.29) In what follows, we shall explain this diagram. 2.3.1.1 Commutative C∗-algebra C0(Ω) and Commutative W ∗-algebra L∞(Ω, ν) Let Ω a locally compact space, for example, it suffices to image Ω as follows. R(= the real line), R2(= plane), Rn(= n-dimensional Euclidean space), [a, b](= interval), finite setΩ(= {ω1, ..., ωn}) (with discrete metric dD) where the discrete metric dD is defined by dD(ω, ω ′) = 1 (ω 6= ω′),= 0 (ω = ω′). Define the continuous functions space C0(Ω) such that C0(Ω) = {f : Ω→ C | f is complex-valued continuous on Ω, lim ω→∞ f(ω) = 0} (2.30) where "limω→∞ f(ω) = 0" means (B) for any positive real ε > 0, there exists a compact set K(⊆ Ω) such that {ω | ω ∈ Ω \K, |f(ω)| > ε} = ∅ 24 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement Therefore, if Ω is compact, the, the condition "limω→∞ f(ω) = 0" is not needed, and thus, C0(Ω) is usually denoted by C(Ω). In this note, even if Ω is compact, we often denote C(Ω) by C0(Ω). Defining the norm ‖ * ‖C0(Ω) in a complex vector space C0(Ω) such that ‖f‖C0(Ω) = max ω∈Ω |f(ω)| (2.31) we get the Banach space ( C0(Ω), ‖ * ‖C0(Ω) ) . Let Ω be a locally compact space, and consider the σ-finite measure space (Ω,BΩ, ν), where, BΩ is the Borel field, i.e., the smallest σ-field that contains all open sets. Further, assume that (C) for any open set U ⊆ Ω, it holds that 0 < ν(U) 5∞ ♠Note 2.1. Without loss of generality, we can assume that Ω is compact by the Stone-Čech compactification. Also, we can assume that ν(Ω) = 1. Define the Banach space Lr(Ω, ν) (where, r = 1, 2,∞) by the all complex-valued measurable functions f : Ω→ C such that ‖f‖Lr(Ω,ν) <∞ The norm ‖f‖Lr(Ω,ν) is defined by ‖f‖Lr(Ω,ν) =  [∫ Ω |f(ω)|r ν(dω) ]1/r (when r = 1, 2) ess.sup ω∈Ω |f(ω)| (when r =∞) (2.32) where ess.supω∈Ω|f(ω)| = sup{a ∈ R | ν({ω ∈ Ω : |f(ω)| = a }) > 0} Lr(Ω, ν) is often denoted by Lr(Ω) or Lr(Ω,BΩ, ν). Remark 2.11. [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Consider a Hilbert space H such that H = L2(Ω, ν) For each f ∈ L∞(Ω), define Tf ∈ B(L2(Ω, ν)) such that L2(Ω, ν) 3 φ −→ Tf (φ) = f * φ ∈ L2(Ω, ν) 25 Ishikawa's Homepage 2.3 Classical basic structure[C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Then, under the identification: L∞(Ω) 3 f ←→ identification Tf ∈ B(L2(Ω, ν)) (2.33) we see that f ∈ L∞(Ω) ⊆ B(L2(Ω, ν)) and further, we have the classical basic structure: [C0(Ω) ⊆ L∞(Ω) ⊆ B(L2(Ω, ν))] (2.34) This will be shown in what follows. Riese theorem (cf. [87]) says that C0(Ω) ∗ = M(Ω)(= the set of all complex-valued measures on Ω ) (2.35) Therefore, for any F ∈ C0(Ω), ρ ∈ C0(Ω)∗ = M(Ω), we have the bi-linear form which is written by the several ways such as ρ(F ) = C0(Ω)∗ ( ρ, F ) C0(Ω) = M(Ω) ( ρ, F ) C0(Ω) = ∫ Ω F (ω)ρ(dω) (2.36) Also, the dual norm is calculated as follows. ‖ρ‖C0(Ω)∗ = sup{|ρ(F ) | ‖F‖C0(Ω) = 1} = sup ||F ||C0(Ω)=1 | ∫ Ω F (ω)ρ(dω)| = sup Ξ,Γ∈BΩ ( |Re(ρ(Ξ))−Re(ρ(Ξc))|2 + |Im(ρ(Γ))− Im(ρ(Γc))|2 )1/2 =‖ρ‖M(Ω) (2.37) where, Ξc is the complement of Ξ, and Re(z)="the real part of the complex number z", Im(z)="the imaginary part of the complex number z". Further, we see that L1(Ω, ν)∗ = L∞(Ω, ν) in the same sense, L1(Ω, ν) = L∞(Ω, ν)∗ Also, it is clear that C0(Ω) ⊆ L∞(Ω, ν) For any f ∈ L∞(Ω, ν), there exist fn ∈ C0(Ω), n = 1, 2, .. such that ν({ω ∈ Ω | limn→∞ fn(ω) 6= f(ω)} = 0 |fn(ω)| ≤ ‖f‖L∞(Ω,ν) (∀ω ∈ Ω,∀n = 1, 2, 3, ...) 26 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement Therefore, we see lim n→∞ | 〈 φ, (f − fn)φ 〉 L2(Ω,ν) | ≤ lim n→∞ ∫ Ω |fn(ω)− f(ω)| * |φ(ω)|2ν(dω) = 0 (∀φ ∈ L2(Ω, ν)) Hence, the weak closure of C0(Ω) is equal to L ∞(Ω, ν) Then, we have the classical basic structure: [C0(Ω) ⊆ L∞(Ω) ⊆ B(L2(Ω, ν))] (2.38) Theorem 2.12. [Gelfand theorem (cf. [80]) ] Consider a general basic structure: [A ⊆ A ⊆ B(H)] where it is assumed that A is commutative. Then, there exists a measure space (Ω,BΩ, ν) (where Ω is a locally compact space) such that A = C0(Ω), A = L ∞(Ω, ν), B(H) = B(L2(Ω, ν)) where Ω is called a spectrum. 2.3.2 Classical basic structure[C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] and State space Consider the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))]. Then, we see the following diagram: (D): Classical basic structure and State space M p +1(Ω) (≈Ω) C∗-pure state ⊂ M+1(Ω) (probability measure) C∗-mixed state ⊂ M(Ω) xdual C0(Ω) ⊆−−−−−−−→ subalgebra weak-closure L∞(Ω) ⊆−−−−−−→ subalgebra B(L2(Ω))y pre-dual (2.39) L1+1(Ω, ν) (probability density function) W ∗-mixed state ⊂ L1(Ω, ν) 27 Ishikawa's Homepage 2.3 Classical basic structure[C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] In the above, the mixed state space Sm(C0(Ω) ∗) is characterized as Sm(C0(Ω) ∗) ={ρ ∈M(Ω) : ρ ≥ 0, ||ρ||M(Ω) = 1} ={ρ ∈M(Ω) : ρ is a probability measure on Ω } =:M+1(Ω) (2.40) Also, the pure state space Sp(C0(Ω) ∗) is Sp(C0(Ω) ∗) ={ρ = δω0 ∈ Sp(C0(Ω)∗) : δω0 is the point measure at ω0(∈ Ω), ω0 ∈ Ω} ≡Mp+1(Ω) (2.41) Here, the point measure δω0 ∈M(Ω) is defined by∫ Ω f(ω)δω0(dω) = f(ω0) (∀f ∈ C0(Ω)) Therefore, M p +1(Ω) = S p(C0(Ω) ∗) 3 δω ←→ identification ω ∈ Ω (2.42) Under this identification, we consider that Sp(C0(Ω) ∗) = Ω Also, it is well known that L1(Ω, ν)∗ = L∞(Ω, ν) Therefore, the W ∗-mixed state space is characterized by L1+1(Ω, ν) = {f ∈ L1(Ω, ν) : f ≥ 0, ∫ Ω f(ω)ν(dω) = 1} = the set of all probability density functions on Ω (2.43) Remark 2.13. [The case that Ω is finite: C0(Ω) = L ∞(Ω, ν), M(Ω) = L1(Ω, ν) ] Let Ω be a finite set {ω1, ω2, ..., ωn} with the discrete metric dD and the counting measure ν. Here, the counting measure ν is defined by ν(D) = ][D](= "the number of the elements of D") 28 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement Then, we see that C0(Ω) = {F : Ω→ C | F is a complex valued function on Ω} = L∞(Ω, ν) And thus, we see that ρ ∈M+1(Ω) ⇐⇒ ρ = n∑ k=1 pkδωk ( n∑ k=1 pk = 1, pk ≥ 0) and f ∈ L1+1(Ω, ν) ⇐⇒ n∑ k=1 f(ωk) = 1. f(ωk) ≥ 0 In this sense, we have the following identifications: M+1(Ω) = L 1 +1(Ω, ν) ( or, M(Ω) = L 1(Ω, ν)) After all, we have the following identification: C0(Ω) = L ∞(Ω) = Cn M(Ω) = L1(Ω) = Cn (2.44) where the norm ‖ * ‖C0(Ω) in the former is defined by ‖z‖C0(Ω) = max k=1,2,...,n |zk| ∀z =  z1 z2 ... xn  ∈ Cn (2.45) and the norm ‖ * ‖M(Ω) in the latter is defined by ‖z‖M(Ω) = n∑ k=1 |zk| ∀z =  z1 z2 ... xn  ∈ Cn (2.46) 29 Ishikawa's Homepage 2.4 State and Observable-the primary quality and the secondary quality- 2.4 State and Observable-the primary quality and the secondary quality- 2.4.1 In the beginning Our present purpose is to learn the following spell (= Axiom 1) by rote. (A): Axiom 1(pure measurement)(cf. This will be able to be read in §2.7) With any system S, a basic structure [A ⊆ A]B(H) can be associated in which measurement theory of that system can be formulated. In [A ⊆ A]B(H), consider a W ∗-measurement MA ( O=(X,F, F ), S[ρ] ) ( or, C∗-measurementMA ( O=(X,F, F ), S[ρ] ) ) . That is, consider • a W ∗-measurement MA ( O, S[ρ] ) ( or, C∗-measurement MA ( O=(X,F, F ), S[ρ] ) ) of an observable O=(X,F, F ) for a state ρ(∈ Sp(A∗) : state space) Then, the probability that a measured value x (∈ X) obtained by the W ∗-measurement MA ( O, S[ρ] ) ( or, C∗-measurement MA ( O=(X,F, F ), S[ρ] ) ) belongs to Ξ (∈ F) is given by ρ(F (Ξ))(≡ A∗(ρ, F (Ξ))A) (if F (Ξ) is essentially continuous at ρ, or see Definition 2.14 ). The "learning by rote" urges us to understand the mathematical definitions of (]1) Basic structure[A ⊆ A]B(H), state space Sp(A∗) (]2) observable O=(X,F, F ), etc. In the previous section, we studied the above (]1), that is, we discussed the following classification: (B) General basic structure[A ⊆ A]B(H) state space [Sp(A∗),Sm(A∗),S p (A∗)] =⇒  Quantum basic structure[C(H) ⊆ B(H)]B(H) state space [Sp(Tr(H)),Sm(Tr(H))=S m (Tr(H))] Classical basic structure[C0(Ω) ⊆ L∞(Ω, ν)]B(L2(Ω,ν)) state space [Ω,M+1(Ω),L∞(Ω,ν)] In this section, we shall study the above (]2), i.e., "Observable" 30 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement Recall the famous words: "the primary quality" and "the secondary quality" due to John Locke, an English philosopher and physician regarded as one of the most influential of Enlightenment thinkers and known as the "Father of Classical Liberalism". We think the following correspondence:{ [state] ←→ [the primary quality] [observable] ←→ [the secondary quality] (2.47) And thus, we think • These (i.e., "state" and "observable") are the concepts which form the basis of dualism. Also, the following table (which may include my fiction ) promotes the better understanding of quantum language as well as the other world-views( i.e., the conventional philosophies). Table 2.1: Observable * State * System in world-views (cf. Table 3.1) World descriptionQuantum language observable state system Plato idea / / Aristotle / eidos hyle Locke secondary quality primary quality / Newton / state point mass statistics / parameter population quantum mechanics observable state(≈ wave function) particle ♠Note 2.2. It may be understandable to consider "observable" ="the partition of word"="the secondary quality" (2.48) For example, Chapter 1 (Figure 1.2) says that ( fc, fh ) is the partition between "cold" and "hot". 1 fc fh 0 10 20 30 40 50 60 70 80 90 100 Chapter 1 (Figure 1.2): Cold or hot? Also, "measuring instrument" is the instrument that choose a word among words. In this sense, we consider that "observable"= "measurement instrument". Also, The reason that John Locke's 31 Ishikawa's Homepage 2.4 State and Observable-the primary quality and the secondary quality- sayings "primary quality (e.g., length, weight, etc.)" and "secondary quality (e.g., sweet, dark, cold, etc.)" is that these words form the basis of dualism. 2.4.2 Dualism (in philosophy) and duality (in mathematics) The following question may be significant: (C1) Why did philosophers continue persisting in dualism? As the typical answer, we may consider that (C2) "I" is the special existence, and thus, we would like to draw a line between "I" and "matter". But, we think that this is only quibbling. We want to connect the question (C1) with the following mathematical question: (C3) Why do mathematicians investigate "dual space"? Of course, the question "why?" is non-sense in mathematics. If we have to answer this, we have no answer except the following (D): (D) If we consider the dual space A∗, calculation progresses deeply. Thus, we want to consider the relation between the dualism and the dual space such as{ [the primary quality] ←→ the state in the dual space A∗ [the secondary quality] ←→ the observable in C∗ algebra A (or, W ∗-algebra A) (2.49) Thus, we consider that the answer to the (C1) is also "calculation progresses deeply". 2.4.3 Essentially continuous In §2.1.2, we introduced the following diagram: (E):General basic structure and state space Sp(A∗) C∗−purestate ⊂ Sm(A∗) C∗-mixed state ⊂ A∗xdual A ⊆−−−−−−−−−−−−−→ subalgebra*weak-closure A ⊆−−−−−−→ subalgebra B(H)y pre-dual (2.50) S m (A∗) W ∗-mixed state ⊂ A∗ 32 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement In the above diagram, we introduce the following definition. Definition 2.14. [Essentially continuous (cf. ref. [31] ) ] An element F (∈ A) is said to be essentially continuous at ρ0(∈ Sm(A∗)), if there uniquely exists a complex number α such that (F1) if ρn (∈ S m (A∗)) weakly converges to ρ0(∈ Sm(A∗)) (That is, limn→∞ A∗ ( ρn, G ) A = A∗ ( ρ0, G ) A (∀G ∈ A(⊆ A) ), then limn→∞ A∗ ( ρn, F ) A = α Then, the value ρ0(F ) (= A∗ ( ρ0, F ) A) is defined by the α Of course, for any ρ0(∈ Sm(A∗)), F (∈ A) is essentially continuous at ρ0. This "essentially continuous" is chiefly used in th case that ρ0(∈ Sp(A∗)). Remark 2.15. [Essentially continuous in quantum system and classical system] [I]: Consider the quantum basic structure [C(H) ⊆ B(H)]B(H). Then, we see (C(H))∗ = T(H) = B(H)∗ Thus, we have ρ ∈ Sp(C(H)∗) ⊆ Tr(H), F ∈ C(H) = B(H), which implies that ρ(G) = C(H)∗ ( ρ, F ) ) B(H) = Tr(H) ( ρ, F ) ) B(H) (2.51) Thus, we see that "essentially continuous" ⇔ "continuous" in quantum case. [II]: Next, consider the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))]. A function F (∈ L∞(Ω, ν)) is essentially continuous at ω0 (∈ Ω = Sp(C0(Ω)∗)), if and only if it holds that (F2) if ρn(∈ L1+1(Ω, ν) satisfies that lim n→∞ ∫ Ω G(ω)ρn(ω)ν(dω) = G(ω0) (∀G ∈ C0(Ω)) then there uniquely exists a complex number α such that lim n→∞ ∫ Ω F (ω)ρn(ω)ν(dω) = α (2.52) Then, the value of F (ω) is defined by α, that is, F (ω0) = α. 33 Ishikawa's Homepage 2.4 State and Observable-the primary quality and the secondary quality- 0 (Ω, ν)ω1 ω2 Figure 2.1: not essentially continuous at ω1, essentially continuous at ω2 2.4.4 The definition of "observable (=measuring instrument)" In this section, we introduce "observable", which is also said to be "measuring instrument" or "POVM (=positive operator valued measure space)". Definition 2.16. [Set ring, set field, σ-field] Let X be a set ( or locally compact space). The F ( ⊆ 2X = P(X) = {A | A ⊆ X}, the power set of X ) (or, the pair (X,F)) is called a ring ( of sets), if it satisfies that (a) : ∅(="empty set") ∈ F, (b) : Ξi ∈ F (i = 1, 2, . . .) =⇒ n∪ i=1 Ξi ∈ F, n∩ i=1 Ξi ∈ F (c) : Ξ1,Ξ2 ∈ F =⇒ Ξ1 \ Ξ2 ∈ F ( where, Ξ1 \ Ξ2 = {x | x ∈ Ξ1, x /∈ Ξ2}) Also, if X ∈ F holds, the ring F(or, the pair (X,F)) is called a field (of sets). And further, (d) if the formula (b) holds in the case that n =∞, a field F is said to be σ-field. And the pair (X,F) is called a measurable space. The following definition is most important. In this note, we mainly devote ourselves to the W ∗-observable. Definition 2.17. [Observable,measured value space] Consider the basic structure [A ⊆ A ⊆ B(H)] (G1):C ∗observable A triplet O=(X,R, F ) is called a C∗-observable (or, C∗-measuring instrument ) in A, if it satisfies as follows. (i) (X,R) is a ring of sets. 34 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement (ii) a map F : R→ A satisfies that (a) 0 5 F (Ξ) ≤ I (∀Ξ ∈ R), F (∅) = 0, (b) for any ρ(∈ Sp(A∗)), there exists a probability space (X,R, Pρ) such that (where, R is the smallest σ-field such that R ⊆ R) such that A∗ ( ρ, F (Ξ) ) A = Pρ(Ξ) (∀Ξ ∈ R) (2.53) Also, X [resp. (X,F, Pρ)] is called a measured value space [resp. sample probability space ]. (G2):W ∗observable A triplet O=(X,F, F ) is called a W ∗-observable (or, W ∗-measuring instrument ) in A, if it satisfies as follows. (i) (X,F) is a σ-field. (ii) a map F : F → A satisfies that (a) 0 5 F (Ξ) (∀Ξ ∈ F), F (∅) = 0, F (X) = I (b) for any ρ(∈ Sm(A∗)), there exists a probability space (X,F, Pρ) such that A∗ ( ρ, F (Ξ) ) A = Pρ(Ξ) (∀Ξ ∈ F) (2.54) The observable O=(X,F, F ) is called a projective observable, if it holds that F (Ξ)2 = F (Ξ) (∀Ξ ∈ F). In this note, we aways assume Hypothesis 2.19 below: Definition 2.18. Let ρ ∈ Sm(A∗), and (X,F, F ) be a W ∗-observable in A. Fρ = {Ξ ∈ F | F (Ξ) is essentially continuous at ρ }. The probability space (X,F, Pρ) is called its sample probability space, if it holds that (]1) F is the smallest σ-field that contains Fρ. (]2) A∗ ( ρ, F (Ξ) ) A = Pρ(Ξ) (∀Ξ ∈ Fρ) (2.55) Concerning the C∗-observable, the sample probability space clearly exists. On the other hand, concerning the W ∗-observable, we have to say something as follows. As mentioned in Remark 2.15, in quantum cases ( thus, A∗ = Tr(H) = A∗ ), the (]1) and (]2) clearly hold. 35 Ishikawa's Homepage 2.4 State and Observable-the primary quality and the secondary quality- However, in the classical cases, we do not know whether the existence of the sample probability space follows from the definition of the W ∗-observable. Thus, in this note, we do not add the condition (]) in the definition of the W ∗-observable. Hypothesis 2.19. [Sample probability space]. In the above situation, the existence of the sample probability space is always assumed. 36 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement 2.5 Examples of classical observables We shall mention several examples of classical observables. The observables introduced in Example 2.20-Example 2.23 are characterized as a C∗observable as well as a W ∗observable. In what follows (except Example 2.20), consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Example 2.20. [Existence observable ] Consider the basic structure: [A ⊆ A ⊆ B(H)] Define the observable O(exi) ≡ (X, {∅, X}, F (exi)) in W ∗-algebra A such that: F (exi)(∅) ≡ 0, F (exi)(X) ≡ I (2.56) which is called the existence observable (or, null observable). Consider any observable O = (X,F, F ) in A. Note that {∅, X} ⊆ F. And we see that F (∅) = 0, F (X) = I Thus, we see that (X, {∅, X}, F (exi)) = (X, {∅, X}, F ), and therefore, we say that any observable O = (X,F, F ) includes the existence observable O(exi). ♠Note 2.3. The above is associated with Berkley's words: (]1) To be is to be perceived (by George Berkeley(1685-1753)) which is peculiar to dualism: This is opposite to Einstein's saying in monism : (]2) The moon is there whether one looks at it or not. (i.e., Physics holds without observers.) in Einstein and Tagore's conversation. (cf. Note 12.2)。 Example 2.21. [The resolution of the identity I; The word's partition] Let [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] be the classical basic structure. We find the similarity between an observable O and the resolution of the identity I in what follows. Consider an observable O ≡ (X,F, F ) in L∞(Ω) such that X is a countable set (i.e., X ≡ {x1, x2, ...}) and F = P(X) = {Ξ | Ξ ⊆ X}, i.e., the power set of X. Then, it is clear that 37 Ishikawa's Homepage 2.5 Examples of classical observables (i) F ({xk}) ≥ 0 for all k = 1, 2, ... (ii) ∑∞ k=1[F ({xk})](ω) = 1 (∀ω ∈ Ω), which imply that the [F ({xk}) : k = 1, 2, ...] can be regarded as the resolution of the identity element I. Thus we say that • An observable O ( ≡ (X,F, F ) ) in L∞(Ω) can be regarded as " the resolution of the identity I 0 1 [F ({x1})](ω) [F ({x2})](ω) [F ({x3})](ω) Ω 100 Figure 2.2: O ≡ ({x1, x2, x3}, 2{x1,x2,x3}, F ) In Figure 2.2, assume that Ω = [0, 100] is the axis of temperatures ( ◦C), and put X = {C(="cold"), L (="lukewarm" = "not hot enough"), H(="hot") }. And further, put fx1 = fC, fx2 = fL, fx3 = fH. Then, the resolution {fx1 , fx2 , fx3} can be regarded as the word's partition C(="cold"), L(="lukewarm"="not hot enough"), H(="hot") . Also, putting F(= 2X) = {∅, {x1}, {x2}, {x3}, {x1, x2}, {x2, x3}, {x1, x3}, X} and [F (∅)](ω) = 0, [F (X)](ω) = fx1(ω) + fx2(ω) + fx3(ω) = 1 [F ({x1})](ω) = fx1(ω), [F ({x2})](ω) = fx2(ω), [F ({x3})](ω) = fx3(ω) [F ({x1, x2})](ω) = fx1(ω) + fx2(ω), [F ({x2, x3})](ω) = fx2(ω) + fx3(ω) [F ({x1, x3})](ω) = fx1(ω) + fx3(ω) then, we have the observable (X,F(= 2X), F ) in L∞([0, 100]). 38 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement Example 2.22. [Triangle observable ] Let [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] be the classical basic structure. For example, define the state space Ω by the closed interval [0, 100] (⊆ R). For each n ∈ N10010 = {0, 10, 20, . . . , 100}, define the (triangle) continuous function gn : Ω → R by gn(ω) =  0 (0 5 ω 5 n− 10) ω − n− 10 10 (n− 10 5 ω 5 n) −ω − n+ 10 10 (n 5 ω 5 n+ 10) 0 (n+ 10 5 ω 5 100) (2.57) 1 0 10 20 30 40 50 60 70 80 90 100 g0 g10 g20 g30 g40 g50 g60 g70 g80 g90 g100 Figure 2.3: Triangle observable Putting Y = N10010 and define the triangle observable O4 = (Y, 2Y , F4) such that [F4(∅)](ω) = 0, [F4(Y )](ω) = 1 [F4(Γ)](ω) = ∑ n∈Γ gn(ω) (∀Γ ∈ 2N 100 10 ) Then, we have the triangle observable O4 = (Y (= N10010 ), 2Y , F4) in L∞([0, 100]). Example 2.23. [Normal observable] x y 6 y = 1√ 2πσ2 e− x2 2σ2 σ−σ 2σ−2σ 68.3% 95.4% Figure 2.4: Error function Consider a classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))]. Here, Ω = R(= the real line) or, Ω = interval [a, b] (⊆ R), which is assumed to have Lebesgue measure ν(dω)(= 39 Ishikawa's Homepage 2.5 Examples of classical observables dω). Let σ > 0, which is call a standard deviation. The normal observable OGσ=(R,BR, Gσ) in L∞(Ω, ν) is defined by [Gσ(Ξ)](ω) = 1√ 2πσ2 ∫ Ξ e− (x−ω)2 2σ2 dx (∀Ξ ∈ BR(Borel field),∀ω ∈ Ω(= R or [a, b])) This is the most fundamental observable in statistics. The following examples introduced in Example 2.24 and Example 2.25 are not C∗observables but W ∗observables. This implies that the W ∗-algebraic approach is more powerful than the C∗-algebraic approach. Although the C∗-observable is easy, it is more narrow than the W ∗observable. Thus, throughout this note, we mainly devote ourselves to W ∗-algebraic approach. Example 2.24. [Exact observable ] Consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))]. Let BΩ be the Borel field in Ω, i.e., the smallest σ-field that contains all open sets. For each Ξ ∈ BΩ, define the definition function χΞ : Ω→ R such that χ Ξ (ω) =  1 (ω ∈ Ξ) 0 (ω /∈ Ξ) (2.58) Put [F (exa)(Ξ)](ω) = χΞ(ω) (Ξ ∈ BΩ, ω ∈ Ω). The triplet O(exa) = (Ω,BΩ, F (exa)) is called the exact observable in L∞(Ω, ν). This is the W ∗-observable and not C∗-observable, since [F (exa)(Ξ)](ω) is not always continuous. For the argument about the sample probability space (cf. Definition 2.18 ), see Example 2.33. Example 2.25. [Rounding observable] Define the state space Ω by Ω = [0, 100]. For each n ∈ N10010 ={0, 10, 20, . . . , 100}, define the discontinuous function gn : Ω→ [0, 1] such that gn(ω) =  0 (0 5 ω 5 n− 5) 1 (n− 5 < ω 5 n+ 5) 0 (n+ 5 < ω 5 100) * * * * * * * * * * * * 1 0 10 20 30 40 50 60 70 80 90 100 g0 g10 g20 g50 g80 g90 g100 Figure 2.5: Round observable 40 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement Define the observable ORND = (Y (=N10010 ), 2Y , GRND) in L∞(Ω, ν) such that [GRND(∅)](ω) = 0, [GRND(Y )](ω) = 1 [GRND(Γ)](ω) = ∑ n∈Γ gn(ω) (∀Γ ∈ 2Y = 2N 100 10 ) Recall that gn is not continuous. Thus, this is not C ∗-observable but W ∗-observable. 41 Ishikawa's Homepage 2.6 System quantity - The origin of observable 2.6 System quantity - The origin of observable In classical mechanics, the term "observable" usually means the continuous real valued function on a state space (that is, physical quantity). An observable in measurement theory (= quantum language ) is characterized as the natural generalization of the physical quantity. This will be explained in the following examples. Example 2.26. [System quantity] Let [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] be the classical basic structure. A continuous real valued function f : Ω → R ( or generally, a measurable Rn-valued function f : Ω → Rn ) is called a system quantity (or in short, quantity) on Ω. Define the projective observable O = (R,BR, F ) in L∞(Ω, ν) such that [F (Ξ)](ω) =  1 when ω ∈ f−1(Ξ) 0 when ω /∈ f−1(Ξ) (∀Ξ ∈ BR) Here, note that f(ω) = lim N→∞ N2∑ n=−N2 n N [ F ( [ n N , n+ 1 N ) )] (ω) = ∫ R λ[F (dλ)](ω) (2.59) Thus, we have the following identification: f (system quantity on Ω) ←→ O = (R,BR, F ) (projective observable in L∞(Ω, ν)) (2.60) This O is called the observable representation of a system quantity f . Therefore, we say that (a) An observable in measurement theory is characterized as the natural generalization of the physical quantity. Example 2.27. [Position observable , momentum observable , energy observable ] Consider Newtonian mechanics in the classical basic algebra [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L∞(Ω, ν))]. For simplicity, consider the two dimensional space Ω = Rq × Rp={(q, p) = (position,momentum) | q, p ∈ R} The following quantities are fundamental: (]1) :q : Ω→ R, q(q, p) =q (∀(q, p) ∈ Ω) 42 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement (]2) :p : Ω→ R, p(q, p) =p (∀(q, p) ∈ Ω) (]3) :ẽ : Ω→ R, ẽ(q, p) =[potential energy ] + [kinetic energy ] =U(q) + p2 2m (Hamiltonian) (∀(q, p) ∈ Ω) where, m is the mass of a particle. Under the identification (2.60), the above (]1), (]2) and (]3) is respectively called a position observable, a momentum observable and an energy observable. Example 2.28. [Hermitian matrix is projective observable ] Consider the quantum basic structure in the case that H = Cn, that is, [B(Cn) ⊆ B(Cn) ⊆ B(Cn)] Now, we shall show that an Hermitian matrix A(∈ B(Cn)) can be regarded as a projective observable. For simplicity, this is shown in the case that n = 3. We see (for simplicity, assume that xj 6= xk(if j 6= k) ) A = U∗ x1 0 00 x2 0 0 0 x3 U (2.61) where U (∈ B(C3)) is the unitary matrix and xk ∈ R. Put FA({x1}) = U∗ 1 0 00 0 0 0 0 0 U, FA({x2}) = U∗ 0 0 00 1 0 0 0 0 U, FA({x3}) = U∗ 0 0 00 0 0 0 0 1 U FA(R \ {x1, x2, x3}) = 0 0 00 0 0 0 0 0  , Thus, we get the projective observable OA = (R,BR, FA) in B(C3). Hence, we have the following identification2: A (Hermitian matrix) ←→ OA = (R,BR, FA) (projective observable ) (2.62) 2 For example, in the case that x1 = x2, it suffices to define FA({x1}) = U∗ 1 0 00 1 0 0 0 0 U, FA({x3}) = U∗ 0 0 00 0 0 0 0 1 U FA(R \ {x1, x3}) = 0 0 00 0 0 0 0 1  And, we have the projection observable OA = (R,BR, FA). 43 Ishikawa's Homepage 2.6 System quantity - The origin of observable Let A(∈ B(Cn)) be an Hermitian matrix. Under this identification, we have the quantum measurement MB(Cn)(OA, S[ρ]), where ρ = |ω〉〈ω|, ω =  ω1 ω2 ... ωn  ∈ Cn, ‖ω‖ = 1 Born's quantum measurement theory (or, Axiom 1 (§2.7) ) says that (]) The probability that a measured value x(∈ R) is obtained by the quantum measurement MB(Cn)(OA, S[ρ]) is given by Tr(ρ * FA({x})) ( = 〈ω, FA({x})ω〉 ). (for the trace: "Tr", recall Definition 2.9). Therefore, the expectation of a measured value is given by∫ R x〈ω, FA(dx)ω〉 = 〈ω,Aω〉 (2.63) Also, its variance (δωA) 2 is given by (δωA) 2 = ∫ R (x− 〈ω,Aω〉)2〈ω, FA(dx)ω〉 = 〈Aω,Aω〉 − |〈ω,Aω〉|2 = ||(A− 〈ω,Aω〉)ω||2 (2.64) Example 2.29. [Spectrum decomposition] Let H be a Hilbert space. Consider the quantum basic structure [C(H) ⊆ B(H) ⊆ B(H)]. The spectral theorem (cf. [87]) asserts the following equivalence: ((a)⇔(b)), that is, (a) T is a self-adjoint operator on Hilbert space H (b) There exists a projective observable O = (R,BR, F ) in B(H) such that T = ∫ ∞ −∞ λF (dλ) (2.65) Since the definition of "unbounded self-adjoint operator" is not easy, in this note we regard the (b) as the definition. In the sense of the (b), we consider the identification: self-adjoint operator T ←→ identification spectrum decomposition O = (R,BR, F ) (2.66) 44 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement This quantum identification should be compared to the classical identification (2.60). The above argument can be extended as follows. That is, we have the following equivalence: ((c)⇔(d)), that is, (c) T1, T2 are commutative self-adjoint operators on Hilbert space H (d) There exists a projective observable Ô = (R2,BR2 , G) in B(H) such that T1 = ∫ R2 λ1G(dλ1dλ2), T2 = ∫ R2 λ2G(dλ1dλ2) (2.67) 45 Ishikawa's Homepage 2.7 Axiom 1 - No science without measurement 2.7 Axiom 1 - No science without measurement Measurement theory (= quantum language ) is formulated as follows. • measurement theory (=quantum language) := [Axiom 1] Measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spells (a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } manual to use spells Now we can explain Axiom 1 (measurement). 2.7.1 Axiom 1 for measurement With any system S, a basic structure [A ⊆ A ⊆ B(H)] can be associated in which measurement theory of the system can be formulated. A state (or precisely, pure state) of the systemS is represented by an element of state space Sp(A∗). An observable (= measuring instrument) is represented by a C∗-observable O = (X,F, F ) in A ( or, W ∗-observable O = (X,F, F ) in A ). (A1) An observer takes a measurement of an observable [O] for a state ρ, and gets a measured value x(∈ X). In a basic structure [A ⊆ A ⊆ B(H)], consider a W ∗-measurement MA ( O=(X,F, F ), S[ρ] )( or, C∗-measurement MA ( O=(X,F, F ), S[ρ] ) ) . Preparation 2.30. Consider • a W ∗-measurement MA ( O, S[ρ] ) ( or, C∗-measurement MA ( O=(X,F, F ), S[ρ] ) ) of an observable O=(X,F, F ) for a state ρ(∈ Sp(A∗) : state space) Note that (A2) { W ∗-measurement MA ( O, S[ρ] ) * * * O is W ∗observable , ρ ∈ Sp(A∗) C∗-measurement MA ( O, S[ρ] ) * * * O is C∗observable , ρ ∈ Sp(A∗) In this lecture, we mainly devote ourselves to W ∗-measurements. 46 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement (B): Axiom 1(measurement) pure type (This can be read under the preparation to this section ) With any system S, a basic structure [A ⊆ A]B(H) can be associated in which measurement theory of that system can be formulated. In [A ⊆ A]B(H), consider a W ∗-measurement MA ( O=(X,F, F ), S[ρ] ) ( or, C∗-measurementMA ( O=(X,F, F ), S[ρ] ) ) . That is, consider • a W ∗-measurement MA ( O, S[ρ] ) ( or, C∗-measurement MA ( O=(X,F, F ), S[ρ] ) ) of an observable O=(X,F, F ) for a state ρ(∈ Sp(A∗) : state space) Then, the probability that a measured value x (∈ X) obtained by the W ∗-measurement MA ( O, S[ρ] ) ( or, C∗-measurement MA ( O=(X,F, F ), S[ρ] ) ) belongs to Ξ (∈ F) is given by ρ(F (Ξ))(≡ A∗(ρ, F (Ξ))A) (if F (Ξ) is essentially continuous at ρ, or see Definition 2.14 ). This axiom is a kind of generalization (or, a linguistic turn) of Born's probabilistic interpretation of quantum mechanics. 3 That is, (the law proposed by Born) quantum mechanics (Born's quantum measurement ) (physics) −−−−−−−−→ linguistic turn (a kind of spell) measurement theory(Axiom 1) (metaphysics, language) (2.68) ♠Note 2.4. The above axiom is due to Max Born (1926). There are many opinions for the term "probability". For example, Einstein sent Born the following letter (1926): (]1) Quantum mechanics is certainly imposing. But an inner voice tells me that it is not yet the real thing. The theory says a lot, but does not really bring us any closer to the secret of the "old one." I, at any rate, am convinced that He does not throw dice. From a viewpoint of quantum mechanics, I want to believe that both Born and Einstein are right. That is because I assert that quantum mechanics is not physics. 2.7.2 A simplest example Now we shall describe Example1.2 ( Cold or hot?) in terms of quantum language (i.e., Axiom 1 ). 3 Ref. [6]: Born, M. "Zur Quantenmechanik der Stossprozesse (Vorläufige Mitteilung)", Z. Phys. (37) pp.863–867 (1926). 47 Ishikawa's Homepage 2.7 Axiom 1 - No science without measurement Example 2.31. [(continued from Example1.2) The measurement of "cold or hot" for water in a cup ] Consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Here, Ω = the closed interval [0, 100](⊂ R) with Lebesgue measure ν. The state space Sp(C0(Ω) ∗) is characterized as Sp(C0(Ω) ∗) = {δω ∈M(Ω) | ω ∈ Ω} ≈ Ω = [0, 100] 1 fc fh 0°C 10°C 20°C 30°C 40°C 50°C 60°C 70°C 80°C 90°C 100°C Figure 2.6: Cold? Hot? In Example 1.2, we consider this [C-H]-thermometer O = (fc, fh), where the state space Ω = [0, 100], the measured value space X = {c, h}. That is, fc(ω) =  1 (0 5 ω 5 10) 70−ω 60 (10 5 ω 5 70) 0 (70 5 ω 5 100) , fh(ω) = 1− fc(ω) Then, we have the (cold-hot) observable Och = (X, 2 X , Fch) in L ∞(Ω) such that [Fch(∅)](ω) = 0, [Fch(X)](ω) = 1 [Fch({c})](ω) = fc(ω), [Fch({h})](ω) = fh(ω) Thus, we get a measurement ML∞(Ω)(Och, S[δω ]) ( or in short, ML∞(Ω)(Och, S[ω]). Therefore, for example, putting ω = 55 ◦C, we can, by Axiom 1 (§2.7), represent the statement (A1) in Example 1.2 as follows. (a) the probability that a measured valuex(∈ X={c, h}) obtained by measurement ML∞(Ω)(Och, S[ω(=55)]) belongs to set  ∅ {c} {h} {c, h}  is given by  [Fch(∅)](55) = 0 [Fch({c})](55) = 0.25 [Fch({h})](55) = 0.75 [Fch({c, h})](55) = 1  Or more precisely, (b) When an observer takes a measurement by [[C-H]-instrument] measuring instrumentOch=(X,2X ,Fch) for [water in cup] (system(measuring object)) with [55 ◦C] (state(= ω ∈ Ω) ) , the probability that measured value [ c h ] is obtained is given by [ fc(55) = 0.25 fh(55) = 0.75 ] 48 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement 2.8 Examples: Classical measurements (urn problem, etc.) 2.8.1 linguistic world-view - Wonder of man's linguistic competence The applied scope of physics physics (realistic world-description method) is rather clear. But the applied scope of measurement theory is ambiguous. What we can do in measurement theory (= quantum language) is (a)  (a1): Use the language defined by Axiom 1 ( §2.7) (a2): Trust in man's linguistic competence Thus, some readers may doubt that (b) Is it science? However, it should be noted that the spirit of measurement theory is different from that of physics. 2.8.2 Elementary examples-urn problem, etc. Since measurement theory is a language, we can not master it without exercise. Thus, we present simple examples in what follows. Example 2.32. [ The measurement of the approximate temperature of water in a cup (continued from Example2.22 [triangle observable ])] Consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] where Ω = "the closed interval [0, 100]" with the Lebesgue measure ν. Let testees drink water with various temperature ω ◦C (0 5 ω 5 100). And you ask them "How many degrees( ◦C) is roughly this water?" Gather the data, ( for example, hn(ω) persons say n ◦C (n = 0, 10, 20, . . . , 90, 100). and normalize them, that is, get the polygonal lines. For example, define the state space Ω by the closed interval [0, 100] (⊆ R) with the Lebesgue measure. For each n ∈ N10010 = {0, 10, 20, . . . , 100}, define the (triangle) continuous function gn : Ω→ [0, 1] by gn(ω) =  0 (0 5 ω 5 n− 10) ω − n− 10 10 (n− 10 5 ω 5 n) −ω − n+ 10 10 (n 5 ω 5 n+ 10) 0 (n+ 10 5 ω 5 100) 49 Ishikawa's Homepage 2.8 Examples: Classical measurements (urn problem, etc.) 1 0 10 20 30 40 50 60 70 80 90 100 g0 g10 g20 g30 g40 g50 g60 g70 g80 g90 g100 Figure 2.7: Triangle observable (a) You choose one person from the testees, and you ask him/her "How many degrees( ◦C) is roughly this water?". Then the probability that he/she says [ "about 40 ◦C" "about 50 ◦C" ] is given by [ g40(47) = 0.25 f50(47) = 0.75 ] This is described in terms of Axiom 1 ( §2.7) in what follows. Putting Y = N10010 , define the triangle observable O4 = (Y, 2Y , G4) in L∞(Ω) such that [G4(∅)](ω) = 0, [G4(Y )](ω) = 1 [G4(Γ)](ω) = ∑ n∈Γ gn(ω) (∀Γ ∈ 2N 100 10 ,∀ω ∈ Ω = [0, 100]) Then, we have the triangle observable O4 = (Y (= N10010 ), 2Y , G4) in L∞([0, 100]). And we get a measurement ML∞(Ω)(O 4, S[δω ]). For example, putting ω=47 ◦C, we see, by Axiom 1 ( §2.7), that (b) the probability that a measured value obtained by the measurement ML∞(Ω)(O 4, S[ω(=47)]) is [ about 40 ◦C about 50 ◦C ] is given by [ [G4({40})](47) = 0.3 [G4({50})](47) = 0.7 ] Therefore, we see: statement (a) (ordinary language) −−−−−−→ translation statement (b) (quantum language) (2.69) /// Example 2.33. [Exact measurement] Consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Let BΩ be the Borel field. Then, define the exact observable O (exa) = (X(= Ω),F(= BΩ), F (exa)) in L∞(Ω, ν) such that [F (exa)(Ξ)](ω) = χ Ξ (ω) =  1 (ω ∈ Ξ) 0 (ω /∈ Ξ) (∀Ξ ∈ BΩ) Let δω0 ≈ ω0(∈ Ω). Consider the exact measurement ML∞(Ω,ν)(O(exa), S[δω0 ]). Here, Axiom 1 ( §2.7) says: 50 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement (a) Let D(⊆ Ω) be arbitrary open set such that ω0 ∈ D. Then, the probability that a measured value obtained by the exact measurement ML∞(Ω,ν)(O (exa), S[δω0 ]) belongs to D is given by C0(Ω)∗ ( δω0 , χD ) L∞(Ω,ν) = 1 From the arbitrariness of D, we conclude that (b) a measured value ω0 is, with the probability 1, obtained by the exact measurement ML∞(Ω,ν) (O (exa), S[δω0 ]). Further, put Fω0 = {Ξ ∈ F : ω0 /∈ "the closure of Ξ"\ "the interior of Ξ"} Then, when Ξ ∈ Fω0 , F (Ξ) is continuous at ω0. And, F is the smallest σ-field that contains Fω0 . Therefore, we have the probability space (X,F, Pδω0 ) such that Pδω0 (Ξ) = [F (Ξ)](ω0) (∀Ξ ∈ Fω0) that is, (c) the exact measurement ML∞(Ω,ν)(O (exa), S[δω0 ]) has the sample space (X,F, Pδω0 ) (= (Ω, BΩ, Pδω0 )) Example 2.34. [Urn problem] There are two urns U1 and U2. The urn U1 [resp. U2] contains 8 white and 2 black balls [resp. 4 white and 6 black balls] (cf. Table 2.2, Figure 2.7). Table 2.2: urn problem Urn w*b white ball black ball Urn U1 8 2 Urn U2 4 6 Here, consider the following statement (a): (a) When one ball is picked up from the urn U2, the probability that the ball is white is 0.4. 51 Ishikawa's Homepage 2.8 Examples: Classical measurements (urn problem, etc.) ω1 ω2 Figure 2.8: Urn problem In measurement theory, the statement (a) is formulated as follows: Assuming U1 * * * "the urn with the state ω1" U2 * * * "the urn with the state ω2" define the state space Ω by Ω = {ω1, ω2} with the discrete metric and the counting measure ν (i.e., ν({ω1}) = ν({ω2}) = 1). That is, we assume the identification; U1 ≈ ω1, U2 ≈ ω2, Thus, consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Put "w" = "white", "b" = "black", and put X = {w, b}. And define the observable O ( ≡ (X ≡ {w, b}, 2{w,b}, F ) ) in L∞(Ω) by [F ({w})](ω1) = 0.8, [F ({b})](ω1) = 0.2, [F ({w})](ω2) = 0.4, [F ({b})](ω2) = 0.6. Thus, we get the measurement ML∞(Ω)(O, S[δω2 ]). Here, Axiom 1 ( §2.7) says that (b) the probability that a measured value w is obtained by ML∞(Ω)(O, S[δω2 ]) is given by F ({b})(ω2) = 0.4 Therefore, we see: statement (a) (ordinary language) −−−−−−→ translation statement (b) (quantum language) (2.70) 52 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement ♠Note 2.5. [L∞(Ω, ν), or in short, L∞(Ω)] In the above example, the counting measure ν (i.e., ν({ω1}) = ν({ω2}) = 1) is not absolutely indispensable. For example, even if we assume that ν({ω1}) = 2 and ν({ω2}) = 1/3, we can assert the same conclusion. Thus, in this note, L∞(Ω, ν) is often abbreviated to L∞(Ω). ♠Note 2.6. The statement (a) in Example 2.34 is not necessarily guaranteed, that is, When one ball is picked up from the urn U2, the probability that the ball is white is 0.4. is not guaranteed. What we say is that the statement (a) in ordinary language should be written by the measurement theoretical statement (b) It is a matter of course that "probability" can not be derived from mathematics itself. For example, the following (]1) and (]2) are not guaranteed. (]1) From the set {1, 2, 3, 4, 5}, choose one number. Then, the probability that the number is even is given by 2/5 (]2) From the closed interval [0, 1], choose one number x. Then, the probability that x ∈ [a, b] ⊆ [0, 1] is given by |b− a| The common sense - "probability" can not be derived from mathematics itself - is well known as Bertrand's paradox (cf. §9.11). Thus, it is usual to add the term "at random" to the above (]1) and (]2). In this note, this term "at random" is usually omitted. Example 2.35. [Blood type system] The ABO blood group system is the most important blood type system (or blood group system) in human blood transfusion. Let U1 be the whole Japanese's set and let U2 be the whole Indian's set. Also, assume that the distribution of the ABO blood group system [O:A:B:AB] concerning Japanese and Indians is determined in (Table 2.3). Table 2.3: The ratio of the ABO blood group system J or IABO blood group O A B AB Japanese U1 30% 40% 20% 10% Indian U2 30% 20% 40% 10% Consider the following phenomenon: 53 Ishikawa's Homepage 2.8 Examples: Classical measurements (urn problem, etc.) (a) Choose one person from the the whole Indian's set U2 at random. Then the probability that the person's blood type is  O A B AB  is given by  0.3 0.2 0.4 0.1  In what follows, we shall translate the statement (a) described in ordinary language to quantum language. Put Ω = {ω1, ω2} and consider the discrete metric (Ω, dD). We get consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Therefore, the pure state space is defined by Sp(C0(Ω) ∗) = {δω1 , δω2} Here, consider δω1 * * * "the state of the whole Japanese's set U1(i.e., population)"4 δω2 * * * "the state of the whole India's set U1(i.e., population)", That is, we consider the following identification: (Therefore, image Figure 2.9): U1 ≈ δω1 , U2 ≈ δω2 U1≈δω1 U2≈δω2 Japanese [3:4:2:1] Indian [3:2:4:1] Figure 2.9: Population(=system)≈urn Define the blood type observable OBT = ({O,A,B,AB}, 2{O,A,B,AB}, FBT) in L∞(Ω, ν) such that [FBT({O})](ω1) = 0.3, [FBT({A})](ω1) = 0.4 4 Note that "population" = "system" (cf. Table 2.1 ). 54 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement [FBT({B})](ω1) = 0.2, [FBT({AB})](ω1) = 0.1 (2.71) and, [FBT({O})](ω2) = 0.3, [FBT({A})](ω2) = 0.2 [FBT({B})](ω2) = 0.4, [FBT({AB})](ω2) = 0.1 (2.72) Thus we get the measurement ML∞(Ω,ν)(OBT, S[δω2 ]). Hence, the above (a) is translated to the following statement (in terms of quantum language): (b) The probability that a measured value  O A B AB  is obtained by the measurement ML∞(Ω,ν)(OBT, S[δω2 ]) is given by C0(Ω)∗ ( δω2 , FBT({O}) ) L∞(Ω,ν) = [FBT({O})](ω2) = 0.3 C0(Ω)∗ ( δω2 , FBT({A}) ) L∞(Ω,ν) = [FBT({A})](ω2) = 0.2 C0(Ω)∗ ( δω2 , FBT({B}) ) L∞(Ω,ν) = [FBT({B})](ω2) = 0.4 C0(Ω)∗ ( δω2 , FBT({AB}) ) L∞(Ω,ν) = [FBT({AB})](ω2) = 0.1  ♠Note 2.7. Readers may feel that Example 2.34–Example 2.35 are too easy. However, as mentioned in (a) of Sec. 2.8.1, what we can do is •  to be faithful to Axioms to trust in Man's linguistic competence If some find the other language that is more powerful than quantum language, it will be praised as the greatest discovery in the history of science. That is because this discovery is regarded as beyond the discovery of quantum mechanics. 55 Ishikawa's Homepage 2.9 Simple quantum measurement (Stern=Gerlach experiment ) 2.9 Simple quantum measurement (Stern=Gerlach experiment ) 2.9.1 Stern=Gerlach experiment Example 2.36. [Quantum measurement( Schtern–Gerlach experiment (1922))] Assume that we examine the beam (of silver particles(or simply, electrons) after passing through the magnetic field. Then, as seen in the following figure, we see that all particles are deflected either equally upwards or equally downwards in a 50:50 ratio. See Figure 2.10. S N electron e state ω = [ α1 α2 ] [↑] U© [↓] D© Screen Figure 2.10: Stern–Gerlach experiment (1922) Consider the two dimensional Hilbert space H = C2, And therefore, we get the noncommutative basic algebra B(H), that is, the algebra composed of all 2 × 2 matrices. Thus, we have the quantum basic structure: [C(H) ⊆ B(H) ⊆ B(H)] = [B(C2) ⊆ B(C2) ⊆ B(C2)] since the dimension of H is finite. The spin state of an electron P is represented by ρ(= |ω〉〈ω|), where ω ∈ C2 such that ‖ω‖ = 1. Put ω = [ α1 α2 ] ( where, ||ω||2 = |α1|2 + |α2|2 = 1 ). Define Oz ≡ (Z, 2Z , Fz), the spin observable concerning the z-axis, such that, Z = {↑, ↓} and Fz({↑}) = [ 1 0 0 0 ] , Fz({↓}) = [ 0 0 0 1 ] , (2.73) Fz(∅) = [ 0 0 0 0 ] , Fz({↑, ↓}) = [ 1 0 0 1 ] . 56 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement Here, Born's quantum measurement theory (the probabilistic interpretation of quantum mechanics) says that (]) When a quantum measurementMB(C2)(O, S[ρ]) is taken, the probability that a measured value [ ↑ ↓ ] is obtained is given by  〈ω, F z({↑})ω〉 = |α1|2 〈ω, F z({↓})ω〉 = |α2|2  That is, putting ω (= [ α1 α2 ] ), we says that When the electron with a spin state state ρ progresses in a magnetic field, the probability that the Geiger counter [ U© D© ] sounds is give by  [ α1 α2 ] [1 0 0 0 ] [ α1 α2 ] = |α1|2 [ α1 α2 ] [0 0 0 1 ] [ α1 α2 ] = |α2|2  Also, we can define Ox ≡ (X, 2X , F x), the spin observable concerning the x-axis, such that, X = {↑x, ↓x} and F x({↑x}) = [ 1/2 1/2 1/2 1/2 ] , F x({↓x}) = [ 1/2 −1/2 −1/2 1/2 ] . (2.74) And furthermore, we can define Oy ≡ (Y, 2Y , F y), the spin observable concerning the y-axis, such that, Y = {↑y, ↓y} and F y({↑y}) = [ 1/2 i/2 −i/2 1/2 ] , F y({↓y}) = [ 1/2 −i/2 i/2 1/2 ] , (2.75) where i = √ −1. Here, putting Ŝx = Fx({↑})− Fx({↓}), Ŝy = Fy({↑})− Fy({↓}), Ŝz = Fz({↑})− Fz({↓}) we have the following commutation relation: ŜyŜz − ŜzŜy = 2iŜx, ŜzŜx − ŜxŜz = 2iŜy, ŜxŜy − ŜyŜx = 2iŜz (2.76) 57 Ishikawa's Homepage 2.10 de Broglie paradox in B(C2) 2.10 de Broglie paradox in B(C2) Axiom 1(measurement) includes the paradox ( that is, so called de Broglie paradox "there is something faster than light"). In what follows, we shall explain de Broglie paradox in B(C2), though the original idea is mentioned in B(L2(R)) (cf. §11.3, and refs.[13, 81]). Also, it should be noted that the argument below is essentially the same as the Stern=Gerlach experiment. Example 2.37. [de Broglie paradox in B(C2) ] Let H be a two dimensional Hilbert space, i.e., H = C2. Consider the quantum basic structure: [B(C2) ⊆ B(C2) ⊆ B(C2)] Now consider the situation in the following Figure 2.11. D2(= (|f2〉〈f2|)) (photon detector) D1(= (|f1〉〈f1|)) (photon detector) u= 1√ 2 (f1+f2) −−−−−−−−→ 1√ 2 f1 ? √ −1√ 2 f2 half mirror 1 course1 course2 photon P Figure 2.11: [D2 +D1] = observable O Let us explain this figure in what follows. Let f1, f2 ∈ H such that f1 = [ 1 0 ] ∈ C2, f2 = [ 0 1 ] ∈ C2 Put u = f1 + f2√ 2 Thus, we have the state ρ = |u〉〈u| (∈ Sp(B(C2))). Let U(∈ B(C2)) be an unitary operator such that U = [ 1 0 0 eiπ/2 ] 58 Ishikawa's Homepage Chap. 2 Axiom 1 - measurement and let Φ : B(C2)→ B(C2) be the homomorphism such that Φ(F ) = U∗FU (∀F ∈ B(C2)) Consider the observable Of = ({1, 2}, 2{1,2}, F ) in B(C2) such that F ({1}) = |f1〉〈f1|, F ({2}) = |f2〉〈f2| and thus, define the observable ΦOf = ({1, 2}, 2{1,2},ΦF ) by ΦF (Ξ) = U∗F (Ξ)U (∀Ξ ⊆ {1, 2}) Let us explain Figure 2.11. The photon P with the state u = 1√ 2 (f1 + f2) ( precisely, |u〉〈u| ) rushed into the half-mirror 1 (A1) the f1 part in u passes through the half-mirror 1, and goes along the course 1 to the photon detector D1. (A2) the f2 part in u rebounds on the half-mirror 1 (and strictly saying, the f2 changes to√ −1f2, we are not concerned with it ), and goes along the course 2 to the photon detector D2. Thus, we have the measurement: MB(C2)(ΦOf , S[ρ]) (2.77) And thus, we see: (B) The probability that a [ measured value 1 measured value 2 ] is obtained by the measurement MB(C2)(ΦOf , S[ρ]) is given by[ Tr(ρ * ΦF ({1})) Tr(ρ * ΦF ({2})) ] = [ 〈u,ΦF ({1})u〉 〈u,ΦF ({2})u〉 ] = [ 〈Uu, F ({1})Uu〉 〈Uu, F ({2})Uu〉 ] = [ |〈u, f1〉|2 |〈u, f2〉|2 ] = [ 1 2 1 2 ] This is easy, but it is deep in the following sense. (C) Assume that Detector D1 and Detector D2 are very far. And assume that the photon P is discovered at the detector D1. Then, we are troubled if the photon P is also discovered at the detector D2. Thus, in order to avoid this difficulty, the photon P (discovered at the detector D1) has to eliminate the wave function √ −1√ 2 f2 in an instant. In this sense, the (B) implies that there may be something faster than light 59 Ishikawa's Homepage 2.10 de Broglie paradox in B(C2) This is the de Broglie paradox (cf. [13, 81]). From the view point of quantum language, we give up to solve the paradox, that is, we declare that Stop to be bothered! (Also, see [70]). ♠Note 2.8. The de Broglie paradox (i.e., there may be something faster than light ) always appears in quantum mechanics. For example, the readers should confirm that it appears in Example 2.36 (Schtern-Gerlach experiment). I think that • the de Broglie paradox is the only paradox in quantum mechanics 60 Ishikawa's Homepage Chapter 3 The linguistic Copenhagen interpretation (dualism and idealism) Measurement theory (= quantum language ) is formulated as follows. • measurement theory (=quantum language) := [Axiom 1] Measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } manual to use spells Measurement theory says that • Describe every phenomenon modeled on Axioms 1 and 2 (by a hint of the linguistic interpretation)! Since we dealt with simple examples in the previous chapter, we did not need the linguistic interpretation. In this chapter, we study several more difficult problems with the linguistic interpretation. Also, the linguistic interpretation may be called "the linguistic Copenhagen interpretation" since we believe that it is the true colors of so called Copenhagen interpretation (cf. Section 1.1.1). 3.1 The linguistic Copenhagen interpretation 3.1.1 The review of Axiom 1 ( measurement: §2.7) In the previous chapter, we introduced Axiom 1 (measurement ) as follows. 61 3.1 The linguistic Copenhagen interpretation (A): Axiom 1(measurement) pure type (cf. It was able to read under the preparation to §2.7) ) With any system S, a basic structure [A ⊆ A]B(H) can be associated in which measurement theory of that system can be formulated. In [A ⊆ A]B(H), consider a W ∗-measurement MA ( O=(X,F, F ), S[ρ] ) ( or, C∗-measurementMA ( O=(X,F, F ), S[ρ] ) ) . That is, consider • a W ∗-measurement MA ( O, S[ρ] ) ( or, C∗-measurement MA ( O=(X,F, F ), S[ρ] ) ) of an observable O=(X,F, F ) for a state ρ(∈ Sp(A∗) : state space) Then, the probability that a measured value x (∈ X) obtained by the W ∗-measurement MA ( O, S[ρ] ) ( or, C∗-measurement MA ( O=(X,F, F ), S[ρ] ) ) belongs to Ξ (∈ F) is given by ρ(F (Ξ))(≡ A∗(ρ, F (Ξ))A) (if F (Ξ) is essentially continuous at ρ, or see Definition 2.14 ). Here, note that (B1) the above axiom is a kind of spell (i.e., incantation, magic words, metaphysical statement), and thus, it is impossible to verify them experimentally. In this sense, the above axiom corresponds to "a priori synthetic judgment" in Kant's philosophy (cf. [62]). And thus, we say: (B2) After we learn the spell (= Axiom 1) by rote, we have to exercise and lesson the spell (= Axiom 1). Since quantum language is a language, it may be unable to use well at first. It will make progress gradually, while applying a trial-and-error method. However, (C1) if we would like to make speed of acquisition of a quantum language as quick as possible, we may want the good manual to use the axioms. Here, we think that (C2) the linguistic interpretation = the manual to use the spells (Axiom 1 and 2) 3.1.2 Descartes figure (in the linguistic interpretation) In what follows, let us explain the linguistic interpretation. The concept of "measurement" can be, for the first time, understood in dualism. Let us explain it. The image of "measurement" is as shown in Figure 3.1. 62 Ishikawa's Homepage Chap. 3 The linguistic Copenhagen interpretation (dualism and idealism) • observer (I(=mind)) system (matter)  - [observable] [measured value] a©interfere b©perceive a reaction [state] Figure 3.1:[Descartes Figure]:The image of "measurement(= a©+ b©)" in dualism In the above, (D1) a©: it suffices to understand that "interfere" is, for example, "apply light". b©: perceive the reaction. That is, "measurement" is characterized as the interaction between "observer" and "measuring object". However, (D2) In measurement theory, "interaction" must not be emphasized. Therefore, in order to avoid confusion, it might better to omit the interaction " a© and b©" in Figure 3.1. After all, we think that: (D3) It is clear that there is no measured value without observer (i.e., brain). Thus, we consider that measurement theory is composed of three key-words: measured value (observer,brain, mind) , observable (= measuring instrument ) (thermometer, eye, ear, body, polar star (cf. Note 3.1 later)) , state (matter) , (3.1) and thus, it might be called "trialism" (and not "dualism"). But, according to the custom, it is called "dualism" in this note. 3.1.3 The linguistic interpretation [(E1)-(E7)] The linguistic interpretation is "the manual to use Axiom 1 and 2". Thus, there are various explanations for the linguistic interpretations. However, it is usual to consider that the linguistic interpretation is characterized as the following (E). And the most important is Only one measurement is permitted 63 Ishikawa's Homepage 3.1 The linguistic Copenhagen interpretation (E):The linguistic interpretation (=quantum language interpretation) With Descartes figure 3.1 (and (E1)-(E7)) in mind, describe every phenomenon in terms of Axioms 1 and 2 (E1) Consider the dualism composed of "observer" and "system( =measuring object)". And therefore, "observer" and "system" must be absolutely separated. If it says for a metaphor, we say "Audience should not be up to the stage". (E2) Of course, "matter(=measuring object)" has the space-time. On the other hand, the observer does not have the space-time. Thus, the question: "When and where is a measured value obtained?" is out of measurement theory, Thus, there is no tense in measurement theory. This implies that there is no tense in science. (E3) In measurement theory, "interaction" must not be emphasized. (E4) Only one measurement is permitted. Thus, the state after measurement (or, wave function collapse, the influence of measurement) is meaningless. (cf. Projection Postulate 11.6) (E5) There is no probability without measurement. (E6) State never moves, and so on. Also, since our assertion is quantum language is the final goal of dualistic idealism (="Descartes=Kant philosophy") (cf. 8© in Figure 1.1), we have to assert that (E7) Many of maxims of the philosophers (particularly, the dualistic idealism ) can be regarded as a part of the linguistic interpretation. Some may think that the (E7) is unbelievable. However, (F) Since the purpose of philosophies and that of quantum language are the same, that is, the non-realistic world view, it is natural to consider that 64 Ishikawa's Homepage Chap. 3 The linguistic Copenhagen interpretation (dualism and idealism) maxims of philosophers ≈ the linguistic interpretation Recall the following figure: Figure 3.1. [=Figure 1.1:The location of quantum language in the history of world-description] Parmenides Socrates 0©:Greek philosophy Plato Aristotle Schola-−−−−→ sticism 1© −−→ (monism) Newton (realism) 2© → relativity theory −−−−−−→ 3© → quantum mechanics −−−−−−→ 4© −→ (dualism) Descartes Locke,... Kant (idealism) 6©−→ (linguistic view) linguistic philosophy language−−−−−→ 8© language−−−−−−→ 7©  5©−→ (unsolved) theory of everything (quantum phys.)  10©−→ (=MT) quantum language (language) Figure 1.1: The history of the world-view statistics system theory language−−−−−→ 9© the linguistic view the realistic view In the above, we regard [ 0© −→ 1© −→ 6© −→ 8© −→ 10©] (3.2) as a genealogy of the dualistic idealism. Talking cynically, we say that • Philosophers continued investigating "linguistic interpretation" (="how to use Axioms 1 and 2") without Axioms 1 and 2. For example, "Only one measurement is permitted" and "State never moves" may be related to Parmenides' words; There are no "plurality", but only "one". And therefore, there is no movement. (3.3) 65 Ishikawa's Homepage 3.1 The linguistic Copenhagen interpretation Table 3.1: Trialism (i.e., dualism ) in world-views (cf. Table 2.1) Quantum language measured value observable state (system) Plato / idea (cf. Note 3.1) / Aristotle / / edios (hyle) Thomas Aquinas universale post rem universale ante rem / (universale in re) Descartes I, mind, brain body (cf. Note 3.1) / (matter) Locke / secondary quality primary quality (/) Newton / / state (point mass) statistics sample space / parameter (population) quantum mechanics measured value observable state (particle) Thus, we want to assert that Parmenides (born around BC. 515) is the oldest discoverer of the linguistic interpretation. Also, we propose the following table: ♠Note 3.1. In the above table, Newtonian mechanics may be the most understandable. We regard "Plato idea" as "absolute standard". And, we want to understand that Newton is similar to Aristotle, since their assertions belong to the realistic world view(cf. Figure 1.1). Also, recall the formula (3.1), that is, "observable"="measuring instrument"="body". Thus, as the examples of "observable", we think: eyes, ears, glasses, telescope, compass, etc. If "compass" is accepted, "the polar star" should be also accepted as the example of the observable. In the same sense, "the jet stream to an airplane" is a kind of observable (cf. Section 8.1 (pp.129-135) in [39] ). Also, if it is certain that Descartes is the first discoverer of "I", I have to retract my understanding of Scholasticism in Table 3.1. Although I have no confidence about Scholasticism, the discover of three words ("post rem", "ante rem", "in re") should be remarkable. 66 Ishikawa's Homepage Chap. 3 The linguistic Copenhagen interpretation (dualism and idealism) 3.2 Tensor operator algebra 3.2.1 Tensor product of Hilbert space The linguistic interpretation (§3.1) says "Only one measurement is permitted" which implies "only one measuring object" or "only one state". Thus, if there are several states, these should be regarded as "only one state". In order to do it, we have to prepare "tensor operator algebra". That is, (A) "several states" combine several into one−−−−−−−−−−−−−−→ by tensor operator algebra "one state" In what follows, we shall introduce the tensor operator algebra. Let H,K be Hilbert spaces. We shall define the tensor Hilbert space H ⊗ K as follows. Let {em | m ∈ N ≡ {1, 2, . . .}} be the CONS (i.e, complete orthonormal system ) in H. And, let {fn | n ∈ N ≡ {1, 2, . . .}} be the CONS in K. For each (m,n) ∈ N2, consider the symbol "em ⊗ fn". Here, consider the following "space": H ⊗K = { g = ∑ (m,n)∈N2 αm,nem ⊗ fn ∣∣∣ ||g||H⊗K ≡ [ ∑ (m,n)∈N2 |αm,m|2]1/2 <∞ } (3.4) Also, the inner product 〈*, *〉H⊗K is represented by 〈em1 ⊗ fn1 , em2 ⊗ fn2〉H⊗K ≡ 〈em1 , em2〉H * 〈fn1 , fn2〉K = { 1 (m1, n1) = (m2, n2) 0 (m1, n1) 6= (m2, n2) (3.5) Thus, summing up, we say (B) the tensor Hilbert space H ⊗K is defined by the Hilbert space with the CONS {em ⊗ fn | (m,n) ∈ N2}. For example, for any e = ∑∞ m=1 αmem ∈ H and any f = ∑∞ n=1 βnfm ∈ H, the tensor e ⊗ f is defined by e⊗ f = ∑ (m,n)∈N2 αmβn(em ⊗ fn) Also, the tensor norm ||û||H⊗K (û ∈ H ⊗K) is defined by ||û||H⊗K = |〈û, û〉H⊗K |1/2 67 Ishikawa's Homepage 3.2 Tensor operator algebra Example 3.2. [Simple example:tensor Hilbert space C2⊗C3] Consider the 2-dimensional Hilbert space H = C2 and the 3-dimensional Hilbert space K = C3. Now we shall define the tensor Hilbert space H ⊗K = C2 ⊗ C3 as follows. Consider the CONS {e1, e2} in H such as e1 = [ 1 0 ] , e2 = [ 0 1 ] And, consider the CONS {f1.f2, f3} in K such as f1 = 10 0  , f2 = 01 0  , f2 = 00 1  Therefore, the tensor Hilbert space H ⊗K = C2 ⊗ C3 has the CONS such as e1 ⊗ f1 = [ 1 0 ] ⊗ 10 0  , e1 ⊗ f2 = [10 ] ⊗ 01 0  , e1 ⊗ f3 = [10 ] ⊗ 00 1  , e2 ⊗ f1 = [ 0 1 ] ⊗ 10 0  , e2 ⊗ f2 = [01 ] ⊗ 01 0  , e2 ⊗ f3 = [01 ] ⊗ 00 1  Thus, we see that H ⊗K = C2 ⊗ C3 = C6 That is because the CONS {ei ⊗ fj | i = 1, 2, 3, j = 1, 2} in H ⊗ K can be regarded as {gk | k = 1, 2, ..., 6} such that g1 = e1 ⊗ f1 =  1 0 0 0 0 0  , g2 = e1 ⊗ f2 =  0 1 0 0 0 0  , g3 = e1 ⊗ f3 =  0 0 1 0 0 0  , g4 = e2 ⊗ f1 =  0 0 0 1 0 0  , g5 = e2 ⊗ f2 =  0 0 0 0 1 0  , g6 = e2 ⊗ f3 =  0 0 0 0 0 1  This Example 3.2 can be easily generalized as follows. Theorem 3.3. [Finite tensor Hilbert space ] Cm1 ⊗ Cm2 ⊗ * * * ⊗ ⊗Cmn = C ∑n k=1mk (3.6) 68 Ishikawa's Homepage Chap. 3 The linguistic Copenhagen interpretation (dualism and idealism) Theorem 3.4. [Concrete tensor Hilbert space ] L2(Ω1, ν1)⊗ L2(Ω2, ν2) = L2(Ω1 × Ω2, ν1 ⊗ ν2) (3.7) where, ν1 ⊗ ν2 is the product measure. Definition 3.5. [Infinite tensor Hilbert space ] Let H1, H2, ..., Hk, ... be Hilbert spaces. Then, the infinite tensor Hilbert space ⊗∞ k=1Hk can be defined as follows. For each k(∈ N), consider the CONS {ejk}∞j=1 in a Hilbert space Hk. For any map b : N→ N, define the symbol ⊗∞ k=1 e b(k) k such that ∞⊗ k=1 e b(k) k = e b(1) 1 ⊗ e b(2) 2 ⊗ e b(3) 3 ⊗ * * * Then, we have: { ∞⊗ k=1 e b(k) k ∣∣∣ b : N→ N is a map} (3.8) Hence we can define the infinite Hilbert space ⊗∞ k=1Hk such that it has the CONS (3.8). 3.2.2 Tensor basic structure For each continuous linear operators F ∈ B(H), G ∈ B(K), the tensor operator F ⊗ G ∈ B(H ⊗K) is defined by (F ⊗G)(e⊗ f) = Fe⊗Gf (∀e ∈ H, f ∈ K) Definition 3.6. [Tensor C∗-algebra and Tensor W ∗-algebra ] Consider basic structures [A1 ⊆ A1 ⊆ B(H1)] and [A2 ⊆ A2 ⊆ B(H2)] [I]: The tensor C∗-algebra A1 ⊗A2 is defined by the smallest C∗-algebra Â such that {F ⊗G (∈ B(H1 ⊗H2)) | F ∈ A1, G ∈ A2} ⊆ Â ⊆ B(H1 ⊗H2) [II]: The tensor W ∗-algebra A1 ⊗A2 is defined by the smallest W ∗-algebra Ã such that {F ⊗G (∈ B(H1 ⊗H2)) | F ∈ A1, G ∈ A2} ⊆ Ã ⊆ B(H1 ⊗H2) Here, note that A1 ⊗A2 = A1 ⊗A2. 69 Ishikawa's Homepage 3.2 Tensor operator algebra Theorem 3.7. [Tensor basic structure ] [I]: Consider basic structures [A1 ⊆ A1 ⊆ B(H1)] and [A2 ⊆ A2 ⊆ B(H2)] Then, we have the tensor basic structure: [A1 ⊗A2 ⊆ A1 ⊗A2 ⊆ B(H1 ⊗H2)] [II]: Consider quantum basic structures [C(H1) ⊆ B(H1) ⊆ B(H1)] and [C(H2) ⊆ B(H2) ⊆ B(H2)]. Then, we have tensor quantum basic structure: [C(H1) ⊆ B(H1) ⊆ B(H1)]⊗ [C(H2) ⊆ B(H2) ⊆ B(H2)] =[C(H1 ⊗H2) ⊆ B(H1 ⊗H2) ⊆ B(H1 ⊗H2)] [III]: Consider classical basic structures [C0(Ω1) ⊆ L∞(Ω1, ν1) ⊆ B(L2(Ω1, ν1))] and [C0(Ω2) ⊆ L∞(Ω2, ν2) ⊆ B(L2(Ω2 ν2))]. Then, we have tensor classical basic structure: [C0(Ω1) ⊆ L∞(Ω1 ⊆ ν1) ⊆ B(L2(Ω1, ν1))]⊗ [C0(Ω2) ⊆ L∞(Ω2 ⊆ ν2) ⊆ B(L2(Ω2, ν2))] =[C0(Ω1 × Ω2) ⊆ L∞(Ω1 × Ω2, ν1 ⊗ ν2) ⊆ B(L2(Ω1 × Ω2, ν1 ⊗ ν2))] Theorem 3.8. The ⊗∞ k=1B(Hk) (⊆ B( ⊗∞ k=1Hk)) is defined by the smallest C ∗-algebra that contains F1 ⊗ F2 ⊗ * * * ⊗ Fn ⊗ I ⊗ I ⊗ * * * ( ∈ B( ∞⊗ k=1 Hk) ) (∀Fk ∈ B(Hk), k = 1, 2, ..., n, n = 1, 2, ...) Then, it holds that ∞⊗ k=1 B(Hk) = B( ∞⊗ k=1 Hk) (3.9) Theorem 3.9. The followings hold: (i) : ρk ∈ A∗k =⇒ n⊗ k=1 ρk ∈ ( n⊗ k=1 Ak) ∗ (ii) : ρk ∈ Sm(A∗k) =⇒ n⊗ k=1 ρk ∈ Sm(( n⊗ k=1 Ak) ∗) (iii) : ρk ∈ Sp(A∗k) =⇒ n⊗ k=1 ρk ∈ Sp(( n⊗ k=1 Ak) ∗) ♠Note 3.2. The theory of operator algebra is a deep mathematical theory. However, in this note, we do not use more than the above preparation. 70 Ishikawa's Homepage Chap. 3 The linguistic Copenhagen interpretation (dualism and idealism) 3.3 The linguistic Copenhagen interpretation - Only one measurement is permitted In this section, we examine the linguistic interpretation (§3.1), i.e., "Only one measurement is permitted". "Only one measurement" implies that "only one observable" and "only one state". That is, we see: [only one measurement] =⇒  only one observable (=measuring instrument) only one state (3.10) ♠Note 3.3. Although there may be several opinions, I believe that the standard Copenhagen interpretation also says "only one measurement is permitted". Thus, some think that this spirit is inherited to quantum language. However, our assertion is reverse, namely, the Copenhagen interpretation is due to the linguistics interpretation. That is, we assert that not " Copenhagen interpretation =⇒ Linguistic interpretation " but " Linguistic interpretation =⇒ Copenhagen interpretation " 3.3.1 "Observable is only one" and simultaneous measurement Recall the measurement Example 2.31 (Cold or hot?) and Example 2.32 (Approximate temperature), and consider the following situation: (a) There is a cup in which water is filled. Assume that the temperature is ω ◦C (0 5 ω 5 100). Consider two questions: "Is this water cold or hot?" "How many degrees( ◦C) is roughly the water?" This implies that we take two measurements such that (]1): ML∞(Ω)(Och=({c, h}, 2{c,h}, Fch), S[ω]) in Example2.31 (]2) : ML∞(Ω) (O 4 =(N10010 , 2N 100 10 , G4), S[ω]) in Example2.32 ML∞(Ω)(Och, S[ω]) ML∞(Ω) (O 4, S[ω]) ω ◦C 71 Ishikawa's Homepage 3.3 The linguistic Copenhagen interpretation - Only one measurement is permitted However, as mentioned in the linguistic interpretation, "only one measurement" =⇒"only one observable" Thus, we have the following problem. Problem 3.10. Represent two measurements ML∞(Ω)(Och=({c, h}, 2{c,h}, Fch), S[ω]) and ML∞(Ω)(O 4=(N10010 , 2N 100 10 , G4), S[ω]) by only one measurement. This will be answered in what follows. Definition 3.11. [Product measurable space] For each k = 1, 2, . . . , n, consider a measurable (Xk, Fk). The product space×nk=1Xk of Xk (k = 1, 2, . . . , n) is defined by n × k=1 Xk = {(x1, x2, . . . , xn) | xk ∈ Xk (k = 1, 2, . . . , n)} Similarly, define the product×nk=1 Ξk of Ξk(∈ Fk) (k = 1, 2, . . . , n) by n × k=1 Ξk = {(x1, x2, . . . , xn) | xk ∈ Ξk (k = 1, 2, . . . , n)} Further, the σ-field  nk=1Fk on the product space×nk=1Xk is defined by (])  nk=1Fk is the smallest field including {×nk=1 Ξk | Ξk ∈ Fk (k = 1, 2, . . . , n)} (×nk=1Xk,  nk=1Fk) is called the product measurable space. Also, in the case that (X,F) = (Xk,Fk) (k = 1, 2, . . . , n), the product space ×nk=1Xk is denoted by Xn, and the product measurable space (×nk=1Xk,  nk=1Fk) is denoted by (Xn,Fn). Definition 3.12. [Simultaneous observable , simultaneous measurement] Consider the basic structure [A ⊆ A ⊆ B(H)]. Let ρ ∈ Sp(A∗). For each k = 1, 2, . . . , n, consider a measurement MA (Ok = (Xk,Fk, Fk), S[ρ]) in A. Let (×nk=1Xk,  nk=1Fk) be the product measurable space. An observable Ô = (×k∈K Xk,  nk=1Fk, F ) in A is called the simultaneous observable of {Ok : k = 1, 2, ..., n}, if it satisfies the following condition: F (Ξ1 × Ξ2 × * * * × Ξn) = F1(Ξ1) * F2(Ξ2) * * *Fn(Ξn) (3.11) ( ∀Ξk ∈ Fk (k = 1, 2, . . . , n)) Ô is also denoted by ×nk=1Ok, F = ×nk=1 Fk. Also, the measurement MA(×nk=1Ok, S[ρ]) is called the simultaneous measurement. Here, it should be noted that • the existence of the simultaneous observable×nk=1Ok is not always guaranteed. though it always exists in the case that A is commutative (this is, A = L∞(Ω)). 72 Ishikawa's Homepage Chap. 3 The linguistic Copenhagen interpretation (dualism and idealism) In what follows, we shall explain the meaning of "simultaneous observable". Let us explain the simultaneous measurement. We want to take two measurements MA(O1, S[ρ]) and measurement MA(O2, S[ρ]). That is, it suffices to image the following: (b) state ρ(∈Sp(A∗)) −−−−−→ −→ observable O1=(X1,F1,F1) −−−−−−−→ M A (O1,S[ρ]) measured value x1(∈X1) −→ observable O2=(X2,F2,F2) −−−−−−−→ M A (O2,S[ρ]) measured value x2(∈X2) However, according to the linguistic interpretation (§3.1), two measurements MA(O1, S[ρ]) and MA(O2, S[ρ]) can not be taken. That is, The (b) is impossible Therefore, combining two observables O1 and O2, we construct the simultaneous observable O1 × O2, and take the simultaneous measurement MA(O1 × O2, S[ρ]) in what follows. (c) state ρ(∈Sp(A∗)) −−−−−−−→ simultaneous observable O1×O2 −−−−−−−−−→ M A (O1×O2,S[ρ]) measured value (x1,x2)(∈X1×X2) The (c) is possible if O1 × O2 exists Answer 3.13. [The answer to Problem3.10] Consider the state space Ω such that Ω = [0, 100], the closed interval. And consider two observables, that is, [C-H]-observable Och = (X={c, h}, 2X , Fch) (in Example2.31) and triangle observable O4 = (Y (=N10010 ), 2Y , G4) (in Example2.32). Thus, we get the simultaneous observable Och×O4 = ({c, h}×N10010 , 2{c,h}×N 100 10 , Fch× G4), and we can take the simultaneous measurement ML∞(Ω)(Och × O4, S[ω]). For example, putting ω = 55, we see (d) when the simultaneous measurement ML∞(Ω)(Och × O4, S[55]) is taken, the probability that the measured value  (c, about 50 ◦C) (c, about 60 ◦C) (h, about 50 ◦C) (h, about 60 ◦C)  is obtained is given by  0.125 0.125 0.375 0.375  (3.12) That is because [(Fch ×G4)({(c, about 50 ◦C)})](55) 73 Ishikawa's Homepage 3.3 The linguistic Copenhagen interpretation - Only one measurement is permitted =[Fch({c})](55) * [G4({about 50 ◦C})](55) = 0.25 * 0.5 = 0.125 and similarly, [(Fch ×G4)({(c, about 60 ◦C)})](55) = 0.25 * 0.5 = 0.125 [(Fch ×G4)({(h, about 50 ◦C)})](55) = 0.75 * 0.5 = 0.375 [(Fch ×G4)({(h, about 60 ◦C)})](55) = 0.75 * 0.5 = 0.375 ♠Note 3.4. The above argument is not always possible. In quantum mechanics, a simultaneous observable O1 × O2 does not always exist (See the following Example 3.14 and Heisenberg's uncertainty principle in Sec.4.4). Example 3.14. [The non-existence of the simultaneous spin observables] Assume that the electron P has the (spin) state ρ = |u〉〈u| ∈ Sp(B(C2)), where u = [ α1 α2 ] (where, |u| = (|α1|2 + |α2|2)1/2 = 1) Let Oz = (X(= {↑, ↓}), 2X , F z) be the spin observable concerning the z-axis such that F z({↑}) = [ 1 0 0 0 ] , F z({↓}) = [ 0 0 0 1 ] Thus, we have the measurement MB(C2)(Oz = (X, 2 X , F z), S[ρ]). Let Ox = (X, 2 X , F x) be the spin observable concerning the x-axis such that F x({↑}) = [ 1/2 1/2 1/2 1/2 ] , F x({↓}) = [ 1/2 −1/2 −1/2 1/2 ] Thus, we have the measurement MB(C2)(Ox = (X, 2 X , F x), S[ρ]) Then we have the following problem: (a) Two measurements MB(C2)(Oz = (X, 2 X , F z), S[ρ]) and MB(C2)(Ox = (X, 2 X , F x), S[ρ]) are taken simultaneously? This is impossible. That is because the two observable Oz and Ox do not commute. For example, we see F z({↑})F x({↑}) = [ 1 0 0 0 ] * [ 1/2 1/2 1/2 1/2 ] = [ 1/2 1/2 0 0 ] F x({↑})F z({↑}) = [ 1/2 1/2 1/2 1/2 ] * [ 1 0 0 0 ] = [ 1/2 0 1/2 0 ] And thus, F x({↑})F z({↑}) 6= F z({↑})F x({↑}) /// 74 Ishikawa's Homepage Chap. 3 The linguistic Copenhagen interpretation (dualism and idealism) The following theorem is clear. For completeness, we add the proof to it. Theorem 3.15. [Exact measurement and system quantity] Consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Let O (exa) 0 = (X,F, F (exa)) (i.e., (X,F, F (exa)) = (Ω,BΩ, χ) ) be the exact observable in L∞(Ω, ν). Let O1 = (R,BR, G) be the observable that is induced by a quantity g : Ω→ R as in Example 2.26(system quantity). Consider the simultaneous observable O (exa) 0 ×O1. Let (x, y) (∈ X×R) be a measured value obtained by the simultaneous measurement ML∞(Ω,ν)(O(exa)0 ×O1, S[δω ]). Then, we can surely believe that x = ω, and y = g(ω). Proof. Let D0(∈ BΩ) be arbitrary open set such that ω(∈ D0 ⊆ Ω=X). Also, let D1(∈ BR) be arbitrary open set such that g(ω) ∈ D1. The probability that a measured value (x, y) obtained by the measurement ML∞(Ω,ν)(O (exa) 0 ×O1, S[δω ]) belongs to D0×D1 is given by χD0 (ω)* χ g−1(D1) (ω) = 1. Since D0 and D1 are arbitrary, we can surely believe that x = ω and y = g(ω). 3.3.2 "State does not move" and quasi-product observable We consider that "only one measurement" =⇒"state does not move" That is because (a) In order to see the state movement, we have to take measurement at least more than twice. However, the "plural measurement" is prohibited. Thus, we conclude "state does not move" Review 3.16. [= Example 2.34:urn problem] There are two urns U1 and U2. The urn U1 [resp. U2] contains 8 white and 2 black balls [resp. 4 white and 6 black balls] (cf. Figure 3.2). Table 3.2: urn problem Urn w*b white ball black ball Urn U1 8 2 Urn U2 4 6 Here, consider the following statement (a): (a) When one ball is picked up from the urn U2, the probability that the ball is white is 0.4. 75 Ishikawa's Homepage 3.3 The linguistic Copenhagen interpretation - Only one measurement is permitted ω1(≈ U1) ω2(≈ U2) Figure 3.2: Urn problem In measurement theory, the statement (a) is formulated as follows: Assuming U1 * * * "the urn with the state ω1" U2 * * * "the urn with the state ω2" define the state space Ω by Ω = {ω1, ω2} with discrete metric and counting measure ν. That is, we assume the identification; U1 ≈ ω1, U2 ≈ ω2, Thus, consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Put "w" = "white", "b" = "black", and put X = {w, b}. And define the observable Owb ( ≡ (X ≡ {w, b}, 2{w,b}, Fwb) ) in L∞(Ω) by [Fwb({w})](ω1) = 0.8, [Fwb({b})](ω1) = 0.2, [Fwb({w})](ω2) = 0.4, [Fwb({b})](ω2) = 0.6. (3.13) Thus, we get the measurement ML∞(Ω)(Owb, S[δω2 ]). Here, Axiom 1 ( §2.7) says that (b) the probability that a measured value w is obtained by ML∞(Ω)(Owb, S[δω2 ]) is given by Fwb({b})(ω2) = 0.4 Thus, the above statement (b) can be rewritten in the terms of quantum language as follows. (c) the probability that a measured value [ w b ] is obtained by the measurement ML∞(Ω)(Owb, S[ω2]) is given by[ ∫ Ω [Fwb({w})](ω)δω2(dω) = [Fwb({w})](ω2) = 0.4∫ Ω [Fwb({b})](ω)δω2(dω) = [Fwb({b})](ω2) = 0.6 ] Problem 3.17. (a) [Sampling with replacement]: Pick out one ball from the urn U2, and recognize the color ("white" or "black") of the ball. And the ball is returned to the 76 Ishikawa's Homepage Chap. 3 The linguistic Copenhagen interpretation (dualism and idealism) urn. And again, Pick out one ball from the urn U2, and recognize the color of the ball. Therefore, we have four possibilities such that. (w,w) (w, b) (b, w) (b, b) It is a common sense that the probability that  (w,w) (w, b) (b, w) (b, b)  is given by  0.16 0.24 0.24 0.36  Now, we have the following problem: (a) How do we describe the above fact in term of quantum language? Answer Is suffices to consider the simultaneous measurement ML∞(Ω)(O 2 wb, S[δω2 ]) (= ML∞(Ω)(Owb×Owb, S[δω2 ]) ), where O2wb = ({w, b} × {w, b}, 2{w,b}×{w,b}, F 2wb(= Fwb × Fwb)). The, we calculate as follows. F 2wb({(w,w)})(ω1) = 0.64, F 2wb({(w, b)})(ω1) = 0.16 F 2wb({(b, w)})(ω1) = 0.16, F 2wb({(b, b)})(ω1) = 0.4 and F 2wb({(w,w)})(ω2) = 0.16, F 2wb({(w, b)})(ω2) = 0.24 F 2wb({(b, w)})(ω2) = 0.24, F 2wb({(b, b)})(ω2) = 0.36 Thus, we conclude that (b) the probability that a measured value  (w,w) (w, b) (b, w) (b, b)  is obtained by ML∞(Ω)(Owb×Owb, S[δω2 ]) is given by  [Fwb({w})](ω2) * [Fwb({w})](ω2) = 0.16 [Fwb({w})](ω2) * [Fwb({b})](ω2) = 0.24 [Fwb({b})](ω2) * [Fwb({w})](ω2) = 0.24 [Fwb({b})](ω2) * [Fwb({b})](ω2) = 0.36  Problem 3.18. (a) [Sampling without replacement]: Pick out one ball from the urn U2, and recognize the color ("white" or "black") of the ball. And the ball is not returned to the urn. And again, Pick out one ball from the urn U2, and recognize the color of the ball. Therefore, we have four possibilities such that. (w,w) (w, b) (b, w) (b, b) 77 Ishikawa's Homepage 3.3 The linguistic Copenhagen interpretation - Only one measurement is permitted It is a common sense that the probability that  (w,w) (w, b) (b, w) (b, b)  is given by  12/90 24/90 24/90 30/90  Now, we have the following problem: (a) How do we describe the above fact in term of quantum language? Now, recall the simultaneous observable (Definition3.12) as follows. Let Ok = (Xk, Fk, Fk) (k = 1, 2, . . . , n ) be observables in A. The simultaneous observable Ô = (×nk=1Xk,  nk=1Fk, F ) is defined by F (Ξ1 × Ξ2 × * * * × Ξn) = F1(Ξ1)F2(Ξ2) * * *Fn(Ξn) (∀Ξk ∈ Fk, ∀k = 1, 2, . . . , n) The following definition ("quasi-product observable") is a kind of simultaneous observable: Definition 3.19. [quasi-product observable ] Let Ok = (Xk, Fk, Fk) (k = 1, 2, . . . , n ) be observables in a W ∗-algebra A. Assume that an observable O12...n = (×nk=1Xk,  nk=1Fk, F12...n) satisfies F12...n(X1 × * * * ×Xk−1 × Ξk ×Xk+1 × * * * ×Xn) = Fk(Ξk) (3.14) (∀Ξk ∈ Fk, ∀k = 1, 2, . . . , n) The observable O12...n = (×nk=1Xk,  nk=1Fk, F12...n) is called a quasi-product observable of {Ok | k = 1, 2, . . . , n}, and denoted by qp × k=1,2,...,n Ok = ( n × k=1 Xk,  nk=1Fk, qp × k=1,2,...,n Fk) Of course, a simultaneous observable is a kind of quasi-product observable. Therefore, quasiproduct observable is not uniquely determined. Also, in quantum systems, the existence of the quasi-product observable is not always guaranteed. Answer 3.20. [The answer to Problem 3.17] Define the quasi-product observable Owb qp × Owb = ({w, b} × {w, b}, 2{w,b}×{w,b}, F12(= Fwb qp × Fwb)) of Owb = ({w, b}, 2{w,b}, F ) in L∞(Ω) such that F12({(w,w)})(ω1) = 8× 7 90 , F12({(w, b)})(ω1) = 8× 2 90 F12({(b, w)})(ω1) = 2× 8 90 , F12({(b, b)})(ω1) = 2× 1 90 F12({(w,w)})(ω2) = 4× 3 90 , F12({(w, b)})(ω2) = 4× 6 90 78 Ishikawa's Homepage Chap. 3 The linguistic Copenhagen interpretation (dualism and idealism) F12({(b, w)})(ω2) = 6× 4 90 , F12({(b, b)})(ω2) = 6× 5 90 Thus, we have the (quasi-product) measurement ML∞(Ω)(O12, S[ω]) Therefore, in terms of quantum language, we describe as follows. (b) the probability that a measured value  (w,w) (w, b) (b, w) (b, b)  is obtained dy ML∞(Ω)(Owb qp× Owb, S[δω2 ]) is given by  [F12({(w,w)})](ω2) = 4×390 [F12({(w, b)})](ω2) = 4×690 [F12({(b, w)})](ω2) = 4×690 [F12({(b, b)})](ω2) = 6×590  3.3.3 Only one state and parallel measurement For example, consider the following situation: (a) There are two cups A1 and A2 in which water is filled. Assume that the temperature of the water in the cup Ak (k = 1, 2) is ωk ◦C (0 5 ωk 5 100). Consider two questions "Is the water in the cup A1 cold or hot?" and "How many degrees( ◦C) is roughly the water in the cup A2?". This implies that we take two measurements such that (]1): ML∞(Ω)(Och=({c, h}, 2{c,h}, Fch), S[ω1]) in Example2.31 (]2) : ML∞(Ω) (O 4 =(N10010 , 2N 100 10 , G4), S[ω2]) in Example2.32 ML∞(Ω)(Och, S[ω1]) ω1 ◦C A1 ML∞(Ω) (O 4, S[ω2]) ω2 ◦C A2 However, as mentioned in the above, "only one state" must be demanded. Thus, we have the following problem. 79 Ishikawa's Homepage 3.3 The linguistic Copenhagen interpretation - Only one measurement is permitted Problem 3.21. Represent two measurements ML∞(Ω)(Och=({c, h}, 2{c,h}, Fch), S[ω1]) and ML∞(Ω)(O 4 =(N10010 , 2N 100 10 , G4), S[ω2]) by only one measurement. This will be answered in what follows. Definition 3.22. [Parallel observable] For each k = 1, 2, . . . , n, consider a basic structure [Ak ⊆ Ak ⊆ B(Hk)], and an observable Ok = (Xk,Fk, Fk) in Ak. Define the observable Õ = (×nk=1Xk,  nk=1Fk, F ) in ⊗n k=1Ak such that F (Ξ1 × Ξ2 × * * * × Ξn) = F1(Ξ1)⊗ F2(Ξ2)⊗ * * * ⊗ Fn(Ξn) (3.15) ∀Ξk ∈ Fk (k = 1, 2, . . . , n) Then, the observable Õ = (×nk=1Xk,  nk=1Fk, F ) is called the parallel observable in ⊗n k=1Ak, and denoted by F = ⊗n k=1 Fk, Õ = ⊗n k=1Ok. the measurement of the parallel observable Õ =⊗n k=1Ok, that is, the measurement M ⊗n k=1 Ak (Õ, S[⊗nk=1 ρk]) is called a parallel measurement, and denoted by M⊗n k=1 Ak ( ⊗n k=1Ok, S[ ⊗n k=1 ρk] ) or ⊗n k=1MAk(Ok, S[ρk]). The meaning of the parallel measurement is as follows. Our present purpose is • to take both measurements MA1(O1, S[ρ1]) and MA2(O2, S[ρ2]) Then. image the following: (b)  state ρ1(∈Sp(A∗1)) −−−−−−−→ observable O1 −−−−−−−−→ M A1 (O1,S[ρ1]) measured value x1(∈X1) state ρ2(∈Sp(A∗2)) −−−−−−−→ observable O2 −−−−−−−−→ M A2 (O2,S[ρ2]) measured value x2(∈X2) However, according to the linguistic interpretation (§3.1), two measurements can not be taken. Hence, The (b) is impossible Thus, two states ρ1 and ρ1 are regarded as one state ρ1⊗ρ2, and further, combining two observables O1 and O2, we construct the parallel observable O1 ⊗ O2, and take the parallel measurement MA1⊗A2(O1 ⊗ O2, S[ρ1⊗ρ2]) in what follows. (c) state ρ1⊗ρ2(∈Sp(A∗1)⊗Sp(A∗2)) −→ parallel observable O1⊗O2 −−−−−−−−−−−−−−−→ M A1⊗A2 (O1⊗O2,S[ρ1⊗ρ2]) measured value (x1,x2)(∈X1×X2) 80 Ishikawa's Homepage Chap. 3 The linguistic Copenhagen interpretation (dualism and idealism) The (c) is always possible Example 3.23. [The answer to Problem 3.21 ] Put Ω1 = Ω2 = [0, 100], and define the state space Ω1 × Ω2. And consider two observables, that is, the [C-H]-observable Och = (X={c, h}, 2X , Fch) in C(Ω1) (in Example2.31) and triangle-observable O4 = (Y (=N10010 ), 2Y , G4) in L∞(Ω2) (in Example2.32). Thus, we get the parallel observable Och ⊗ O4 = ({c, h} × N10010 , 2{c,h}×N 100 10 , Fch ⊗ G4) in L∞(Ω1 × Ω2), take the parallel measurement ML∞(Ω1×Ω2)(Och ⊗ O4, S[(ω1,ω2)]). Here, note that δω1 ⊗ δω2 = δ(ω1,ω2) ≈ (ω1, ω2). For example, putting (ω1, ω2) = (25, 55), we see the following. (d) When the parallel measurement ML∞(Ω1×Ω2)(Och ⊗O4, S[(25,55)]) is taken, the probability that the measured value  (c, about 50 ◦C) (c, about 60 ◦C) (h, about 50 ◦C) (h, about 60 ◦C)  is obtained is given by  0.375 0.375 0.125 0.125  That is because [(Fch ⊗G4)({(c, about 50 ◦C)})](25, 55) =[Fch({c})](25) * [G4({about 50 ◦C})](55) = 0.75 * 0.5 = 0.375 Thus, similarly, [(Fch ⊗G4)({(c, about 60 ◦C)})](25, 55) = 0.75 * 0.5 = 0.375 [(Fch ⊗G4)({(h, about 50 ◦C)})](25, 55) = 0.25 * 0.5 = 0.125 [(Fch ⊗G4)({(h, about 60 ◦C)})](25, 55) = 0.25 * 0.5 = 0.125 Remark 3.24. Also, for example, putting (ω1, ω2) = (55, 55), we see: (e) the probability that a measured value  (c, about 50 ◦C) (c, about 60 ◦C) (h, about 50 ◦C) (h, about 60 ◦C)  is obtained by parallel measurement ML∞(Ω1×Ω2)(Och ⊗ O4, S[(55,55)]) is given by  0.125 0.125 0.375 0.375  81 Ishikawa's Homepage 3.3 The linguistic Copenhagen interpretation - Only one measurement is permitted That is because, we similarly, see [Fch({c})](55) * [G4({about 50 ◦C})](55) = 0.25 * 0.5 = 0.125 [Fch({c})](55) * [G4({about 60 ◦C})](55) = 0.25 * 0.5 = 0.125 [Fch({h})](55) * [G4({about 50 ◦C})](55) = 0.75 * 0.5 = 0.375 [Fch({h})](55) * [G4({about 60 ◦C})](55) = 0.75 * 0.5 = 0.375 (3.16) Note that this is the same as Answer 3.13 (cf. Note 3.5 later). The following theorem is clear. But, the assertion is significant. Theorem 3.25. [Ergodic property] For each k = 1, 2, * * * , n, consider a measurement ML∞(Ω)(Ok(:= (Xk,Fk, Fk)), S[δω ]) with the sample probability space (Xk,Fk, P ω k ). Then, the sample probability spaces of the simultaneous measurement ML∞(Ω)(×nk=1Ok, S[δω ]) and the parallel measurement ML∞(Ωn) ( ⊗n k=1Ok, S[⊗nk=1δω ]) are the same, that is, these are the same as the product probability space ( n × k=1 Xk,  nk=1Fk, n⊗ k=1 P ωk ) (3.17) Proof. It is clear, and thus we omit the proof. ( Also, see Note 3.5 later.) Example 3.26. [The parallel measurement is always meaningful in both classical and quantum systems ] The electron P1 has the (spin) state ρ1 = |u1〉〈u1| ∈ Sp(B(C2)) such that u1 = [ α1 β1 ] (where, ‖u1‖ = (|α1|2 + |β1|2)1/2 = 1) Let Oz = (X(= {↑, ↓}), 2X , F z) be the spin observable concerning the z-axis such that F z({↑}) = [ 1 0 0 0 ] , F z({↓}) = [ 0 0 0 1 ] Thus, we have the measurement MB(C2)(Oz = (X, 2 X , F z), S[ρ1]). The electron P2 has the (spin) state ρ2 = |u2〉〈u2| ∈ Sp(B(C2)) such that u = [ α2 β2 ] (where, ‖u2‖ = (|α2|2 + |β2|2)1/2 = 1) Let Ox = (X, 2 X , F x) be the spin observable concerning the x-axis such that F x({↑}) = [ 1/2 1/2 1/2 1/2 ] , F x({↓}) = [ 1/2 −1/2 −1/2 1/2 ] Thus, we have the measurement MB(C2)(Ox = (X, 2 X , F x), S[ρ2]) Then we have the following problem: 82 Ishikawa's Homepage Chap. 3 The linguistic Copenhagen interpretation (dualism and idealism) (a) Two measurements MB(C2)(Oz = (X, 2 X , F z), S[ρ1]) and MB(C2)(Ox = (X, 2 X , F x), S[ρ2]) are taken simultaneously? This is possible. It can be realized by the parallel measurement MB(C2)⊗B(C2)(Oz ⊗ Oz = (X ×X, 2X×X , F z ⊗ F x), S[ρ⊗ρ]) That is, (b) The probability that a measured value  (↑, ↑) (↑, ↓) (↓, ↑) (↓, ↓)  is obtained by the parallel measurement MB(C2)⊗B(C2)(Oz ⊗ Oz, S[ρ⊗ρ]) is given by 〈u, F z({↑})u〉〈u, F x({↑})u〉 = p1p2 〈u, F z({↑})u〉〈u, F x({↓})u〉 = p1(1− p2) 〈u, F z({↓})u〉〈u, F x({↑})u〉 = (1− p1)p2 〈u, F z({↓})u〉〈u, F x({↓})u〉 = (1− p1)(1− p2)  where p1 = |α1|2, p2 = 12(|α1| 2 + α1α2 + α1α2 + |α2|2) ♠Note 3.5. Theorem 3.25 is rather deep in the following sense. For example, "To toss a coin 10 times" is a simultaneous measurement. On the other hand, "To toss 10 coins once" is characterized as a parallel measurement. The two have the same sample space. That is, "spatial average" = "time average" which is called the ergodic property. This means that the two are not distinguished by the sample space and not the measurements (i.e., a simultaneous measurement and a parallel measurement). However, this is peculiar to classical pure measurements. It does not hold in classical mixed measurements and quantum measurement. 83 Ishikawa's Homepage

Chapter 4 Linguistic Copenhagen interpretation of quantum systems Measurement theory (= quantum language ) is formulated as follows. • measurement theory (=quantum language) := [Axiom 1] Measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } manual to use spells Measurement theory says that • Describe every phenomenon modeled on Axioms 1 and 2 (by a hint of the linguistic interpretation)! In this chapter, we devote ourselves to the linguistic interpretation (§3.1) for general (or, quantum) systems. 4.1 Kolmogorov's extension theorem and the linguistic interpretation Kolmogorov's probability theory (cf. [63] ) starts from the following spell: (]) Let (X,F, P ) be a probability space. Then, the probability that a event Ξ (∈ F) happens is given by P (Ξ) And, through trial and error, Kolmogorov found his extension theorem, which says that (]) "Only one probability space is permitted" which surely corresponds to (]) "Only one measurement is permitted" in the linguistic interpretation (§3.1) 85 4.1 Kolmogorov's extension theorem and the linguistic interpretation Therefore, we want to say that (]) Parmenides (born around BC. 515) and Kolmogorov (1903-1987) said about the same thing (cf. Parmenides' words (3.3)). Let Λ be a set (called an index set). For each λ ∈ Λ, consider a set Xλ. For any subsets Λ1 ⊆ Λ2( ⊆ Λ), πΛ1,Λ2 is the natural map such that: πΛ1,Λ2 : × λ∈Λ2 Xλ −→ × λ∈Λ1 Xλ. (4.1) Especially, put πΛ = πΛ,Λ. Consider the basic structure [A ⊆ A ⊆ B(H)] For each λ ∈ Λ, consider an observable (Xλ,Fλ, Fλ) in A. Note that the quasi-product observable O ≡ (×λ∈ΛXλ, ×λ∈ΛFλ, FΛ) of { (Xλ,Fλ, Fλ) | λ ∈ Λ } is characterized as the observable such that: FΛ(π −1 {λ}(Ξλ)) = Fλ(Ξλ) (∀Ξλ ∈ Fλ, ∀λ ∈ Λ), (4.2) though the existence and the uniqueness of a quasi-product observable are not guaranteed in general. The following theorem says something about the existence and uniqueness of the quasi-product observable. Let Λ be a set. For each λ ∈ Λ, consider a set Xλ. For any subset Λ1 ⊆ Λ2( ⊆ Λ), define the natural map πΛ1,Λ2 :×λ∈Λ2 Xλ −→×λ∈Λ1 Xλ by × λ∈Λ2 Xλ 3 (xλ)λ∈Λ2 7→ (xλ)λ∈Λ1 ∈ × λ∈Λ1 Xλ (4.3) The following theorem guarantees the existence and uniqueness of the observable. It should be noted that this is due to the the linguistic interpretation (§3.1), i.e., "only one measurement is permitted". Theorem 4.1. [ Kolmogorov extension theorem in measurement theory ( cf. [28, 30] ) ] Consider the basic structure [A ⊆ A ⊆ B(H)] For each λ ∈ Λ, consider a Borel measurable space (Xλ,Fλ), where Xλ is a separable complete metric space. Define the set P0(Λ) such as P0(Λ) ≡ {Λ ⊆ Λ | Λ is finite }. Assume that the family of the observables { OΛ ≡ (×λ∈ΛXλ,×λ∈Λ Fλ, FΛ ) | Λ ∈ P0(Λ) } in A satisfies the following "consistency condition": 86 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems • for any Λ1, Λ2 ∈ P0(Λ) such that Λ1 ⊆ Λ2, FΛ2 ( π−1Λ1,Λ2(ΞΛ1) ) = FΛ1 ( ΞΛ1 ) (∀ΞΛ1 ∈ × λ∈Λ1 Fλ). (4.4) Then, there uniquely exists the observable ÕΛ ≡ (×λ∈ΛXλ,×λ∈Λ Fλ, FΛ) in A such that: FΛ ( π−1Λ (ΞΛ) ) = FΛ ( ΞΛ ) (∀ΞΛ ∈ × λ∈Λ Fλ, ∀Λ ∈ P0(Λ)). Proof. For the proof, see refs.[28, 30]. Corollary 4.2. [Infinite simultaneous observable ] Consider the basic structure [A ⊆ A ⊆ B(H)] Let Λ be a set. For each λ ∈ Λ, assume that Xλ is a separable complete metric space, Fλ is its Borel field. For each λ ∈ Λ, consider an observable Oλ = (Xλ,Fλ, Fλ) in A such that it satisfies the commutativity condition, that is, Fk1(Ξk1)Fk2(Ξk2) = Fk2(Ξk2)Fk1(Ξk1) (∀Ξk1 ∈ Fk1 , ∀Ξk2 ∈ Fk2 , k1 6= k2) (4.5) Then, a simultaneous observable Ô = (×λ∈ΛXλ,  λ∈ΛFλ, F=×λ∈Λ Fλ) uniquely exists. That is, for any finite set Λ0(⊆ Λ), it holds that F ( (× λ∈Λ0 Ξλ)× ( × λ∈Λ\Λ0 Xλ) ) = × λ∈Λ0 Fλ(Ξλ) (∀Ξλ ∈ Fλ, ∀λ ∈ Λ0) Proof. The proof is a direct consequence of Theorem 4.1. Thus, it is omitted. Remark 4.3. Now we can answer the following question: (B) Why is Kolmogorov's extension theory fundamental in probability theory ? That is, I can assert the following chain: (Linguistic interpretation) Only one measurement is permitted −→ (Kolmogorov's extension theorem 4.1 in quantum language ) The existence of measurement −→ (Kolmogorov's extension theorem) The existence of sample space /// 87 Ishikawa's Homepage 4.2 The law of large numbers in quantum language 4.2 The law of large numbers in quantum language 4.2.1 The sample space of infinite parallel measurement ⊗∞ k=1MA(O = (X,F, F ), S[ρ]) Consider the basic structure [A ⊆ A ⊆ B(H)]( that is, [C(H) ⊆ B(H) ⊆ B(H)], or [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] ) and measurement MA(O = (X,F, F ), S[ρ]), which has the sample probability space (X,F, Pρ) Note that the existence of the infinite parallel observable Õ (= ⊗∞ k=1O) = (X N, ∞k=1F, F (= ⊗∞ k=1 F )) in an infinite tensor W ∗-algebra ⊗∞ k=1A is assured by Kolmogorov's extension theorem (Corollary4.2). For completeness, let us calculate the sample probability space of the parallel measurement M⊗∞ k=1 A (Õ, S[⊗∞k=1 ρ]) in both cases (i.e., quantum case and classical case): Preparation 4.4. [I]: quantum system: The quantum infinite tensor basic structure is defined by [C(⊗∞k=1H) ⊆ B(⊗∞k=1H) ⊆ B(⊗∞k=1H)] Therefore, infinite tensor state space is characterized by Sp(Tr(⊗∞k=1H)) ⊂ Sm(Tr(⊗∞k=1H)) = S m (Tr(⊗∞k=1H)) (4.6) Since Definition 2.17 says that F = Fρ (∀ρ ∈ Sp(Tr(H))), the sample probability space (XN, ∞k=1F, P⊗∞k=1 ρ) of the infinite parallel measurement M⊗∞k=1B(H)(⊗∞k=1O = (XN, ∞k=1F, ⊗k = 1∞F ), S[⊗∞k=1 ρ]) is characterized by P⊗∞ k=1 ρ (Ξ1 × Ξ2 × * * * × Ξn × ( ∞ × k=n+1 X)) = n × k=1 Tr(H) ( ρ, F (Ξk) ) B(H) (4.7) ( ∀Ξk ∈ F = Fρ, ( k = 1, 2, . . . , n), n = 1, 2, 3 * * * ) which is equal to the infinite product probability measure ⊗n k=1 Pρ. [II]: classical system: Without loss of generality, we assume that the state space Ω is compact, and ν(Ω) = 1 (cf. Note 2.1). Then, the classical infinite tensor basic structure is defined by [C0(×∞k=1Ω) ⊆ L∞(×∞k=1Ω,⊗∞k=1ν) ⊆ B(L2(×∞k=1Ω,⊗∞k=1ν))] (4.8) Therefore, the infinite tensor state space is characterized by Sp(C0(×∞k=1Ω)∗) ( ≈ ∞ × k=1 Ω ) (4.9) 88 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems Put ρ = δω. the sample probability space (X N, ∞k=1F, P⊗∞k=1 ρ) of the infinite parallel measurement ML∞(×∞k=1Ω,⊗∞k=1ν)(⊗ ∞ k=1O = (X N, ∞k=1F,⊗k = 1∞F ), S[⊗∞k=1 ρ]) is characterized by P⊗∞ k=1 ρ (Ξ1 × Ξ2 × * * * × Ξn × ( ∞ × k=n+1 X)) = n × k=1 [F (Ξk)](ω) (4.10) ( ∀Ξk ∈ F = Fρ, ( k = 1, 2, . . . , n), n = 1, 2, 3 * * * ) which is equal to the infinite product probability measure ⊗n k=1 Pρ. [III]: Conclusion: Therefore, we can conclude (]) in both cases, the sample probability space (XN, ∞k=1F, P⊗∞k=1 ρ) is defined by the infinite product probability space (XN, ∞k=1F, ⊗∞ k=1 Pρ) Summing up, we have the following theorem ( the law of large numbers ). Theorem 4.5. [The law of large numbers ] Consider the measurement MA(O = (X,F, F ), S[ρ]) with the sample probability space (X,F, Pρ). Then, by Kolmogorov's extension theorem (Corollary4.2), we have the infinite parallel measurement: M⊗∞ k=1 A (⊗∞k=1O = (XN, ∞k=1F,⊗∞k=1F ), S[⊗∞k=1 ρ]) The sample probability space (XN, ∞k=1F, P⊗∞k=1 ρ) is characterized by the infinite probability space (XN, ∞k=1F, ⊗∞ k=1 Pρ). Further, we see (A) for any f ∈ L1(X,Pρ), put Df = { (x1, x2, . . .) ∈ XN | lim n→∞ f(x1) + f(x2) + * * *+ f(xn) n = E(f) } (4.11) ( where, E(f) = ∫ X f(x)Pρ(dx) ) Then, it holds that P⊗∞ k=1 ρ (Df ) = 1 (4.12) That is, we see, almost surely,∫ X f(x)Pρ(dx) (population mean) = limn→∞ f(x1)+f(x2)+***+f(xn) n (sample mean) (4.13) Remark 4.6. [Frequency probability ] In the above, consider the case that f(x) = χ Ξ (x) = { 1 (x ∈ Ξ) 0 (x /∈ Ξ) (Ξ ∈ F) 89 Ishikawa's Homepage 4.2 The law of large numbers in quantum language Then, put Dχ Ξ = { (x1, x2, . . .) ∈ XN | lim n→∞ ][{k | xk ∈ Ξ, 1 ≤ k ≤ n} n = Pρ(Ξ) } (4.14) (where, ][A] is the number of the elements of the set A) Then, it holds that P⊗∞ k=1 ρ (Dχ Ξ ) = 1 (4.15) Therefore, the law of large numbers (Theorem 4.5) says that (]1) the probability in Axiom 1 ( §2.7) can be regarded as "frequency probability" Thus, we have the following opinion: (]2)  G. Galileo * * * the originator of the realistic world view J. Bernoulli * * * the originator of the linguistic world view 4.2.2 Mean, variance, unbiased variance Consider the measurement MA(O = (R,BR, F ), S[ρ]). Let (R,BR, Pρ) be its sample probability space. That is, consider the case that a measured value space X = R. Here, define: population mean(μρO) : E[MA(O = (R,BRF ), S[ρ])] = ∫ R xPρ(dx)(= μ) (4.16) population variance((σρO) 2) : V [MA(O = (R,BRF ), S[ρ])] = ∫ R (x− μ)2Pρ(dx) (4.17) Assume that a measured value (x1, x2, x3, ..., xn)(∈ Rn) is obtained by the parallel measurement ⊗nk=1MA(O, S[ρ]). Put sample distribution(νn) : νn = δx1 + δx2 + * * *+ δxn n ∈M+1(X) sample mean(μn) : E[⊗nk=1MA(O, S[ρ])] = x1 + x2 + * * *+ xn n (= μ) = ∫ R xνn(dx) sample variance(s2n) : V [⊗nk=1MA(O, S[ρ])] = (x1 − μ)2 + (x2 − μ)2 + *+ (x2 − μ)2 n = ∫ R (x− μ)2νn(dx) unbiased variance(u2n) : U [⊗nk=1MA(O, S[ρ])] = (x1 − μ)2 + (x2 − μ)2 + *+ (x2 − μ)2 n− 1 = n n− 1 ∫ R (x− μ)2νn(dx) Under the above preparation, we have: 90 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems Theorem 4.7. [Population mean, population variance, sample mean, sample variance] Assume that a measured value (x1, x2, x3, * * * )(∈ RN) is obtained by the infinite parallel measurement⊗∞ k=1MA(O = (R,BR, F ), S[ρ]). Then, the law of large numbers (Theorem4.5) says that (4.16) = population mean(μρO) = limn→∞ x1 + x2 + * * *+ xn n =: μ = sample mean (4.17) = population variance(σρO) = limn→∞ (x1 − μρO)2 + (x2 − μ ρ O) 2 + * * *+ (xn − μρO)2 n = lim n→∞ (x1 − μ)2 + (x2 − μ)2 + * * *+ (xn − μ)2 n =: sample variance Example 4.8. [Spectrum decomposition] Consider the quantum basic structure [C(H) ⊆ B(H) ⊆ B(H)] Let A be a self-adjoint operator on H, which has the spectrum decomposition (i.e., projective observable) OA = (R,BR, FA) such that A = ∫ R λFA(dλ) That is, under the identification: self-adjoint operator: A ←→ identification spectrum decomposition:OA = (R,BR, FA) the self-adjoint operator A is regarded as the projective observable OA = (R,BR, FA). Fix the state ρu = |u〉〈u| ∈ Sp(Tr(H)). Consider the measurement MB(H)(OA, S[|u〉〈u|]). Then, we see population mean(μρuOA) : E[MB(H)(OA, S[|u〉〈u|])] = ∫ R λ〈u, FA(dλ)u〉 = 〈u,Au〉 (4.18) population variance((σρuOA) 2) : V [MB(H)(OA, S[|u〉〈u|])] = ∫ R (λ− 〈u,Au〉)2〈u, FA(dλ)u〉 = ‖(A− 〈u,Au〉)u‖2 (4.19) 4.2.3 Robertson's uncertainty principle Now we can introduce Robertson's uncertainty principle as follows. Theorem 4.9. [Robertson's uncertainty principle (parallel measurement) (cf. [77]) ] Consider the quantum basic structure [C(H) ⊆ B(H) ⊆ B(H)]. Let A1 and A2 be unbounded selfadjoint operators on a Hilbert space H, which respectively has the spectrum decomposition: OA1 = (R,BR, FA1) to OA1 = (R,BR, FA1) 91 Ishikawa's Homepage 4.2 The law of large numbers in quantum language Thus, we have two measurements MB(H)(OA1 , S[ρu]) and MB(H)(OA2 , S[ρu]), where ρu = |u〉〈u| ∈ Sp(C(H)∗). To take two measurements means to take the parallel measurement: MB(Cn)(OA1 , S[ρu]) ⊗ MB(Cn)(OA2 , S[ρu]), namely, MB(H)⊗B(H)(OA1 ⊗ OA2 , S[ρu⊗ρu]) Then, the following inequality (i.e., Robertson's uncertainty principle ) holds that σρuA1 * σ ρu A2 = 1 2 |〈u, (A1A2 − A2A1)u〉| (∀|u〉〈u| = ρu, ‖u‖H = 1) where σρuA1 and σ ρu A2 are shown in (4.19), namely,{ σρuA1 = [〈A1u,A1u〉 − |〈u,A1u〉| 2] 1/2 = ‖(A1 − 〈u,A1u〉)u‖ σρuA2 = [〈A2u,A2u〉 − |〈u,A2u〉| 2] 1/2 = ‖(A2 − 〈u,A2u〉)u‖ Therefore, putting [A1, A2] ≡ A1A2 − A2A1, we rewrite Robertson's uncertainty principle as follows: ‖A1u‖ * ‖A2u‖ ≥ ‖(A1 − 〈u,A1u〉)u‖ * ‖(A2 − 〈u,A2u〉)u‖ ≥ |〈u, [A1, A2]u〉|/2 (4.20) For example, when A1(= Q) [resp. A2(= P ) ] is the position observable [resp. momentum observable ] (i.e., QP − PQ = ~ √ −1), it holds that σρuQ * σ ρu P = 1 2 ~ Proof. Robertson's uncertainty principle (4.20) is essentially the same as Schwarz inequality, that is, |〈u, [A1, A2]u〉| = |〈u, (A1A2 − A2A1)u〉| = ∣∣∣〈u,((A1 − 〈u,A1u〉)(A2 − 〈u,A2u〉)− (A2 − 〈u,A2u〉)(A1 − 〈u,A1u〉))u〉∣∣∣ ≤2‖(A1 − 〈u,A1u〉)u‖ * ‖(A2 − 〈u,A2u〉)u‖ 92 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems 4.3 Heisenberg's uncertainty principle 4.3.1 Why is Heisenberg's uncertainty principle famous? Heisenberg's uncertainty principle is as follows. Proposition 4.10. [Heisenberg's uncertainty principle (cf. [19]:1927) ] (i) The position x of a particle P can be measured exactly. Also similarly, the momentum p of a particle P can be measured exactly. However, the position x and momentum p of a particle P can not be measured simultaneously and exactly, namely, the both errors ∆x and ∆p can not be equal to 0. That is, the position x and momentum p of a particle P can be measured simultaneously and approximately, (ii) And, ∆x and ∆p satisfy Heisenberg's uncertainty principle as follows. ∆x *∆p + ~(= Plank constant/2π+1.5547× 10−34Js). (4.21) This was discovered by Heisenberg's thought experiment due to γ-ray microscope. It is (A) one of the most famous statements in the 20-th century. But, we think that it is doubtful in the following sense. ♠Note 4.1. I think, strictly speaking, that Heisenberg's uncertainty principle(Proposition 4.10) is meaningless. That is because, for example, (]) The approximate measurement and "error" in Proposition 4.10 are not defined. This will be improved in Theorem 4.15 in the framework of quantum mechanics. That is, Heisenberg's thought experiment is an excellent idea before the discovery of quantum mechanics. Some may ask that If it be so, why is Heisenberg's uncertainty principle (Proposition 4.10) famous? I think that Heisenberg's uncertainty principle (Proposition 4.10) was used as the slogan for advertisement of quantum mechanics in order to emphasize the difference between classical mechanics and quantum mechanics. And, this slogan was completely successful. This kind of slogan is not rare in the history of science. For example, recall "cogito proposition (due to Descartes)", that is, I think, therefore I am. which is also meaningless (cf. §8.4). However, it is certain that the cogito proposition built the foundation of modern science. 93 Ishikawa's Homepage 4.3 Heisenberg's uncertainty principle ♠Note 4.2. Heisenberg's uncertainty principle(Proposition 4.10) may include contradiction (cf. ref. [23]), if we think as follows (]) it is "natural" to consider that ∆x = |x− x|, ∆p = |p− p|, where{ Position: [x : exact measured value (=true value), x : measured value] Momentum: [p : exact measured value (=true value), p : measured value] However, this is in contradiction with Heisenberg's uncertainty principle (4.21). That is because (4.21) says that the exact measured value (x, p) can not be measured. As seen in Remark 4.23, note that the concept of "true vale" is nonsense. 4.3.2 The mathematical formulation of Heisenberg's uncertainty principle In this section, we shall propose the mathematical formulation of Heisenberg's uncertainty principle 4.10. Consider the quantum basic structure: [C(H) ⊆ B(H) ⊆ B(H)] Let Ai (i = 1, 2) be arbitrary self-adjoint operator on H. For example, it may satisfy that [A1, A2](:= A1A2 − A2A1) = ~ √ −1I Let OAi = (R,B, FAi) be the spectral representation of Ai, i.e., Ai = ∫ R λFAi(dλ), which is regarded as the projective observable in B(H). Let ρ0 = |u〉〈u| be a state, where u ∈ H and ‖u‖ = 1. Thus, we have two measurements: (B1) MB(H)(OA1 :=(R,B, FA1), S[ρu]) by (4.18)−−−−−−−−−→ expectation 〈u,A1u〉 (B2) MB(H)(OA2 :=(R,B, FA2), S[ρu]) by (4.18)−−−−−−−−−→ expectation 〈u,A2u〉 (∀ρu = |u〉〈u| ∈ Sp(C(H)∗)) However, since it is not always assumed that A1A2−A2A1 = 0, we can not expect the existence of the simultaneous observable OA1 × OA2 , namely, 94 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems • in general, two observables OA1 and OA2 can not be simultaneously measured That is, (B3) the measurement MB(H)(OA1 × OA2 , S[ρu]) is impossible, Thus, we have the question: Then, what should be done? In what follows, we shall answer this. Let K be another Hilbert space, and let s be in K such that ‖s‖ = 1. Thus, we also have two observables OA1⊗I :=(R,B, FA1⊗I) and OA2⊗I :=(R,B, FA2⊗I) in the tensor algebra B(H ⊗K). Put the tensor state ρus = |u⊗ s〉〈u⊗ s| And we have the following two measurements: (C1) MB(H⊗K)(OA1⊗I , S[ρus]) by (4.18)−−−−−−−−−→ expectation 〈u⊗ s, (A1 ⊗ I)(u⊗ s)〉 = 〈u,A1u〉 (C2) MB(H⊗K)(OA2⊗I , S[ρus]) by (4.18)−−−−−−−−−→ expectation 〈u⊗ s, (A2 ⊗ I)(u⊗ s)〉 = 〈u,A2u〉 It is a matter of course that (C1)=(B1) (C2)=(B2) and (C3) MB(H⊗K)(OA1⊗I × OA2⊗I , S[ρus]) is impossible. Thus, overcoming this difficulty, we prepare the following idea: Preparation 4.11. Let Âi (i = 1, 2) be arbitrary self-adjoint operator on the tensor Hilbert space H ⊗K, where it is assumed that [Â1, Â2](:= Â1Â2 − Â2Â1) = 0 (i.e., the commutativity) (4.22) Let OÂi = (R,B, FÂi) be the spectral representation of Âi, i.e.Âi = ∫ R λFÂi(dλ), which is regarded as the projective observable in B(H ⊗ K). Thus, we have two measurements as follows: 95 Ishikawa's Homepage 4.3 Heisenberg's uncertainty principle (D1) MB(H⊗K)(OÂ1 , S[ρus]) by (4.18)−−−−−−→ expectation 〈u⊗ s, Â1(u⊗ s)〉 (D2) MB(H⊗K)(OÂ2 , S[ρus]) by (4.18)−−−−−−→ expectation 〈u⊗ s, Â2(u⊗ s)〉 Note, by the commutative condition (4.22), that the two can be measured by the simultaneous measurement MB(H⊗K)(OÂ1 × OÂ2 , S[ρus]), where OÂ1 × OÂ2 = (R 2,B2, FÂ1 × FÂ2). Again note that any relation between Ai ⊗ I and Âi is not assumed. However, • we want to regard this simultaneous measurement as the substitute of the above two (C1) and (C2). That is, we want to regard (D1) and (D2) as the substitute of (C1) and (C2) For this, we have to prepare Hypothesis 4.9 below. Putting Ni := Âi − Ai ⊗ I (and thus, Âi = Ni + Ai ⊗ I) (4.23) we define the ∆ρus Ni and ∆ ρus Ni such that ∆u⊗s Ni =‖Ni(u⊗ s)‖ = ‖(Âi − Ai ⊗ I)(u⊗ s)‖ (4.24) ∆ u⊗s Ni =‖(Ni − 〈u⊗ s, Ni(u⊗ s)〉)(u⊗ s)‖ =‖((Âi − Ai ⊗ I)− 〈u⊗ s, (Âi − Ai ⊗ I)(u⊗ s)〉)(u⊗ s)‖ where the following inequality: ∆ρus Ni ≥ ∆ρusNi (4.25) is common sense. By the commutative condition (4.22), (4.23) implies that [N1, N2] + [N1, A2 ⊗ I] + [A1 ⊗ I, N2] = −[A1 ⊗ I, A2 ⊗ I] (4.26) Here, we should note that the first term (or, precisely, |〈u⊗ s, [the first term](u⊗ s)〉| ) of (4.26) can be, by the Robertson uncertainty relation (cf. Theorem4.9), estimated as follows: 2∆ ρus N1 *∆ρusN2 ≥ |〈u⊗ s, [N1, N2](u⊗ s)〉| (4.27) 96 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems 4.3.2.1 Average value coincidence conditions; approximately simultaneous measurement However, it should be noted that In the above, any relation between Ai ⊗ I and Âi is not assumed. Thus, we think that the following hypothesis is natural. Hypothesis 4.12. [Average value coincidence conditions ]. We assume that 〈u⊗ s, Ni(u⊗ s)〉 = 0 (∀u ∈ H, i = 1, 2) (4.28) or equivalently, 〈u⊗ s, Âi(u⊗ s)〉 = 〈u,Aiu〉 (∀u ∈ H, i = 1, 2) (4.29) That is, the average measured value of MB(H⊗K)(OÂi , S[ρus]) =〈u⊗ s, Âi(u⊗ s)〉 =〈u,Aiu〉 =the average measured value of MB(H)(OAi , S[ρu]) (∀u ∈ H, ||u||H = 1, i = 1, 2) Hence, we have the following definition. Definition 4.13. [Approximately simultaneous measurement] Let A1 and A2 be (unbounded) self-adjoint operators on a Hilbert space H. The quartet (K, s, Â1, Â2) is called an approximately simultaneous observable of A1 and A2, if it satisfied that (E1) K is a Hilbert space. s ∈ K, ‖s‖K = 1, Â1 and Â2 are commutative self-adjoint operators on a tensor Hilbert space H ⊗ K that satisfy the average value coincidence condition (4.28), that is, 〈u⊗ s, Âi(u⊗ s)〉 = 〈u,Aiu〉 (∀u ∈ H, i = 1, 2) (4.30) Also, the measurement MB(H⊗K)(OÂ1 × OÂ2 , S[ρus]) is called the approximately simultaneous measurement of MB(H)(OA1 , S[ρu]) and MB(H)(OA2 , S[ρu]). Thus, under the average coincidence condition, we regard (D1) and (D2) as the substitute of (C1) and (C2) 97 Ishikawa's Homepage 4.3 Heisenberg's uncertainty principle And (E2) ∆ ρus N1 (= ‖(Â1−A1⊗ I)(u⊗ s)‖) and ∆ρusN2 (= ‖(Â2−A2⊗ I)(u⊗ s)‖) are called errors of the approximate simultaneous measurement measurement MB(H⊗K)(OÂ1 × OÂ2 , S[ρus]) Lemma 4.14. Let A1 and A2 be (unbounded) self-adjoint operators on a Hilbert space H. And let (K, s, Â1, Â2) be an approximately simultaneous observable of A1 and A2. Then, it holds that ∆ρus Ni = ∆ ρus Ni (4.31) 〈u⊗ s, [N1, A2 ⊗ I](u⊗ s)〉 = 0 (∀u ∈ H) (4.32) 〈u⊗ s, [A1 ⊗ I, N2](u⊗ s)〉 = 0 (∀u ∈ H) (4.33) The proof is easy, thus, we omit it. Under the above preparations, we can easily get "Heisenberg's uncertainty principle" as follows. ∆ρus N1 *∆ρus N2 (= ∆ ρus N1 *∆ρusN2 ) ≥ 1 2 |〈u, [A1, A2]u〉| (∀u ∈ H such that ||u|| = 1) (4.34) Summing up, we have the following theorem: Theorem 4.15. [The mathematical formulation of Heisenberg's uncertainty principle] Let A1 and A2 be (unbounded) self-adjoint operators on a Hilbert space H. Then. we have the followings: (i) There exists an approximately simultaneous observable(K, s, Â1, Â2) of A1 and A2, that is, s ∈ K, ‖s‖K = 1, Â1 and Â2 are commutative self-adjoint operators on a tensor Hilbert space H⊗K that satisfy the average value coincidence condition (4.28). Therefore, the approximately simultaneous measurement MB(H⊗K)(OÂ1 × OÂ2 , S[ρus]) exists. (ii) And further, we have the following inequality (i.e., Heisenberg's uncertainty principle). ∆ρus N1 *∆ρus N2 (= ∆ ρus N1 *∆ρusN2 ) = ‖(Â1 − A1 ⊗ I)(u⊗ s)‖ * ‖(Â2 − A2 ⊗ I)(u⊗ s)‖ ≥ 1 2 |〈u, [A1, A2]u〉| (∀u ∈ H such that ||u|| = 1) (4.35) (iii) In addition, if A1A2 − A2A1 = ~ √ −1, we see that ∆ρus N1 *∆ρus N2 ≥ ~/2 (∀u ∈ H such that ||u|| = 1) (4.36) 98 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems Proof. For the proof of (i) and (ii), see • Ref. [23]: S. Ishikawa, Rep. Math. Phys. Vol.29(3), 1991, pp.257–273, As shown in the above (4.34), the proof (ii) is easy (cf. [30, 71]), but the proof (i) is not easy (cf. [7, 30]). 4.3.3 Without the average value coincidence condition Now we have the complete form of Heisenberg's uncertainty relation as Theorem 4.15, To be compared with Theorem 4.15, we should note that the conventional Heisenberg's uncertainty relation (= Proposition 4.10) is ambiguous. Wrong conclusions are sometimes derived from the ambiguous statement (= Proposition 4.10). For example, in some books of physics, it is concluded that EPR-experiment (Einstein, Podolosky and Rosen [14], or, see the following section) conflicts with Heisenberg's uncertainty relation. That is, [I ] Heisenberg's uncertainty relation says that the position and the momentum of a particle can not be measured simultaneously and exactly. On the other hand, [II ] EPR-experiment says that the position and the momentum of a certain "particle"can be measured simultaneously and exactly ( Also, see Note 4.3. ) Thus someone may conclude that the above [I] and [II] includes a paradox, and therefore, EPR-experiment is in contradiction with Heisenberg's uncertainty relation. Of course, this is a misunderstanding. This "paradox"was solved in [23, 30]. Now we shall explain the solution of the paradox. [Concerning the above [I]] Put H = L2(Rq). Consider two-particles system in H ⊗H = L2(R2(q1,q2)). In the EPR problem, we, for example, consider the state ue ( ∈ H ⊗ H = L2(R2(q1,q2))) ( or precisely, |ue〉〈ue| ) such that: ue(q1, q2) = √ 1 2πεσ e− 1 8σ2 (q1−q2−a)2− 1 8ε2 (q1+q2−b)2 * eiφ(q1,q2) (4.37) where ε is assumed to be a sufficiently small positive number and φ(q1, q2) is a real-valued function. Let A1 : L 2(R2(q1,q2))→ L 2(R2(q1,q2)) and A2 : L 2(R2(q1,q2))→ L 2(R2(q1,q2)) be (unbounded) self-adjoint operators such that A1 = q1, A2 = ~∂ i∂q1 . (4.38) 99 Ishikawa's Homepage 4.3 Heisenberg's uncertainty principle Then, Theorem 4.15 says that there exists an approximately simultaneous observable(K, s, Â1, Â2) of A1 and A2. And thus, the following Heisenberg's uncertainty relation (= Theorem 4.15) holds, ‖Â1ue − A1ue‖ * ‖Â2ue − A2ue‖ ≥ ~/2 (4.39) [Concerning the above [II]] However, it should be noted that, in the above situation we assume that the state ue is known before the measurement. In such a case, we may take another measurement as follows: Put K = C, s = 1. Thus, (H ⊗H) ⊗K = H ⊗H, u ⊗ s = u ⊗ 1 = u. Define the self-adjoint operators Â1 : L 2(R2(q1,q2)) → L 2(R2(q1,q2)) and Â2 : L 2(R2(q1,q2)) → L2(R2(q1,q2)) such that Â1 = b− q2, Â2 = A2 = ~∂ i∂q1 (4.40) Note that these operators commute. Therefore, (]) we can take an exact simultaneous measurement of Â1 and Â2 (for the state ue). And moreover, we can easily calculate as follows: ‖Â1ue − A1ue‖ = [ ∫∫ R2 ∣∣∣((b− q2)− q1)√ 1 2πεσ e− 1 8σ2 (q1−q2−a)2− 1 8ε2 (q1+q2−b)2 * eiφ(q1,q2) ∣∣∣2dq1dq2]1/2 = [ ∫∫ R2 ∣∣∣((b− q2)− q1)√ 1 2πεσ e− 1 8σ2 (q1−q2−a)2− 1 8ε2 (q1+q2−b)2 ∣∣∣2dq1dq2]1/2 = √ 2ε, (4.41) and ‖Â2ue − A2ue‖ = 0. (4.42) Thus we see ‖Â1ue − A1ue‖ * ‖Â2ue − A2ue‖ = 0. (4.43) However it should be again noted that, the measurement (]) is made from the knowledge of the state ue. [[I] and [II] are consistent ] The above conclusion (4.43) does not contradict Heisenberg's uncertainty relation (4.39), since the measurement (]) is not an approximate simultaneous measurement of A1 and A2. In other words, the (K, s, Â1, Â2) is not an approximately simultaneous observable of A1 and A2. Therefore, we can conclude that 100 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems (F) Heisenberg's uncertainty principle is violated without the average value coincidence condition (cf. Remark 3 in ref.[23], or p.316 in [30]). ♠Note 4.3. Some may consider that the formulas (4.41) and (4.42) imply that the statement [II] is true. However, it is not true. This is answered in Remark 8.15. Also, we add the following remark. Remark 4.16. Calculating the second term (precisely , 〈u⊗s,"the second term"(u⊗s)〉) and the third term (precisely , 〈u⊗ s,"the third term"(u⊗ s)〉) in (4.26), we get, by Robertson's uncertainty principle (4.20), 2∆ ρus N1 * σ(A2;u) ≥ |〈u⊗ s, [N1, A2 ⊗ I](u⊗ s)〉| (4.44) 2∆ ρus N2 * σ(A1;u) ≥ |〈u⊗ s, [A⊗I, N2](u⊗ s)〉| (4.45) (∀u ∈ H such that ||u|| = 1) and, from (4.26), (4.27), (4.44),(4.45), we can get the following inequality ∆ρus N1 *∆ρus N2 + ∆ρus N2 * σ(A1;u) + ∆ρusN1 * σ(A2;u) ≥∆ρusN1 *∆ ρus N2 + ∆ ρus N2 * σ(A1;u) + ∆ ρus N1 * σ(A2;u) ≥1 2 |〈u, [A1, A2]u〉| (∀u ∈ H such that ||u|| = 1) (4.46) Since we do not assume the average value coincidence condition, it is a matter of course that this (4.46) is more rough than Heisenberg's uncertainty principle (4.35) If a certain interpretation is adopted such that ∆ρus N1 and ∆ρus N2 mean "error:ε(A1, u)" and "disturbance:η(A2, u)", respectively, then the inequality (4.46), i.e., ε(A1, u)η(A2, u) + ε(A1, u)σ(A2, u) + σ(A1, u)η(A2, u) ≥ 1 2 |〈u, [A1, A2]u〉| is called Ozawa's inequality (cf. [72]). He asserted that this inequality is a faithful description of Heisenberg's thought experiment ( due to γ-ray microscope ). 101 Ishikawa's Homepage 4.4 EPR-paradox (1935) and faster-than-light 4.4 EPR-paradox (1935) and faster-than-light 4.4.1 EPR-paradox Next, let us explain EPR-paradox (Einstein–Poolside–Rosen: [14, 81]). Consider Two electrons P1 and P2 and their spins. The tensor Hilbert space H = C2 ⊗ C2 is defined in what follows. That is, e1 = [ 1 0 ] , e2 = [ 0 1 ] (i.e., the complete orthonormal system {e1, e2} in the C2), C2 ⊗ C2 = { ∑ i,j=1,2 αijei ⊗ ej | αij ∈ C, i, j = 1, 2} Put u = ∑ i,j=1,2 αijei ⊗ ej and v = ∑ i,j=1,2 βijei ⊗ ej. And the inner product 〈u, v〉C2⊗C2 is defined by 〈u, v〉 C2⊗C2 = ∑ i,j=1,2 αi,j * βi,j Therefore, we have the tensor Hilbert space H = C2 ⊗ C2 with the complete orthonormal system {e1 ⊗ e1, e1 ⊗ e2, e2 ⊗ e1, e2 ⊗ e2}. For each F ∈ B(C2) and G ∈ B(C2), define the F ⊗G ∈ B(C2 ⊗ C2) (i.e., linear operator F ⊗G : C2 ⊗ C2 → C2 ⊗ C2 ) such that (F ⊗G)(u⊗ v) = Fu⊗Gv Let us define the entangled state ρ = |s〉〈s| of two particles P1 and P2 such that s = 1√ 2 (e1 ⊗ e2 − e2 ⊗ e1) Here, we see that 〈s, s〉 C2⊗C2 = 1 2 〈e1 ⊗ e2 − e2 ⊗ e1, e1 ⊗ e2 − e2 ⊗ e1〉C2⊗C2 = 1 2 (1 + 1) = 1, and thus, ρ is a state. Also, assume that two particles P1 and P2 are far. Let O = (X, 2X , F z) in B(C2) (where X = {↑, ↓} ) be the spin observable concerning the z-axis such that F z({↑}) = [ 1 0 0 0 ] , F z({↓}) = [ 0 0 0 1 ] 102 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems The parallel observable O⊗ O = (X2, 2X × 2X , F z ⊗ F z) in B(C2 ⊗ C2) is defined by (F z ⊗ F z)({(↑, ↑)}) = F z({↑})⊗ F z({↑}) = [ 1 0 0 0 ] ⊗ [ 1 0 0 0 ] (F z ⊗ F z)({(↓, ↑)}) = F z({↓})⊗ F z({↑}) = [ 0 0 0 1 ] ⊗ [ 1 0 0 0 ] (F z ⊗ F z)({(↑, ↓)}) = F z({↑})⊗ F z({↓}) = [ 1 0 0 0 ] ⊗ [ 0 0 0 1 ] (F z ⊗ F z)({(↓, ↓)}) = F z({↓})⊗ F z({↓}) = [ 0 0 0 1 ] ⊗ [ 0 0 0 1 ] Thus, we get the measurement MB(C2⊗C2)(O⊗O, S[ρ]) The, Born's quantum measurement theory says that When the parallel measurementmeasurement MB(C2⊗C2)(O⊗ O, S[s]) is taken, the probability that the measured value  (↑, ↑) (↓, ↑) (↑, ↓) (↓, ↓)  is obtained is given by  〈s, (F z ⊗ F z)({(↑, ↑)})s〉 C2⊗C2 = 0 〈s, (F z ⊗ F z)({(↓, ↑)})s〉 C2⊗C2 = 0.5 〈s, (F z ⊗ F z)({(↑, ↓)})s〉 C2⊗C2 = 0.5 〈s, (F z ⊗ F z)({(↓, ↓)})s〉 C2⊗C2 = 0  That is because, F z({↑})e1 = e1, F z({↓})e2 = e2, F z({↑})e2 = F z({↓})e1 = 0 For example, 〈s, (F z ⊗ F z)({(↑, ↓)})s〉 C2⊗C2 = 1 2 〈(e1 ⊗ e2 − e2 ⊗ e1), (F z({↑})⊗ F z({↓})(e1 ⊗ e2 − e2 ⊗ e1)〉C2⊗C2 = 1 2 〈(e1 ⊗ e2 − e2 ⊗ e1), e1 ⊗ e2〉C2⊗C2 = 1 2 Here, it should be noted that we can assume that the x1 and the x2 (in (x1, x2) ∈ { (↑z, ↑z), (↑z, ↓z), (↓z, ↑z), (↓z, ↓z)}) are respectively obtained in Tokyo and in New York (or, in the earth and in the polar star). (b) (probability12 ) ↑z Tokyo ↓z New York or (c) (probability12 ) ↓z Tokyo ↑z New York This fact is, figuratively speaking, explained as follows: 103 Ishikawa's Homepage 4.4 EPR-paradox (1935) and faster-than-light • Immediately after the particle in Tokyo is measured and the measured value ↑z [resp. ↓z] is observed, the particle in Tokyo informs the particle in New York "Your measured value has to be ↓z [resp. ↑z]". Therefore, the above fact implies that quantum mechanics says that there is something faster than light. This is essentially the same as the de Broglie paradox (cf. [81]). That is, • if we admit quantum mechanics, we must also admit the fact that there is something faster than light (i.e., so called "non-locality"). ♠Note 4.4. EPR-paradox is closely related to the fact that quantum syllogism does not hold in general. This will be discussed in Chapter 8. The Bohr-Einstein debates were a series of public disputes about quantum mechanics between Albert Einstein and Niels Bohr. Although there may be several opinions, I regard this debates as Einstein (realistic view) ←→ v.s. Bohr (linguistic view) For the further argument, see Section 10.7 (Leibniz-Clarke debates). ♠Note 4.5. [Shut up and calculate]. The above argument may suggest that there is something faster than light. However, when faster-than-light appears, our standing point is Stop being bothered This is not only our opinion but also most physicists'. In fact, in Mermin's book [70], he said (a) "Most physicists, I think it is fair to say, are not bothered." (b) If I were forced to sum up in one sentence what the Copenhagen interpretation says to me, it would be "Shut up and calculate" If it is so, we want to assert that the linguistic interpretation (§3.1) is the true colors of "the Copenhagen interpretation". That is because I also consider that (c) If I were forced to sum up in one sentence what the linguistic interpretation says to me, it would be "Shut up and calculate." 104 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems 4.5 Bell's inequality should be reconsidered This section is extracted from the following paper: Ref. [52]; Ishikawa,S., Bell's inequality should be reconsidered in quantum language , JQIS, Vol. 7, No.4 , 140-154, 2017, DOI: 10.4236/jqis.2017.74011 (http://www.scirp.org/Journal/PaperInformation.aspx?PaperID=80813) 4.5.1 Bell's inequality in mathematics Bell's inequality is important in the relation of "the hidden variable". J. Bell showed that, if Bell's inequality is violated, then the hidden variable does not exist. However, it should be noted that even if Bell's inequality is violated, it does not imply that quantum mechanics is wrong. In this section I would like to mention some of the things about Bell's inequality, though I am not concerned with "the hidden variable". Firstly, let us mention Bell's inequality in mathematics. Theorem 4.17. [The conventional Bell's inequality (cf. refs. [76, 10, 81])] The mathematical Bell's inequality is as follows: Let (Θ,B, P ) be a probability space. Let (f1, f2, f3, f4) : Θ → X4(≡ {−1, 1}4) be a measurable functions. Define the correlation functions Rij(i = 1, 2, j = 3, 4) by ∫ Θ fi(θ)fj(θ)P (dθ). Then, the following mathematical Bell's inequality ( or precisely, CHSH inequality (cf. ref. [10])) holds: |R13 − R14|+ |R23 + R24| ≤ 2 (4.47) Proof. It is easy as follows. "the left-hand side of the above eq.(4.47)" ≤ ∫ Θ |f3(θ)− f4(θ)|P (dθ) + ∫ Θ |f3(θ) + f4(θ)|P (dθ) ≤ 2 This completes the proof. This theorem is too easy, but we must remember the linguistic interpretation: (]) There is no probability (or, no probability space ) without measurements. Thus, in this section, we discuss "What is the probability space in Theorem 4.17?". 105 Ishikawa's Homepage 4.5 Bell's inequality should be reconsidered 4.5.2 Bell's inequality holds in both classical and quantum systems Now let us consider a kind of generalization of the quasi-product observable (cf. Definition 3.19) as follows. Definition 4.18. [Combinable, Combined observable(cf. ref. [26])] Let {S1, S2, ..., Sj} be a family (i.e., a set of sets) such that Sl ⊆ {1, 2, ..., n} (∀l = 1, 2, ..., j}). For each l ∈ {1, 2, ..., j}, consider an observable Ol = (×s∈Sl Xs,  s∈SlFs, Fl) in a W ∗-algebra A, and define a natural map πl :×k=1,2,...,nXk →×s∈Sl Xs such that × k=1,2,...,n Xk 3 (xk)k=1,2,...n 7→ (xk)k∈Sl ∈ × k∈Sl Xk Here, the {Ol : l = 1, 2, ..., j} is said to be combinable, if there exists an observable O = (×k=1,2,...,nXk,  k=1,2,...,nFk, F ) in A such that F (π−1l (× s∈Sl Ξs)) = Fl(× s∈Sl Ξs) (Ξs ∈ Fs, s ∈ Sl) Also, the observable O is called a combined observable of {Ol : l = 1, 2, ..., j} Note that, for each l, a measurement MA(Ol, S[ρ0]) is included in MA(O, S[ρ0]). In this section we devote ourselves to the following simple combined observable. Example 4.19. [Combined observable ] Let [A,A]B(H) be a basic structure. Put X = {−1, 1}. Let O1 = (X,P(X), F1), O2 = (X,P(X), F2), O3 = (X,P(X), F3), O4 = (X,P(X), F3) be observables in A. Consider four observables: O13 = (X 2,P(X2), F13), O14 = (X 2,P(X2), F14), O23 = (X 2,P(X2), F23), O24 = (X 2,P(X2), F24) in A such that F13({x} ×X) = F14({x} ×X) = F1({x}) F23({x} ×X) = F24({x} ×X) = F2({x}) F13(X × {x}) = F23(X × {x}) = F3({x}) F14(X × {x}) = F24(X × {x}) = F4({x}) (4.48) for any x ∈ {−1, 1}. The four observables O13, O14, O23 and O24 are said to be combinable if there exists an observable O = (X4,P(X4), F ) in A such that F13({(x1, x3)}) = F ({x1} ×X × {x3} ×X), F14({(x1, x4)}) = F ({x1} ×X ×X × {x4}) F23({(x2, x3)}) = F (X × {x2} × {x3} ×X), F24({(x2, x4)}) = F (X × {x2} ×X × {x4}) (4.49) 106 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems for any (x1, x2, x3, x4) ∈ X4. The observable O is said to be a combined observable of Oij (i = 1, 2, j = 3, 4). Also, the measurement MA(O = (X 4,P(X4), F ), S[ρ0]) is called the combined measurement of MA(O13, S[ρ0]), MA(O14, S[ρ0]), MA(O23, S[ρ0]) and MA(O24, S[ρ0]). Remark 4.20. (i): Note that the formula (4.49) implies (4.48). The condition (4.48) is not needed. (ii): Syllogism (i.e., [[A⇒ B] ∧ [B ⇒ C]]⇒ [A⇒ C] ) does not hold in quantum systems but in classical systems (cf. Section 8.7). A certain combined observable plays an important role in the proof of the classical syllogism (cf. ref. [26]). The following theorem is all of our insistence concerning Bell's inequality. We assert that this is the true Bell's inequality. Theorem 4.21. [Bell's inequality in quantum language] Let [A,A]B(H) be a basic structure. Put X = {−1, 1}. Fix the pure state ρ0 ( ∈ Sp(A∗) ) . And consider the four measurements MA(O13 = (X 2,P(X2), F13), S[ρ0]), MA(O14 = (X 2,P(X2), F14), S[ρ0]), MA(O23 = (X 2,P(X2), F23), S[ρ0]) and MA(O24 = (X 2,P(X2), F24), S[ρ0]). Or equivalently, consider the parallel measurement ⊗i=1,2,j=3,4MA(Oij = (X2,P(X2), Fij), S[ρ0]). Define four correlation functions (i = 1, 2, j = 3, 4) such that Rij = ∑ (u,v)∈X×X u * v ρ0(Fij({(u, v)})) Assume that four observables O13 = (X 2,P(X2), F13), O14 = (X 2,P(X2), F14), O23 = (X2,P(X2), F23) and O24 = (X 2,P(X2), F24) are combinable, that is, we have the combined observable O = (X4,P(X4), F ) in A such that it satisfies the formula (4.49). Then we have a combined measurement MA(O = (X 4,P(X4), F ), S[ρ0]) of MA(O13, S[ρ0]), MA(O14, S[ρ0]), MA(O23, S[ρ0]) and MA(O24, S[ρ0]). And further, we have Bell's inequality in quantum language as follows. |R13 −R14|+ |R23 +R24| 5 2 (4.50) Proof. Clearly we see, i = 1, 2, j = 3, 4, Rij = ∑ (x1,x2,x3,x4)∈X×X×X×X xi * xj ρ0(F ({(x1, x2, x3, x4)})) (4.51) ( for example, R13 = ∑ (x1,x2,x3,x4)∈X×X×X×X x1 * x3 ρ0(F ({(x1, x2, x3, x4)})) ) . Therefore, we see that |R13 −R14|+ |R23 +R24| 107 Ishikawa's Homepage 4.5 Bell's inequality should be reconsidered = ∑ (x1,x2,x3,x4)∈X×X×X×X [ |x1 * x3 − x1 * x4|+ |x2 * x3 + x2 * x4| ] ρ0(F ({(x1, x2, x3, x4)})) = ∑ (x1,x2,x3,x4)∈X×X×X×X [ |x3 − x4|+ |x3 + x4| ] ρ0(F ({(x1, x2, x3, x4)})) ≤ 2 This completes the proof. As the corollary of this theorem, we have the followings: Corollary 4.22. Consider the parallel measurement ⊗i=1,2,j=3,4MA(Oij = (X2,P(X2), Fij), S[ρ0]) as in Theorem 4.21. Let x = ( (x113, x 2 13), (x 1 14, x 2 14), (x 1 23, x 2 23), (x 1 24, x 2 24) ) ∈ X8(≡ {−1, 1}8) be a measured value of the parallel measurement ⊗i=1,2,j=3,4MA(Oij = (X2,P(X2), Fij), S[ρ0]). Let N be sufficiently large natural number. Consider N -parallel measurement ⊗N n=1 [ ⊗i=1,2,j=2,3 MA(Oij := (X 2,P(X2), Fij), S[ρ0]) ]. Let {xn}Nn=1 be the measured value. That is, {xn}Nn=1 =  ( (x1,113 , x 2,1 13 ), (x 1,1 14 , x 2,1 14 ), (x 1,1 23 , x 2,1 23 ), (x 1,1 24 , x 2,1 24 ) ) ( (x1,213 , x 2,2 13 ), (x 1,2 14 , x 2,2 14 ), (x 1,2 23 , x 2,2 23 ), (x 1,2 24 , x 2,2 24 ) ) ... ... ...( (x1,N13 , x 2,N 13 ), (x 1,N 14 , x 2,N 14 ), (x 1,N 23 , x 2,N 23 ), (x 1,N 24 , x 2,N 24 ) )  ∈ (X8)N Here, note that the law of large numbers says: for sufficiently large N , Rij ≈ 1 N N∑ n=1 x1,nij x 2,n ij (i = 1, 2, j = 3, 4). Then, it holds, by the formula (4.50), that | N∑ n=1 x1,n13 x 2,n 13 N − N∑ n=1 x1,n14 x 2,n 14 N |+ | N∑ n=1 x1,n23 x 2,n 23 N + N∑ n=1 x1,n24 x 2,n 24 N | ≤ 2, (4.52) which is also called Bell's inequality in quantum language. Remark 4.23. [(i):The conventional Bell's inequality (cf. refs. [10, 76, 81])] From the mathematical point of view, the formulas (4.47) and (4.50) are the same. However, the probability space (X4,P(X4), ρ0(F (*))) in Theorem 4.21 is visible and concrete. [(ii): "true value" (or, "hidden value")] In Theorem 4.21, we have the combined measurement MA(O = (X 4,P(X4), F ), S[ρ0]). Thus, some may consider that 108 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems • the true value (x1, x2, x3, x4) (of observables Ok, k = 1, 2, 3, 4 in Example 4.19 ) can be obtained by the measurement MA(O = (X 4,P(X4), F ), S[ρ0]). No-Go theorem (cf. [76] ) is usually mentioned in terms of Einstein's world view. However, • If No-Go theorem is mentioned in terms of Bohr's world view, we think that No-Go theorem is the existence theorem of the combined observable. 4.5.3 "Bell's inequality" is violated in classical systems as well as quantum systems In the previous section, we show that Theorem 4.21 (or Corollary 4.22) says (F1) Under the combinable condition (cf. Example 4.19), Bell's inequality (4.50) (or, (4.52)) holds in both classical systems and quantum systems. Or, equivalently, (F2) If Bell's inequality (4.50) (or, (4.52)) is violated, then the combined observable does not exist, and thus, we cannot obtain the measured value ( by the combined measurement). Remark 4.24. This is similar to the following elementary statement in quantum mechanics: (F′2) We have no simultaneous measurement (= combined measurement ) of the position observable Q and the momentum observable P , and thus we cannot obtain the measured value ( by the simultaneous measurement), which may be, from Einstein's point of view, represented that "true value (or, hidden variable) of the position and momentum" does not exist. Since the error ∆ is usually defined by ∆ = |rough measured value − true value|, it is not easy to define the errors ∆Q and ∆P in Heisenberg's uncertainty principle ∆Q *∆P ≥ ~/2 (cf. Note 4.2 ). As seen in Section 4.3, this definition was completed and Heisenberg's uncertainty principle was proved (cf. Corollary 1 in ref. [23]). Also, according to the maxim of dualism: "To be is to be perceived" due to G. Berkeley, we think that it is not necessary to name that does not exist (or equivalently, that is not measured ). The above statement (F2) makes us expect that (G) Bell's inequality (4.50) (or, (4.52)) is violated in classical systems as well as quantum systems without the combinable condition. This (G) was already shown in my previous paper [31]. However, I received a lot of questions concerning (G) from the readers. Thus, in this section, we again explain the (G) precisely. 109 Ishikawa's Homepage 4.5 Bell's inequality should be reconsidered 4.5.3.1 Bell test experiment In order to show the (G), three steps ([Step:I] ∼[Step:III]) are prepared in what follows. [Step: I]. Put X = {−1, 1}. Define complex numbers ak(= αk + βk √ −1 ∈ C : the complex field) (k = 1, 2, 3, 4) such that |ak| = 1. Define the probability space (X2,P(X2), νaiaj) such that (i = 1, 2, j = 3, 4) νaiaj({(1, 1)})= νaiaj({(−1,−1)})= (1− αiαj − βiβj)/4 νaiaj({(−1, 1)})= νaiaj({(1,−1)})= (1 + αiαj + βiβj)/4 (4.53) The correlation R(ai, aj) (i = 1, 2, j = 3, 4) is defined as follows: R(ai, aj) ≡ ∑ (x1,x2)∈X×X x1 * x2νaiaj({(x1, x2)}) = −αiαj − βiβj (4.54) Now we have the following problem: (H) Find a measurement MA(Oaiaj := (X 2, P(X2), Faiaj), S[ρ0]) (i = 1, 2, j = 3, 4) such that ρ0(Faiaj(Ξ)) = νaiaj(Ξ) (∀Ξ ∈ P(X2)) (4.55) and Fa1a3({x1} ×X) = Fa1a4({x1} ×X) Fa2a3({x2} ×X) = Fa2a4({x2} ×X) Fa1a3(X × {x3}) = Fa2a3(X × {x3}) Fa1a4(X × {x4}) = Fa2a4(X × {x4}) (∀xk ∈ X(≡ {−1, 1}), k = 1, 2, 3, 4) which is the same as the condition (4.48) [Step: II]. Let us answer this problem (H) in the two cases (i.e., classical case and quantum case), that is, •  (i):the case of quantum systems: [A = B(C2)⊗B(C2)(≡ B(C2 ⊗ C2)), A = B(C2)⊗B(C2)] (ii):the case of classical systems: [A = C0(Ω)⊗ C0(Ω)(≡ C0(Ω× Ω)), A = L∞(Ω)⊗ L∞(Ω) ] (i):the case of quantum system: [A = B(C2)⊗B(C2)] Put e1 = [ 1 0 ] , e2 = [ 0 1 ] (∈ C2). 110 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems For each ak (k = 1, 2, 3, 4), define the observable Oak ≡ ( X,P(X), Gak ) in B(C2) such that Gak({1}) = 1 2 [ 1 āk ak 1 ] , Gak({−1}) = 1 2 [ 1 −āk −ak 1 ] . where āk = αk − βk √ −1. Then, we have four observable: Ôai = (X,P(X), Gai ⊗ I), Ôaj = (X,P(X), I ⊗Gaj) (i = 1, 2, j = 3, 4) (4.56) and further, Oaiaj = (X 2,P(X2), Faiaj := Gai ⊗Gaj) (i = 1, 2, j = 3, 4) (4.57) in B(C2)⊗B(C2), where it should be noted that Faiaj is separated by Gai and Gaj . Further define the singlet state ρ0 = |ψs〉〈ψs| ( ∈ Sp(B(C2 ⊗ C2)∗) ) , where ψs = (e1 ⊗ e2 − e2 ⊗ e1)/ √ 2 Thus we have the measurement MB(C2⊗C2)(Oaiaj , S[ρ0]) in B(C2) ⊗ B(C2) (i = 1, 2, j = 3, 4). The followings are clear: for each (x1, x2) ∈ X2(≡ {−1, 1}2), ρ0(Faiaj({(x1, x2)})) = 〈ψs, (Gai({x1})⊗Gaj({x2}))ψs〉 = νaiaj({(x1, x2)}) (i = 1, 2, j = 3, 4) (4.58) For example, we easily see: ρ0(Faibj({(1, 1)})) = 〈ψs, (Gai({1})⊗Gaj({1}))ψs〉 = 1 8 〈(e1 ⊗ e2 − e2 ⊗ e1), ( [ 1 āi ai 1 ] ⊗ [ 1 āj aj 1 ] )(e1 ⊗ e2 − e2 ⊗ e1)〉 =18〈( [ 1 0 ] ⊗ [ 0 1 ] − [ 0 1 ] ⊗ [ 1 0 ] ), ( [ 1 āi ai 1 ] ⊗ [ 1 āj aj 1 ] )( [ 1 0 ] ⊗ [ 0 1 ] − [ 0 1 ] ⊗ [ 1 0 ] )〉 = 1 8 〈( [ 1 0 ] ⊗ [ 0 1 ] − [ 0 1 ] ⊗ [ 1 0 ] ), ( [ 1 ai ] ⊗ [ āj 1 ] − [ āi 1 ] ⊗ [ 1 aj ] )〉 = 1 8 (2− aāj − āiaj) = (1− αiαj − βiβj)/4 = νaiaj({(1, 1)}). Therefore, the measurement MB(C2⊗C2)(Oaiaj , S[ρ0]) satisfies the condition (H). (ii):the case of classical systems: [A = C0(Ω)⊗ C0(Ω) = C0(Ω× Ω)] Put ω0(= (ω ′ 0, ω ′′ 0)) ∈ Ω × Ω、ρ0 = δω0 (∈ Sp(C0(Ω× Ω) ∗), i.e., the point measure at ω0) ). Define the observable Oaiaj := (X 2,P(X2), Faiaj) in L ∞(Ω× Ω) such that [Faiaj({(x1, x2)})](ω) = νaiaj({(x1, x2)}) (∀(x1, x2) ∈ X2, i = 1, 2, j = 3, 4, ∀ω ∈ Ω× Ω) 111 Ishikawa's Homepage 4.5 Bell's inequality should be reconsidered Thus, we have four observables Oaiaj = (X 2,P(X2), Faiaj) (i = 1, 2, j = 3, 4) (4.59) in L∞(Ω × Ω) ( though the variables are not separable (cf. the formula (4.57) ). Then, it is clear that the measurement MC0(Ω×Ω)(Oaiaj , S[δω0 ]) satisfies the condition (H). (ii)′:the case of classical systems: [A = C0(Ω)⊗ C0(Ω) = C0(Ω× Ω)] It is easy to show a lot of different answers from the above (ii). For example, as a slight generalization of (9), define the probability measure νtaiaj (0 ≤ t ≤ 1) such that νtaiaj({(1, 1)})= ν t aiaj ({(−1,−1)})= (1− t(αiαj + βiβj))/4 νtaiaj({(−1, 1)})= ν t aiaj ({(1,−1)})= (1 + t(αiαj + βiβj))/4 (4.60) And consider the real-valued continuous function t(∈ C0(Ω × Ω)) such that 0 ≤ t(ω′, ω′′) ≤ 1 (∀ω = (ω′, ω′′) ∈ Ω × Ω). And assume that t(ω0) = 1 for some ω0(= (ω′0, ω′′0)) ∈ Ω × Ω、 ρ0 = δω0 (∈ Sp(C0(Ω× Ω) ∗), i.e., the point measure at ω0) ). Define the observable Oaiaj := (X2,P(X2), Faiaj) in L ∞(Ω× Ω) such that [Faiaj({(x1, x2)})](ω) = νt(ω)aiaj({(x1, x2)}) (∀(x1, x2) ∈ X 2, i = 1, 2, j = 3, 4, ∀ω ∈ Ω× Ω) (4.61) Thus, we have four observables Oaiaj = (X 2,P(X2), Faiaj) (i = 1, 2, j = 3, 4) in L∞(Ω × Ω) ( though the variables are not separable (cf. the formula (4.57) ). Then, it is clear that the measurement ML∞(Ω×Ω)(Oaiaj , S[δω0 ]) satisfies the condition (H). [Step: III]. As defined by (9), consider four complex numbers ak(= αk + βk √ −1; k = 1, 2, 3, 4) such that |ak| = 1. Thus we have four observables Oa1a3 := (X 2,P(X2), Fa1a3), Oa1a4 := (X 2,P(X2), Fa1a4), Oa2a3 := (X 2,P(X2), Fa2a3), Oa2a4 := (X 2,P(X2), Fa2a4), in A. Thus, we have the parallel measurement ⊗i=1,2,j=3,4 MA(Oaiaj := (X2,P(X2), Faiaj), S[ρ0]) in ⊗ i=1,2,j=3,4 A. Thus, putting a1 = √ −1, a2 = 1, a3 = 1 + √ −1√ 2 , a4 = 1− √ −1√ 2 , 112 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems we see, by (10), that |R(a1, a3)−R(a1, a4)| + |R(a2, a3) +R(a2, a4)| = 2 √ 2 (4.62) Further, assume that the measured value is x(∈ X8). That is, x = ( (x113, x 2 13), (x 1 14, x 2 14), (x 1 23, x 2 23), (x 1 24, x 2 24) ) ∈ × i,j=1,2 X2(≡ {−1, 1}8) LetN be sufficiently large natural number. ConsiderN -parallel measurement ⊗N n=1 [⊗i=1,2,j=3,4 MA(Oaiaj := (X 2,P(X2), Faiaj), S[ρ0]) ]. Assume that its measured value is {xn}Nn=1. That is, {xn}Nn=1 =  ( (x1,113 , x 2,1 13 ), (x 1,1 14 , x 2,1 14 ), (x 1,1 23 , x 2,1 23 ), (x 1,1 24 , x 2,1 24 ) ) ( (x1,213 , x 2,2 13 ), (x 1,2 14 , x 2,2 14 ), (x 1,2 23 , x 2,2 23 ), (x 1,2 24 , x 2,2 24 ) ) ... ... ...( (x1,N13 , x 2,N 13 ), (x 1,N 14 , x 2,N 14 ), (x 1,N 23 , x 2,N 23 ), (x 1,N 24 , x 2,N 24 ) )  ∈ ( × i=1,2,j=3,4 X2 )N (≡ {−1, 1}8N ) Then, the law of large numbers says that R(ai, aj) ≈ 1 N N∑ n=1 x1,nij x 2,n ij (i = 1, 2, j = 3, 4) This and the formula (18) say that | N∑ n=1 x1,n13 x 2,n 13 N − N∑ n=1 x1,n14 x 2,n 14 N |+ | N∑ n=1 x1,n23 x 2,n 23 N + N∑ n=1 x1,n24 x 2,n 24 N | ≈ 2 √ 2 (4.63) Therefore, Bell's inequality (4.50) (or, (4.52)) is violated in classical systems as well as quantum systems. Remark 4.25. For completeness, note that the observables Oaiaj (i = 1, 2, j = 3, 4) in the classical L∞(Ω × Ω) are not combinable in spite that these commute. Also, note that the formulas (4.60) and (4.61) imply that [Fa1a3({x} ×X)](ω) = [Fa1a4({x} ×X)](ω) = 1/2, [Fa2a3({x} ×X)](ω) = [Fa2a4({x} ×X)](ω) = 1/2, [Fa1a3(X × {x})](ω) = [Fa2a3(X × {x})](ω) = 1/2, [Fa1a4(X × {x})](ω) = [Fa2a4(X × {x})](ω) = 1/2 (∀x ∈ X, ∀ω ∈ Ω× Ω), which is similar as (4.48). 113 Ishikawa's Homepage 4.5 Bell's inequality should be reconsidered 4.5.4 Conclusion In Bohr-Einstein debates (refs. [14, 5]), Einstein's standing-point (that is, "the moon is there whether one looks at it or not" (i.e., physics holds without observers) ) is on the side of the realistic world view in Figure 1. On the other hand, we think that Bohr's standing point (that is, "to be is to be perceived" (i.e., there is no science without measurements )) is on the side of the linguistic world view in Figure 1.1. In this paper, contrary to Bell's spirit (which inherits Einstein's spirit), we try to discuss Bell's inequality in Bohr's spirit (i.e., in the framework of quantum language). And we show Theorem 4.21 ( Bell's inequality in quantum language), which says the statement (F2), that is, (I1) (≡ (F2)): [ from Bohr's standing-point]: If Bell's inequality (4.50) (or, (4.52)) is violated, then the combined observable does not exist, and thus, we cannot obtain the measured value (by the measurement of the combined observable). Also, recall that Bell's original argument (which is under the influence of Bohr-Einstein debates) says, roughly speaking, that (I2) [ from Einstein's standing-point]: If the mathematical Bell's inequality (4.47) is violated in Bell test experiment (the quantum case of Section 4.5.3), then hidden variables do not exist. It should be note that the concept of "hidden variable" is independent of measurements, thus, the (I2) is a philosophical statement in Einstein's spirit, or precisely, the (I2) may says that quantum mechanical phenomenon (i.e., Bell test experiment) cannot be described in Einstein's spirit. On the other hand, our (I1) is not related Einstein's spirit, that is, it is a statement in Bohr's spirit (i.e., there is no science without measurements). It is sure that Bell's answer (I2) is philosophically attractive, however, we believe in the scientific superiority of our answer (I1). For example, consider the following problem: (J) [Problem]: Why is Bell's inequality violated in the Bell test experiment ( mentioned in Section 4.5.3)? It is sure that everybody agrees to the answer (I1) and not (I2). Thus, the scientific superiority of our answer (I1) is clear. That is, we think that Bell's (I2) is a philosophical view of the scientific (I1). If so, we can, for the first time, understand Bell's inequality from the practical point of view. 114 Ishikawa's Homepage Chap. 4 Linguistic Copenhagen interpretation of quantum systems That is, Theorem 4.21 is the true Bell's inequality. And we conclude that whether or not Bell's inequality holds does not depend on whether classical systems or quantum systems (in Sections 4.5.3), but depend on whether the combined measurement exists or not (in Section 4.5.2). Thus, Bell's inequality is violated even in classical systems (in Section 4.5.3). Remark 4.26. Note that the great disputes in the history of the world view (cf. Figure 1.1 in Section 1.1) are always formed as follows: Einstein,... realistic world view (monistic realism) ←→ v.s. Bohr,... linguistic world view (dualistic idealism) For example, Table 4.1 : The realistic world view vs the linguistic world view Dispute R vs. L the realistic world view the linguistic world view Greek philosophy Aristotle Plato Problem of universals Nominalisme(William of Ockham) Realismus(Anselmus) Space*times Clarke( Newton) Leibniz Quantum mechanics Einstein (cf. [14]) Bohr (cf. [5]) (cf. Note 10.7 in Chapter 10 or ref. [49]). 115 Ishikawa's Homepage

Chapter 5 Fisher statistics (I) Measurement theory (= quantum language ) is formulated as follows. • measurement theory (=quantum language) := [Axiom 1] Measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } manual to use spells Measurement theory says that • Describe every phenomenon modeled on Axioms 1 and 2 (by a hint of the linguistic interpretation)! In this chapter, we study Fisher statistics in terms of Axiom 1 ( measurement: §2.7). We shall emphasize the reverse relation between measurement and inference (such as "the two sides of a coin"). The readers can read this chapter without the knowledge of statistics. 5.1 Statistics is, after all, urn problems 5.1.1 Population(=system)↔state Example 5.1. The density functions of the whole Japanese male's height and the whole American male's height is respectively defined by fJ and fA. That is,∫ β α fJ(x)dx = A Japanese male's population whose height is from α(cm) to β(cm) A Japanese male's overall population 117 5.1 Statistics is, after all, urn problems ∫ β α fA(x)dx = An American male's population whose height is from α(cm) to β(cm) An American male's overall population Let the density functions fJ and fA be regarded as the probability density functions fJ and fA such as (A) From [ the set of all Japanese males the set of all American males ] , choose a person (at random). Then, the probability that his height is from α(cm) to β(cm) is given by[ [Fh([α, β))](ωJ) = ∫ β α fJ(x)dx [Fh([α, β))](ωA) = ∫ β α fA(x)dx ] Now, let us represent the statements (A1) and (A2) in terms of quantum language: Define the state space Ω by Ω = {ωJ , ωA} with the discrete metric dD and the counting measure ν such that ν({ωJ}) = 1, ν({ωA}) = 1( It does not matter, even if ν({ωJ}) = a, ν({ωA}) = b (a, b > 0) ) . U1≈δωJ U2≈δωA All Japanese males in this urn U1 All American males in this urn U2 Figure 5.1: Population≈urn(↔state) Thus, we have the classical basic structure: Classical basic structure[C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] The pure state space is defined by Sp(C0(Ω) ∗) = {δωJ , δωA} ≈ {ωJ , ωA} = Ω Here, we consider that δωJ * * * "the state of the set U1 of all Japanese males", δωA * * * "the state of the set U2 of all American males", and thus, we have the following identification (that is, Figure 5.1): U1 ≈ δωJ , U2 ≈ δωA The observable Oh = (R,B, Fh) in L∞(Ω, ν) is already defined by (A). Thus, we have the measurement ML∞(Ω)(Oh, S[δω ]) (ω ∈ Ω = {ωJ , ωA}). The statement(A) is represented in terms of quantum language by 118 Ishikawa's Homepage Chap. 5 Fisher statistics (I) (B) The probability that a measured value obtained by the measurement [ ML∞(Ω)(Oh, S[ωJ ]) ML∞(Ω)(Oh, S[ωA]) ] belongs to an interval [α, β) is given by  C0(Ω)∗(δωJ , Fh([α, β)))L∞(ω,ν) = [Fh([α, β))](ωJ) C0(Ω) ∗ ( δωA , Fh([α, β)) ) L∞(ω,ν) = [Fh([α, β))](ωA)  Therefore, we get: statement (A) (ordinary language) −−−−−−→ translation statement (B) (quantum language) 5.1.2 Normal observable and student t-distribution Consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] where Ω = R (=the real line) with the Lebesgue measure ν. Let σ > 0 be a standard deviation, which is assumed to be fixed. Define the measured value space X by R (i.e., X = R ). Define the normal observable OGσ = (X(= R),BR, Gσ) in L∞(Ω, ν) such that [Gσ(Ξ)](ω) = 1√ 2πσ ∫ Ξ exp [ − 1 2σ2 (x− ω)2 ] dx (5.1) (∀Ξ ∈ BX(= BR), ∀ω ∈ Ω(= R)) where BR is the Borel field. For example, 1√ 2πσ2 ∫ σ −σ e− x2 2σ2 dx = 0.683..., 1√ 2πσ2 ∫ 2σ −2σ e− x2 2σ2 dx = 0.954..., 1√ 2πσ2 ∫ 1.96σ −1.96σ e− x2 2σ2 dx+0.95 x y 6 y = 1√ 2πσ2 e− x2 2σ2 σ−σ 2σ−2σ 68.3% 95.4% Figure 5.2: Error function 119 Ishikawa's Homepage 5.1 Statistics is, after all, urn problems Next, consider the parallel observable ⊗n k=1OGσ = (Rn,BRn , ⊗n k=1Gσ) in L ∞(Ωn, ν⊗n) and restrict it on K = {(ω, ω, . . . , ω) ∈ Ωn | ω ∈ Ω}(⊆ Ωn) This is essentially the same as the simultaneous observable On = (Rn,BRn ,×nk=1Gσ) in L∞(Ω). That is, [( n × k=1 Gσ)(Ξ1 × Ξ2 × * * * × Ξn)](ω) = n × k=1 [Gσ(Ξk)](ω) = n × k=1 1√ 2πσ ∫ Ξk exp [ − 1 2σ2 (xk − ω)2 ] dxk (5.2) (∀Ξk ∈ BX(= BR), ∀ω ∈ Ω(= R)) Then, for each (x1, x2, * * * , xn) ∈ Xn(= Rn), define xn = x1 + x2 + * * *+ xn n U2n = (x1 − xn)2 + (x2 − xn)2 + * * *+ (xn − xn)2 n− 1 and define the map ψ : Rn → R such that ψ(x1, x2, . . . , xn) = xn − ω Un/ √ n Then, we have the observable OTσn = (X(= R),BR, T σ n ) in L ∞(R) such that [T σn (Ξ)](ω) = [ Gσ ( {(x1, x2, ..., xn) ∈ Rn | xn − ω Un/ √ n ∈ Ξ} )] (ω) (∀Ξ ∈ F) (5.3) The observable OTσn = (X(= R),BR, T σ n ) in L ∞(R) is called the student t observable . Here, putting fσn (x) = {Γ(n/2)√ (n− 1)πΓ((n− 1)/2) (1 + x2 n− 1 )−n/2 (Γ is Gamma function) (5.4) we see that [T σn (Ξ)](ω) = ∫ Ξ fσn (x)dx (∀Ξ ∈ F) (5.5) which is independent of ω and σ. Also note that lim n→∞ fσn (x) = lim n→∞ Γ(n/2)√ (n− 1)πΓ((n− 1)/2) (1 + x2 n− 1 )−n/2 = 1√ 2π e− x2 2 thus, if n ≥ 30, it can be regarded as the normal distribution N(0, 1)( that is, mean 0, the standard deviation 1). 120 Ishikawa's Homepage Chap. 5 Fisher statistics (I) 5.2 The reverse relation between Fisher ( =inference) and Born ( =measurement) In this section, we consider the reverse relation between Fisher ( =inference) and Born ( =measurement) 5.2.1 Inference problem ( Statistical inference ) Before we mention Fisher's maximum likelihood method, we exercise the following problem: Problem 5.2. [Urn problem( =Example2.34), A simplest example of Fisher's maximum likelihood method] There are two urns U1 and U2. The urn U1 [resp. U2] contains 8 white and 2 black balls [resp. 4 white and 6 black balls]. - [∗] U1(≈ ω1) U2(≈ ω2) Figure 5.3: Pure measurement (Fisher's maximum likelihood method) Here consider the following procedures (i) and (ii). (i) One of the two (i.e., U1 or U2) is chosen and is settled behind a curtain. Note, for completeness, that you do not know whether it is U1 or U2. (ii) Pick up a ball out of the unknown urn behind the curtain. And you find that the ball is white. Here, we have the following problem: (iii) Infer the urn behind the curtain, U1 or U2? The answer is easy, that is, the urn behind the curtain is U1. That is because the urn U1 has more white balls than U2. The above problem is too easy, but it includes the essence of Fisher maximum likelihood method. 5.2.2 Fisher's maximum likelihood method in measurement theory We begin with the following notation: 121 Ishikawa's Homepage 5.2 The reverse relation between Fisher ( =inference) and Born ( =measurement) Notation 5.3. [MA(O, S[∗])]: Consider the measurement MA (O=(X,F, F ), S[ρ]) formulated in the basic structure [A ⊆ A ⊆ B(H)]. Here, note that (A1) In most cases that the measurement MA (O=(X,F, F ), S[ρ]) is taken, it is usual to think that the state ρ (∈ Sp(A∗)) is unknown. That is because (A2) the measurement MA(O, S[ρ]) may be taken in order to know the state ρ. Therefore, when we want to stress that we do not know the state ρ The measurement MA (O=(X,F, F ), S[ρ]) is often denoted by (A3) MA (O=(X,F, F ), S[∗]) Further, consider the subset K(⊆ Sp(A∗)). When we know that the state ρ belongs to K, MA (O=(X,F, F ), S[∗]) is denoted by MA(O, S[∗]((K))). Therefore, it suffices to consider that MA(O, S[∗]) = MA(O, S[∗]((S p(A∗)))) Using this notation MA(O, S[∗]), we characterize our problem (i.e., inference) as follows. Problem 5.4. [Inference problem] (a) Assume that a measured value obtained by MA(O=(X,F, F ), S[∗]((K))) belongs to Ξ(∈ F). Then, infer the unknown state [∗] (∈ Ω) or, (b) Assume that a measured value (x, y) obtained by MA(O=(X × Y,F  G, H), S[∗]((K))) belongs to Ξ× Y (Ξ ∈ F). Then, infer the probability that y ∈ Γ. Before we answer the problem, we emphasize the reverse relation between "inference" and "measurement". The measurement is "the view from the front", that is, (B1) (observable[O], state[ω(∈ Ω)]) measurement−−−−−−−−−−−→ ML∞(Ω)(O,S[ω]) measured value[x(∈ X)] On the other hand, the inference is "the view from the back", that is, (B2) (observable[O],measured value[x ∈ Ξ(∈ F)]) inference−−−−−−−−−→ ML∞(Ω)(O,S[∗]) state [ω(∈ Ω)] In this sense, we say that 122 Ishikawa's Homepage Chap. 5 Fisher statistics (I) the inference problem is the reverse problem of measurement Therefore, it suffices to image Fig. 5.4. (measuring object) unknown state −−−−−−−→ (measurement){ }} { observable (measuring instrument) −−−−−−−−−→ probabilistic measured value (output)} {{ } (observer) 6 inference Figure 5.4: The image of inference In order to answer the above problem 5.4, we shall describe Fisher maximum likelihood method in terms of measurement theory. Theorem 5.5. [(Answer to Problem 5.4(b)): Fisher's maximum likelihood method(the general case)] Consider the basic structure [A ⊆ A ⊆ B(H)] Assume that a measured value(x, y) obtained by a measurement MA(O=(X×Y,F  G, H), S[∗]((K))) belongs to Ξ× Y (Ξ ∈ F). Then, there is reason to infer that the probability P (Γ) that y ∈ Γ is equal to P (Γ) = ρ0(H(Ξ× Γ)) ρ0(H(Ξ× Y )) (∀Γ ∈ G) where, ρ0 ∈ K is determined by. ρ0(H(Ξ× Y )) = max ρ∈K ρ(H(Ξ× Y )) (5.6) Proof. Assume that ρ1, ρ2 ∈ K and ρ1(H(Ξ × Y )) < ρ2(H(Ξ × Y )). By Axiom 1 ( measurement: §2.7) (i) the probability that a measured value(x, y) obtained by a measurement MA(O, S[ρ1]) belongs to Ξ× Y is equal to ρ1(H(Ξ× Y )) (ii) the probability that a measured value(x, y) obtained by a measurement MA(O, S[ρ2]) belongs to Ξ× Y is equal to ρ2(H(Ξ× Y )) 123 Ishikawa's Homepage 5.2 The reverse relation between Fisher ( =inference) and Born ( =measurement) Since we assume that ρ1(H(Ξ × Y )) < ρ2(H(Ξ × Y )), we can conclude that "(i) is more rare than (ii)". Thus, there is a reason to infer that [∗] = ω2. Therefore, the ρ0 in (5.6) is reasonable. Since the probability that a measured value(x, y) obtained by MA(O, S[ρ0]) belongs to Ξ× Γ is given by ρ0(H(Ξ× Γ)), we complete the proof of Theorem 5.5. Theorem 5.6. [(Answer to 5.4(a)): Fisher's maximum likelihood method in classical case ] (i): Consider a measurement ML∞(Ω)(O =(X,F, F ), S[∗]((K))). Assume that we know that a measured value obtained by a measurement ML∞(Ω)(O, S[∗]((K))) belongs to Ξ (∈ F). Then, there is a reason to infer that the unknown state state [∗] is ω0 (∈ Ω) such that [F (Ξ)](ω0) = max ω∈Ω [F (Ξ)](ω) 0 1 Ω ω0 [F (Ξ)](ω) Figure 5.5: Fisher maximum likelihood method (ii): Assume that a measured value x0 (∈ X) is obtained by a measurement ML∞(Ω)(O =(X,F, F ), S[∗]((K))). Define the likelihood function f(x, ω) by f(x, ω) = inf ω1∈K [ lim Ξ3x,[F (Ξ)](ω1)6=0,Ξ→{x} [F (Ξ)](ω) [F (Ξ)](ω1) ] (5.7) Then, there is a reason to infer that [∗] = ω0(∈ K) such that f(x0, ω0) = 1. Proof. Consider Theorem 5.5 in the case that [A ⊆ A ⊆ B(H)] = [C0(Ω) ⊆ L∞(Ω) ⊆ B(L2(Ω)] Thus, in the measurement ML∞(Ω)(O=(X × Y,F  G, H), S[∗]((K))), consider the case that Fixed O1=(X,F, F ), any O2=(Y,G, G), O=O1 × O2 = (X × Y,F  G, F ×G), ρ0 = δω0 Then, we see P (Γ) = [H(Ξ)](ω0)× [G(Γ)](ω0) [H(Ξ)](ω0)× [G(Y )](ω0) = [G(Γ)](ω0) (∀Γ ∈ G) (5.8) And, from the arbitrariness of O2, there is a reason to infer that [∗] = δω0( ≈ identification ω0) 124 Ishikawa's Homepage Chap. 5 Fisher statistics (I) ♠Note 5.1. The linguistic interpretation says that the state after measurement is non-sense. In this sense, the readers may consider that (]1) Theorem 5.6 is also non-sense However, we say that (]2) in the sense of (5.8), Theorem 5.6 should be accepted. or (]3) as far as classical system, it suffices to believe in Theorem 5.6 Answer 5.7. [The answer to Problem 5.2 by Fisher's maximum likelihood method] You do not know which the urn behind the curtain is, U1 or U2. Assume that you pick up a white ball from the urn. The urn is U1 or U2? Which do you think? - [∗] U1≈ω1 U2≈ω2 Figure 5.6: Pure measurement (Fisher's maximum likelihood method) Answer: Consider the measurement ML∞(Ω)(O= ({w, b}, 2{w,b}, F ), S[∗]), where the observable Owb = ({w, b}, 2{w,b}, Fwb) in L∞(Ω) is defined by [Fwb({w})](ω1) = 0.8, [Fwb({b})](ω1) = 0.2 [Fwb({w})](ω2) = 0.4, [Fwb({b})](ω2) = 0.6 (5.9) Here, we see: max{[Fwb({w})](ω1), [Fwb({w})](ω2)} = max{0.8, 0.4} = 0.8 = Fwb({w})](ω1) 125 Ishikawa's Homepage 5.2 The reverse relation between Fisher ( =inference) and Born ( =measurement) Then, Fisher's maximum likelihood method (Theorem 5.6) says that [∗] = ω1 Therefore, there is a reason to infer that the urn behind the curtain is U1. ♠Note 5.2. As seen in Figure 5.4 , inference (Fisher maximum likelihood method) is the reverse of measurement (i.e., Axiom 1 due to Born). Here note that (a) Born's discovery "the probabilistic interpretation of quantum mechanics" in [6] (1926) (b) Fisher's great book "Statistical Methods for Research Workers" (1925) Thus, it is surprising that Fisher and Born investigated the same thing in the different fields in the same age. 126 Ishikawa's Homepage Chap. 5 Fisher statistics (I) 5.3 Examples of Fisher's maximum likelihood method All examples mentioned in this section are easy for the readers who studied the elementary of statistics. However, it should be noted that these are consequence of Axiom 1 ( measurement: §2.7). Example 5.8. [Urn problem] Each urn U1, U2, U3 contains many white balls and black ball such as: Table 5.1: urn problem w*b Urn Urn U1 Urn U2 Urn U3 white ball 80% 40% 10% black ball 20% 60% 90% Here, (i) one of three urns is chosen, but you do not know it. Pick up one ball from the unknown urn. And you find that its ball is white. Then, how do you infer the unknown urn, i.e., U1, U2 or U3? Further, (ii) And further, you pick up another ball from the unknown urn (in (i)). And you find that its ball is black. That is, after all, you have one white ball and one black ball. Then, how do you infer the unknown urn, i.e., U1, U2 or U3? In what follows, we shall answer the above problems (i) and (ii) in terms of measurement theory. Consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Put δωj(≈ ωj)←→ [the state such that urn Uj is chosen] (j = 1, 2, 3) Thus, we have the state space Ω ( ={ω1, ω2, ω3} ) with the counting measure ν. Further, define the observable O = ({w, b}, 2{w,b}, F ) in C(Ω) such that F ({w})(ω1) = 0.8, F ({w})(ω2) = 0.4, F ({w})(ω3) = 0.1 F ({b})(ω1) = 0.2, F ({b})(ω2) = 0.6, F ({b})(ω3) = 0.9 127 Ishikawa's Homepage 5.3 Examples of Fisher's maximum likelihood method Answer to (i): Consider the measurement ML∞(Ω)(O, S[∗]), by which a measured value "w" is obtained. Therefore, we see [F ({w})](ω1) = 0.8 = max ω∈Ω [F ({w})](ω) = max{0.8, 0.4, 0.1} Hence, by Fisher's maximum likelihood method (Theorem5.6) we see that [∗] = ω1 Thus, we can infer that the unknown urn is U1. Answer to (ii): Next, consider the simultaneous measurement ML∞(Ω)(×2k=1O = (X2, 2X 2 , F=×2k=1 F ), S[∗]), by which a measured value (w, b) is obtained. Here, we see [F ({(w, b)})](ω) = [F ({w})](ω) * [F ({b})](ω) thus, [F ({(w, b)})](ω1) = 0.16, [F ({(w, b)})](ω2) = 0.24, [F ({(w, b)})](ω3) = 0.09 Hence, by Fisher's maximum likelihood method (Theorem5.6), we see that [∗] = ω2 Thus, we can infer that the unknown urn is U2. Example 5.9. [Normal observable(i): Ω = R] As mentioned before, we again discuss the normal observable in what follows. Consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] (where, Ω = R) Fix σ > 0, and consider the normal observable OGσ = (R,BR, Gσ) in L∞(R) (where Ω = R) such that [Gσ(Ξ)](μ) = 1√ 2πσ ∫ Ξ exp[− 1 2σ2 (x− μ)2]dx (∀Ξ ∈ BR, ∀μ ∈ Ω = R) Thus, the simultaneous observable ×3k=1OGσ (in short, O3Gσ) = (R3,BR3 , G3σ) in L∞(R) is defined by 128 Ishikawa's Homepage Chap. 5 Fisher statistics (I) [G3σ(Ξ1 × Ξ2 × Ξ3)](μ) = [Gσ(Ξ1)](μ) * [Gσ(Ξ2)](μ) * [Gσ(Ξ3)](μ) = 1 ( √ 2πσ)3 ∫∫∫ Ξ1×Ξ2×Ξ3 exp[− (x1 − μ) 2 + (x2 − μ)2 + (x3 − μ)2 2σ2 ] × dx1dx2dx3 (∀Ξk ∈ BR, k = 1, 2, 3, ∀μ ∈ Ω = R) Thus, we get the measurement ML∞(R)(O 3 Gσ , S[∗]) Now we consider the following problem: (a) Assume that a measured value (x01, x 0 2, x 0 3) (∈ R3) is obtained by the measurement ML∞(R)(O3Gσ , S[∗]). Then, infer the unknown state [∗](∈ R). Answer(a) Put Ξi = [x 0 i − 1 N , x0i + 1 N ] (i = 1, 2, 3) Assume that N is sufficiently large. Fisher's maximum likelihood method (Theorem5.6) says that the unknown state[ ∗ ] = μ0 is found in what follows. [G3σ(Ξ1 × Ξ2 × Ξ3)](μ0) = max μ∈R [G3σ(Ξ1 × Ξ2 × Ξ3)](μ) Since N is sufficiently large, we see 1 ( √ 2πσ)3 exp[− (x 0 1 − μ0)2 + (x02 − μ0)2 + (x03 − μ0)2 2σ2 ] = max μ∈R [ 1 ( √ 2πσ)3 exp[− (x 0 1 − μ)2 + (x02 − μ)2 + (x03 − μ)2 2σ2 ] ] That is, (x01 − μ0)2 + (x02 − μ0)2 + (x03 − μ0)2 = min μ∈R { (x01 − μ)2 + (x02 − μ)2 + (x03 − μ)2 } Therefore, solving d dμ {* * * } = 0, we conclude that μ0 = x01 + x 0 2 + x 0 3 3 [Normal observable(ii)] Next consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] (where, Ω = R× R+) and consider the case: 129 Ishikawa's Homepage 5.3 Examples of Fisher's maximum likelihood method • we know that the length of the pencil μ is satisfied that 10cm μ L cm ≤30. And we assume that (]) the length of the pencil μ and the roughness σ of the ruler are unknown. That is, assume that the state space Ω = [10, 30] × R+ ( ={μ ∈ R | 10 5 μ 5 30} × {σ ∈ R | σ > 0} ) Define the observable O = (R,BR, G) in L∞([10, 30]× R+) such that [G(Ξ)](μ, σ) = [Gσ(Ξ)](μ) (∀Ξ ∈ BR, ∀(μ, σ) ∈ Ω = [10, 30]× R+) Therefore, the simultaneous observable O3 = (R3,BR3 , G 3) in C([10, 30]× R+) is defined by [G3(Ξ1 × Ξ2 × Ξ3)](μ, σ) = [G(Ξ1)](μ, σ) * [G(Ξ2)](μ, σ) * [G(Ξ3)](μ, σ) = 1 ( √ 2πσ)3 ∫ Ξ1×Ξ2×Ξ3 exp[− (x1 − μ) 2 + (x2 − μ)2 + (x3 − μ)2 2σ2 ]dx1dx2dx3 (∀Ξk ∈ BR, k = 1, 2, 3, ∀(μ, σ) ∈ Ω = [10, 30]× R+) Thus, we get the simultaneous measurement ML∞([10,30]×R+)(O 3, S[∗]). Here, we have the following problem: (b) When a measured value (x01, x 0 2, x 0 3) ( ∈ R3) is obtained by the measurement ML∞([10,30]×R+) (O3, S[∗]), infer the unknown state [∗](= (μ0, σ0) ∈ [10, 30] × R+), i.e., the length μ0 of the pencil and the roughness σ0 of the ruler. Answer (b) By the same way of (a), Fisher's maximum likelihood method (Theorem5.6) says that the unknownstate [ ∗ ] = (μ0, σ0) such that 1 ( √ 2πσ0)3 exp[− (x 0 1 − μ0)2 + (x02 − μ0)2 + (x03 − μ0)2 2σ20 ] = max (μ,σ)∈[10,30]×R+ { 1 ( √ 2πσ)3 exp[− (x 0 1 − μ)2 + (x02 − μ)2 + (x03 − μ)2 2σ2 ] } (5.10) Thus, solving ∂ ∂μ {* * * } = 0, ∂ ∂σ {* * * } = 0 we see μ0 =  10 (when (x01 + x 0 2 + x 0 3)/3 < 10 ) (x01 + x 0 2 + x 0 3)/3 (when 10 5 (x01 + x02 + x03)/3 5 30 ) 30 (when 30 < (x01 + x 0 2 + x 0 3)/3 ) (5.11) σ0 = √ {(x01 − μ)2 + (x02 − μ)2 + (x03 − μ)2}/3 130 Ishikawa's Homepage Chap. 5 Fisher statistics (I) where μ = (x01 + x 0 2 + x 0 3)/3 Example 5.10. [Fisher's maximum likelihood method for the simultaneous normal measurement]. Consider the simultaneous normal observable OnG = (Rn,BnR, Gn) in L∞(R × R+) (such as defined in formula (5.2)). This is essentially the same as the simultaneous observable On = (Rn,BRn ,×nk=1Gσ) in L∞(R× R+). That is, [( n × k=1 Gσ)(Ξ1 × Ξ2 × * * * × Ξn)](ω) = n × k=1 [Gσ(Ξk)](ω) = n × k=1 1√ 2πσ ∫ Ξk exp [ − 1 2σ2 (xk − μ)2 ] dxk (∀Ξk ∈ BX(= BR), ∀ω = (μ, σ) ∈ Ω(= R× R+)) Assume that a measured value x = (x1, x2, . . . , xn)(∈ Rn) is obtained by the measurement ML∞(R×R+)(O n = (Rn,BnR, Gnσ),S[∗]). The likelihood function Lx(μ, σ)(= L(x, (μ, σ)) is equal to Lx(μ, σ) = 1 ( √ 2πσ)n exp[− ∑n k=1(xk − μ)2 2σ2 ] or, in the sense of (5.7), Lx(μ, σ) = 1 ( √ 2πσ)n exp[− ∑n k=1(xk−μ)2 2σ2 ] 1 ( √ 2πσ(x))n exp[− ∑n k=1(xk−μ(x))2 2σ(x)2 ] (5.12) (∀x = (x1, x2, . . . , xn) ∈ Rn, ∀ω = (μ, σ) ∈ Ω = R× R+). Therefore, we get the following likelihood equation: ∂Lx(μ, σ) ∂μ = 0, ∂Lx(μ, σ) ∂σ = 0 (5.13) which is easily solved. That is, Fisher's maximum likelihood method (Theorem5.6) says that the unknown state [∗] = (μ, σ) (∈ R× R+) is inferred as follows. μ = μ(x) = x1 + x2 + . . .+ xn n , (5.14) σ = σ(x) = √∑n k=1(xk − μ(x))2 n (5.15) 131 Ishikawa's Homepage 5.4 Moment method: useful but artificial 5.4 Moment method: useful but artificial Let us explain the moment method (cf. [30]), which as well as Fisher's maximum likelihood method are frequently used. Consider the measurement MA ( O ≡ (X,F, F ), S[ρ] ) , and its parallel measurement⊗nk=1MA ( O ≡ (X,F, F ), S[ρ] ) (= M⊗A (⊗n k=1 O := (X n,Fn, ⊗n k=1 F ), S[⊗nk=1ρ] ) . Assume that the measured value (x1, x2, ..., xn)(∈ Xn) is obtained by the parallel measurement. Assume that n is sufficiently large. By the law of large numbers (Theorem 4.5), we can assure that M+1(X) 3 νn ( ≡ δx1 + δx2 + * * *+ δxn n ) + ρ(F (*)) ∈M+1(X) (5.16) Thus, (A) in order to infer the unknown state ρ(∈ Sp(A∗)), it suffices to solve the equation (5.16) For example, we have several methods to solve the equation (5.16) as follows. (B1) Solve the following equation: ‖νn(*)− ρ(F (*))‖M(X) = min{‖νn(*)− ρ1(F (*))‖M(X) | ρ1(∈ Sp(A∗))} (5.17) (B2) For some f1, f2, * * * , fn ∈ C(X) (= the set of all continuous functions on X), it suffices to find ρ(∈ Sp(A∗)) such that ∆(ρ) = minρ1(∈Sp(A∗)) ∆(ρ1), where ∆(ρ) = n∑ k=1 ∣∣∣ ∫ X fk(ξ)νn(dξ)− ∫ X fk(ξ)ρ(F (dξ)) ∣∣∣ = n∑ k=1 ∣∣∣fk(x1) + fk(x2) + * * *+ fk(xn) n − ∫ X fk(ξ)ρ(F (dξ)) ∣∣∣ (B3) In the cases of the classical measurement ML∞(Ω) ( O ≡ (X,F, F ), S[ρ] ) (putting ρ = δω), it suffices to solve 0 = n∑ k=1 ∣∣∣fk(x1) + fk(x2) + * * *+ fk(xn) n − ∫ X fk(ξ)[F (dξ)](ω) ∣∣∣ (5.18) or, it suffices to solve f1(x1)+f1(x2)+***+f1(xn) n − ∫ X f1(ξ)[F (dξ)](ω) = 0 f2(x1)+f2(x2)+***+f2(xn) n − ∫ X f2(ξ)[F (dξ)](ω) = 0 . . . . . . . . . . . . fm(x1)+fm(x2)+***+fm(xn) n − ∫ X fm(ξ)[F (dξ)](ω) = 0 132 Ishikawa's Homepage Chap. 5 Fisher statistics (I) (B4) Particularly, in the case that X = {ξ1, ξ2, * * * , ξm} is finite, define f1, f2, * * * , fm ∈ C(X) by fk(ξ) = χ{ξk}(ξ) = { 1 (ξ = ξk) 0 (ξ 6= ξk) and, it suffices to find the ρ(= δω) such that n∑ k=1 ∣∣∣χ{ξk}(x1) + χ{ξk}(x2) + * * *+ χ{ξk}(xn) n − ∫ X χ{ξk}(ξ)ρ(F (dξ)) ∣∣∣ = n∑ k=1 ∣∣∣][{xm : ξk = xm}] n − [F ({ξk}](ω)) ∣∣∣ = 0 The above methods are all the moment method. Note that (C1) It is desirable that n is sufficiently large, but the moment method may be valid even when n = 1. (C2) The choice of fk is artificial ( on the other hand, Fisher' maximum likelihood method is natural). Problem 5.11. [=Problem5.2: Urn problem: by the moment method] You do not know which the urn behind the curtain is, U1 or U2. Assume that you pick up a white ball from the urn. The urn is U1 or U2? Which do you think? - [∗] U1≈ω1 U2≈ω2 Figure 5.7: Inference(by moment method) Answer: Consider the measurement ML∞(Ω)(O= ({w, b}, 2{w,b}, F ), S[∗]). Here, recall that the observable Owb = ({w, b}, 2{w,b}, Fwb) in L∞(Ω) is defined by [Fwb({w})](ω1) = 0.8, [Fwb({b})](ω1) = 0.2 133 Ishikawa's Homepage 5.4 Moment method: useful but artificial [Fwb({w})](ω2) = 0.4, [Fwb({b})](ω2) = 0.6 Since a measured value "w" is obtained, the approximate sample space ({w, b}, 2{w,b}, ν1) is obtained as ν1({w}) = 1, ν1({b}) = 0 [when the unknown state [∗] is ω1] (5.17) = |1− 0.8|+ |0− 0.2| [when the unknown state [∗] is ω2] (5.17) = |1− 0.4|+ |0− 0.6| Thus, by the moment method, we can infer that [∗] = ω1, that is, the urn behind the curtain is U1. [II] The above may be too easy. Thus, we add the following problem. Problem 5.12. [Sampling with replacement]: As mentioned in the above, assume that "white ball" is picked. and the ball is returned to the urn. And further, we pick "black ball", and it is returned to the urn. Repeat this, after all, assume that we get "w", "b", "b", "w", "b", "w", "b", Then, we have the following problem: (a) Which the urn behind the curtain is U1 or U2? Answer: Consider the simultaneous measurement ML∞(Ω)(×7k=1O= ({w, b}7, 2{w,b} 7 , ×7k=1F ), S[∗]). And assume that the measured value is (w, b, b, w, b, w, b). Then, [when [∗] is ω1] (5.17) = |3/7− 0.8|+ |4/7− 0.2| = 52/70 [when [∗] is ω2] (5.17) = |3/7− 0.4|+ |4/7− 0.6| = 10/70 Thus, by the moment method, we can infer that [∗] = ω2, that is, the urn behind the curtain is U2. 134 Ishikawa's Homepage Chap. 5 Fisher statistics (I) Example 5.13. [The most important example of moment method] Putting Ω = R × R+ = {ω = (μ, σ) | μ ∈ R, σ > 0} with Lebesgue measure ν, Consider the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Assume that the observable OG = (X(= R),BR, G) in L∞(Ω, ν) satisfies that∫ R ξ[G(dξ)](μ, σ) = μ, ∫ R (ξ − μ)2[G(dξ)](μ, σ) = σ2 (∀ω = (μ, σ) ∈ Ω(= R× R+)) Here, assume that a measured value (x1, x2, x3)(∈ R3) is obtained by the simultaneous measurement×3k=1 ML∞(Ω)(OG, S[∗]). That is, we have the 3-sample distribution ν3 such that ν3 = δx1 + δx2 + δx3 3 ∈M+1(R) Put f1(ξ) = ξ, f2(ξ) = ξ 2. Then, by the moment method (5.18), we see: 0 = 2∑ k=1 ∣∣∣ ∫ R ξkν3(dξ)− ∫ R ξk[G(dξ)](ω) ∣∣∣ = 2∑ k=1 ∣∣∣(x1)k + (x2)k + (xn)k 3 − ∫ R ξk[G(dξ)](μ, σ) ∣∣∣ = ∣∣∣x1 + x2 + x3 3 − μ ∣∣∣ + ∣∣∣(x1)2 + (x2)2 + (x3)2 3 − (σ2 + μ2) ∣∣∣ Thus, we get: μ = x1 + x2 + xn 3 σ2 = (x1) 2 + (x2) 2 + (x3) 2 3 − μ2 = (x1 − x1+x2+xn3 ) 2 + (x2 − x1+x2+xn3 ) 2 + (x3 − x1+x2+xn3 ) 2 3 which is the same as the (5.11) concerning the normal measurement. ♠Note 5.3. Consider the measurement ML∞(Ω)(O=(X, 2X , F ), S[∗]), where X = {x1, x2, ..., xn} is finite. Then, we see that "Fisher's maximum likelihood method"="moment method" . [Answer] Assume that a measured valuexm(∈ X) is obtained by the measurementMA(O=(X, 2 X , F ), S[∗]) [Fisher's maximum likelihood method]: 135 Ishikawa's Homepage 5.4 Moment method: useful but artificial (a) Find ω0(∈ Ω) such that [F ({xm})](ω0) = max ω∈Ω [F ({xm})](ω) [Moment method]: (b) Since we get the approximate sample probability space (X, 2X , δxm), we see |0− [F ({x1})](ω)|+ * * *+ |0− [F ({xm−1})](ω)|+ |1− [F ({xm})](ω)| + |0− [F ({xm+1})](ω)|+ * * *+ |0− [F ({xn})](ω)| =[F ({x1})](ω) + * * *+ [F ({xm−1})](ω) + [F ({xm})](ω) + [F ({xm+1})](ω) + * * *+ [F ({xn})](ω) =1− 2[F ({xm})](ω) Thus, it suffice to find ω0(∈ Ω) such that 1− 2[F ({xm})](ω0) = min ω (1− 2[F ({xm})](ω)) Thus, Fisher's maximum likelihood method and the moment method are the same in this case. 136 Ishikawa's Homepage Chap. 5 Fisher statistics (I) 5.5 Monty Hall problem - Non-Bayesian approach - Monty Hall problem is as follows1. Problem 5.14. [Monty Hall problem ] You are on a game show and you are given the choice of three doors. Behind one door is a car, and behind the other two are goats. You choose, say, door 1, and the host, who knows where the car is, opens another door, behind which is a goat. For example, the host says that ([) the door 3 has a goat. And further, he now gives you the choice of sticking with door 1 or switching to door 2? What should you do? ? ? ? door door door No. 1 No. 2 No. 3 Figure 5.8: Monty Hall problem Answer: Put Ω = {ω1, ω2, ω3} with the discrete topology dD and the counting measure ν. Thus consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Assume that each state δωm(∈ Sp(C(Ω)∗)) means δωm ⇔ the state that the car is behind the door m (m = 1, 2, 3) Define the observable O1 ≡ ({1, 2, 3}, 2{1,2,3}, F1) in L∞(Ω) such that [F1({1})](ω1) = 0.0, [F1({2})](ω1) = 0.5, [F1({3})](ω1) = 0.5, [F1({1})](ω2) = 0.0, [F1({2})](ω2) = 0.0, [F1({3})](ω2) = 1.0, [F1({1})](ω3) = 0.0, [F1({2})](ω3) = 1.0, [F1({3})](ω3) = 0.0, (5.19) 1This section is extracted from the followings: (a) Ref. [30]: S. Ishikawa, "Mathematical Foundations of Measurement Theory," Keio University Press Inc. 2006. (b) Ref. [34]: S. Ishikawa, "Monty Hall Problem and the Principle of Equal Probability in Measurement Theory," Applied Mathematics, Vol. 3 No. 7, 2012, pp. 788-794. doi: 10.4236/am.2012.37117. 137 Ishikawa's Homepage 5.5 Monty Hall problem - Non-Bayesian approach - where it is also possible to assume that F1({2})(ω1) = α, F1({3})(ω1) = 1−α (0 < α < 1). The fact that you say "the door 1" clearly means that you take a measurement ML∞(Ω)(O1, S[∗]). Here, we assume that a) "a measured value 1 is obtained by the measurement ML∞(Ω)(O1, S[∗])" ⇔ The host says "Door 1 has a goat" b) "measured value 2 is obtained by the measurement ML∞(Ω)(O1, S[∗]) " ⇔ The host says "Door 2 has a goat" c) "measured value 3 is obtained by the measurement ML∞(Ω)(O1, S[∗]) " ⇔ The host says "Door 3 has a goat" Recall that, in Problem 5.14, the host said "Door 3 has a goat". This implies that you get the measured value "3" by the measurement ML∞(Ω)(O1, S[∗]). Therefore, Theorem 5.6 (Fisher's maximum likelihood method) says that you should pick door number 2. That is because we see that max{[F1({3})](ω1), [F1({3})](ω2), [F1({3})](ω3)} = max{0.5, 1.0, 0.0} = 1.0 = [F1({3})](ω2) and thus, there is a reason to infer that wquaualweigh[∗] = δω2 . Thus, you should switch to door 2. This is the first answer to Problem 5.14 (Monty-Hall problem). ♠Note 5.4. Examining the above example, the readers should understand that the problem "What is measurement?" is an unreasonable demand. Thus, we abandon the realistic approach, and accept the metaphysical approach. Also, for a Bayesian approach to Monty Hall problem, see Chapter 9 and Chapter 19. Remark 5.15. [The answer by the moment method] In the above, a measured value "3" is obtained by the measurement ML∞(Ω)(O=({1, 2, 3}, 2{1,2,3}, F ), S[∗]). Thus, the approximate sample space ({1, 2, 3}, 2{1,2,3}, ν1) is obtained such that ν1({1}) = 0, ν1({2}) = 0, ν1({3}) = 1. Therefore, [when the unknown [∗] is ω1] (5.17) = |0− 0|+ |0− 0.5|+ |1− 0.5| = 1, 138 Ishikawa's Homepage Chap. 5 Fisher statistics (I) [when the unknown [∗] is ω2] (5.17) = |0− 0|+ |0− 0|+ |1− 1| = 0 [when the unknown [∗] is ω3] (5.17) = |0− 0|+ |0− 1|+ |1− 0| = 2. Thus, we can infer that [∗] = ω2. That is, you should change to the Door 2. 139 Ishikawa's Homepage 5.6 The two envelope problem -Non-Bayesian approach - 5.6 The two envelope problem-Non-Bayesian approach - This section is extracted from the following: Ref. [47]: S. Ishikawa; The two envelopes paradox in non-Bayesian and Bayesian statistics ( arXiv:1408.4916v4 [stat.OT] 2014 ) Also, for a Bayesian approach to the two envelope problem, see Chapter 9. 5.6.1 Problem(the two envelope problem) The following problem is the famous "two envelope problem( cf. [68] )". Problem 5.16. [The two envelope problem] The host presents you with a choice between two envelopes (i.e., Envelope A and Envelope B). You know one envelope contains twice as much money as the other, but you do not know which contains more. That is, Envelope A [resp. Envelope B] contains V1 dollars [resp. V2 dollars]. You know that (a) V1 V2 = 1/2 or, V1 V2 = 2 Define the exchanging map x : {V1, V2} → {V1, V2} by x = { V2, ( if x = V1), V1 ( if x = V2) You choose randomly (by a fair coin toss) one envelope, and you get x1 dollars (i.e., if you choose Envelope A [resp. Envelope B], you get V1 dollars [resp. V2 dollars] ). And the host gets x1 dollars. Thus, you can infer that x1 = 2x1 or x1 = x1/2. Now the host says "You are offered the options of keeping your x1 or switching to my x1". What should you do? Envelope A Envelope B Figure 5.9: Two envelope problem [(P1):Why is it paradoxical?]. You get α = x1. Then, you reason that, with probability 1/2, x1 is equal to either α/2 or 2α dollars. Thus the expected value (denoted Eother(α) at this 140 Ishikawa's Homepage Chap. 5 Fisher statistics (I) moment) of the other envelope is Eother(α) = (1/2)(α/2) + (1/2)(2α) = 1.25α (5.20) This is greater than the α in your current envelope A. Therefore, you should switch to B. But this seems clearly wrong, as your information about A and B is symmetrical. This is the famous two-envelope paradox (i.e., "The Other Person's Envelope is Always Greener" ). 5.6.2 Answer: the two envelope problem 5.16 Consider the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] where the locally compact space Ω is arbitrary, that is, it may be R+ = {ω | ω ≥ 0} or the one point set {ω0} or Ω = {2n | n = 0,±1,±2, . . .}. Put X = R+ = {x | x ≥ 0}. Consider two continuous (or generally, measurable ) functions V1 : Ω→ R+ and V2 : Ω→ R+. such that V2(ω) = 2V1(ω) or, 2V2(ω) = V1(ω) (∀ω ∈ Ω) For each k = 1, 2, define the observable Ok = (X(= R+),F(= BR+ : the Borel field), Fk) in L∞(Ω, ν) such that [Fk(Ξ)](ω) = { 1 ( if Vk(ω) ∈ Ξ) 0 ( if Vk(ω) /∈ Ξ) (∀ω ∈ Ω,∀Ξ ∈ F = BR+ i.e., the Bore field in X(= R+) ) Further, define the observable O = (X,F, F ) in L∞(Ω, ν) such that F (Ξ) = 1 2 ( F1(Ξ) + F2(Ξ) ) (∀Ξ ∈ F) (5.21) That is, [F (Ξ)](ω) =  1 ( if V1(ω) ∈ Ξ, V2(ω) ∈ Ξ) 1/2 ( if V1(ω) ∈ Ξ, V2(ω) /∈ Ξ) 1/2 ( if V1(ω) /∈ Ξ, V2(ω) ∈ Ξ) 0 ( if V1(ω) /∈ Ξ, V2(ω) /∈ Ξ) (∀ω ∈ Ω,∀Ξ ∈ F = BX i.e., Ξ is a Borel set in X(= R+) ) Fix a state ω(∈ Ω), which is assumed to be unknown. Consider the measurement ML∞(Ω,ν)(O = (X,F, F ), S[ω]). Axiom 1 (§2.7) says that 141 Ishikawa's Homepage 5.6 The two envelope problem -Non-Bayesian approach - (A1) the probability that a measured value { V1(ω) V2(ω) } is obtained by the measurement ML∞(Ω,ν)(O = (X,F, F ), S[ω]) is given by { 1/2 1/2 } If you switch to { V2(ω) V1(ω) } , your gain is { V2(ω)− V1(ω) = ω V1(ω)− V2(ω) = −ω } . Therefore, the expectation of switching is (V2(ω)− V1(ω))/2 + (V1(ω)− V2(ω))/2 = 0 That is, it is wrong "The Other Person's envelope is Always Greener". Remark 5.17. The condition (a) in Problem 5.16 is not needed. This condition plays a role to confuse the essence of the problem. 5.6.3 Another answer: the two envelope problem 5.16 For the preparation of the following section (§ 5.6.4), consider the state space Ω such that Ω = R+ with Lebesgue measure ν. Thus, we start from the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Also, putting Ω = {(ω, 2ω) | ω ∈ R+}, we consider the identification: Ω 3 ω ←→ (identification) (ω, 2ω) ∈ Ω (5.22) Further, define V1 : Ω(≡ R+)→ X(≡ R+) and V2 : Ω(≡ R+)→ X(≡ R+) such that V1(ω) = ω, V2(ω) = 2ω (∀ω ∈ Ω) And define the observable O = (X(= R+),F(= BR+ : the Borel field), F ) in L ∞(Ω, ν) such that [F (Ξ)](ω) =  1 ( if ω ∈ Ξ, 2ω ∈ Ξ) 1/2 ( if ω ∈ Ξ, 2ω /∈ Ξ) 1/2 ( if ω /∈ Ξ, 2ω ∈ Ξ) 0 ( if ω /∈ Ξ, 2ω /∈ Ξ) (∀ω ∈ Ω,∀Ξ ∈ F) Fix a state ω(∈ Ω), which is assumed to be unknown. Consider the measurement ML∞(Ω,ν)(O = (X,F, F ), S[ω]). Axiom 1 ( measurement: §2.7) says that 142 Ishikawa's Homepage Chap. 5 Fisher statistics (I) (A2) the probability that a measured value { x = V1(ω) = ω x = V2(ω) = 2ω } is obtained by ML∞(Ω,ν)(O = (X,F, F ), S[ω]) is given by { 1/2 1/2 } If you switch to { V2(ω) V1(ω) } , your gain is { V2(ω)− V1(ω) V1(ω)− V2(ω) } . Therefore, the expectation of switching is (V2(ω)− V1(ω))/2 + (V1(ω)− V2(ω))/2 = 0 That is, it is wrong "The Other Person's envelope is Always Greener". Remark 5.18. The readers should note that Fisher's maximum likelihood method is not used in the two answers ( in §5.6.2 and §5.6.3). If we try to apply Fisher's maximum likelihood method to Problem 5.16 ( Two envelope problem), we get into a dead end. This is shown below. 5.6.4 Where do we mistake in (P1) of Problem 5.16? Now we can answer to the question: Where do we mistake in (P1) of Problem 5.16? Let us explain it in what follows. Assume that (a) a measured value α is obtained by the measurement ML∞(Ω,ν)(O = (X,F, F ), S[∗]) Then, we get the likelihood function f(α, ω) such that f(α, ω) ≡ inf ω1∈Ω [ lim Ξ→{x},[F (Ξ)](ω1)6=0 [F (Ξ)](ω) [F (Ξ)](ω1) ] = { 1 (ω = α/2 or α) 0 ( elsewhere ) 6 α (α 2 , α) (α, 2α) X(= R+) Ω(≈ Ω = R+) Figure 5.10: Two envelope problem 143 Ishikawa's Homepage 5.6 The two envelope problem -Non-Bayesian approach - Therefore, Fisher's maximum likelihood method says that (B1) unknown state [∗] is equal to α/2 or α( If [∗] = α/2 [resp. [∗] = α ], then the switching gain is (α/2− α) [resp. (2α− α)] ) . However, Fisher's maximum likelihood method does not say (B2)  "the probability that [∗] = α/2"=1/2 "the probability that [∗] = α"=1/2 "the probability that [∗] is otherwise"=0 Therefore, we can not calculate ( such as (5.20)): (α/2− α)× 1 2 + (2α− α)× 1 2 = 1.25α (C1) Thus, the sentence "with probability 1/2" in [(P1):Why is it paradoxical?] is wrong. Hence, we can conclude that (C2) If "state space" is specified, there will be no method of a mistake. since the state space is not declared in [(P1):Why is it paradoxical?]. After all, we see (D) If "state space" is specified, there will be no room to make a mistake. since the state space is not declared in [(P1):Why is it paradoxical ?]. Remark 5.19. The condition (b) in Problem 5.16 is indispensable. Without this condition, we can not difine the observable O = (X,F, F ) by the formula (5.23), and thus we can not solve Problem 5.16. However, it is usual to assume the principle of equal weight (i.e., no information is interpreted as a fair coin toss ), or more precisely, (]) the principle that, in the absence of any reason to expect one event rather than another, all the possible events should be assigned the same probability Under this hypothesis, the condition (b) may be often omitted. Also, we will again discuss the principle of equal weight in Chapters 9 and 18. 144 Ishikawa's Homepage Chap. 5 Fisher statistics (I) ♠Note 5.5. The readers may think that (]1) the answer of Problem 5.16 is a direct consequence of the fact that the information about A and B is symmetrical (as mentioned in [(P1): Why is it paradoxical?] in Problem 5.16). That is, it suffices to point out the symmetry. This answer (]1) may not be wrong. But we think that the (]1) is not sufficient. That is because (]2) in the above answer (]1), the problem "What kind of theory (or, language, world view) is used?" is not clear. On the other hand, the answer presented in Section 5.6.2 is based on quantum language. This is quite important. For example, someone may paradoxically assert that it is impossible to decide "Geocentric model vs. Heliocentrism", since motion is relative. However, we can say, at least, that (]3) Heliocentrism is more handy (than Geocentric model) under Newtonian mechanics. That is, I think that (]4) Geocentric model may not be wrong under Aristotle's world view. Therefore, I think that the true meaning of the Copernican revolution is Aristotle's world view −−−−−−−−−−−−−−−−−→ (the Copernican revolution) Newtonian mechanical world view (5.23) and not Geocentric model −−−−−−−−−−−−−−−−−→ (the Copernican revolution) Heliocentrism (5.24) Thus, this (5.24) is merely one of the symbolic events in the Copernican revolution (5.23). The readers should recall my only one assertion in this note, i.e., Figure 1.1 (The history of the world views). 145 Ishikawa's Homepage

Chapter 6 The confidence interval and statistical hypothesis testing The standard university course of statistics is as follows: 1© Inference (maximum likelihood method) (moment method) −→ 2© confidence interval −→ 3© statistical hypothesis testing −→ 4© ANOVA (Analysis of Variance) In the previous chapter, we are concerned with 1© (inference) in quantum language. In this chapter, we devote ourselves to 2© and 3© (confidence interval and statistical hypothesis testing). This chapter is extracted from Ref. [41]: S. Ishikawa; A quantum linguistic characterization of the reverse relation between confidence interval and hypothesis testing ( arXiv:1401.2709 [math.ST] 2014 ) 6.1 Review: classical quantum language(Axiom 1) Firstly, we review classical measurement theory as follows. 147 6.1 Review: classical quantum language(Axiom 1) (A): Axiom 1(measurement) classical pure type (cf. This can be read under the preparation to §2.7) ) With any classical system S, a basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] can be associated in which measurement theory of that classical system can be formulated. In [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))], consider a W ∗-measurement ML∞(Ω,ν) ( O=(X,F, F ), S[δω ] ) ( or, C∗-measurement ML∞(Ω) ( O=(X,F, F ), S[δω ] ) ) . That is, consider • a W ∗-measurement ML∞(Ω,ν) ( O, S[δω ] ) ( or, C∗-measurement ML∞(Ω) ( O=(X,F, F ), S[δω ] ) ) of an observable O=(X,F, F ) for a state δω(∈Mp(Ω) : state space) Then, the probability that a measured value x (∈ X) obtained by the W ∗-measurement ML∞(Ω,ν) ( O, S[δω ] ) ( or, C∗-measurement ML∞(Ω) ( O=(X,F, F ), S[δω ] ) ) belongs to Ξ (∈ F) is given by δω(F (Ξ))(≡ [F (Ξ)](ω) = M(Ω)(δω, F (Ξ))L∞(Ω.ν)) (if F (Ξ) is essentially continuous at δω, or see Definition 2.14 ). In this chapter, we devote ourselves to the simultaneous normal measurement as follows. Example 6.1. [Normal observable]. Let R be the real axis. Define the state space Ω = R×R+, where R+ = {σ ∈ R|σ > 0} with the Lebesgue measure ν. Consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] The normal observable OG = (R,BR, G) in L∞(Ω(≡ R× R+)) is defined by [G(Ξ)](ω) = 1√ 2πσ ∫ Ξ exp[− (x− μ) 2 2σ2 ]dx (6.1) (∀Ξ ∈ BR(= the Borel field in R)), ∀ω = (μ, σ) ∈ Ω = R× R+). Example 6.2. [Simultaneous normal observable]. Let n be a natural number. Let OG = (R,BR, G) be the normal observable in L∞(R × R+). Define the n-th simultaneous normal observable OnG = (Rn,BnR, Gn) in L∞(R× R+) such that [Gn(×nk=1Ξk)](ω) =×nk=1[G(Ξk)](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ ×nk=1Ξk exp[− ∑n k=1(xk − μ)2 2σ2 ]dx1dx2 * * * dxn (6.2) 148 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing (∀Ξk ∈ BR(k = 1, 2, . . . , n), ∀ω = (μ, σ) ∈ Ω = R× R+). Thus, we have the simultaneous normal measurement ML∞(R×R+)(O n G = (Rn,BnR, Gn), S[(μ,σ)]). Consider the maps μ : Rn → R, SS : Rn → R and σ : Rn → R such that μ(x) = μ(x1, x2, . . . , xn) = x1 + x2 + * * *+ xn n (∀x = (x1, x2, . . . , xn) ∈ Rn) (6.3) SS(x) = SS(x1, x2, . . . , xn) = n∑ k=1 (xk − μ(x))2 (∀x = (x1, x2, . . . , xn) ∈ Rn) (6.4) σ(x) = σ(x1, x2, . . . , xn) = √∑n k=1(xk − μ(x))2 n (∀x = (x1, x2, . . . , xn) ∈ Rn) (6.5) Therefore, we get and calculate (by the formulas of Gauss integrals ( in § 7.4)) two image observables μ(OnG) = (R,BR, Gn ◦μ−1) and SS(OnG) = (R+,BR+ , Gn ◦SS −1 ) in L∞(R×R+) as follows. [(Gn ◦ μ−1)(Ξ1)](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ {x∈Rn : μ(x)∈Ξ1} exp[− ∑n k=1(xk − μ)2 2σ2 ]dx1dx2 * * * dxn = √ n√ 2πσ ∫ Ξ1 exp[− n(x− μ) 2 2σ2 ]dx (6.6) (∀Ξ1 ∈ BR, ∀ω = (μ, σ) ∈ Ω ≡ R× R+). and, [(Gn ◦ SS−1)(Ξ2)](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ {x∈Rn : SS(x)∈Ξ2} exp[− ∑n k=1(xk − μ)2 2σ2 ]dx1dx2 * * * dxn = ∫ Ξ2/σ2 pχ 2 n−1(x)dx (6.7) ( ∀Ξ2 ∈ BR+ , ∀ω = (μ, σ) ∈ Ω ≡ R× R+). where pχ 2 n−1(x) is the probability density function of χ 2-distribution with (n − 1) degree of freedom. That is, pχ 2 n−1(x) = x(n−1)/2−1e−x/2 2(n−1)/2Γ((n− 1)/2) (x > 0) (6.8) where, Γ is the Gamma function. 149 Ishikawa's Homepage 6.2 The reverse relation between confidence interval method and statistical hypothesis testing 6.2 The reverse relation between confidence interval method and statistical hypothesis testing In what follows, we shall mention the reverse relation (such as "the two sides of a coin") between confidence interval method and statistical hypothesis testing. We devote ourselves to the classical systems, i.e., the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] 6.2.1 The confidence interval method Consider an observable O = (X,F, F ) in L∞(Ω). Let Θ be a locally compact space (called the second state space), which has the semi-metric dxΘ (∀x ∈ X) such that, (]) for each x ∈ X, the map dxΘ : Θ2 → [0,∞) satisfies (i):dxΘ(θ, θ) = 0, (ii):dxΘ(θ1, θ2) = d x Θ(θ2, θ1), (ii):d x Θ(θ1, θ3) ≤ dxΘ(θ1, θ2) + dxΘ(θ2, θ3). Further, consider two maps E : X → Θ and π : Ω→ Θ. Here, E : X → Θ and π : Ω→ Θ is respectively called an estimator and a system quantity. Theorem 6.3. [Confidence interval method ]. Let a positive number α be 0 < α 1, for example, α = 0.05. For any state ω( ∈ Ω), define the positive number δ1−αω ( > 0) such that: δ1−αω = inf{δ > 0 : [F ({x ∈ X : dxΘ(E(x), π(ω)) < δ})](ω) ≥ 1− α} (6.9) Then we say that: (A) the probability, that the measured value x obtained by the measurement ML∞(Ω) ( O := (X,F, F ), S[ω0] ) satisfies the following condition (6.10), is more than or equal to 1− α (e.g., 1− α = 0.95). dxΘ(E(x), π(ω0)) ≤ δ1−αω0 (6.10) And further, put D1−α,Θx = {π(ω)(∈ Θ) : dxΘ(E(x), π(ω)) ≤ δ1−αω }. (6.11) which is called the (1− α)-confidence interval. Here, we see the following equivalence: (6.10) ⇐⇒ D1−α,Θx 3 π(ω0). (6.12) 150 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing x0 E π E(x0) π(ω0) * ω0 D1−α,Θx0 Θ ΩX Figure 6.1 Confidence interval D1−α,Θx0 Remark 6.4. [(B1):The meaning of confidence interval]. Consider the parallel measurement⊗J j=1 ML∞(Ω) ( O := (X,F, F ), S[ω0] ) , and assume that a measured value x = (x1, x2, . . . , xJ)(∈ XJ) is obtained by the parallel measurement. Recall the formula (6.12). Then, it surely holds that lim J→∞ Num[{j | D1−α,Θxj 3 π(ω0)] J ≥ 1− α(= 0.95) (6.13) where Num[A] is the number of the elements of the set A. Hence Theorem 6.3 can be tested by numerical analysis (with random number). Similarly, Theorem 6.5 ( mentioned later ) can be tested. [(B2)] Also, note that (6.9) = δ1−αω = inf{δ > 0 : [F ({x ∈ X : dxΘ(E(x), π(ω)) < δ})](ω) ≥ 1− α} = inf{η > 0 : [F ({x ∈ X : dxΘ(E(x), π(ω)) ≥ η})](ω) ≤ α} (6.14) 6.2.2 Statistical hypothesis testing Next, we shall explain the statistical hypothesis testing, which is characterized as the reverse of the confident interval method. Theorem 6.5. [Statistical hypothesis testing]. Let α be a real number such that 0 < α 1, for example, α = 0.05. For any state ω( ∈ Ω), define the positive number ηαω ( > 0) such that: ηαω = inf{η > 0 : [F ({x ∈ X : dxΘ(E(x), π(ω)) ≥ η})](ω) ≤ α} (6.15) ( by the (6.14), note that δ1−αω = η α ω) Then we say that: 151 Ishikawa's Homepage 6.2 The reverse relation between confidence interval method and statistical hypothesis testing (C) the probability, that the measured value x obtained by the measurement ML∞(Ω) ( O := (X,F, F ), S[ω0] ) satisfies the following condition (6.16), is less than or equal to α (e.g., α = 0.05). dxΘ(E(x), π(ω0)) ≥ ηαω0 . (6.16) Further, consider a subset HN of Θ, which is called a "null hypothesis". Put Rα,ΘHN = ∩ ω∈Ω such that π(ω)∈HN {E(x)(∈ Θ) : dxΘ(E(x), π(ω)) ≥ ηαω}. (6.17) which is called the (α)-rejection region of the null hypothesis HN . Then we say that: (D) the probability, that the measured value x obtained by the measurement ML∞(Ω) ( O := (X,F, F ), S[ω0] ) (where π(ω0) ∈ HN) satisfies the following condition (6.18), is less than or equal to α (e.g., α = 0.05). RαHN 3 E(x). (6.18) x0 E π E(x0) π(ω0) * ω0 RαHN Θ ΩX Figure 6.2: Rejection region RαHN (when HN = {π(ω0)} Corollary 6.6. [The reverse relation between Confidence interval and statistical hypothesis testing ]. Let 0 < α 1. Consider an observable O = (X,F, F ) in L∞(Ω), and the second state space Θ (i.e., locally compact space with a semi-metric dxΘ(x ∈ X) ). And consider the estimator E : X → Θ and the system quantity π : Ω→ Θ. Define δ1−αω by (6.9), and define ηαω by (6.15) ( and thus, δ1−αω = η α ω). (E) [Confidence interval method]. for each x ∈ X, define (1− α)-confidence interval by D1−α,Θx = {π(ω)(∈ Θ) : dxΘ(E(x), π(ω)) < δ1−αω } (6.19) Also, D1−α,Ωx = {ω(∈ Ω) : dxΘ(E(x), π(ω)) < δ1−αω } (6.20) 152 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing Here, assume that a measured value x(∈ X) is obtained by the measurement ML∞(Ω) ( O := (X,F, F ), S[ω0] ) . Then, we see that (E1) the probability that D1−α,Θx 3 π(ω0) or, in the same sense D1−α,Ωx 3 ω0 is more than 1− α. (F) [statistical hypothesis testing]. Consider the null hypothesis HN(⊆ Θ). Assume that the state ω0(∈ Ω) satisfies: π(ω0) ∈ HN(⊆ Θ) Here, put, Rα;ΘHN = ∩ ω∈Ω such that π(ω)∈HN {E(x)(∈ Θ) : dxΘ(E(x), π(ω)) ≥ ηαω}. (6.21) or, Rα;XHN = E −1(Rα;ΘHN ) = ∩ ω∈Ω such that π(ω)∈HN {x(∈ X) : dxΘ(E(x), π(ω)) ≥ ηαω}. (6.22) which is called the (α)-rejection region of the null hypothesis HN . Assume that a measured value x(∈ X) is obtained by the measurement ML∞(Ω) ( O := (X,F, F ), S[ω0] ) . Then, we see that (F1) the probability that "E(x) ∈ Rα;ΘHN " or, in the same sense, "x ∈ R α;X HN " (6.23) is less than α. 153 Ishikawa's Homepage 6.3 Confidence interval and statistical hypothesis testing for population mean 6.3 Confidence interval and statistical hypothesis testing for population mean Consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Fix a positive number α such that 0 < α 1, for example, α = 0.05. 6.3.1 Preparation (simultaneous normal measurement) Example 6.7. Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]) in L ∞(R×R+). Here, the simultaneous normal observable OnG = (Rn,BnR, Gn) is defined by [Gn(×nk=1Ξk)](ω) =×nk=1[G(Ξk)](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ ×nk=1Ξk exp[− ∑n k=1(xk − μ)2 2σ2 ]dx1dx2 * * * dxn (6.24) (∀Ξk ∈ BR(k = 1, 2, . . . , n), ∀ω = (μ, σ) ∈ Ω = R× R+). Therefore, the state space Ω and the measured value space X are defined by Ω = R× R+ X = Rn Also, the second state space Θ is defined by Θ = R The estimator E : Rn → Θ(≡ R) and the system quantityπ : Ω → Θ are respectively defined by E(x) = E(x1, x2, . . . , xn) = μ(x) = x1 + x2 + * * *+ xn n Ω = R× R+ 3 ω = (μ, σ) 7→ π(ω) = μ ∈ Θ = R Also, the semi-metric d (1) Θ in Θ is defined by d (1) Θ (θ1, θ2) = |θ1 − θ2| (∀θ1, θ2 ∈ Θ = R) 154 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing 6.3.2 Confidence interval Our present problem is as follows. Problem 6.8. [Confidence interval]. Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]). Assume that a measured valuex ∈ X = Rn is obtained by the measurement. Let 0 < α 1. Then, find the D1−α;Θx (⊆ Θ) (which may depend on σ) such that • the probability that μ ∈ D1−α;Θx is more than 1− α. Here, the more D1−α;Θx (⊆ Θ) is small, the more it is desirable. Consider the following semi-distance d (1) Ω in the state space R× R+: d (1) Ω ((μ1, σ1), (μ2, σ2)) = |μ1 − μ2| (6.25) For any ω = (μ, σ)( ∈ Ω = R× R+), define the positive number δ1−αω ( > 0) such that: δ1−αω = inf{η > 0 : [F (E−1(Balld(1)Ω (ω; η))](ω) ≥ 1− α} where Ball d (1) Ω (ω; η) = {ω1( ∈ Ω) : d(1)Ω (ω, ω1) ≤ η} = [μ− η, μ+ η]× R+ Hence we see that E−1(Ball d (1) Ω (ω; η)) = E−1([μ− η, μ+ η]× R+) ={(x1, . . . , xn) ∈ Rn : μ− η ≤ x1 + . . .+ xn n ≤ μ+ η} (6.26) Thus, [Gn(E−1(Ball d (1) Ω (ω; η))](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ μ−η≤x1+...+xn n ≤μ+η exp[− ∑n k=1(xk − μ)2 2σ2 ]dx1dx2 * * * dxn = 1 ( √ 2πσ)n ∫ * * * ∫ −η≤x1+...+xn n ≤η exp[− ∑n k=1(xk) 2 2σ2 ]dx1dx2 * * * dxn = √ n√ 2πσ ∫ η −η exp[− nx 2 2σ2 ]dx = 1√ 2π ∫ √nη/σ − √ nη/σ exp[− x 2 2 ]dx (6.27) Solving the following equation: 1√ 2π ∫ −z(α/2) −∞ exp[− x 2 2 ]dx = 1√ 2π ∫ ∞ z(α/2) exp[− x 2 2 ]dx = α 2 (6.28) 155 Ishikawa's Homepage 6.3 Confidence interval and statistical hypothesis testing for population mean we define that δ1−αω = σ√ n z( α 2 ) (6.29) Then, for any x ( ∈ Rn), we get D1−α,Ωx ( the (1− α)-confidence interval of x ) as follows: D1−α,Ωx = {ω(∈ Ω) : dΩ(E(x), ω) ≤ δ1−αω } = {(μ, σ) ∈ R× R+ : |μ− μ(x)| = |μ− x1 + . . .+ xn n | ≤ σ√ n z( α 2 )} (6.30) Also, D1−α,Θx = {π(ω)(∈ Θ) : dΩ(E(x), ω) ≤ δ1−αω } = {μ ∈ R : |μ− μ(x)| = |μ− x1 + . . .+ xn n | ≤ σ√ n z( α 2 )} which depends on σ. R R+ D1−α,Ωx 6 μ(x) Figure 6.3: Confidence interval D1−α,Ωx for the semi-distance d (1) Ω 6.3.3 Statistical hypothesis testing[null hypothesisHN = {μ0}(⊆ Θ = R)] Problem 6.9. [Statistical hypothesis testing]. Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]). Assume the null hypothesis HN such that HN = {μ0}(⊆ Θ = R)) Let 0 < α 1. Then, find the rejection region Rα;ΘHN (⊆ Θ) (which may depend on σ) such that • the probability that a measured value x(∈ Rn) obtained by ML∞(R×R+) (OnG = (Rn,BnR, Gn), S[(μ0,σ)]) satisfies that E(x) ∈ Rα;ΘHN 156 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing is less than α. Here, the more the rejection region Rα;ΘHN is large, the more it is desirable. Define the null hypothesis HN such that HN = {μ0}(⊆ Θ(= R)) For any ω = (μ, σ)( ∈ Ω = R× R+), define the positive number ηαω ( > 0) such that: ηαω = inf{η > 0 : [F (E−1(BallCd(1)Θ (π(ω); η))](ω) ≤ α} where BallC d (1) Θ (π(ω); η) = {θ( ∈ Θ) : d(1)Θ (μ, θ) ≥ η} = ( (−∞, μ− η] ∪ [μ+ η,∞) ) Hence we see that E−1(BallC d (1) Θ (π(ω); η)) = E−1 ( (−∞, μ− η] ∪ [μ+ η,∞) ) ={(x1, . . . , xn) ∈ Rn : x1 + . . .+ xn n ≤ μ− η or μ+ η ≤ x1 + . . .+ xn n } ={(x1, . . . , xn) ∈ Rn : | (x1 − μ) + . . .+ (xn − μ) n | ≥ η} (6.31) Thus, [Gn(E−1(BallC d (1) Θ (π(ω); η))](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ | (x1−μ)+...+(xn−μ) n |≥η exp[− ∑n k=1(xk − μ)2 2σ2 ]dx1dx2 * * * dxn = 1 ( √ 2πσ)n ∫ * * * ∫ |x1+...+xn n |≥η exp[− ∑n k=1(xk) 2 2σ2 ]dx1dx2 * * * dxn = √ n√ 2πσ ∫ x≥η exp[− nx 2 2σ2 ]dx = 1√ 2π ∫ x≥ √ nη/σ exp[− x 2 2 ]dx (6.32) Solving the following equation: 1√ 2π ∫ −z(α/2) −∞ exp[− x 2 2 ]dx = 1√ 2π ∫ ∞ z(α/2) exp[− x 2 2 ]dx = α 2 (6.33) we define that ηαω = σ√ n z( α 2 ) (6.34) 157 Ishikawa's Homepage 6.3 Confidence interval and statistical hypothesis testing for population mean Therefore, we get RαHN ( the (α)-rejection region of HN(= {μ0} ⊆ Θ(= R)) ) as follows: Rα,Θ{μ0} = ∩ π(ω)=μ∈{μ0} {E(x)(∈ Θ = R) : d(1)Θ (E(x), π(ω)) ≥ η α ω} = {E(x)(= x1 + . . .+ xn n ) ∈ R : μ(x)− μ0 = x1 + . . .+ xn n − μ0 ≥ σ√ n z( α 2 )} (6.35) Remark 6.10. Note that the Rα,Θ{μ0} ( the (α)-rejection region of {μ0} ) depends on σ. Thus, putting Rα{μ0}×R+ = {(μ(x), σ) ∈ R× R+ : |μ0 − μ(x)| = |μ0 − x1 + . . .+ xn n | ≥ σ√ n z( α 2 )} (6.36) we see that Rα{μ0}×R+="the slash part in Figure 6.4". R σ Rα{μ0}×R+ 6 μ0 Figure 6.4: Rejection region Rα{μ0} (which depends on σ) 6.3.4 Statistical hypothesis testing[null hypothesisHN = (−∞, μ0](⊆ Θ(= R))] Our present problem was as follows Problem 6.11. [Statistical hypothesis testing]. Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]). Assume the null hypothesis HN such that HN = (−∞, μ0](⊆ Θ = R)) Let 0 < α 1. Then, find the rejection region Rα;ΘHN (⊆ Θ) (which may depend on σ) such that • the probability that a measured value x(∈ Rn) obtained by ML∞(R×R+) (OnG = 158 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing (Rn,BnR, Gn), S[(μ0,σ)]) satisfies that E(x) ∈ Rα;ΘHN is less than α. Here, the more the rejection region Rα;ΘHN is large, the more it is desirable. [Rejection region of HN = (−∞, μ0] ⊆ Θ(= R)]. Consider the simultaneous measurement ML∞(R×R+) (O n N = (Rn,BnR, Gn), S[(μ,σ)]) in L∞(R × R+). Thus, we consider that Ω = R × R, X = Rn. Assume that the real σ in a state ω = (μ, σ) ∈ Ω is fixed and known. Put Θ = R The formula (6.3) urges us to define the estimator E : Rn → Θ(≡ R) such that E(x) == μ(x) = x1 + x2 + * * *+ xn n (6.37) And consider the quantity π : Ω→ Θ such that Ω = R× R+ 3 ω = (μ, σ) 7→ π(ω) = μ ∈ Θ = R Consider the following semi-distance d (2) Θ in Θ(= R): d (2) Θ ((θ1, θ2) =  |θ1 − θ2| θ0 ≤ θ1, θ2 |θ2 − θ0| θ1 ≤ θ0 ≤ θ2 |θ1 − θ0| θ2 ≤ θ0 ≤ θ1 0 θ1, θ2 ≤ θ0 (6.38) Define the null hypothesis HN such that HN = (−∞, μ0](⊆ Θ(= R)) For any ω = (μ, σ)( ∈ Ω = R× R+), define the positive number ηαω ( > 0) such that: ηαω = inf{η > 0 : [F (E−1(BallCd(2)Θ (π(ω); η))](ω) ≤ α} where BallC d (2) Θ (π(ω); η) = {θ( ∈ Θ) : d(2)Θ (μ, θ) ≥ η} = ( (−∞, μ− η] ∪ [μ+ η,∞) ) Hence we see that E−1(BallC d (2) Θ (π(ω); η)) = E−1 ( [μ+ η,∞) ) ={(x1, . . . , xn) ∈ Rn : μ+ η ≤ x1 + . . .+ xn n } 159 Ishikawa's Homepage 6.3 Confidence interval and statistical hypothesis testing for population mean ={(x1, . . . , xn) ∈ Rn : (x1 − μ) + . . .+ (xn − μ) n ≥ η} (6.39) Thus, [Gn(E−1(BallC d (2) Θ (π(ω); η))](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ (x1−μ)+...+(xn−μ) n ≥η exp[− ∑n k=1(xk − μ)2 2σ2 ]dx1dx2 * * * dxn = 1 ( √ 2πσ)n ∫ * * * ∫ x1+...+xn n ≥η exp[− ∑n k=1(xk) 2 2σ2 ]dx1dx2 * * * dxn = √ n√ 2πσ ∫ |x|≥η exp[− nx 2 2σ2 ]dx = 1√ 2π ∫ |x|≥ √ nη/σ exp[− x 2 2 ]dx (6.40) Solving the following equation: 1√ 2π ∫ −z(α/2) −∞ exp[− x 2 2 ]dx = 1√ 2π ∫ ∞ z(α/2) exp[− x 2 2 ]dx = α (6.41) we define that ηαω = σ√ n z(α) (6.42) Then, we get Rα,ΘHN ( the (α)-rejection region of HN(= (−∞, μ0] ⊆ Θ(= R)) ) as follows: Rα,Θ(−∞,μ0] = ∩ π(ω)=μ∈(−∞,μ0] {E(x)(∈ Θ = R) : d(2)Θ (E(x), π(ω)) ≥ η α ω} = {E(x)(= x1 + . . .+ xn n ) ∈ R : x1 + . . .+ xn n − μ0 ≥ σ√ n z(α)} (6.43) Thus, in a similar way of Remark 6.10, we see that Rα(−∞,μ0]×R+="the slash part in Figure 6.5", where Rα(−∞,μ0]×R+ = {(E(x)(= x1 + . . .+ xn n ), σ) ∈ R× R+ : x1 + . . .+ xn n − μ0 ≥ σ√ n z(α)} (6.44) 160 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing R σ Rα(−∞,μ0]×R+ 6 μ0 Figure 6.5: Rejection region Rα,Θ(−∞,μ0] (which depends on σ) 161 Ishikawa's Homepage 6.4 Confidence interval and statistical hypothesis testing for population variance 6.4 Confidence interval and statistical hypothesis testing for population variance 6.4.1 Preparation (simultaneous normal measurement) Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]) in L∞(R × R+). Here, recall that the simultaneous normal observable OnG = (Rn,BnR, Gn) is defined by [Gn(×nk=1Ξk)](ω) =×nk=1[G(Ξk)](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ ×nk=1Ξk exp[− ∑n k=1(xk − μ)2 2σ2 ]dx1dx2 * * * dxn (6.45) (∀Ξk ∈ BR(k = 1, 2, . . . , n), ∀ω = (μ, σ) ∈ Ω = R× R+). where, note that Ω = R× R+ X = Rn The second state space Θ is Θ = R+ Putting μ(x) = x1 + x2 + * * *+ xn n we define the estimator E : Rn → Θ(≡ R+) by E(x) = E(x1, x2, . . . , xn) = √ (x1 − μ(x))2 + (x2 − μ(x))2 + * * *+ (xn − μ(x))2 n and the system quantity π : Ω→ Θ by Ω = R× R+ 3 ω = (μ, σ) 7→ π(ω) = σ ∈ Θ = R+ 162 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing 6.4.2 Confidence interval Our present problem is as follows. Problem 6.12. [Confidence interval for population variance]. Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]). Assume that a measured valuex ∈ X = Rn is obtained by the measurement. Let 0 < α 1. Then, find the D1−α;Θx (⊆ Θ) (which may depend on μ) such that • the probability that σ ∈ D1−α;Θx is more than 1− α Here, the more D1−α;Θx (⊆ Θ) is small, the more it is desirable. Consider the following semi-distance d (1) Θ in Θ(= R+): d (1) Θ (θ1, θ2) = | ∫ σ2 σ1 1 σ dσ| = | log σ1 − log σ2| (6.46) For any ω = (μ, σ)( ∈ Ω = R× R+), define the positive number δ1−αω ( > 0) such that: δ1−αω = inf{η > 0 : [F (E−1(Balld(1)Θ (ω; η))](ω) ≥ 1− α} = inf{η > 0 : [F (E−1(BallC d (1) Θ (ω; η))](ω) ≤ α} (6.47) where BallC d (1) Θ (ω; η) = BallC d (1) Θ ((μ;σ), η) = R× {σ′ : | log(σ′/σ)| ≥ η} = R× ( (0, σe−η] ∪ [σeη,∞) ) (6.48) Then, E−1(BallC d (1) Θ (ω; η)) = E−1 ( R× ( (0, σe−η] ∪ [σeη,∞) )) ={(x1, . . . , xn) ∈ Rn : (∑n k=1(xk − μ(x))2 n )1/2 ≤ σe−η or σeη ≤ (∑n k=1(xk − μ(x))2 n )1/2 } (6.49) Hence we see, by the Gauss integral (6.7), that [Gn(E−1(BallC d (1) Θ (ω; η))](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ E−1 ( R× ( (0,σe−η ]∪[σeη ,∞) )) exp[− ∑n k=1(xk − μ)2 2σ2 ]dx1dx2 * * * dxn = ∫ ne−2η 0 pχ 2 n−1(x)dx+ ∫ ∞ ne2η pχ 2 n−1(x)dx = 1− ∫ ne2η ne−2η pχ 2 n−1(x)dx (6.50) 163 Ishikawa's Homepage 6.4 Confidence interval and statistical hypothesis testing for population variance Using the chi-squared distribution pχ 2 n−1(x) (with n− 1 degrees of freedom) in (6.8), define the δ1−αω such that 1− α = ∫ ne2δ1−αω ne−2δ 1−α ω pχ 2 n−1(x)dx (6.51) where it should be noted that the δ1−αω depends on only α and n. Thus, put δ1−αω = δ 1−α n (6.52) Hence we get, for any x ( ∈ X), the D1−α,Ωx ( the (1− α)-confidence interval of x ) as follows: D1−α,Ωx = {ω(∈ Ω) : d (1) Θ (E(x), π(ω)) ≤ δ 1−α n } = {(μ, σ) ∈ R× R+ : σe−δ 1−α n ≤ (∑n k=1(xk − μ(x))2 n )1/2 ≤ σeδ 1−α n } (6.53) Recalling (6.4), i.e., σ(x) = (∑n k=1(xk−μ(x))2 n )1/2 = (SS(x) n ) 1/2 , we conclude that D1−α,Ωx = {(μ, σ) ∈ R× R+ : σ(x)e−δ 1−α n ≤ σ ≤ σ(x)eδ 1−α n } = {(μ, σ) ∈ R× R+ : e−2δ 1−α n n SS(x) ≤ σ2 ≤ e 2δ1−αn n SS(x)} (6.54) And D1−α,Θx = {σ ∈ R+ : σ(x)e−δ 1−α n ≤ σ ≤ σ(x)eδ 1−α n } = {(μ, σ) ∈ R× R+ : e−2δ 1−α n n SS(x) ≤ σ2 ≤ e 2δ1−αn n SS(x)} R R+ D1−α,Ωx 6 σ(x)eδ 1−α n I σ(x)e−δ 1−α n Figure 6.6: Confidence interval D1−α,Ωx for the semi-distance d (1) Θ 164 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing 6.4.3 Statistical hypothesis testing[null hypothesisHN = {σ0} ⊆ Θ = R+] Our present problem is as follows. Problem 6.13. [Statistical hypothesis testing]. Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]). Assume the null hypothesis HN such that HN = {σ0}(⊆ Θ = R)) Let 0 < α 1. Then, find the rejection region Rα;ΘHN (⊆ Θ) (which may depend on μ) such that • the probability that a measured valuex(∈ Rn) obtained by ML∞(R×R+) (OnG = (Rn,BnR, Gn), S[(μ0,σ)]) satisfies that E(x) ∈ Rα;ΘHN is less that α. Here, the more the rejection region Rα;ΘHN is large, the more it is desirable. For any ω = (μ, σ)( ∈ Ω = R× R+), define the positive number ηαω ( > 0) such that: ηαω = inf{η > 0 : [F (E−1(BallCd(1)Θ (ω; η))](ω) ≤ α} Recall that ηαω = δ 1−α ω = δ 1−α n (= η α n) Hence we get the Rα,ΘHN ( the (α)-rejection region of HN = {σ0} ⊆ Θ = R+ ) as follows: Rα,ΘHN = R α,Θ {σ0} = ∩ π(ω)=σ∈{σ0} {E(x)(∈ Θ) : d(1)Θ (E(x), π(ω)) ≥ η α ω} = {E(x)(∈ Θ = R+) : d(1)Θ (E(x), σ0) ≥ η α n} = {σ(x)(∈ Θ = R+) : σ(x) ≤ σ0e−η α n or σ0e ηαn ≤ σ(x)} (6.55) where σ(x) = (∑n k=1(xk−μ(x))2 n )1/2 . Thus, in a similar way of Remark 6.10, we see that RαR×{σ0}="the slash part in Figure 6.7", where RαR×{σ0} = {(μ, σ(x)) ∈ R× R+ : σ(x) ≤ σ0e −ηαn or σ0e ηαn ≤ σ(x)} (6.56) 165 Ishikawa's Homepage 6.4 Confidence interval and statistical hypothesis testing for population variance μ R+ RαR×{σ0} 6 σ0e ηαn  σ0 I σ0e−η α n Figure 6.7: Rejection region RαR×{σ0} 6.4.4 Statistical hypothesis testing[null hypothesisHN = (0, σ0] ⊆ Θ = R+] Our present problem is as follows. Problem 6.14. [Statistical hypothesis testing]. Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]). Assume the null hypothesis HN such that HN = (0, σ0](⊆ Θ = R)) Let 0 < α 1. Then, find the rejection region Rα;ΘHN (⊆ Θ) (which may depend on μ) such that • the probability that a measured valuex(∈ Rn) obtained by ML∞(R×R+) (OnG = (Rn,BnR, Gn), S[(μ0,σ)]) satisfies that E(x) ∈ Rα;ΘHN is less that α. Here, the more the rejection region Rα;ΘHN is large, the more it is desirable. Consider the following semi-distance d (2) Θ in Θ(= R+): d (2) Θ (σ1, σ2) =  | ∫ σ2 σ1 1 σ dσ| = | log σ1 − log σ2| (σ0 ≤ σ1, σ2) | ∫ σ2 σ0 1 σ dσ| = | log σ0 − log σ2| (σ1 ≤ σ0 ≤ σ2) | ∫ σ1 σ0 1 σ dσ| = | log σ0 − log σ1| (σ2 ≤ σ0 ≤ σ1) 0 (σ1, σ2 ≤ σ0) (6.57) For any ω = (μ, σ)( ∈ Ω = R× R+), define the positive number ηαω ( > 0) such that: ηαω = inf{η > 0 : [F (E−1(BallCd(2)Θ (ω; η))](ω) ≤ α} (6.58) 166 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing where BallC d (2) Θ (ω; η) = BallC d (2) Θ ((μ; σ), η) = R× [σeη,∞) (6.59) Then, E−1(BallC d (2) Θ (ω; η)) = E−1 ( [σeη,∞) ) ={(x1, . . . , xn) ∈ Rn : σeη ≤ σ(x) = (∑n k=1(xk − μ(x))2 n )1/2 } (6.60) Hence we see, by the Gauss integral (6.7), that [Gn(E−1(BallC d (2) Θ (ω; η))](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ σ0eη≤σ(x) exp[− ∑n k=1(xk − μ)2 2σ2 ]dx1dx2 * * * dxn = ∫ ∞ ne2ησ2 σ2 pχ 2 n−1(x)dx ≤ ∫ ∞ ne2η pχ 2 n−1(x)dx (6.61) Solving the following equation, define the (ηαn) ′(> 0) such that α = ∫ ∞ ne2(η α n ) ′ pχ 2 n−1(x)dx (6.62) Hence we get the Rα,ΘHN ( the (α)-rejection region of HN = (0, σ0] ) as follows: Rα,ΘHN = R α,Θ (0,σ0] = ∩ π(ω)∈(0,σ0] {E(x)(∈ Θ = R+) : d(2)Θ (E(x), π(ω)) ≥ η α ω} = ∩ π(ω)∈(0,σ0] {E(x)(∈ Θ) : d(2)Θ (E(x), π(ω)) ≥ (η α n) ′} = {σ(= σ(x)) ∈ R+ : σ0e(η α n ) ′ ≤ σ(x)} (6.63) where σ(x) = (∑n k=1(xk−μ(x))2 n )1/2 . Thus, in a similar way of Remark 6.10, we see that RαR×(0,σ0]="the slash part in Figure 6.8", where RαR×(0,σ0] = {(μ, σ(x)) ∈ R× R+ : σ0e (ηαn ) ′ ≤ σ(x)} (6.64) 167 Ishikawa's Homepage 6.4 Confidence interval and statistical hypothesis testing for population variance μ R+ RαR×(0,σ0] 6 σ0e (ηαn ) ′  σ0 I σ0e−(η α n ) ′ Figure 6.8: Rejection region RαR×(0,σ0] 168 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing 6.5 Confidence interval and statistical hypothesis testing for the difference of population means 6.5.1 Preparation (simultaneous normal measurement) Consider the parallel measurementML∞((R×R+)×(R×R+)) (O n G⊗OmG = (Rn×Rm ,BnR BmR , Gn⊗ Gm), S[(μ1,σ1,μ2,σ2)]) (in L ∞((R× R+)× (R× R+))) of two normal measurements. Assume that σ1 and σ2 are fixed and known. Thus, this parallel measurement is represented by ML∞(R×R) (O n Gσ1 ⊗ OmGσ1 = (R n × Rm ,BnR BmR , Gσ1n ⊗ Gσ2m), S[(μ1,μ2)]) in L∞(R × R). Here, recall the normal observable (6.1), i.e., [Gσ(Ξ)](μ) = 1√ 2πσ ∫ Ξ exp[− (x− μ) 2 2σ2 ]dx (∀Ξ ∈ BR(=Borel field in R)), ∀μ ∈ R). (6.65) Therefore, we have the state space Ω = R2 = {ω = (μ1, μ2) : μ1, μ2 ∈ R}. Put Θ = R with the distance d (1) Θ (θ1, θ2) = |θ1 − θ2| and consider the quantity π : R2 → R by π(μ1, μ2) = μ1 − μ2 (6.66) The estimator E : X(= X × Y = Rn × Rm)→ Θ(= R) is defined by E(x1, . . . , xn, y1, . . . , ym) = ∑n k=1 xk n − ∑m k=1 yk m (6.67) For any ω = (μ1, μ2)( ∈ Ω = R × R), define the positive number ηαω(= δ1−αω ) ( > 0) such that: ηαω(= δ 1−α ω ) = inf{η > 0 : [F (E−1(BallCd(1)Θ (π(ω); η))](ω) ≥ α} where BallC d (1) Θ (π(ω); η) = (−∞, μ1 − μ2 − η] ∪ [μ1 − μ2 + η,∞). Define the null hypothesis HN (⊆ Θ = R) such that HN = {θ0} Now let us calculate the ηαω as follows: E−1(BallC d (1) Θ (π(ω); η)) = E−1((−∞, μ1 − μ2 − η] ∪ [μ1 − μ2 + η,∞)) ={(x1, . . . , xn, y1, . . . , ym) ∈ Rn × Rm : | ∑n k=1 xk n − ∑m k=1 yk m − (μ1 − μ2)| ≥ η} ={(x1, . . . , xn, y1, . . . , ym) ∈ Rn × Rm : | ∑n k=1(xk − μ1) n − ∑m k=1(yk − μ2) m | ≥ η} (6.68) 169 Ishikawa's Homepage 6.5 Confidence interval and statistical hypothesis testing for the difference of population means Thus, [(Nσ1 n ⊗Nσ2m)(E−1(BallCd(1)Θ (π(ω); η))](ω) = 1 ( √ 2πσ1)n( √ 2πσ2)m × ∫ * * * ∫ | ∑n k=1 (xk−μ1) n − ∑m k=1 (yk−μ2) m |≥η exp[− ∑n k=1(xk − μ1)2 2σ21 − ∑m k=1(yk − μ2)2 2σ22 ]dx1dx2 * * * dxndy1dy2 * * * dym = 1 ( √ 2πσ1)n( √ 2πσ2)m ∫ * * * ∫ | ∑n k=1 xk n − ∑m k=1 yk m |≥η exp[− ∑n k=1 xk 2 2σ21 − ∑m k=1 yk 2 2σ22 ]dx1dx2 * * * dxndy1dy2 * * * dym =1− 1√ 2π( σ21 n + σ22 m )1/2 ∫ η −η exp[− x 2 2( σ21 n + σ22 m ) ]dx (6.69) Using the z(α/2) in (6.33), we get that ηαω = δ 1−α ω = ( σ21 n + σ22 m )1/2z( α 2 ) (6.70) 6.5.2 Confidence interval Our present problem is as follows Problem 6.15. [ Confidence interval for the difference of population means]. Let σ1 and σ2 be positive numbers which are assumed to be fixed. Consider the parallel measurement ML∞(R×R) (OnGσ1 ⊗O m Gσ1 = (Rn×Rm ,BnR BmR , Gσ1n⊗Gσ2m), S[(μ1,μ2)]). Assume that a measured value x = (x, y) = (x1, . . . , xn, y1, . . . , ym) ( ∈ Rn × Rm) is obtained by the measurement. Let 0 < α 1. Then, find the confidence interval D1−α;Θ(x,y) (⊆ Θ) (which may depend on σ1 and σ2) such that • the probability that μ1 − μ2 ∈ D1−α;Θ(x,y) is more than 1− α. Here, the more the confidence interval D1−α;Θ(x,y) is small, the more it is desirable. Therefore, for any x = (x, y) = (x1, . . . , xn, y1, . . . , ym) ( ∈ Rn × Rm), we get D1−αx ( the (1− α)-confidence interval of x ) as follows: D1−α,Ωx = {ω(∈ Ω) : dΘ(E(x), π(ω)) ≤ δ 1−α ω } = {(μ1, μ2) ∈ R× R : | ∑n k=1 xk n − ∑m k=1 yk m − (μ1 − μ2)| ≤ ( σ21 n + σ22 m )1/2z( α 2 )} (6.71) 170 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing 6.5.3 Statistical hypothesis testing[rejection region: null hypothesisHN = {μ0} ⊆ Θ = R] Our present problem is as follows Problem 6.16. [Statistical hypothesis testing for the difference of population means]. Consider the parallel measurement ML∞(R×R) (O n Gσ1 ⊗ OmGσ1 = (R n × Rm ,BnR BmR , Gσ1n ⊗ Gσ2m), S[(μ1,μ2)]). Assume that π(μ1, μ2) = μ1 − μ2 = θ0 ∈ Θ = R that is, assume the null hypothesisHN such that HN = {θ0}(⊆ Θ = R)) Let 0 < α 1. Then, find the rejection region Rα;ΘHN (⊆ Θ) (which may depend on μ) such that • the probability that a measured value(x, y)(∈ Rn×Rm) obtained by ML∞(R×R) (OnGσ1 ⊗ OmGσ1 = (R n × Rm ,BnR BmR , Gσ1n ⊗Gσ2m), S[(μ1,μ2)]) satisfies E(x, y) = x1 + x2 + * * *+ xn n − y1 + y2 + * * *+ ym m ∈ Rα;ΘHN is less than α. Here, the more the rejection region Rα;ΘHN is large, the more it is desirable. By the formula (6.70), we see that the rejection regionRαx ( (α)-rejection region of HN = {θ0}(⊆ Θ) ) is defined by Rα,ΘHN = ∩ ω=(μ1,μ2)∈Ω(=R2) such that π(ω)=μ1−μ2∈HN (={θ0}) {E(x)(∈ Θ) : d(1)Θ (E(x), π(ω)) ≥ η α ω} = {μ(x)− μ(y) ∈ Θ(= R) : |μ(x)− μ(y)− θ0| ≥ ( σ21 n + σ22 m )1/2z( α 2 )} (6.72) or, Rα,XHN = ∩ ω=(μ1,μ2)∈Ω(=R2) such that π(ω)=μ1−μ2∈HN (={θ0}) {x(∈ Rn × Rm) : d(1)Θ (E(x), π(ω)) ≥ η α ω} = {x(∈ Rn × Rm) : |μ(x)− μ(y)− θ0| ≥ ( σ21 n + σ22 m )1/2z( α 2 )} (6.73) Here, μ(x) = ∑n k=1 xk n , μ(y) = ∑m k=1 yk m 171 Ishikawa's Homepage 6.5 Confidence interval and statistical hypothesis testing for the difference of population means 6.5.4 Statistical hypothesis testing[rejection region: null hypothesisHN = (−∞, θ0] ⊆ Θ = R] Our present problem is as follows Problem 6.17. [Statistical hypothesis testing for the difference of population means]. Consider the parallel measurement ML∞(R×R) (O n Gσ1 ⊗ OmGσ1 = (R n × Rm ,BnR BmR , Gσ1n ⊗ Gσ2m), S[(μ1,μ2)]). Assume that π(μ1, μ2) = μ1 − μ2 = (−∞, θ0] ⊆ Θ = R that is, assume the null hypothesisHN such that HN = (−∞, θ0](⊆ Θ = R)) Let 0 < α 1. Then, find the rejection region Rα;ΘHN (⊆ Θ) (which may depend on μ) such that • the probability that a measured value(x, y)(∈ Rn×Rm) obtained by ML∞(R×R) (OnGσ1 ⊗ OmGσ1 = (R n × Rm ,BnR BmR , Gσ1n ⊗Gσ2m), S[(μ1,μ2)]) satisfies E(x, y) = x1 + x2 + * * *+ xn n − y1 + y2 + * * *+ ym m ∈ Rα;ΘHN is less than α. Here, the more the rejection region Rα;ΘHN is large, the more it is desirable. Since the null hypothesis HN is assumed as follows: HN = (−∞, θ0], it suffices to define the semi-distance d (1) Θ in Θ(= R) such that d (1) Θ (θ1, θ2) =  |θ1 − θ2| (∀θ1, θ2 ∈ Θ = R such that θ0 ≤ θ1, θ2) max{θ1, θ2} − θ0 (∀θ1, θ2 ∈ Θ = R such that min{θ1, θ2} ≤ θ0 ≤ max{θ1, θ2}) 0 (∀θ1, θ2 ∈ Θ = R such that θ1, θ2 ≤ θ0) (6.74) Then, we can easily see that Rα,ΘHN = ∩ ω=(μ1,μ2)∈Ω(=R2) such that π(ω)=μ1−μ2∈HN (=(−∞,θ0]) {E(x)(∈ Θ) : d(1)Θ (E(x), π(ω)) ≥ η α ω} = {μ(x)− μ(y) ∈ R : μ(x)− μ(y)− θ0 ≥ ( σ21 n + σ22 m )1/2z(α)} (6.75) 172 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing 6.6 Student t-distribution of population mean 6.6.1 Preparation Example 6.18. [Student t-distribution]. Consider the simultaneous measurement ML∞(R×R+) (OnG = (Rn,BnR, Gn), S[(μ,σ)]) in L∞(R × R+). Thus, we consider that Ω = R × R+, X = Rn. Put Θ = R with the semi-distance dxΘ(∀x ∈ X) such that dxΘ(θ1, θ2) = |θ1 − θ2| σ′(x)/ √ n (∀x ∈ X = Rn, ∀θ1, θ2 ∈ Θ = R) (6.76) where σ′(x) = √ n n−1σ(x). The quantity π : Ω(= R× R+)→ Θ(= R) is defined by Ω(= R× R+) 3 ω = (μ, σ) 7→ π(μ, σ) = μ ∈ Θ(= R) (6.77) Also, define the estimator E : X(= Rn)→ Θ(= R) such that E(x) = E(x1, x2, . . . , xn) = μ(x) = x1 + x2 + * * *+ xn n (6.78) Define the null hypothesis HN (⊆ Θ = R)) such that HN = {μ0} (6.79) Thus, for any ω = (μ0, σ)( ∈ Ω = R× R+), we see that [Gn({x ∈ X(= Rn) : dxΘ(E(x), π(ω)) ≥ η})](ω) =[Gn({x ∈ X : |μ(x)− μ0| σ′(x)/ √ n ≥ η})](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ η≤ |μ(x)−μ0| σ′(x)/ √ n exp[− ∑n k=1(xk − μ0)2 2σ2 ]dx1dx2 * * * dxn = 1 ( √ 2π)n ∫ * * * ∫ η≤ |μ(x)| σ′(x)/ √ n exp[− ∑n k=1(xk) 2 2 ]dx1dx2 * * * dxn =1− ∫ η −η ptn−1(x)dx (6.80) where ptn−1 is the t-distribution with n − 1 degrees of freedom. Solving the equation 1 − α =∫ ηαω −ηαω ptn−1(x)dx, we get δ1−αω = η α ω = t(α/2) 173 Ishikawa's Homepage 6.6 Student t-distribution of population mean 6.6.2 Confidence interval Our present problem is as follows Problem 6.19. [Confidence interval]. Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]). Assume that a measured valuex ∈ X = Rn is obtained by the measurement. Let 0 < α 1. Then, find the confidence interval D1−α;Θx (⊆ Θ) (which does not depend on σ) such that • the probability that μ ∈ D1−α;Θx is more than 1− α Here, the more the confidence interval D1−α;Θx is small, the more it is desirable. Therefore, for any x ( ∈ X), we get D1−α,Θx ( the (1− α)-confidence interval of x ) as follows: D1−αx = {π(ω)(∈ Θ) : ω ∈ Ω, dxΘ(E(x), π(ω)) ≤ δ1−αω } = {μ ∈ Θ(= R) : μ(x)− σ ′(x)√ n t(α/2) ≤ μ ≤ μ(x) + σ ′(x)√ n t(α/2)} (6.81) D1−α,Ωx = {ω = (μ, σ)(∈ Ω) : ω ∈ Ω, dxΘ(E(x), π(ω)) ≤ δ1−αω } = {ω = (μ, σ)(∈ Ω) : μ(x)− σ ′(x)√ n t(α/2) ≤ μ ≤ μ(x) + σ ′(x)√ n t(α/2)} (6.82) 6.6.3 Statistical hypothesis testing[null hypothesisHN = {μ0}(⊆ Θ = R)] Our present problem was as follows Problem 6.20. [Statistical hypothesis testing]. Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]). Assume that μ = μ0 That is, assume the null hypothesis HN such that HN = {μ0}(⊆ Θ = R)) Let 0 < α 1. Then, find the rejection region Rα;ΘHN (⊆ Θ) (which does not depend on σ) such that • the probability that a measured valuex(∈ Rn) obtained by ML∞(R×R+) (OnG = (Rn,BnR, Gn), S[(μ0,σ)]) satisfies E(x) ∈ Rα;ΘHN 174 Ishikawa's Homepage Chap. 6 The confidence interval and statistical hypothesis testing is less than α. Here, the more the rejection region Rα;ΘHN is large, the more it is desirable. The rejection regionRα,ΘHN ( (α)-rejection region of null hypothesis HN(= {μ0}) ) is calculated as follows: Rα,ΘHN = ∩ ω=(μ,σ)∈Ω(=R×R+) such that π(ω)=μ∈HN (={μ0}) {E(x)(∈ Θ) : dxΘ(E(x), π(ω)) ≥ ηαω} = {μ(x) ∈ Θ(= R) : |μ(x)− μ0| σ′(x)/ √ n ≥ t(α/2)} = {μ(x) ∈ Θ(= R) : μ0 ≤ μ(x)− σ′(x)√ n t(α/2) or μ(x) + σ′(x)√ n t(α/2) ≤ μ0} (6.83) Also, Rα,XHN = ∩ ω=(μ,σ)∈Ω(=R×R+) such that π(ω)=μ∈HN (={μ0}) {x ∈ X : dxΘ(E(x), π(ω)) ≥ ηαω} = {x ∈ X = Rn : |μ(x)− μ0| σ′(x)/ √ n ≥ t(α/2)} = {x ∈ X = Rn : μ0 ≤ μ(x)− σ′(x)√ n t(α/2) or μ(x) + σ′(x)√ n t(α/2) ≤ μ0} (6.84) 6.6.4 Statistical hypothesis testing[null hypothesis HN = (−∞, μ0](⊆ Θ = R )] Our present problem was as follows Problem 6.21. [Statistical hypothesis testing]. Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]). Assume that μ ∈ (−∞, μ0] That is, assume the null hypothesis HN such that HN = (−∞, μ0](⊆ Θ = R)) Let 0 < α 1. Then, find the rejection region Rα;ΘHN (⊆ Θ) (which does not depend on σ) such that • the probability that a measured valuex(∈ Rn) obtained by ML∞(R×R+) (OnG = (Rn,BnR, Gn), S[(μ0,σ)]) satisfies E(x) ∈ Rα;ΘHN 175 Ishikawa's Homepage 6.6 Student t-distribution of population mean is less than α. Here, the more the rejection region Rα;ΘHN is large, the more it is desirable. Since the null hypothesis HN is assumed as follows: HN = (−∞, μ0], it suffices to define the semi-distance dxΘ in Θ(= R) such that dxΘ(θ1, θ2) =  |θ1−θ2| σ′(x)/ √ n (∀θ1, θ2 ∈ Θ = R such that μ0 ≤ θ1, θ2) max{θ1,θ2}−μ0 σ′(x)/ √ n (∀θ1, θ2 ∈ Θ = R such that min{θ1, θ2} ≤ μ0 ≤ max{θ1, θ2}) 0 (∀θ1, θ2 ∈ Θ = R such that θ1, θ2 ≤ μ0) (6.85) for any x ∈ X = Rn. Then, (α)-rejection regionRα,ΘHN is calculated as follows. Rα,ΘHN = ∩ ω=(μ,σ)∈Ω(=R×R+) such that π(ω)=μ∈HN (=(−∞,μ0]) {E(x)(∈ Θ) : dxΘ(E(x), π(ω)) ≥ ηαω} = {μ(x) ∈ Θ(= R) : μ0 ≤ μ(x)− σ′(x)√ n t(α)} (6.86) Also, Rα,XHN = ∩ ω=(μ,σ)∈Ω(=R×R+) such that π(ω)=μ∈HN (=(−∞,μ0]) {x(∈ X = Rn) : dxΘ(E(x), π(ω)) ≥ ηαω} = {x(∈ X = Rn) : μ0 ≤ μ(x)− σ′(x)√ n t(α)} (6.87) Remark 6.22. There are many ideas of statistical hypothesis testing. The most natural idea is the likelihood-ratio, which is discussed in (a) Ref. [30]: S. Ishikawa, "Mathematical Foundations of Measurement Theory," Keio University Press Inc. 2006. (b) Ref. [33]: S. Ishikawa, "A Measurement Theoretical Foundation of Statistics," Applied Mathematics, Vol. 3, No. 3, 2012, pp. 283-292. doi: 10.4236/am.2012.33044 Also, we think that the arguments concerning "null hypothesis vs. alternative hypothesis" and "one-sided test and two-sided test" are practical and not theoretical. 176 Ishikawa's Homepage Chapter 7 ANOVA( = Analysis of Variance) The standard university course of statistics is as follows: 1© Inference (likelihood method, moment method) −→ 2© confidence interval −→ 3© statistical hypothesis testing −→ 4© ANOVA In the previous chapters, we studied 1©, 2© and 3©. In this chapter, we devote ourselves to 4©(ANOVA). This chapter is extracted from the following. Ref. [42]: S. Ishikawa, ANOVA (analysis of variance) in the quantum linguistic formulation of statistics ( arXiv:1402.0606 [math.ST] 2014 ) 7.1 Zero way ANOVA (Student t-distribution) In the previous chapter, we introduced the statistical hypothesis testing for student tdistribution, which is characterized as "zero" way ANOVA (analysis of variance ). In this section, we review "zero" way ANOVA (analysis of variance ). Consider the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] where Ω = R× R+ = {(μ, σ) | μ is real, σ is positive real} Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]) ( in L∞(R× R+)). For completeness, recall that 177 7.1 Zero way ANOVA (Student t-distribution) [Gn(×nk=1Ξk)](ω) =×nk=1[G(Ξk)](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ ×nk=1Ξk exp[− ∑n k=1(xk − μ)2 2σ2 ]dx1dx2 * * * dxn (7.1) (∀Ξk ∈ BR(k = 1, 2, . . . , n), ∀ω = (μ, σ) ∈ Ω = R× R+). And recall the state space Ω = R × R+, the measured value space X = Rn, the second state space(=parameter space) Θ = R. Also, recall the estimator E : X(= Rn) → Θ(= R) defined by E(x) = E(x1, x2, . . . , xn) = μ(x) = x1 + x2 + * * *+ xn n (7.2) and the system quantity π : Ω(= R× R+)→ Θ(= R) defined by Ω(= R× R+) 3 ω = (μ, σ) 7→ π(μ, σ) = μ ∈ Θ(= R) (7.3) The essence of "studentized" is to define the semi-metric dxΘ(∀x ∈ X) in the second state space Θ(= R)such that dxΘ(θ (1), θ(2)) = |θ(1) − θ(2)|√ nσ(x) = |θ(1) − θ(2)|√ SS(x) (∀x ∈ X = Rn, ∀θ(1), θ(2) ∈ Θ = R) (7.4) where SS(x) = SS(x1, x2, . . . , xn) = n∑ k=1 (xk − μ(x))2 (∀x = (x1, x2, . . . , xn) ∈ Rn) Thus, as mentioned in the previous chapter, our problem is characterized as follows. Problem 7.1. [The zero-way ANOVA]. Consider the simultaneous normal measurement ML∞(R×R+) (O n G = (Rn,BnR, Gn), S[(μ,σ)]) Here, assume that μ = μ0 That is, the null hypothesis HN is defined by HN = {μ0} (⊆ Θ = R)). Consider 0 < α 1. Then, find the largest Rα;ΘHN (⊆ Θ) (independent of σ) such that (A1) the probability that a measured value x(∈ Rn) (obtained by ML∞(R×R+)(OnG = (X(≡ Rn),BnR, Gn), S[(μ0,σ)])) satisfies E(x) ∈ Rα;ΘHN (7.5) is less than α. 178 Ishikawa's Homepage Chap. 7 ANOVA( = Analysis of Variance) We see, for any ω = (μ0, σ)( ∈ Ω = R× R+), [Gn({x ∈ X : dxΘ(E(x), π(ω)) ≥ η})](ω) =[Gn({x ∈ X : |μ(x)− μ0|√ SS(x) ≥ η})](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ η √ n−1≤ |μ(x)−μ0|√ SS(x)/ √ n−1 exp[− ∑n k=1(xk − μ0)2 2σ2 ]dx1dx2 * * * dxn = 1 ( √ 2π)n ∫ * * * ∫ η2n(n−1)≤ n(μ(x)) 2 SS(x)/(n−1) exp[− ∑n k=1(xk) 2 2 ]dx1dx2 * * * dxn (7.6) (A2) by the formula of Gauss integrals ( Formula 7.8(A)(§7.4)), we see = ∫ ∞ η2n(n−1) pF(1,n−1)(t)dt = α ( e.g., α = 0.05) (7.7) where pF(1,n−1) is the probability density function of F -distribution with (1, n − 1) degree of freedom. Note that the probability density function pF(n1,n2)(t) of F -distribution with (n1, n2) degree of freedom is defined by pF(n1,n2)(t) = 1 B(n1/2, n2/2) (n1 n2 )n1/2 t(n1−2)/2 (1 + n1t/n2)(n1+n2)/2 (t ≥ 0) (7.8) where B(*, *) is the Beta function. The α-point: F n2n1,α (> 0) is defined by∫ ∞ F n2 n1,α pF(n1,n2)(t)dt = α (0 < α 1. e.g., α = 0.05) (7.9) Thus, it suffices to solve the following equation: η2n(n− 1) = F 1n−1,α (7.10) Therefore, (ηαω) 2 = F 1n−1,α n(n− 1) (7.11) Then, the rejection regionRα;ΘHN ( (or R α;X HN ) is calculated as Rα;ΘHN = ∩ ω=(μ,σ)∈Ω(=R×R+) such that π(ω)=μ∈HN (={μ0}) {E(x)(∈ Θ) : dxΘ(E(x), π(ω)) ≥ ηαω} 179 Ishikawa's Homepage 7.1 Zero way ANOVA (Student t-distribution) = {μ(x) ∈ Θ(= R) : |μ(x)− μ0|√ SS(x) ≥ ηαω} = {μ(x) ∈ Θ(= R) : |μ(x)− μ0| σ(x) ≥ ηαω √ n} = { μ(x) ∈ Θ(= R) : |μ(x)− μ0| σ(x) ≥ √ F 1n−1,α n− 1 } = { μ(x) ∈ Θ(= R) : μ0 ≤ μ(x)− σ(x) √ F 1n−1,α n− 1 or μ(x) + σ(x) √ F 1n−1,α n− 1 ≤ μ0 } (7.12) and, Rα;XHN = E −1(Rα;ΘHN ) = { x ∈ X(= Rn) : μ0 ≤ μ(x)− σ(x) √ F 1n−1,α n− 1 or μ(x) + σ(x) √ F 1n−1,α n− 1 ≤ μ0 } (7.13) ♠Note 7.1. (i): It should be noted that the mathematical part is only the (A2). (ii): Also, note that (]) F -distribution with (1, n− 1) degree of freedom = the student t-distribution with (n− 1) degree of freedom Thus, we conclude that (7.12) = (6.83) (7.13) = (6.84) 180 Ishikawa's Homepage Chap. 7 ANOVA( = Analysis of Variance) 7.2 The one way ANOVA For each i = 1, 2, * * * , a, a natural number ni is determined. And put, n = ∑a i=1 ni. Consider the parallel simultaneous normal observable OnG = (X(≡ Rn),BnR, Gn) ( in L∞(Ω(≡ (Ra × R+)) ) such that [Gn(Ξ)](ω) = 1 ( √ 2πσ)n ∫ * * * ∫ Ξ exp[− ∑a i=1 ∑ni k=1(xik − μi)2 2σ2 ] a × i=1 ni× k=1 dxik (7.14) (∀ω = (μ1, μ2, . . . , μa, σ) ∈ Ω = Ra × R+, Ξ ∈ BnR) That is, consider ML∞(Ra×R+)(O n G = (X(≡ Rn),BnR, Gn), S[(μ=(μ1,μ2,*** ,μa),σ)]) Put ai as follows. αi = μi − ∑a i=1 μi a (∀i = 1, 2, . . . , a) (7.15) and put, Θ = Ra Thus,, the system quantity π : Ω→ Θ is defined as follows. Ω = Ra × R+ 3 ω = (μ1, μ2, . . . , μa, σ) 7→ π(ω) = (α1, α2, . . . , αa) ∈ Θ = Ra (7.16) Define the null hypothesis HN(⊆ Θ = Ra) as follows. HN = {(α1, α2, . . . , αa) ∈ Θ = Ra : α1 = α2 = . . . = αa = α} = {( a{ }} { 0, 0, . . . , 0)} (7.17) Here, note the following equivalence: "μ1 = μ2 = . . . = μa"⇔ "α1 = α2 = . . . = αa = 0"⇔ "(7.17)" Hence, our problem is as follows. Problem 7.2. [The one-way ANOVA]. Put n = ∑a i=1 ni. Consider the parallel simultaneous normal measurement ML∞(Ra×R+)(O n G = (X(≡ Rn), BnR, Gn), S[(μ=(μ1,μ2,*** ,μa),σ)]) Here, assume 181 Ishikawa's Homepage 7.2 The one way ANOVA that μ1 = μ2 = * * * = μa that is, π(μ1, μ2, * * * , μa) = (0, 0, * * * , 0) Namely, assume that the null hypothesis is HN = {(0, 0, * * * , 0)} (⊆ Θ = R)). Consider 0 < α 1. Then, find the largest Rα;ΘHN (⊆ Θ) (independent of σ) such that (A1) the probability that a measured value x(∈ Rn) (obtained by ML∞(Ra×R+)(OnG = (X(≡ Rn),BnR, Gn), S[(μ=(μ1,μ2,*** ,μa),σ)])) satisfies E(x) ∈ Rα;ΘHN is less than α. Consider the weighted Euclidean norm ‖θ(1) − θ(2)‖Θ in Θ = Ra as follows. ‖θ(1) − θ(2)‖Θ = √√√√ a∑ i=1 ni ( θ (1) i − θ (2) i )2 (∀θ(`) = (θ(`)1 , θ (`) 2 , . . . , θ (`) a ) ∈ Ra, ` = 1, 2) Also, put X = Rn 3 x = ((xik)k=1,2,...,ni)i=1,2,...,a xi* = ∑ni k=1 xik ni , x** = ∑a i=1 ∑ni k=1 xik ni , (7.18) Theorem 5.6 (Fisher's maximum likelihood method) urges us to calculate σ(x)(= √ SS(x) n ) as follows. For x ∈ X = Rn, SS(x) = SS(((xik) k=1,2,...,ni)i=1,2,...,a ) = a∑ i=1 ni∑ k=1 (xik − xi*)2 = a∑ i=1 ni∑ k=1 (xik − ∑ni k=1 xik ni )2 = a∑ i=1 ni∑ k=1 ((xik − μi)− ∑ni k=1(xik − μi) ni )2 182 Ishikawa's Homepage Chap. 7 ANOVA( = Analysis of Variance) =SS(((xik − μi) k=1,2,...,ni)i=1,2,...,a ) (7.19) For each x ∈ X = Rn, define the semi-norm dxΘ in Θ such that dxΘ(θ (1), θ(2)) = ‖θ(1) − θ(2)‖Θ√ SS(x) (∀θ(1), θ(2) ∈ Θ)). (7.20) Further, define the estimator E : X(= Rn)→ Θ(= Ra) as follows. E(x) =E((xik)i=1,2,...,a,k=1,2,...,n) = (∑ni k=1 x1k n − ∑a i=1 ∑ni k=1 xik n , ∑ni k=1 x2k n − ∑a i=1 ∑ni k=1 xik n , . . . , ∑ni k=1 xak n − ∑a i=1 ∑ni k=1 xik n ) = (∑ni k=1 xik n − ∑a i=1 ∑ni k=1 xik n ) i=1,2,...,a = (xi* − x**)i=1,2,...,a (7.21) Thus, we get ‖E(x)− π(ω)‖2Θ =|| (∑ni k=1 xik n − ∑a i=1 ∑ni k=1 xik n ) i=1,2,...,a − (αi)i=1,2,...,a||2Θ =|| (∑ni k=1 xik n − ∑a i=1 ∑ni k=1 xik n − (μi − ∑a i=1 μi a ) ) i=1,2,...,a ||2Θ remarking the null hypothesis HN (i.e., μi − ∑a k=1 μi a = αi = 0(i = 1, 2, . . . , a)), =|| (∑ni k=1 xik n − ∑a i=1 ∑ni k=1 xik n ) i=1,2,...,a ||2Θ = a∑ i=1 ni(xi* − x**)2 (7.22) Therefore, for any ω = ((μik)i=12,...,a, k=1,2,...,n, σ)( ∈ Ω = Rn × R+), define the positive real ηαω ( > 0) such that ηαω = inf{η > 0 : [Gn(E−1(BallCdxΘ(π(ω); η))](ω) ≥ α} (7.23) where BallCdxΘ(π(ω); η) = {θ ∈ Θ : d x Θ(π(ω), θ) > η} (7.24) Recalling the null hypothesis HN (i.e., μi − ∑a k=1 μi a = αi = 0(i = 1, 2, . . . , a)) , calculate η α ω as follows. E−1(BallCdxΘ(π(ω); η)) = {x ∈ X = R n : dxΘ(E(x), π(ω)) > η} ={x ∈ X = Rn : ‖E(x)− π(ω)‖ 2 Θ SS(x) = ∑a i=1 ni(xi* − x**)2∑a i=1 ∑ni k=1(xik − xi*)2 > η 2} (7.25) 183 Ishikawa's Homepage 7.2 The one way ANOVA For any ω = (μ1, μ2, . . . , μa, σ) ∈ Ω = Ra × R+ such that π(ω)(= (α1, α2, . . . , αa)) ∈ HN(= {0, 0, . . . , 0)}), we see [Gn(E−1(BallCdxΘ(π(ω); η)))(ω) = 1 ( √ 2πσ)n ∫ * * * ∫ ∑a i=1 ni(x i*−x**)2∑a i=1 ∑ni k=1 (xik−x i*)2>η 2 exp[− ∑a i=1 ∑ni k=1(xik − μi)2 2σ2 ] a × i=1 ni× k=1 dxik = 1 ( √ 2π)n ∫ * * * ∫ ( ∑a i=1 ni(x i*−x**)2/(a−1) ( ∑a i=1 ∑ni k=1 (xik−x i*)2)/(n−a)> η2(n−a) (a−1) exp[− ∑a i=1 ∑ni k=1(xik) 2 2 ] a × i=1 ni× k=1 dxik (A2) By the formula of Gauss integrals (Formula 7.8(B)(§7.4)), we see = ∫ ∞ η2(n−a) (a−1) pF(a−1,n−a)(t)dt = α ( e.g., α=0.05) (7.26) where, pF(a−1,n−a) is a probability density function of the F -distribution with p F (a−1,n−a) degree of freedom. Therefore, it suffices to solve the following equation η2(n− a) (a− 1) = F a−1n−a,α(= "α-point") (7.27) This is solved, (ηαω) 2 = F a−1n−a,α(a− 1)/(n− a) (7.28) Then, we get Rα;Θx (or, R α;X x ; the (α)-rejection region of HN = {(0.0. . . . , 0)}(⊆ Θ = Ra) ) as follows: Rα;ΘHN = ∩ ω=((μi)ai=1,σ)∈Ω(=Ra×R+) such that π(ω)=(μ)ai=1∈HN={(0,0,...,0)} {E(x)(∈ Θ) : dxΘ(E(x), π(ω)) ≥ ηαω} = {E(x)(∈ Θ) : ( ∑a i=1 ni(xi* − x**)2)/(a− 1) ( ∑a i=1 ∑ai k=1(xik − xi*)2))/(n− a) ≥ F a−1 n−a,α} (7.29) Thus, Rα;Xx = E −1(Rα;ΘHN ) = {x ∈ X : ( ∑a i=1 ni(xi* − x**)2)/(a− 1) ( ∑a i=1 ∑ni k=1(xik − xi*)2)/(n− a) ≥ F a−1 n−a,α} (7.30) ♠Note 7.2. It should be noted that the mathematical part is only the (A2). 184 Ishikawa's Homepage Chap. 7 ANOVA( = Analysis of Variance) 7.3 The two way ANOVA 7.3.1 Preparation As one of generalizations of the simultaneous normal observable (7.14), we consider a kind of observable OabnG = (X(≡ Rabn),BabnR , Gabn) in L∞(Ω(≡ (Rab × R+)). [Gabn(Ξ)](ω) = 1 ( √ 2πσ)abn ∫ * * * ∫ Ξ exp[− ∑a i=1 ∑b j=1 ∑n k=1(xijk − μij)2 2σ2 ] n × k=1 b × j=1 a × i=1 dxijk (∀ω = ((μij)i=1,2,...,a,j=1,2,...,b, σ) ∈ Ω = Rab × R+, Ξ ∈ BabnR ) (7.31) Therefore, consider the parallel simultaneous normal measurement: ML∞(Rab×R+)(O abn G = (X(≡ Rabn),BabnR , Gabn), S[(μ=(μij | i=1,2,*** ,a,j=1,2,*** ,b),σ)]) Here, μij = μ(= μ** = ∑a i=1 ∑b j=1 μij ab ) + αi(= μi* − μ** = ∑b j=1 μij b − ∑a i=1 ∑b j=1 μij ab ) + βj(= μ*j − μ** = ∑a i=1 μij a − ∑a i=1 ∑b j=1 μij ab ) + (αβ)ij(= μij − μi* − μ*j + μ**) (7.32) And put, X = Rabn 3 x = (xijk)i=1,2,...,a, j=1,2,...,b, k=1,2,...,n xij* = ∑n k=1 xijk n , xi** = ∑b j=1 ∑n k=1 xijk bn , x*j* = ∑a i=1 ∑n k=1 xijk an , x*** = ∑a i=1 ∑b j=1 ∑n k=1 xijk abn (7.33) 7.3.2 The null hypothesis: μ1* = μ2* = * * * = μa* = μ** Now put, Θ = Ra (7.34) 185 Ishikawa's Homepage 7.3 The two way ANOVA define the system quantity π1 : Ω(= Rab × R+)→ Θ(= Ra) by Ω = Rab × R+ 3 ω = ((μij)i=1,2,...,a,j=1,2,...,b, σ) 7→ π1(ω) = (αi)ai=1(= (μi* − μ**)ai=1) ∈ Θ = Ra (7.35) Define the null hypothesis HN(⊆ Θ = Ra) such that HN = {(α1, α2, . . . , αa) ∈ Θ = Ra : α1 = α2 = . . . = αa = α} (7.36) = {( a{ }} { 0, 0, . . . , 0)} (7.37) Here, "(7.36)⇔(7.37)" is derived from aα = a∑ i=1 αi = a∑ i=1 (μi* − μ**) = ∑a i=1 ∑b j=1 μij b − a∑ i=1 ∑a i=1 ∑b j=1 μij ab = 0 (7.38) Also, define the estimator E : X(= Rabn)→ Θ(= Ra) by E(x) = (∑b j=1 ∑n k=1 xijk bn − ∑a i=1 ∑b j=1 ∑n k=1 xijk abn ) i=1,2,...,a = ( xi** − x*** ) i=1,2,...,a (7.39) Now we have the following problem: Problem 7.3. [The two-way ANOVA]. Consider the parallel simultaneous normal measurement: ML∞(Rab×R+)(O abn G = (X(≡ Rabn),BabnR , Gabn), S[(μ=(μij | i=1,2,*** ,a,j=1,2,*** ,b),σ)]) where we assume that μ1* = μ2* = * * * = μa* = μ** that is, π1(μ1, μ2, * * * , μa) = (0, 0, * * * , 0) namely, consider the null hypothesis HN = {(0, 0, * * * , 0)} (⊆ Θ = Ra)). Let 0 < α 1. Then, find the largest Rα;ΘHN (⊆ Θ)(independent of σ) such that (A1) the probability that a measured value x(∈ Rabn) obtained by ML∞(Rab×R+)(OabnG = (X(≡ Rabn),BabnR , Gabn), S[(μ=(μij | i=1,2,*** ,a,j=1,2,*** ,b),σ)]) satisfies that E(x) ∈ Rα;ΘHN is less than α. 186 Ishikawa's Homepage Chap. 7 ANOVA( = Analysis of Variance) Further, ‖θ(1) − θ(2)‖Θ = √√√√ a∑ i=1 ( θ (1) i − θ (2) i )2 (∀θ(`) = (θ(i)1 , θ (`) 2 , . . . , θ (`) a ) ∈ Ra, ` = 1, 2) Motivated by Theorem 5.6 (Fisher's maximum likelihood method), define and calculate σ(x) ( =√ SS(x)/(abn) ) as follows. SS(x) = SS((xijk)i=1,2,...,a, j=1,2,...,b,k=1,2,...,n) := a∑ i=1 b∑ j=1 n∑ k=1 (xijk − xij*)2 = a∑ i=1 b∑ j=1 n∑ k=1 (xijk − ∑n k=1 xijk n )2 = a∑ i=1 b∑ j=1 n∑ k=1 ((xijk − μij)− ∑n k=1(xijk − μij) n )2 =SS(((xijk − μij)i=1,2,...,a, j=1,2,...,b)k=1,2,*** ,n) (7.40) Define the semi-distance dxΘ ( in Θ = Ra) such that dxΘ(θ (1), θ(2)) = ‖θ(1) − θ(2)‖Θ√ SS(x) (∀θ(1), θ(2) ∈ Θ = Ra,∀x ∈ X = Rabn) (7.41) Define the estimator E : X(= Rabn)→ Θ(= Ra) such that E(x) = (∑b j=1 ∑n k=1 xijk bn − ∑a i=1 ∑b j=1 ∑n k=1 xijk abn ) i=1,2,...,a = ( xi** − x*** ) i=1,2,...,a Therefore, ‖E(x)− π(ω)‖2Θ =|| (∑b j=1 ∑n k=1 xijk bn − ∑a i=1 ∑b j=1 ∑n k=1 xijk abn ) i=1,2,...,a − ( αi ) i=1,2,...,a ||2Θ =|| (∑b j=1 ∑n k=1 xijk bn − ∑a i=1 ∑b j=1 ∑n k=1 xijk abn ) i=1,2,...,a − (∑b j=1 μij b − ∑a i=1 ∑b j=1 μij ab ) i=1,2,...,a ||2Θ =|| (∑n k=1 ∑b j=1(xijk − μij) bn − ∑a i=1 ∑b j=1 ∑n k=1(xijk − μij) abn ) i=1,2,...,a ||2Θ and thus, if the null hypothesis HN is assumed (i.e., μi* − μ** = αi = 0 (∀i = 1, 2, . . . , a) ) =|| (∑n k=1 ∑b j=1 xijk bn − ∑a i=1 ∑b j=1 ∑n k=1 xijk abn ) i=1,2,...,a ||2Θ = a∑ i=1 (xij* − x***)2 (7.42) 187 Ishikawa's Homepage 7.3 The two way ANOVA Thus, for any ω = (μ1, μ2)( ∈ Ω = R× R), define the positive number ηαω ( > 0) such that: ηαω = inf{η > 0 : [G(E−1(BallCdxΘ(π(ω); η))](ω) ≥ α} (7.43) Assume the null hypothesis HN . Now let us calculate the η α ω as follows: E−1(BallCdxΘ(π(ω); η)) = {x ∈ X = R abn : dxΘ(E(x), π(ω)) > η} ={x ∈ X = Rabn : abn ∑a i=1 ∑b j=1(xij* − x***)2∑a i=1 ∑b j=1 ∑n k=1(xijk − xij*)2 > η} (7.44) That is, for any ω = ((μij)i=1,2,...,a, j=1,2,...,b, , σ) ∈ Ω such that π(ω)(= (α1, α2, . . . , αa)) ∈ HN(= {0, 0, . . . , 0)}), [Gabn(E−1(BallCdxΘ(π(ω); η)))(ω) = 1 ( √ 2πσ)abn ∫ * * * ∫ E−1(BallC dx Θ (π(ω);η)) exp[− ∑a i=1 ∑b j=1 ∑n k=1(xijk − μij)2 2σ2 ] n × k=1 b × j=1 a × i=1 dxijk = 1 ( √ 2πσ)abn ∫ * * * ∫ abn ∑a i=1 ∑b j=1 (x ij*−x***)2∑a i=1 ∑b j=1 ∑n k=1 (xijk−x ij*)2>η 2 exp[− ∑a i=1 ∑b j=1 ∑n k=1(xijk − μij)2 2σ2 ] n × k=1 b × j=1 a × i=1 dxijk = 1 ( √ 2π)abn ∫ * * * ∫ ∑a i=1 ∑b j=1 (x ij*−x***)2) (a−1)∑a i=1 ∑b j=1 ∑n k=1 (xijk−x ij*)2 ab(n−1) > η2(ab(n−1)) abn(a−1) exp[− ∑a i=1 ∑b j=1 ∑n k=1(xijk) 2 2 ] n × k=1 b × j=1 a × i=1 dxijk (7.45) (A2) using the formula of Gauss integrals derived in Kolmogorov's probability theory, we finally get as follows. = ∫ ∞ η2(n−1) n(a−1) pF(a−1,ab(n−1))(t)dt = α (e.g., α = 0.05) (7.46) where pF(a−1,ab(n−1)) is the F -distribution with (a − 1, ab(n − 1)) degrees of freedom. Thus, it suffices to calculate the α-point F a−1ab(n−1),α Thus, we see (ηαω) 2 = F a−1ab(n−1),α * n(a− 1)/(n− 1) (7.47) 188 Ishikawa's Homepage Chap. 7 ANOVA( = Analysis of Variance) Therefore, we get Rα;Θx (or, R α;X x ; the (α)-rejection region of HN = {(0.0. . . . , 0)}(⊆ Θ = Ra) ) as follows: Rα;ΘHN = ∩ ω=((μi)ai=1,σ)∈Ω(=Ra×R+) such that π(ω)=(αi)ai=1∈HN={(0,0,...,0)} {E(x)(∈ Θ) : dxΘ(E(x), π(ω)) ≥ ηαω} = {E(x)(∈ Θ) : ( ∑a i=1 ∑b j=1(xij* − x***)2)/(a− 1) ( ∑a i=1 ∑b j=1 ∑n k=1(xijk − xij*)2)/(ab(n− 1)) ≥ F a−1ab(n−1),α} (7.48) Thus, Rα;XHN = E −1(Rα;ΘHN ) = {x(∈ X) : ( ∑a i=1 ∑b j=1(xij* − x***)2)/(a− 1) ( ∑a i=1 ∑b j=1 ∑n k=1(xijk − xij*)2)/(ab(n− 1)) ≥ F a−1ab(n−1),α} (7.49) ♠Note 7.3. It should be noted that the mathematical part is only the (A2). 7.3.3 Null hypothesis: μ*1 = μ*2 = * * * = μ*b = μ** Our present problem is as follows Problem 7.4. [The two-way ANOVA]. Consider the parallel simultaneous normal measurement: ML∞(Rab×R+)(O abn G = (X(≡ Rabn),BabnR , Gabn), S[(μ=(μij | i=1,2,*** ,a,j=1,2,*** ,b),σ)]) where the null hypothesis μ*1 = μ*2 = * * * = μ*b = μ** is assumed. Let 0 < α 1. Then, find the largest Rα;ΘHN (⊆ Θ)(independent of σ) such that (B)′ the probability that a measured value x(∈ Rabn) obtained by ML∞(Rab×R+)(OabnG = (X(≡ Rabn),BabnR , Gabn), S[(μ=(μij | i=1,2,*** ,a,j=1,2,*** ,b),σ)]) satisfies that E(x) ∈ Rα;ΘHN is less than α. 189 Ishikawa's Homepage 7.3 The two way ANOVA Since a and b have the same role, by the similar way of §7.3.2, we can easily solve Problem 7.4. 7.3.4 Null hypothesis: (αβ)ij = 0 (∀i = 1, 2, . . . , a, j = 1, 2, . . . , b ) Now, put Θ = Rab (7.50) And, define the system quantityπ : Ω→ Θ by Ω = Rab × R+ 3 ω = ((μij)i=1,2,...,a, j=1,2,...,b, σ) 7→ π(ω) = ((αβ)ij)i=1,2,...,a, j=1,2,...,b ∈ Θ = Rab (7.51) Here, recall: (αβ)ij = μij − μi* − μ*j + μ** (7.52) Also, the estimator E : X(= Rabn)→ Θ(= Rab) is defined by E((xijk)i=1,...,a, j=1,2,...b, k=1,2,...,n) = (∑n k=1 xijk n − ∑b j=1 ∑n k=1 xijk bn − ∑b j=1 ∑n k=1 xijk an + ∑a i=1 ∑b j=1 ∑n k=1 xijk abn ) i=1,2,...,a j=1,2,...b, = ( xij* − xi** − x*j* + x*** ) i=1,2,...,a j=1,2,...b, (7.53) Our present problem is as follows Problem 7.5. [The two way ANOVA]. Consider the parallel simultaneous normal measurement: ML∞(Rab×R+)(O abn G = (X(≡ Rabn),BabnR , Gabn), S[(μ=(μij | i=1,2,*** ,a,j=1,2,*** ,b),σ)]) The null hypothesis HN(⊆ Θ = Rab) is defined by HN = {((αβ)ij)i=1,2,...,a, j=1,2,...,b ∈ Θ = Rab : (αβ)ij = 0, (∀i = 1, 2, . . . , a, j = 1, 2, . . . , b)} (7.54) That is, (αβ)ij = μij − μi* − μ*j + μ** = 0 (i = 1, 2, * * * , a, j = 1, 2, * * * , b) (7.55) Let 0 < α 1. Then, find the largest Rα;ΘHN (⊆ Θ)(independent of σ) such that 190 Ishikawa's Homepage Chap. 7 ANOVA( = Analysis of Variance) (C1) the probability that a measured value x(∈ Rabn) obtained by ML∞(Rab×R+)(OabnG = (X(≡ Rabn),BabnR , Gabn), S[(μ=(μij | i=1,2,*** ,a,j=1,2,*** ,b),σ)]) satisfies that E(x) ∈ Rα;ΘHN is less than α. Now, ‖θ(1) − θ(2)‖Θ = √√√√ a∑ i=1 b∑ j=1 ( θ (`) ij − θ (`) ij )2 (7.56) (∀θ(`) = (θ(`)ij )i=1,2,...,a, j=1,2,...,b ∈ Rab, ` = 1, 2) and, define the semi-distance dxΘ in Θ by dxΘ(θ (1), θ(2)) = ‖θ(1) − θ(2)‖Θ√ SS(x) (∀θ(1), θ(2) ∈ Θ,∀x ∈ X) (7.57) E((xijk − μij)i=1,...,a, j=1,2,...b, k=1,2,...,n) = (∑n k=1(xijk − μij) n − ∑b j=1 ∑n k=1(xijk − μij) bn − ∑b j=1 ∑n k=1(xijk − μij) an + ∑a i=1 ∑b j=1 ∑n k=1(xijk − μij) abn ) i=1,2,...,a j=1,2,...b, = ( (xij* − μij)− (xi** − μi*)− (x*j* − μ*j) + (x*** − μ**) ) i=1,2,...,a j=1,2,...b, = ( xij* − xi** − x*j* + x*** ) i=1,2,...,a j=1,2,...b (Remark:null hypothesis (αβ)ij = 0) (7.58) Therefore, E((xijk)i=1,...,a, j=1,2,...b, k=1,2,...,n) = E((xijk − μij)i=1,...,a, j=1,2,...b, k=1,2,...,n) (7.59) Thus, for each i = 1, ..., a, j = 1, 2, ...b, Eij(xijk − μij) = ∑n k=1(xijk − μij) n − ∑b j=1 ∑n k=1(xijk − μij) bn − ∑b j=1 ∑n k=1(xijk − μij) an + ∑a i=1 ∑b j=1 ∑n k=1(xijk − μij) abn =Eij(x)− (αβ)ij 191 Ishikawa's Homepage 7.3 The two way ANOVA =xij* − xi** − x*j* + x*** − (αβ)ij (7.60) And, we see: ‖E(x)− π(ω)‖2Θ =|| ( Eij(x)− (αβ)ij ) i=1,2,...,a j=1,2,...b ||2Θ (7.61) Recalling that the null hypothesis HN (i.e., (αβ)ij = 0 (∀i = 1, 2, . . . , a, j = 1, 2, . . . , b) ), we see = a∑ i=1 b∑ j=1 (xij* − xi** − x*j* + x***)2 (7.62) Thus, for each ω = (μ, σ)( ∈ Ω = Rab × R), define the positive real ηαω ( > 0) such that ηαω = inf{η > 0 : [G(E−1(BallCdxΘ(π(ω); η))](ω) ≥ α} (7.63) Recalling the null hypothesisHN (i.e., (αβ)ij = 0 (∀i = 1, 2, . . . , a, j = 1, 2, . . . , b) ), calculate the ηαωas follows. E−1(BallCdxΘ(π(ω); η)) = {x ∈ X = R abn : dxΘ(E(x), π(ω)) > η} ={x ∈ X = Rabn : abn ∑a i=1 ∑b j=1(xij* − xi** − x*j* + x***)2∑a i=1 ∑b j=1 ∑n k=1(xijk − xij*)2 > η2} (7.64) Thus, for any ω = ((μij)i=1,2,...,a, j=1,2,...,b, , σ) ∈ Ω = Rab × R+ such that π(ω) ∈ HN(⊆ Rab) (i.e., (αβ)ij = 0 (∀i = 1, 2, . . . , a, j = 1, 2, . . . , b) ), we see: [Gabn(E−1(BallCdxΘ(π(ω); η)))(ω) = 1 ( √ 2πσ)abn ∫ * * * ∫ E−1(BallC dx Θ (π(ω);η)) exp[− ∑a i=1 ∑b j=1 ∑n k=1(xijk − μij)2 2σ2 ] n × k=1 b × j=1 a × i=1 dxijk = 1 ( √ 2πσ)abn ∫ * * * ∫ {x∈X : dxΘ(E(x),π(ω)≥η} exp[− ∑a i=1 ∑b j=1 ∑n k=1(xijk − μij)2 2σ2 ] n × k=1 b × j=1 a × i=1 dxijk = 1 ( √ 2π)abn ∫ * * * ∫ ∑a i=1 ∑b j=1 (x ij*−xi**−x*j*+x***)2∑a i=1 ∑b j=1 ∑n k=1 (xijk−x ij*)2 > η2 abn exp[− ∑a i=1 ∑b j=1 ∑n k=1(xijk) 2 2 ] n × k=1 b × j=1 a × i=1 dxijk 192 Ishikawa's Homepage Chap. 7 ANOVA( = Analysis of Variance) = 1 ( √ 2π)abn ∫ * * * ∫ ∑a i=1 ∑b j=1 (x ij*−xi**−x*j*+x***)2 (a−1)(b−1)∑a i=1 ∑b j=1 ∑n k=1 (xijk−x ij*)2 ab(n−1) > η2(ab(n−1)) abn(a−1)(b−1) exp[− ∑a i=1 ∑b j=1 ∑n k=1(xijk) 2 2 ] n × k=1 b × j=1 a × i=1 dxijk (7.65) (C2) Then, by the formula of Gauss integrals 7.8(D) (§7.4) , we see = ∫ ∞ η2(n−1) n(a−1)(b−1) pF((a−1)(b−1),ab(n−1))(t)dt = α( e.g., α = 0.05) (7.66) where pF((a−1)(b−1),ab(n−1)) is a probability density function of the F -distribution with ((a−1)(b− 1), ab(n− 1)) degrees of freedom. Hence, it suffices to the following equation: η2(n− 1) n(a− 1)(b− 1) = F (a−1)(b−1) ab(n−1),α (= "α-point") (7.67) thus, we see, (ηαω) 2 = F (a−1)(b−1) ab(n−1),α n(a− 1)(b− 1)/(n− 1) (7.68) Therefore, we get the (α)-rejection region Rα;Θx (or, R α;X x ; HN = {((αβ)ij)i=1,2,*** ,a,j=1,2,*** ,b : (αβ)ij = 0 (i = 1, 2, * * * , a, j = 1, 2, * * * , b)}(⊆ Θ = Rab) ): Rα;ΘHN = ∩ ω=((μij)ai=1 b j=1,σ)∈Ω(=Ra×R+) such that π(ω)=(αβ)ij∈HN {E(x)(∈ Θ) : dxΘ(E(x), π(ω)) ≥ ηαω} = {E(x)(∈ Θ) : ( ∑a i=1 ∑b j=1(xij* − x***)2)/((a− 1)(b− 1)) ( ∑a i=1 ∑b j=1 ∑n k=1(xijk − xij*)2)/(ab(n− 1)) ≥ F (a−1)(b−1)ab(n−1),α } (7.69) Also, Rα;XHN = E −1(Rα;ΘHN ) = {x(∈ X) : ( ∑a i=1 ∑b j=1(xij* − x***)2)/((a− 1)(b− 1)) ( ∑a i=1 ∑b j=1 ∑n k=1(xijk − xij*)2)/(ab(n− 1)) ≥ F (a−1)(b−1)ab(n−1),α } (7.70) ♠Note 7.4. It should be noted that the mathematical part is only the (C2). 193 Ishikawa's Homepage 7.4 Supplement(the formulas of Gauss integrals) 7.4 Supplement(the formulas of Gauss integrals) 7.4.1 Normal distribution, chi-squared distribution, Student t-distribution, F -distribution Definition 7.6. [Fdistribution ]. Let t ≥ 0, and n1 and n2 be natural numbers. The probability density function pF(n1,n2)(t) of F -distribution with the degree of freedom(n1, n2) is defined by pF(n1,n2)(t) = 1 B(n1/2, n2/2) (n1 n2 )n1/2 t(n1−2)/2 (1 + n1t/n2)(n1+n2)/2 (t ≥ 0) (7.71) where, B(*, *) is the Beta function, that is, for x, y > 0, B(x, y) = ∫ 1 0 tx−1(1− t)y−1dt Note that F -distribution with degree of freedom(1, n− 1) = Student t-distribution with the degree of freedom(n− 1) Define two maps μ : Rn → R and SS : Rn → R as follows. μ(x) = μ(x1, x2, * * * , xn) = ∑n k=1 xk n SS(x) = SS(x1, x2, * * * , xn) = n∑ k=1 (xk − μ(x))2 (∀x = (x1, x2, * * * , xn) ∈ Rn) Formula 7.7. [Gauss integral(normal distribution and chi-squared distribution)]. This was already mentioned in (6.6) and (6.7). Formula 7.8. [Gauss integral(F -distribution )]. For c ≥ 0, (A): 1 ( √ 2π)n ∫ * * * ∫ c≤ n(μ(x)) 2 SS(x)/(n−1) exp[− ∑n k=1(xk) 2 2 ]dx1dx2 * * * dxn = ∫ ∞ c pF(1,n−1)(t)dt (7.72) (B): For n = ∑a i=1 ni, 1 ( √ 2π)n ∫ * * * ∫ ( ∑a i=1 ni(x i*−x**)2/(a−1) ( ∑a i=1 ∑ni k=1 (xik−x i*)2)/(n−a)>c exp[− ∑a i=1 ∑ni k=1(xik) 2 2 ] a × i=1 ni× k=1 dxik 194 Ishikawa's Homepage Chap. 7 ANOVA( = Analysis of Variance) = ∫ ∞ c pF(a−1,n−a)(t)dt (7.73) (C): 1 ( √ 2π)abn ∫ * * * ∫ ∑a i=1 ∑b j=1 (x ij*−x***)2 (a−1)∑a i=1 ∑b j=1 ∑n k=1 (xijk−x ij*)2 ab(n−1) >c exp[− ∑a i=1 ∑b j=1 ∑n k=1(xijk) 2 2 ] n × k=1 b × j=1 a × i=1 dxijk = ∫ ∞ c pF(a−1,ab(n−1))(t)dt (7.74) Or, equivalently, (D): 1 ( √ 2π)abn ∫ * * * ∫ ∑a i=1 ∑b j=1 (x ij*−xi**−x*j*+x***)2 (a−1)(b−1)∑a i=1 ∑b j=1 ∑n k=1 (xijk−x ij*)2 ab(n−1) >c exp[− ∑a i=1 ∑b j=1 ∑n k=1(xijk) 2 2 ] n × k=1 b × j=1 a × i=1 dxijk = ∫ ∞ c pF((a−1)(b−1),ab(n−1))(t)dt (7.75) 195 Ishikawa's Homepage

Chapter 8 Practical logic–Do you believe in syllogism?– For examle, consider three kinds of syllogisms as follows. One is the the (natural) logic inherent in our ordinary language such as (]1) Since Socrates is a man and all men are mortal, it follows that Socrates is mortal. Another is the mathematical syllogism such as (]2) "A⇒ B" and "B ⇒ C" imply "A⇒ C" (where "A⇒ B" is defined by "¬A ∨B") It is certain that pure logic (=mathematical logic) is merely a kind of rule in mathematics or meta-mathematics. Thus, mathematical syllogism (]2) is not guaranteed to be applicable to our world such as (]1). However, many philosophers ( e.g. Aristotle) might consciously or unconsciously propose the interpretation such that the two (]1) and (]2) are closely related. The other is "practical logic" that means the logic in measurement theory. In this chapter, we prove the (]1) in classical measurement theory. Also, we point out that syllogism does not hold in quantum systems 1 8.1 Marginal observable and quasi-product observable Definition 8.1. [(=Definition 3.19):quasi-product product observable ] Let Ok = (Xk, Fk, Fk) (k = 1, 2, . . . , n ) be observables in a W ∗-algebra A. Assume that an observable O12...n = 1 This chapter is mostly extracted from the following: (]) Ref. [26]: S. Ishikawa, "Fuzzy Inferences by Algebraic Method," Fuzzy Sets and Systems, Vol. 87, No. 2, 1997, pp. 181-200. doi:10.1016/S0165-0114(96)00035-8 197 8.1 Marginal observable and quasi-product observable (×nk=1Xk,  nk=1Fk, F12...n) satisfies F12...n(X1 × * * * ×Xk−1 × Ξk ×Xk+1 × * * * ×Xn) = Fk(Ξk). (8.1) (∀Ξk ∈ Fk,∀k = 1, 2, . . . , n) The observable O12...n = (×nk=1Xk,  nk=1Fk, F12...n) is called a quasi-product observable of {Ok | k = 1, 2, . . . , n}, and denoted by qp × k=1,2,...,n Ok = ( n × k=1 Xk,  nk=1Fk, qp × k=1,2,...,n Fk). Of course, a simultaneous observable is a kind of quasi-product observable. Therefore, quasiproduct observable is not uniquely determined. Also, in quantum systems, the existence of the quasi-product observable is not always guaranteed. Definition 8.2. [Image observable, marginal observable] Consider the basic structure [A ⊆ A ⊆ B(H)]. And consider the observable O = (X, F, F ) in A. Let (Y,G) be a measurable space, and let f : X → Y be a measurable map. Then, we can define the image observable f(O) = (X, F, F ◦ f−1) in A, where F ◦ f−1 is defined by (F ◦ f−1)(Γ) = F (f−1(Γ)) (∀Γ ∈ G). [Marginal observable] Consider the basic structure [A ⊆ A ⊆ B(H)]. And consider the observable O12...n = (×nk=1Xk,  nk=1Fk, F12...n) in A. For any natural number j such that 1 5 j 5 n, define F (j)12...n such that F (j) 12...n(Ξj) = F12...n(X1 × * * * ×Xj−1 × Ξj ×Xj+1 × * * * ×Xn) (∀Ξj ∈ Fj). Then we have the observable O (j) 12...n = (Xj, Fj, F (j) 12...n) in A. The O (j) 12...n is called a marginal observable of O12...n ( or, precisely, (j)-marginal observable ). Consider a map Pj :×nk=1Xk → Xj such that n × k=1 3 (x1, x2, ..., xj, ..., xn) 7→ xj ∈ Xj. Then, the marginal observable O (j) 12...n is characterized as the image observable Pj(O12...n). The above can be easily generalized as follows. For example, define O (12) 12...n = (X1×X2, F1F2, F (12) 12...n) such that F (12) 12...n(Ξ1 × Ξ2) = F (12) 12...n(Ξ1 × Ξ2 ×X3 × * * * ×Xn) (∀Ξ1 ∈ F1,∀Ξ2 ∈ F2). Then, we have the (12)-marginal observable O (12) 12...n = (X1×X2, F1 F2, F (12) 12...n). Of course, we also see that F12...n = F (12...n) 12...n . 198 Ishikawa's Homepage Chap. 8 Practical logic–Do you believe in syllogism?– The following theorem is often used: Theorem 8.3. Consider the basic structure [A ⊆ A ⊆ B(H)] Let A be a C∗-algebra. Let O1 ≡ (X1,F1, F1) and O2 ≡ (X2,F2, F2) be W ∗-observables in A such that at least one of them is a projective observable. ( So, without loss of generality, we assume that O2 is projective, i.e., F2 = (F2) 2 ) . Then, the following statements are equivalent: (i) There exists a quasi-product observable O12 ≡ (X1×X2,F1×F2, F1 qp ×F2) with marginal observables O1 and O2. (ii) O1 and O2 commute, that is, F1(Ξ1)F2(Ξ2) = F2(Ξ2)F1(Ξ1) (∀Ξ1 ∈ F1,∀Ξ2 ∈ F2). Furthermore, if the above statements (i) and (ii) hold, the uniqueness of the quasi-product observable O12 of O1 and O2 is guaranteed. Proof. See refs. [12, 26, 30]. 199 Ishikawa's Homepage 8.2 Properties of quasi-product observables 8.2 Properties of quasi-product observables Consider the measurement MA(O12=(X1 ×X2,F1  F2, F12), S[ρ]) with the sample probability space (X1 ×X2,F1  F2, A∗ ( ρ, F12(*) ) A). Put RepΞ1×Ξ2ρ [O12] = [ A∗ ( ρ, F12(Ξ1 × Ξ2) ) A A ∗ ( ρ, F12(Ξ1 × Ξc2) ) A A∗ ( ρ, F12(Ξ c 1 × Ξ2) ) A A ∗ ( ρ, F12(Ξ c 1 × Ξc2) ) A ] (∀Ξ1 ∈ F1, ∀Ξ2 ∈ F2) where, Ξc is the complement of Ξ {x ∈ X | x /∈ Ξ}. Also, note that A∗ ( ρ, F12(Ξ1 × Ξ2) ) A + A∗ ( ρ, F12(Ξ1 × Ξc2) ) A = A∗ ( ρ, F (1) 12 ](Ξ1) ) A A∗ ( ρ, F12(Ξ c 1 × Ξc2) ) A + A∗ ( ρ, F12(Ξ c 1 × Ξ2) ) A = A∗ ( ρ, F (1) 12 (Ξ c 1) ) A A∗ ( ρ, F12(Ξ c 1 × Ξc2) ) A + A∗ ( ρ, F12(Ξ1 × Ξc2) ) A = A∗ ( ρ, F (2) 12 (Ξ c 2) ) A A∗ ( ρ, F12(Ξ1 × Ξc2) ) A + A∗ ( ρ, F12(Ξ c 1 × Ξc2) ) A = A∗ ( ρ, F (2) 12 (Ξ c 2) ) A We have the following lemma. Lemma 8.4. [The condition of quasi-product observables] Consider the general basic structure [A ⊆ A ⊆ B(H)]. Let O1 = (X1,F1, F1) and O2 = (X2,F2, F2) be observables in C(Ω). Let O12 = (X1×X2,F1× F2, F12=F1 qp × F2) be a quasi-product observable of O1 and O2. That is, it holds that F1 = F (1) 12 , F2 = F (2) 12 Then, putting α Ξ1×Ξ2 ρ = A∗ ( ρ, F12(Ξ1 × Ξ2) ) A = ρ(F12(Ξ1 × Ξ2)), we see RepΞ1×Ξ2ρ [O12] = [ A∗ ( ρ, F12(Ξ1 × Ξ2) ) A A ∗ ( ρ, F12(Ξ1 × Ξc2) ) A A∗ ( ρ, F12(Ξ c 1 × Ξ2) ) A A ∗ ( ρ, F12(Ξ c 1 × Ξc2) ) A ] = [ α Ξ1×Ξ2 ρ ρ(F1(Ξ1))− α Ξ1×Ξ2 ρ ρ(F2(Ξ2))− α Ξ1×Ξ2 ρ 1 + α Ξ1×Ξ2 ρ − ρ(F1(Ξ1))− ρ(F2(Ξ2)) ] (8.2) and max{0, ρ(F1(Ξ1)) + ρ(F2(Ξ2))− 1} 5 α Ξ1×Ξ2 ρ 5 min{ρ(F1(Ξ1)), ρ(F2(Ξ2))} (∀Ξ1 ∈ F1,∀Ξ2 ∈ F2,∀ρ ∈ Sp(A∗)) (8.3) 200 Ishikawa's Homepage Chap. 8 Practical logic–Do you believe in syllogism?– Reversely, for any α Ξ1×Ξ2 ρ satisfying (8.3), the observable O12 defined by (8.2) is a quasiproduct observable of O1 and O2. Also, it holds that ρ(F (Ξ1 × Ξc2)) = 0 ⇐⇒ α Ξ1×Ξ2 ρ = ρ(F1(Ξ1)) =⇒ ρ(F1(Ξ1)) 5 ρ(F2(Ξ2)) (8.4) Proof. Though this lemma is easy, we add a brief proof for completeness. 0 5 ρ(F ((Ξ′1×Ξ′2))) 5 1, (∀Ξ′1 ∈ F1,Ξ′2 ∈ F2) we see, by (8.2) that 0 5 αΞ1×Ξ2ρ 5 1 0 5 1 + αΞ1×Ξ2ρ − ρ(F1(Ξ1))− ρ(F2(Ξ2)) 5 1 0 5 ρ(F2(Ξ2))− α Ξ1×Ξ2 ρ 5 1 0 5 ρ(F1(Ξ1))− α Ξ1×Ξ2 ρ 5 1 which clearly implies (8.3). Conversely. if α satisfies (8.3),then we easily see (8.2),Also, (8.4) is obvious. This completes the proof. Let O12 = (X1×X2,F1F2, F12=F1 qp ×F2) be a quasi-product observable of O1 = (X1,F1, F1) and O2 = (X2,F2, F2) in A. Consider the measurement MA(O12 =(X1×X2,F1F2, F12=F1 qp ×F2), S[ρ])). And assume that a measured value(x1, x2) (∈ X1 ×X2) is obtained. And assume that we know that x1 ∈ Ξ1. Then, the probability (i.e., the conditional probability) that x2 ∈ Ξ2 is given by P = ρ(F12(Ξ1 × Ξ2)) ρ(F1(Ξ1)) = ρ(F12(Ξ1 × Ξ2)) ρ(F12(Ξ1 × Ξ2)) + ρ(F12(Ξ1 × Ξc2)) And further, it is, by (8.3), estimated as follows. max{0, ρ(F1(Ξ1)) + ρ(F2(Ξ2))− 1} ρ(F12(Ξ1 × Ξ2)) + ρ(F12(Ξ1 × Ξc2)) 5 P 5 min{ρ(F1(Ξ1)), ρ(F2(Ξ2))} ρ(F12(Ξ1 × Ξ2)) + ρ(F12(Ξ1 × Ξc2)) Example 8.5. [Example of tomatoes] Let Ω = {ω1, ω2, ...., ωN} be a set of tomatoes, which is regarded as a compact Hausdorff space with the discrete topology. Consider the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] 201 Ishikawa's Homepage 8.2 Properties of quasi-product observables Consider yes-no observables ORD ≡ (XRD, 2XRD , FRD) and OSW ≡ (XSW, 2XSW , FSW) in C(Ω) such that: XRD = {yRD, nRD} and XSW = {ySW, nSW}, where we consider that "yRD" and "nRD" respectively mean "RED" and "NOT RED". Similarly, "ySW" and "nSW" respectively mean "SWEET" and "NOT SWEET". For example, the ω1 is red and not sweet, the ω2 is red and sweet, etc. as follows. ω1 yRD nSW ω2 yRD ySW ω3 nRD ySW * * * * * * * * * ωK nRD nSW Figure 8.1: Tomatoes ( Red or Sweet? ) Next, consider the quasi-product observable as follows. O12 = (XRD ×XSW, 2XRD×XSW , F=FRD qp ×FSW) That is, Rep{(yRD,ySW)}ωk [O12] = [ [F ({(yRD, ySW)})](ωk) [F ({(yRD, nSW)})](ωk) [F ({(nRD, ySW)})](ωk) [F ({(nRD, nSW)})](ωk) ] = [ α{(yRD,ySW)} [FRD({yRD})]− α{(yRD,ySW)} [FSW({ySW})]− α{(yRD,ySW)} 1 + α{(yRD,ySW)} − [FRD({yRD})]− [FSW({ySW})] ] where α{(yRD,ySW)}(ωk) satisfies the (8.3). When we know that a tomato ωk is red, the probability P that the tomato ωk is sweet is given by P = [F ({(yRD, ySW)})](ωk) [F ({(yRD, ySW)})](ωk) + [F ({(yRD, nSW)})](ωk) = [F ({(yRD, ySW)})](ωk) [FRD({yRD})](ωk) Since [F ({(yRD, ySW)})](ωk) = α{(yRD,ySW)}(ωk), the conditional probability P is estimated by max{0, [F1({yRD})](ωk) + [F2({ySW})](ωk)− 1} [FRD({yRD})](ωk) 5 P 5 min[F1({ySW})](ωk), [F2({ySW})](ωk)} [FRD({yRD})](ωk) 202 Ishikawa's Homepage Chap. 8 Practical logic–Do you believe in syllogism?– 8.3 Implication-the definition of "⇒" 8.3.1 Implication and contraposition In Example 8.5, consider the case that [F ({(yRD, nSW)})](ω) = 0. In this case, we see [F ({(yRD, ySW)})](ω) [F ({(yRD, ySW)})](ω) + [F ({(yRD, nSW)})](ω) = 1 Therefore, when we know that a tomato ω is red, the probability, that the tomato ω is sweet, is equal to 1. That is, "[F ({(yRD, nSW)})](ω) = 0" ⇐⇒ [ "Red" =⇒ "Sweet" ] Motivated by the above argument, we have the following definition. Definition 8.6. [Implication] Consider the general basic structure [A ⊆ A ⊆ B(H)] Let O12 = (X1 × X2, F1  F2, F12=F1 qp ×F2) be a quasi-observable in A Let ρ ∈ Sp(A∗), Ξ1 ∈ F1, Ξ2 ∈ F2. Then, if it holds that ρ(F 12(Ξ1 × (Ξc2))) = 0 this is denoted by [O (1) 12 ; Ξ1] =⇒ M A (O12,S[ρ]) [O (2) 12 ; Ξ2] (8.5) Of course, this (8.5) should be read as follows. (A) Assume that a measured value (x1, x2)(∈ X1×X2) is obtained by a measurementML∞(Ω)(O12, S[ω]). When we know that x1 ∈ Ξ1, then we can assure that x2 ∈ Ξ2. The above argument is generalized as follows. Let O12...n = (×nk=1Xk,  nk=1Fk, F12...n = qp × k=1,2,...,n Fk) be a quasi-product observable in A. Let Ξ1 ∈ Fi and Ξ2 ∈ Fj. Then, the condition A∗ ( ρ, F (ij) 12...n(Ξi × (Ξcj)) ) A = 0 (where, Ξc = X \ Ξ) is denoted by [O (i) 12...n; Ξi] =⇒ M A (O12...n,S[ρ]) [O (j) 12...n; Ξj] (8.6) 203 Ishikawa's Homepage 8.3 Implication-the definition of "⇒" Theorem 8.7. [Contraposition] Let O12 = (X1×X2, F1×F2, F12=F1 qp ×F2) be a quasi-product observable in A. Let ρ ∈ Sp(A∗). Let Ξ1 ∈ F1 and Ξ2 ∈ F2. If it holds that [O (1) 12 ; Ξ1] =⇒ M A (O12,S[ρ]) [O (2) 12 ; Ξ2] (8.7) then we see: [O (1) 12 ; Ξ c 1] ⇐= M A (O12,S[ρ]) [O (2) 12 ; Ξ c 2] Proof. The proof is easy, but we add it. Assume the condition (8.7). That is, A∗ ( ρ, F12(Ξ1 × (X2 \ Ξ2)) ) A = 0 Since Ξ1 × Ξ2c = (Ξc1)c × Ξc2 we see A∗ ( ρ, F12((Ξ c 1) c × Ξc2) ) A = 0 Therefore, we get [O (1) 12 ; Ξ c 1] ⇐= M A (O12,S[ρ]) [O (2) 12 ; Ξ c 2] 204 Ishikawa's Homepage Chap. 8 Practical logic–Do you believe in syllogism?– 8.4 Cogito- I think, therefore I am- This section is published in the following: • ref. [57]: S. Ishikawa; Leibniz-Clarke correspondence, Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. are clarified in quantum language Open Journal of philosophy, Vol. 8, No.5 , 466-480, 2018, DOI: 10.4236/ojpp.2018.85032 (https://www.scirp.org/Journal/PaperInformation.aspx?PaperID=87862) • ref. [58]; S. Ishikawa; Leibniz-Clarke correspondence, Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. are clarified in quantum language; [Revised version] ; Keio Reseach report; 2018; KSTS/RR-18/001, 1-15 (https://philpapers.org/rec/ISHLCB) (http://www.math.keio.ac.jp/academic/research_pdf/report/2018/18001.pdf) Recall the following figure. • observer (I(=mind)) system (matter)  - [observable] [measured value] a©interfere b©perceive a reaction [state] [Descartes Figure 8.2 (=Figure 3.1) ]:The image of "measurement(= a©+ b©)" in dualism The following example may be rather unnatural, but this is indispensable for the wellunderstanding of dualism. Example 8.8. [Brain death(cf. ref. p.89 in [39])] Consider the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Let ωn (∈ Ω = {ω1, ω2, . . . , ωN}) be the state of Peter. Let O12 = (X1 × X2, 2X1×X2 , F12=F1 qp ×F2) be the brain death observable in L∞(Ω) such that X1 = {T, T} X2 = {L,L}, where T = "think", T = "not think", L = "live", L = "not live". For each ωn (n = 1, 2, . . . , N), O12 satisfies the condition in Table 8.2. [Table 8.2 ]: Brain death observable O12 = (X1 ×X2, 2X1×X2 , F12) 205 Ishikawa's Homepage 8.4 Cogito- I think, therefore I am- F1F2 [F2({L})](ωn) [F2({L})](ωn) [F1({T})](ωn) (1 + (−1)n)/2 (=[F12({T}×{L})](ωn)) 0 (=[F12({T}×{L})](ωn)) [F1({T})](ωn) 0 (=[F12({T}×{L})](ωn)) (1− (−1)n)/2 (=[F12({T}×{L})](ωn)) Since [F12({T} × {L})](ωn) = 0, the following formula holds: [O (1) 12 ; {T}] =⇒ ML∞(Ω)(O12,S[ωn]) [O (2) 12 ; {L}] Of course, this implies that (A1) Peter thinks, therefore, Peter lives. This is the same as the statement concerning brain death. Note that in the above example, we see that observer←→doctor, system←→Peter, The above (A1) should not be confused with the following famous Descartes' saying (= cogito proposition): (A2) "I think, therefore I am". in which the following identification may be assumed: observer←→I, system←→I And thus, the above is not a statement in dualism (=measurement theory). In order to propose Figure 8.2 (i.e., dualism) ( that is, in order to establish the concept "I" in science), he started from the ambiguous statement "I think, therefore I am". Summing up, we want to say the following irony: (B) Descartes proposed the dualism (i.e., Figure 8.2 ) by the cogito proposition (A2) which is not understandable in dualism. ♠Note 8.1. It is not true to consider that every phenomena can be describe in terns of quantum language. Although readers may think that the following can be described in measurement theory, but we believe that it is impossible. For example, the followings can not be written by quantum language: 206 Ishikawa's Homepage Chap. 8 Practical logic–Do you believe in syllogism?–  1© : tense-past, present, future - 2© : Heidegger's saying"In-der-Welt-sein" 3© : the measurement of a measurement, 4© : Bergson's subjective time 5© : observer's space-time, 6© : Only the present exists ( due to Augustinus(354-430)) If we want to understand the above words, we have to propose the other scientific languages ( except quantum language). We have to recall Wittgenstein's sayings The limits of my language mean the limits of my world 207 Ishikawa's Homepage 8.5 Combined observable -Only one measurement is permitted - 8.5 Combined observable -Only one measurement is permitted - 8.5.1 Combined observable - only one observable The linguistic interpretation says that "Only one measurement is permitted" ⇒ "only one observable"⇒ "the necessity of the combined observable" Thus, we prepare the following theorem. Theorem 8.9. [The existence theorem of classical combined observable(cf.refs.[26, 30])] Consider the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] And consider observables O12=(X1 × X2,F1  F2, F12) and O23= (X2 × X3, F2  F3, F23) in L∞(Ω, ν). Here, for simplicity, assume that Xi={x1i , x2i , . . . , x ni i } (i = 1, 2, 3) is finite, Also, assume that Fi = 2 Xi . Further assume that O (2) 12 = O (2) 23 (That is, F12(X1 × Ξ2) = F23(Ξ2 ×X3) (∀Ξ2 ∈ 2X2)) Then, we have the observable O123=(X1 ×X2 ×X3,F1 × F2 × F3, F123) in L∞(Ω) such that O (12) 123 = O12, O (23) 123 = O23 That is, F (12) 123 (Ξ1 × Ξ2 ×X3) = F12(Ξ1 × Ξ2), F (23) 123 (X1 × Ξ2 × Ξ3) = F23(Ξ2 × Ξ3) (8.8) (∀Ξ1 ∈ F1, ∀Ξ2 ∈ F2,∀Ξ3 ∈ F3)) The O123 is called the combined observable of O12 and O23. Also, for the general definition of "combined observable", see Definition 4.18. Proof. O123 = (X1 ×X2 ×X3, F1 × F2 × F3, F123) is, for example, defined by [F123({(x1, x2, x3)})](ω) =  [F12({(x1, x2)})](ω) * [F23({(x2, x3)})](ω) [F12(X1 × {x2})](ω) ([F12(X1 × {x2})](ω) 6= 0 and ) 0 ([F12(X1 × {x2})](ω) = 0 and ) 208 Ishikawa's Homepage Chap. 8 Practical logic–Do you believe in syllogism?– (∀ω ∈ Ω, ∀(x1, x2, x3) ∈ X1 ×X2 ×X3) This clearly satisfies (8.8). Counter example 8.10. [Counter example in quantum systems] Theorem 8.9 does not hold in the quantum basic structure [C(H) ⊆ B(H) ⊆ B(H)] For example, put H = Cn, and consider the three Hermitian (n × n)-matrices T1, T2, T3 in B(H) such that T1T2 = T2T1, T2T3 = T3T2, T1T3 6= T3T1 (8.9) For each k = 1, 2, 3, define the spectrum decomposition Ok = (Xk,Fk, Fk) in H (which is regarded as a projective observable) such that Tk = ∫ Xk xkFk(dxk) (8.10) where Xk = R,Fk = BR. From the commutativity, we have the simultaneous observables O12=O1 × O2 = (X1 ×X2,F1  F2, F12 = F1 × F2) and O23=O2 × O3 = (X2 ×X3,F2  F3, F23 = F2 × F3) It is clear that O (2) 12 = O (2) 23 (that is, F12(X1 × Ξ2) = F2(Ξ2) = F23(Ξ2 ×X3) (∀Ξ2 ∈ F2)) However, it should be noted that there does not exist the observable O123=(X1×X2×X3,F1 F2  F3, F123) in B(H) such that O (12) 123 = O12, O (23) 123 = O23 That is because, if O123 exists, Theorem 8.3 says that O1 and O3 commute, and it is in contradiction with the (8.9). Therefore, the combined observable O123 of O12 and O23 does not exist. 209 Ishikawa's Homepage 8.5 Combined observable -Only one measurement is permitted - 8.5.2 Combined observable and Bell's inequality Now we consider the following problem: Problem 8.11. [combined observable and Bell's inequality (cf. [39])] Consider the basic structure [A ⊆ A ⊆ B(H)] Put X1 = X2 = X3 = X4 = {−1, 1}. Let O13=(X1×X3, 2X1×2X3 , F13), O14=(X1×X4, 2X1× 2X4 , F14), O23= (X2 ×X3, 2X2 × 2X3 , F23) and O24= (X2 ×X3, 2X2 × 2X4 , F24) be observables in L∞(Ω) such that O (1) 13 = O (1) 14 , O (2) 23 = O (2) 24 , O (3) 13 = O (3) 23 , O (4) 14 = O (4) 24 Define the probability measure νab on {−1, 1}2 by the formula (4.53). Assume that there exists a state ρ0 ∈ Sp(A∗) such that A∗ ( ρ0, F13({(x1, x3)}) ) A = νa1b1({(x1, x3)}, A∗ ( ρ0, F14({(x1, x4)}) ) A = νa1b2({(x1, x4)} A∗ ( ρ0, F23({(x2, x3)}) ) A = νa2b1({(x2, x3)}, A∗ ( ρ0, F24({(x2, x4)}) ) A = νa2b2({(x2, x4)} Now we have the following problem: (a) Does the observable O1234=(×4k=1Xk,×4k=1 Fk, F1234) in A satisfying the following (]) exist? (]) O (13) 1234 = O13, O (14) 1234 = O14, O (23) 1234 = O23, O (24) 1234 = O24 In what follows, we show that the above observable O1234 does not exist. Assume that the observable O1234=(×4k=1Xk, ×4k=1 Fk, F1234) exists. Then, it suffices to show the contradiction. Define C13(ρ0), C14(ρ0), C23(ρ0) and C24(ρ0) such that C13(ρ0) = ∫ ×4k=1Xk x1 * x3 A∗ ( ρ0, F1234( 4 × k=1 dxk) ) A ( = ∫ X1×X3 x1 * x3 νa1b1(dx1dx3) ) C14(ρ0) = ∫ ×4k=1Xk x1 * x4 A∗ ( ρ0, F1234( 4 × k=1 dxk) ) A ( = ∫ X1×X4 x1 * x4 νa1b2(dx1dx4) ) C23(ρ0) = ∫ ×4k=1Xk x2 * x3 A∗ ( ρ0, F1234( 4 × k=1 dxk) ) A ( = ∫ X2×X3 x2 * x3 νa2b1(dx2dx3) ) C24(ρ0) = ∫ ×4k=1Xk x2 * x4 A∗ ( ρ0, F1234( 4 × k=1 dxk) ) A ( = ∫ X2×X4 x2 * x4 νa2b2(dx2dx4) ) 210 Ishikawa's Homepage Chap. 8 Practical logic–Do you believe in syllogism?– Then, we can easily get the following Bell's inequality: (cf. Bell's inequality (4.47)). |C13(ρ0)− C14(ρ0)|+ |C23(ρ0) + C24(ρ0)| 5 ∫ ×4k=1Xk |x1| * |x3 − x4| +|x2| * |x3 + x4| [ F1234( 4 × k=1 dxk) ] (ρ0) 5 2 (since xk ∈ {−1, 1}) (8.11) However, the formula (4.62) says that this (8.11) must be 2 √ 2. Thus, by contradiction, we says that O1234 satisfying (a) does not exist. Thus we can not take a measurement MA(O1234, S[ρ0]). However, it should be noted that (b) instead of MA(O1234, S[ρ0]). we can take a parallel measurement M⊗4k=1A (O13⊗O14⊗O23⊗ O24, S[⊗4k=1ρ0]). In this case, we easily see that (8.11) = 2 √ 2 as the formula (4.62). That is, (c) in the case of a parallel measurement, Bell's inequality is broken in both quantum and classical systems. ♠Note 8.2. In the above argument, Bell's inequality is used in the framework of measurement theory. This is of course true. Also as seen in Section 4.5.3, J.S. Bell asserted (cf. [4]) that (]) Problem 8.11 is related to the theory of "hidden variables". 211 Ishikawa's Homepage 8.6 Syllogism and its variations 8.6 Syllogism and its variations Next, we shall discuss practical syllogism (i.e., measurement theoretical theorem concerning implication (Definition8.6) ). Before the discussion, we note that (]) Since Theorem8.9 ( The existence of the combined observable) does not hold in quantum system, ( cf. Counter Example8.10), syllogism does not hold. On the other hand, in classical system, we can expect that syllogism holds. This will be proved in the following theorem. Theorem 8.12. [Practical syllogism in classical systems] Consider the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Let O123 = (X1 × X2 × X3, F1 × F2 × F3, F123= qp ×k=1,2,3Fk) be an observable in L∞(Ω) Fix ω ∈ Ω, Ξ1 ∈ F1, Ξ2 ∈ F2, Ξ3 ∈ F3 Then, we see the following (i) – (iii). (i).(practical syllogism) [O (1) 123; Ξ1] =⇒ ML∞(Ω)(O123,S[ω]) [O (2) 123; Ξ2], [O (2) 123; Ξ2] =⇒ ML∞(Ω)(O123,S[ω]) [O (3) 123; Ξ3] implies RepΞ1×Ξ3ω [O (13) 123 ] = [ [F (13) 123 (Ξ1 × Ξ3)](ω) [F (13) 123 (Ξ1 × Ξc3)](ω) [F (13) 123 (Ξ c 1 × Ξ3)](ω) [F (13) 123 (Ξ c 1 × Ξc3)](ω) ] = [ [F (1) 123(Ξ1)](ω) 0 [F (3) 123(Ξ3)](ω)− [F (1) 123(Ξ1)](ω) 1− [F (3) 123(Ξ3)](ω) ] That is, it holds: [O (1) 123; Ξ1] =⇒ ML∞(Ω)(O123,S[ω]) [O (3) 123; Ξ3] (8.12) (ii). [O (1) 123; Ξ1] ⇐= ML∞(Ω)(O123,S[ω]) [O (2) 123; Ξ2], [O (2) 123; Ξ2] =⇒ ML∞(Ω)(O123,S[ω]) [O (3) 123; Ξ3] implies RepΞ1×Ξ3ω [O (13) 123 ] = [ [F (13) 123 (Ξ1 × Ξ3)](ω) [F (13) 123 (Ξ1 × Ξc3)](ω) [F (13) 123 (Ξ c 1 × Ξ3)](ω) [F (13) 123 (Ξ c 1 × Ξc3)](ω) ] 212 Ishikawa's Homepage Chap. 8 Practical logic–Do you believe in syllogism?– = [ α Ξ1×Ξ3 [F (1) 123(Ξ1)](ω)− αΞ1×Ξ3 [F (3) 123(Ξ3)](ω)− αΞ1×Ξ3 1− αΞ1×Ξ3 − [F (1) 123(Ξ1)]− [F (3) 123(Ξ3)] ] where max{[F (2)123(Ξ2)](ω), [F (1) 123(Ξ1)](ω) + [F (3) 123(Ξ3)](ω)− 1} 5 α Ξ1×Ξ3 (ω) 5 min{[F (1)123(Ξ1)](ω), [F (3) 123(Ξ3)](ω)} (8.13) (iii). [O (1) 123; Ξ1] =⇒ ML∞(Ω)(O123,S[ω]) [O (2) 123; Ξ2], [O (2) 123; Ξ2] ⇐= ML∞(Ω)(O123,S[ω]) [O (3) 123; Ξ3] implies RepΞ1×Ξ3ω [O (13) 123 ] = [ [F (13) 123 (Ξ1 × Ξ3)](ω) [F (13) 123 (Ξ1 × Ξc3)](ω) [F (13) 123 (Ξ c 1 × Ξ3)](ω) [F (13) 123 (Ξ c 1 × Ξc3)](ω) ] = [ αΞ1×Ξ3 (ω) [F (1) 123(Ξ1)](ω)− αΞ1×Ξ3 (ω) [F (3) 123(Ξ3)](ω)− αΞ1×Ξ3 (ω) 1− αΞ1×Ξ3 (ω)− [F (1) 123(Ξ1)](ω)− [F (3) 123(Ξ3)](ω) ] where max{0, [F (1)123(Ξ1)](ω) + [F (3) 123(Ξ3)](ω)− [F (2) 123(Ξ2)](ω)} 5 α Ξ1×Ξ3 (ω) 5 min{[F (1)123(Ξ1)](ω), [F (3) 123(Ξ3)](ω)} Proof. (i): By the condition, we see 0 = [F (12) 123 (Ξ1 × Ξc2)](ω) = [F123(Ξ1 × Ξc2 × Ξ3)](ω) + [F123(Ξ1 × Ξc2 × Ξc3)](ω) 0 = [F (23) 123 (Ξ2 × Ξc3)](ω) = [F123(Ξ1 × Ξ2 × Ξc3)](ω) + [F123(Ξc1 × Ξ2 × Ξc3)](ω) Therefore, 0 = [F123(Ξ1 × Ξc2 × Ξ3)](ω) = [F123(Ξ1 × Ξc2 × Ξc3)](ω) 0 = [F123(Ξ1 × Ξ2 × Ξc3)](ω) = [F123(Ξc1 × Ξ2 × Ξc3)](ω) Hence, [F (13) 123 (Ξ1 × Ξc3)](ω) = [F123(Ξ1 × Ξ2 × Ξc3)](ω) + [F (13) 123 (Ξ1 × Ξc2 × Ξc3)](ω) = 0 Thus, we get, (8.12). For the proof of (ii) and (iii), see refs. [26, 30]. 213 Ishikawa's Homepage 8.6 Syllogism and its variations Example 8.13. [Continued from Example 8.5] Let O1 = OSW = (XSW, 2 XSW , FSW) and O3 = ORD = (XRD, 2 XRD , FRD) be as in Example 8.5. Putting XRP = {yRP, nRP}, consider the new observable O2 = ORP = (XRP, 2 XRP , FRP). Here, "yRP" and "nRP" respectively means "ripe" and "not ripe". Put Rep[O1] = [ [FSW({ySW})](ωk), [FSW({nSW})](ωk) ] Rep[O2] = [ [FRP({yRP})](ωk), [FRP({nRP})](ωk) ] Rep[O3] = [ [FRD({yRD})](ωk), [FRD({nRD})](ωk) ] Consider the following quasi-product observable: O12 = (XSW ×XRP, 2XSW×XRP , F12=FSW qp ×FRP) O23 = (XRP ×XRD, 2XRP×XRD , F23=FRP qp ×FRD) Let ωk ∈ Ω. And assume that [O (1) 123; {ySW}] =⇒ ML∞(Ω)(O123,S[ωk]) [O (2) 123; {yRP}], [O (2) 123; {yRP}] =⇒ ML∞(Ω)(O123,S[ωk]) [O (3) 123; {yRD}] (8.14) Then, by Theorem 8.12(i), we get: Rep[O13] = [ [F13({ySW} × {yRD})](ωk) [F13({ySW} × {nRD})](ωk) [F13({nSW} × {yRD})](ωk) [F13({nSW} × {nRD})](ωk) ] = [ [FSW({ySW})](ωk) 0 [FRD({yRD})](ωk)− [FSW({ySW})](ωk) 1− [FRD({yRD})](ωk) ] Therefore, when we know that the tomato ωk is sweet by measurement ML∞(Ω)(O123, S[ωk]), the probability that ωk is red is given by [F13({ySW} × {yRD})](ωk) [F13({ySW} × {yRD})](ωk) + [F13({ySW} × {nRD})](ωk) = [FRD({yRD})](ωk) [FRD({yRD})](ωk) = 1 (8.15) Of course, (8.14) means "Sweet" =⇒ "Ripe" "Ripe" =⇒ "Red" Therefore, by (8.12), we get the following conclusion. "Sweet" =⇒ "Red" However, it is not useful in the market. What we want to know is such as "Red" =⇒ "Sweet" This will be discussed in the following example. 214 Ishikawa's Homepage Chap. 8 Practical logic–Do you believe in syllogism?– Example 8.14. [Continued from Example 8.5] Instead of (8.14), assume that O {y1} 1 ⇐= ML∞(Ω)(O12,S[δωn ]) O {y2} 2 , O {y2} 2 =⇒ ML∞(Ω)(O23,S[δωn ]) O {y3} 3 . (8.16) When we observe that the tomato ωn is "RED", we can infer, by the fuzzy inference ML∞(Ω)(O13, S[δωn ]), the probability that the tomato ωn is "SWEET" is given by Q = [F13({ySW}×{yRD})](ωn) [F13({ySW}×{yRD})](ωn) + [F13({nSW}×{yRD})](ωn) which is, by (8.3), estimated as follows: max { [FRP({yRP})](ωn) [FRD({yRD})](ωn) , [FSW({ySW})] + [FRD({yRD})]− 1 [FRD({yRD})](ωn) } ≤ Q ≤ min{ [FSW({ySW})](ωn) [FRD({yRD})](ωn) , 1}. (8.17) Note that (8.16) implies (and is implied by) "RIPE" =⇒ "SWEET" and "RIPE" =⇒ "RED" . And note that the conclusion (8.17) is somewhat like "RED" =⇒ "SWEET" . Therefore, this estimation (8.17) may be useful in marckets. /// 215 Ishikawa's Homepage 8.7 EPR-paradox says that syllogism does not hold in quantum systems 8.7 EPR-paradox says that syllogism does not hold in quantum systems Remark 8.15. [Syllogism does not hold in quantum system (cf. ref. [36] ) ] Concerning EPR's paper[14], we shall add some remark as follows. Let A and B be particles with the same masses m. Consider the situation described in the following figure:  A B Figure 8.3: The case that "the velocity of A"= −"the velocity of B". The position qA (at time t0) of the particle A can be exactly measured, and moreover, the velocity of vB (at time t0) of the particle B can be exactly measured. Thus, we may conclude that (A) the position and momentum (at time t0) of the particle A are respectively and exactly equal to qA and −mvB ? (As mentioned in Section 4.4.3, this is not in contradiction with Heisenberg' uncertainty principle). However, we have the following question: Is the conclusion (A) true? Now we shall describe the above arguments in quantum system: A quantum two particles system S is formulated in a tensor Hilbert space H = H1 ⊗H1 = L2(Rq1)⊗ L2(Rq2) = L2(R2(q1,q2)). The state u0 ( ∈ H = H1 ⊗H1 = L 2(R2(q1,q2))) ( or precisely, ρ0 = |u0〉〈u0| ) of the system S is assumed to be u0(q1, q2) = √ 1 2πεσ e− 1 8σ2 (q1−q2−2a)2− 1 8ε2 (q1+q2)2 (8.18) where a positive number ε is sufficiently small. For each k = 1, 2, define the self-adjoint operators Qk : L 2(R2(q1,q2))→ L 2(R2(q1,q2)) and Pk : L 2(R2(q1,q2))→ L 2(R2(q1,q2)) by Q1 = q1, P1 = ~∂ i∂q1 Q2 = q2, P2 = ~∂ i∂q2 (8.19) (]1) Let O1 = (R3,BR3 , F1) be the observable representation of the self-adjoint operator (Q1⊗ P2) × (I ⊗ P2). And consider the measurement MB(H)(O1 = (R3,BR3 , F1), S[|u0〉〈u0|]). Assume that the measured value (x1, p2, p2)(∈ R3). That is, (x1, p2) (the position of A1, the momentum of A2) =⇒ MB(H)(O1,S[ρ0]) p2 the momentum of A2 216 Ishikawa's Homepage Chap. 8 Practical logic–Do you believe in syllogism?– (]2) Let O2 = (R2,BR2 , F2) be the observable representation of (I⊗P2)×(P1⊗I). And consider the measurement MB(H)(O2 = (R2,BR2 , F2), S[|u0〉〈u0|]). Assume that the measured value (p2,−p2)(∈ R3). That is, p2 the momentum of A2 =⇒ MB(H)(O2,S[ρ0]) −p2 the momentum of A1 (]3) Therefore, by (]1) and (]2), "syllogism" may say that −p2 the momentum of A1 ( that is, the momentum of A1 is equal to −p2 ) Hence, some assert that (B) The (A) is true But, the above argument ( particularly, "syllogism") is not true, thus, The (A) is not true That is because (]4) (Q1 ⊗ P2)× (I ⊗ P2) and (I ⊗ P2)× (P1 ⊗ I) ( Therefore, O1 and O2 ) do not commute, and thus, the simultaneous observable does not exist. Thus, we can not test the (]3) experimentally. Remark 8.16. After all, we think that EPR-paradox says the following two: (C1) syllogism does not necessarily hold in quantum systems, (C2) there is something faster than light. We think that (C1) is not serious. Thus, we do not need to investigate how to understand the fact (C1). On the other hand, (C2) is serious. Although we have to make efforts to understand the "fact (C2)", this is the problem in physics (i.e., in 5© in Figure 1.1). Recall that the spirit of quantum language (i.e., in 10© in Figure 1.1) is "Stop being bothered." 217 Ishikawa's Homepage

Chapter 9 Mixed measurement theory (⊃Bayesian statistics) Quantum language (= measurement theory ) is classified as follows. (]) measurement theory (=quantum language)  pure type (]1) { classical system : Fisher statistics quantum system : usual quantum mechanics mixed type (]2) { classical system : including Bayesian statistics, Kalman filter quantum system : quantum decoherence In this chapter, we study mixed measurement theory, which includes Bayesian statistics. 9.1 Mixed measurement theory(⊃Bayesian statistics) 9.1.1 Axiom(m) 1 (mixed measurement) In the previous chapters, we studied Axiom 1 (pure measurement: §2.7), that is, pure measurement theory (=quantum language) := [(pure)Axiom 1] pure measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spells (a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } manual to use spells (9.1) In this chapter, we shall study "Axiom(m) 1 (mixed measurement)" in mixed measurement theory, that is, mixed measurement theory (=quantum language) := [(mixed)Axiom(m) 1] mixed measurement (cf. §9.1 ) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spells (a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } manual to use spells (9.2) 219 9.1 Mixed measurement theory(⊃Bayesian statistics) In the previous chapters, we mainly discussed pure measurements listed in Review 9.1, especially W ∗-measurement (A1). Review 9.1. [=Preparation 2.30]. (A1) W ∗-measurement MA ( O= (X,F, F ), S[ρ] ) , where O= (X,F, F ) is a W ∗-observable in A, and ρ(∈ Sp(A∗)) is a pure state. Here, "W ∗-measurement MA ( O, S[ρ] ) " is also denoted by "measurementW ∗ MA ( O. S[ρ] ) " , or "measurement MA ( O. S[ρ] ) " , (A2) C ∗-measurement MA ( O= (X,F, F ), S[ρ] ) , where O= (X,F, F ) is a C∗-observable in A, and ρ(∈ Sp(A∗)) is a pure state. Here, "C∗-measurement MA ( O, S[ρ] ) " is also denoted by "measurementC ∗ MA ( O. S[ρ] ) " , or "measurement MA ( O. S[ρ] ) " . In this chapter, we introduce four "mixed measurements" as follows. Preparation 9.2. (B1) W ∗-mixed measurement MA ( O= (X,F, F ), S[∗](w0) ) , where O= (X,F, F ) is a W ∗observable in A, and w0(∈ S m (A∗)) is a W ∗-mixed state. Here, "W ∗-mixed measurement MA ( O, S[∗](w0) ) " is also denoted by "W ∗-mixed measurementW ∗ MA ( O. S[∗](w0) ) ", or "mixed measurement MA ( O. S[∗](w0) ) " (B2) C ∗-mixed measurement MA ( O= (X,F, F ), S[∗](ρ0) ) , where O= (X,F, F ) is a W ∗observable in A, and ρ0(∈ Sm(A∗)) is a C∗-mixed state. Here, "C∗-mixed measurement MA ( O, S[∗](ρ0) ) " is also denoted by "C∗-mixed measurementW ∗ MA ( O. S[∗](ρ0) ) ", or "mixed measurement MA ( O. S[∗](ρ0) ) " Although we mainly devote ourselves to the above two, we add the followings. (B3) W ∗-mixed measurement MA ( O= (X,F, F ), S[∗](w0) ) , where O= (X,F, F ) is a C∗observable in A, and w0(∈ S m (A∗)) is a W ∗-mixed state. Here, "W ∗-mixed measurement MA ( O, S[∗](w0) ) " is also denoted by "W ∗-mixed measurementC ∗ MA ( O. S[∗](w0) ) ", or "mixed measurement MA ( O. S[∗](w0) ) " (B4) C ∗-mixed measurement MA ( O= (X,F, F ), S[∗](ρ0) ) , where O= (X,F, F ) is a C∗220 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) observable in A, and ρ0(∈ Sm(A∗)) is a C∗-mixed state. Here, "C∗-mixed measurement MA ( O, S[∗](ρ0) ) " is also denoted by "C∗-mixed measurementC ∗ MA ( O. S[∗](ρ0) ) ", or "mixed measurement MA ( O.S[∗](ρ0) ) " We now give Axiom(m) 1 for mixed measurements. We will discuss (C1) mainly, and (C2) when necessary. (C):Axiom(m) 1 (mixed measurement) Let O= (X,F, F ) be an observable in A (C1): Let w0 ∈ S m (A∗). The probability that a measured value obtained by W ∗-mixed measurement MA ( O= (X,F, F ), S[∗](w0) ) belongs to Ξ (∈ F) is given by A∗ (w0, F (Ξ))A ( ≡ w0(F (Ξ)) ) (C2): Let ρ0 ∈ Sm(A∗). The probability that a measured value obtained by C∗-mixed measurement MA ( O= (X,F, F ), S[∗](ρ0) ) belongs to Ξ (∈ F) is given by A∗(ρ0, F (Ξ))A ( ≡ ρ(F (Ξ)) ) As we learned Axiom 1 by rote in pure measurement theory, we have to learn Axiom(m) 1 by rote, and exercise a lot of examples The practices will be done in this chapter. Remark 9.3. In the above Axiom(m) 1, (C1) and (C2) are not so different. (]1) In the quantum case, (C1)=(C2) clearly holds, since S m(Tr(H)) = S m (Tr(H)) in (2.17). (]2) In the classical case, we see L1+1(Ω.ν) 3 w0 ρ0(D)= ∫ D w0(ω)ν(dω)−−−−−−−−−−−−→ ρ0 ∈M+1(Ω) Therefore, in this case, we consider that ML∞(Ω.ν) ( O=(X,F, F ), S[∗](w0) ) = ML∞(Ω.ν) ( O=(X,F, F ), S[∗](ρ0) ) Hence, (C1) and (C2) are not so different. In oder to avoid the confusion, we use the following notation: W ∗-state w0 (∈ S m (A∗) is written by Roman alphabet (e.g., w0, w, v, ...) C∗-state ρ0 (∈ Sm(A∗) is written by Greek alphabet (e.g., ρ0, ρ, ...) /// 221 Ishikawa's Homepage 9.2 Simple examples in mixed measurement theory 9.2 Simple examples in mixed measurement theory Recall the following wise sayings: experience is the best teacher, or custom makes all things Thus, we exercise the following problem. Review 9.4. [Answer 5.7 to Problem 5.2 by Fisher's maximum likelihood method] You do not know the urn behind the curtain. Assume that you pick up a white ball from the urn. Which urn do you think is more likely, U1 or U2 ? - [∗] U1≈ω1 U2≈ω2 Figure 9.1 (= Figure 5.6: ): Pure measurement (Fisher's maximum likelihood method) Answer Consider the state space Ω = {ω1, ω2} with the discrete topology and the measure ν such that ν({ω1}) = 1, ν({ω2}) = 1 (9.3) In the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))], consider the measurement ML∞(Ω)(O= ({W, B}, 2{W,B}, FWB), S[∗]), where the observable OWB = ({W,B}, 2{W,B}, FWB) in L∞(Ω) is defined by [FWB({W})](ω1) = 0.8, [FWB({B})](ω1) = 0.2 [FWB({W})](ω2) = 0.4, [FWB({B})](ω2) = 0.6. (9.4) Here, we see: max{[FWB({W})](ω1), [FWB({W})](ω2)} = max{0.8, 0.4} = 0.8 = FWB({W})](ω1). Then, Fisher's maximum likelihood method (Theorem 5.6) says that [∗] = ω1. Therefore, there is a reason to infer that the urn behind the curtain is U1. Thus, we exercise the following problem. 222 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) Problem 9.5. [mixed measurement ML∞(Ω,ν)(O = (X,F, F ), S[∗](w))] 100p% 100(1-p)% [∗] U1≈ω1 U2≈ω2 Figure 9.2: Mixed measurement (Urn problem) (]1) Assume an unfair coin-tossing (Tp,1−p) such that (0 5 p 5 1): That is,{ the possibility that "head" appears is 100p% the possibility that "tail" appears is 100(1− p)% If "head" [resp. "tail"] appears, put an urn U1(≈ω1) [resp. U2(≈ω2)] behind the curtain. Assume that you do not know which urn is behind the curtain, U1 or U2). The unknown urn is denoted by [∗](∈ {ω1, ω2}). This situation is represented by w ∈ L1+1(Ω, ν) (with the counting measure ν), that is, w(ω) = { p ( if ω = ω1 ) 1− p ( if ω = ω2 ) (]2) Consider the "measurement" such that a ball is picked out from the unknown urn. This "measurement" is denoted by ML∞(Ω,ν)(O, S[∗](w)), and called a mixed measurement. Then, we have the following problems: (a) Calculate the probability that a white ball is picked from the unknown urn behind the curtain ! And further, (b) when a white ball is picked, calculate the probability that the unknown urn behind the curtain is U1 ! We would like to remark • the term "subjective probability" is not used in the above problem. Answer: Assume that the state spaceΩ = {ω1, ω2} is defined by the discrete metric with the 223 Ishikawa's Homepage 9.2 Simple examples in mixed measurement theory following measure ν: ν({ω1}) = 1, ν({ω2}) = 1. (9.5) Thus, we start from the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))], (9.6) in which we consider the mixed measurement ML∞(Ω)(O= ({W, B}, 2{W,B}, F ), S[∗](w)). Here, the observable OWB = ({W,B}, 2{W,B}, FWB) in L∞(Ω) is defined by [FWB({W})](ω1) = 0.8, [FWB({B})](ω1) = 0.2 [FWB({W})](ω2) = 0.4, [FWB({B})](ω2) = 0.6. (9.7) Also, the mixed state w0 ∈ L1+1(Ω, ν) is defined by w0(ω1) = p, w0(ω2) = 1− p. (9.8) Then, by Axiom(m) 1, we see (a): the probability that a measured value x (∈ {W,B}) is obtained by ML∞(Ω)(O= ({W, B}, 2{W,B}, F ), S[∗](w)) is given by P ({x}) = L1(Ω) ( w0, F ({x}) ) L∞(Ω) = ∫ Ω [F ({x})](ω) * w0(ω)ν(dω) = p[F ({x})](ω1) + (1− p)[F ({x})](ω2) = { 0.8p+ 0.4(1− p) (when x = W ) 0.2p+ 0.6(1− p) (when x = B) (9.9) The question (b) will be answered in Answer 9.13. ♠Note 9.1. The following question is natural. That is, (]1) In the above (i), why is "the possibility that [ ∗ ] = ω1 is 100p% * * * " replaced by "the probability that [ ∗ ] = ω1 is 100p% * * * " ? However, the linguistic interpretation says that (]2) there is no probability without measurements. This is the reason why the term "probability" is not used in (i). However, from the practical point of view, we are not sensitive to the difference between "probability" and "possibility". 224 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) Example 9.6. [Mixed spin measurement MB(C2)(O = (X = {↑, ↓}, 2X , F z), S[∗](w))] Consider the quantum basic structure: [B(C2) ⊆ B(C2) ⊆ B(C2)] And consider a particle P1 with spin state ρ1 = |a〉〈a| ∈ Sp(B(C2)), where a = [ α1 α2 ] ∈ C2 ( ‖a‖ = (|α1|2 + |α2|2)1/2 = 1) And consider another particle P2 with spin state ρ2 = |b〉〈b| ∈ Sp(B(C2)), where b = [ β1 β2 ] ∈ C2 ( ‖b‖ = (|β1|2 + |β2|2)1/2 = 1) Here, assume that • the "probability" that the "particle" P is { a particle P1 a particle P2 } is given by { p 1− p } That is, state ρ1 (Particle P1) −−−−−−−−→ "probability" p unknown state [∗] (Particle P ) ←−−−−−−−−−− "probability" 1−p state ρ2 (Particle P2) Here, the unknown state [∗] of Particle P is represented by the mixed statew (∈ Sm(Tr(C2))) such that w = pρ1 + (1− p)ρ2 = p|a〉〈a|+ (1− p)|b〉〈b| Therefore, we have the mixed measurement MB(C2)(Oz = (X, 2 X , F z), S[∗](w)) of the z-axis spin observable Oz = (X,F, F z), where F z({↑}) = [ 1 0 0 0 ] , F z({↓}) = [ 0 0 0 1 ] And we say that (a) the probability that a measured value { ↑ ↓ } is obtained by the mixed measurement MB(C2)(Oz = (X, 2 X , F z), S[∗](w)) is given by Tr(C2) ( w,F z({↑}) ) B(C2) = p|α1|2 + (1− p)|β1|2 Tr(C2) ( w,F z({↓}) ) B(C2) = p|α2|2 + (1− p)|β2|2  Remark 9.7. As seen in the above, we say that 225 Ishikawa's Homepage 9.2 Simple examples in mixed measurement theory (a) Pure measurement theory is fundamental. Adding the concept of "mixed state", we can construct mixed measurement theory as follows. mixed measurement theory ML∞(Ω)(O, S[∗](w)) := pure measurement theory ML∞(Ω)(O, S[∗]) + mixed state w Therefore, There is no mixed measurement without pure measurement That is, in quantum language, there is no confrontation between "frequency probability" and "subjective probability". The reason that a coin-tossing is used in Problem 9.5 is to emphasize that the naming of "subjective probability" is improper. 226 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) 9.3 St. Petersburg two envelope problem This section is extracted from the following: Ref. [47]: S. Ishikawa; The two envelopes paradox in non-Bayesian and Bayesian statistics ( arXiv:1408.4916v4 [stat.OT] 2014 ) Now, we shall review the St. Petersburg two envelope problem (cf. [9]1). Problem 9.8. [The St. Petersburg two envelope problem] The host presents you with a choice between two envelopes (i.e., Envelope A and Envelope B). You are told that each of them contains an amount determined by the following procedure, performed separately for each envelope: (]) a coin was flipped until it came up heads, and if it came up heads on the k-th trial, 2k is put into the envelope. This procedure is performed separately for each envelope. You choose randomly (by a fair coin toss) one envelope. For example, assume that the envelope is Envelope A. And therefore, the host get Envelope B. You find 2m dollars in the envelope A. Now you are offered the options of keeping A (=your envelope) or switching to B (= host's envelope ). What should you do? Envelope A Envelope B Figure 9.2: Two envelope problem [(P2):Why is it paradoxical?]. You reason that, before opening the envelopes A and B, the expected values E(x) and E(y) in A and B is infinite respectively. That is because 1× 1 2 + 2× 1 22 + 22 × 1 23 + * * * =∞ For any 2m, if you knew that A contained x = 2m dollars, then the expected value E(y) in B would still be infinite. Therefore, you should switch to B. But this seems clearly wrong, as your information about A and B is symmetrical. This is the famous St. Petersburg two-envelope paradox (i.e., "The Other Person's Envelope is Always Greener" ). 1 D.J. Chalmers, "The St. Petersburg Two-Envelope Paradox," Analysis, Vol.62, 155-157, (2002) 227 Ishikawa's Homepage 9.3 St. Petersburg two envelope problem 9.3.1 (P2): St. Petersburg two envelope problem: classical mixed measurement Define the state space Ω such that Ω = {ω = 2k | k = 1, 2, * * * }, with the discrete metric and the counting measure ν. And define the exact observable O = (X,F, F ) in L∞(Ω, ν) such that X = Ω, F = 2X ≡ {Ξ | Ξ ⊆ X} [F (Ξ)](ω) = χ Ξ (ω) ≡ { 1 (ω ∈ Ξ) 0 (ω /∈ Ξ) (∀Ξ ∈ F,∀ω ∈ Ω) Define the mixed state w (∈ L1+1(Ω, ν), i.e., the probability density function on Ω) such that w0(ω) = 2 −k (∀ω = 2k ∈ Ω). Consider the mixed measurement ML∞(Ω,ν)(O = (X,F, F ), S[∗](w0)). Axiom (m) 1(C1) (§9.1) says that (A) the probability that a measured value 2k is obtained by ML∞(Ω)(O = (X,F, F ), S[∗](w0)) is given by 2−k. Therefore, the expectation of the measured value is calculated as follows. E = ∞∑ k=1 2k * 2−k =∞ Note that you knew that A contained x = 2m dollars (and thus, E = ∞ > 2m). There is a reason to consider that the switching to B is an advantage. Remark 9.9. After you get a measured value 2m from the envelope A, you can guess (also see Bayes theorem later) that the probability density function w0 changes to the new w1 such that w1(2 m) = 1, w1(2 k) = 0(k 6= m). Thus, now your information about A : w1 and B : w0 is not symmetrical. Hence, in this case, it is true: "The Other Person's envelope is Always Greener". ♠Note 9.2. There are various criterions except the expectaion. For example, consider the criterion such that (]) "the probability that the switching is disadvantageous" < 12 Under this criterion, it is reasonable to judge that{ m = 1 =⇒ switching to B m = 2, 3, ... =⇒ keeping A 228 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) 9.4 Bayesian statistics is to use Bayes theorem Although there may be several opinions for the question "What is Bayesian statistics?", we think that Bayesian statistics is to use Bayes theorem Thus, let us start from Bayes theorem. The following is clear. Theorem 9.10. [The conditional probability]. Consider the mixed measurement MA ( O= (X × Y,F  G, H), S[∗](w) ) , which is formulated in the basic structure [A ⊆ A ⊆ B(H)] Assume that a measured value (x, y) (∈ X×Y ) is obtained by the mixed measurementMA ( O= (X × Y,F  G, H), S[∗](w) ) belongs to Ξ× Y (∈ F). Then, the probability that y ∈ Γ is given by A∗ (w,H(Ξ× Γ))A A∗ (w,H(Ξ× Y ))A (∀Γ ∈ G) Proof. This is due to the property (or, common sense) of conditional probability. In the classical case, this is rewritten as follows. Theorem 9.11. [Bayes' Theorem (in classical mixed measurement)]. Consider the simultaneous measurement MA ( O= (X×Y,F  G, F ×G), S[∗](w0) ) formulated in the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))]. Here the observable O12=(X × Y,F  G, F ×G) is defined by the simultaneous observable of the two observables O1=(X,F, F ) and O2=(Y,G, G). That is, (F ×G)(Ξ× Γ) = F (Ξ) *G(Γ) (∀Ξ ∈ F, ∀Γ ∈ G). (9.10) Assume that (a) a measured value (x, y) (∈ X × Y ) obtained by the mixed measurement ML∞(Ω) ( O12= 229 Ishikawa's Homepage 9.4 Bayesian statistics is to use Bayes theorem (X × Y,F  G, F ×G), S[∗](w0) ) belongs to Ξ× Y (where, Ξ ∈ F). Then, the probability such that "y ∈ Γ" is given by L1(Ω)(w0, H(Ξ× Γ))L∞(Ω) L1(Ω)(w0, H(Ξ× Y ))L∞(Ω) ( = ∫ Ω [F (Ξ)](ω) * [G(Γ)](ω) * w0(ω)ν(dω)∫ Ω [F (Ξ)](ω) * w0(ω)ν(dω) ) . (9.11) Here, putting (b) wnew(ω) = [F (Ξ)](ω)*w0(ω)∫ Ω[F (Ξ)](ω)*w0(ω)ν(dω) ( ∀ω ∈ Ω). we see: (9.23) = ∫ Ω [G(Γ)](ω)wnew(ω)ν(dω) (∀Γ ∈ G). (9.12) Remark 9.12. [How to understand Bayes' Theorem] Bayes' theorem 9.11 is usually read as follows. (b′) If a measured value x (∈ X) obtained by the mixed measurement ML∞(Ω) ( O1= (X,F, F ), S[∗](w0) ) belongs to Ξ (∈ F), then, the following state collapse happens: w0 pre-state −−−→ x ∈ Ξ wnew post-state The above (d) superficially contradicts the linguistic interpretation, which says A state never moves. In this sense, the above (b) or (b′) (i.e., Bayes' theorem) is convenient and makeshift. Answer 9.13. [Bayes' Theorem (=Problem 9.5 and the answer to (c2)) ] Assume that the state space Ω = {ω1, ω2} is defined by the discrete metric with the following measure ν: ν({ω1}) = 1, ν({ω2}) = 1. (9.13) Thus, we start from the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))], (9.14) in which we consider the mixed measurement ML∞(Ω)(O= ({W, B}, 2{W,B}, F ), S[∗](w)). Here, the observable OWB = ({W,B}, 2{W,B}, FWB) in L∞(Ω) is defined by [FWB({W})](ω1) = 0.8, [FWB({B})](ω1) = 0.2, 230 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) [FWB({W})](ω2) = 0.4, [FWB({B})](ω2) = 0.6. (9.15) Also, the mixed state w0 ∈ L1+1(Ω, ν) is defined by w0(ω1) = p, w0(ω2) = 1− p. (9.16) Then, by Axiom(m) 1, we see (a): the probability that a measured value x (∈ {W,B}) is obtained by ML∞(Ω)(O= ({W, B}, 2{W,B}, F ), S[∗](w)) is given by P ({x}) = L1(Ω) ( w0, F ({x}) ) L∞(Ω) = ∫ Ω [F ({x})](ω) * w0(ω)ν(dω) = p[F ({x})](ω1) + (1− p)[F ({x})](ω2) = { 0.8p+ 0.4(1− p) (when x = W ) 0.2p+ 0.6(1− p) (when x = B) (9.17) [ W ∗-algebraic answer to Problem 9.5(c2) in Sec. 9.1.2] Since "white ball" is obtained by a mixed measurement ML∞(Ω)(O, S[∗](w0)), a new mixed state wnew(∈ L1+1(Ω)) is given by wnew(ω) = [F ({W})](ω)w0(ω)∫ Ω [F ({W})](ω)w0(ω)ν(dω) =  0.8p 0.8p+ 0.4(1− p) (when ω = ω1) 0.4(1− p) 0.8p+ 0.4(1− p) (when ω = ω2) [ C∗-algebraic answer to Problem 9.5 (c2) in Sec. 9.1.2] Since "white ball" is obtained by a mixed measurement ML∞(Ω)(O, S[∗](ρ0)), a new mixed state ρnew(∈M+1(Ω)) is given by ρnew = F ({W})ρ0∫ Ω [F ({W})](ω)ρ0(dω) = 0.8p 0.8p+ 0.4(1− p) δω1 + 0.4(1− p) 0.8p+ 0.4(1− p) δω2 . 231 Ishikawa's Homepage 9.5 Two envelope problem (Bayes' method) 9.5 Two envelope problem (Bayes' method) This section is extracted from the following: ref. [47]: S. Ishikawa; The two envelopes paradox in non-Bayesian and Bayesian statistics ( arXiv:1408.4916v4 [stat.OT] 2014 ) Problem 9.14. [ (=Problem5.16): the two envelope problem ] The host presents you with a choice between two envelopes (i.e., Envelope A and Envelope B). You know one envelope contains twice as much money as the other, but you do not know which contains more. That is, Envelope A [resp. Envelope B] contains V1 dollars [resp. V2 dollars]. You know that (a) V1 V2 = 1/2 or, V1 V2 = 2 Define the exchanging map x : {V1, V2} → {V1, V2} by x = { V2, ( if x = V1), V1 ( if x = V2) You choose randomly (by a fair coin toss) one envelope, and you get x1 dollars (i.e., if you choose Envelope A [resp. Envelope B], you get V1 dollars [resp. V2 dollars] ). And the host gets x1 dollars. Thus, you can infer that x1 = 2x1 or x1 = x1/2. Now the host says "You are offered the options of keeping your x1 or switching to my x1". What should you do? Envelope A Envelope B Figure 9.4: Two envelope problem [(P1):Why is it paradoxical?]. You get α = x1. Then, you reason that, with probability 1/2, x1 is equal to either α/2 or 2α dollars. Thus the expected value (denoted Eother(α) at this moment) of the other envelope is Eother(α) = (1/2)(α/2) + (1/2)(2α) = 1.25α (9.18) This is greater than the α in your current envelope A. Therefore, you should switch to B. But this seems clearly wrong, as your information about A and B is symmetrical. This is the famous two-envelope paradox (i.e., "The Other Person's Envelope is Always Greener" ). 232 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) 9.5.1 (P1): Bayesian approach to the two envelope problem Consider the state space Ω such that Ω = R+(= {ω ∈ R | ω ≥ 0}) with Lebesgue measure ν. Thus, we start from the classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Also, putting Ω = {(ω, 2ω) | ω ∈ R+}, we consider the identification: Ω 3 ω ←→ (identification) (ω, 2ω) ∈ Ω (9.19) Further, define V1 : Ω(≡ R+)→ X(≡ R+) and V2 : Ω(≡ R+)→ X(≡ R+) such that V1(ω) = ω, V2(ω) = 2ω (∀ω ∈ Ω) And define the observable O = (X(= R+),F(= BR+ : the Borel field), F ) in L ∞(Ω, ν) such that [F (Ξ)](ω) =  1 ( if ω ∈ Ξ, 2ω ∈ Ξ) 1/2 ( if ω ∈ Ξ, 2ω /∈ Ξ) 1/2 ( if ω /∈ Ξ, 2ω ∈ Ξ) 0 ( if ω /∈ Ξ, 2ω /∈ Ξ) (∀ω ∈ Ω,∀Ξ ∈ F) 6 α (α 2 , α) (α, 2α) X(= R+) Ω(≈ Ω = R+) Figure 9.5: Two envelope problem Recalling the identification : Ω 3 (ω, 2ω)←→ ω ∈ Ω = R+, assume that ρ0(D) = ∫ D w0(ω)dω (∀D ∈ BΩ = BR+) where the probability density function w0 : Ω(≈ R+)→ R+ is assumed to be continuous positive function. That is, the mixed state ρ0(∈ M+1(Ω(= R+))) has the probability density function w0. Axiom(m) 1(§9.1) says that 233 Ishikawa's Homepage 9.5 Two envelope problem (Bayes' method) (A1) The probability P (Ξ) (Ξ ∈ BX = BR+) that a measured value obtained by the mixed measurement ML∞(Ω,dω)(O = (X,F, F ), S[∗](ρ0)) belongs to Ξ(∈ BX = BR+) is given by P (Ξ) = ∫ Ω [F (Ξ)](ω)ρ0(dω) = ∫ Ω [F (Ξ)](ω)w0(ω)dω = ∫ Ξ w0(x/2) 4 + w0(x) 2 dx (∀Ξ ∈ BR+) (9.20) Therefore, the expectation is given by∫ R+ xP (dx) = 1 2 ∫ ∞ 0 x * ( w0(x/2)/2 + w0(x) ) dx = 3 2 ∫ R+ xw0(x)dx Further, Theorem 9.11 ( Bayes' theorem ) says that (A2) When a measured value α is obtained by the mixed measurement ML∞(Ω,dω)(O = (X,F, F ), S[∗](ρ0)), then the post-state ρpost(∈M+1(Ω)) is given by ραpost = w0(α/2) 2 h(α/2) 2 + w0(α) δ(α 2 ,α) + w0(α) w0(α/2) 2 + w0(α) δ(α,2α) (9.21) Hence, (A3) if [∗] = { δ(α 2 ,α) δ(α,2α) } , then you change { α −→ α 2 α −→ 2α } , and thus you get the switching gain{ α 2 − α(= −α 2 ) 2α− α(= α) } . Therefore, the expectation of the switching gain is calculated as follows:∫ R+ ( (−α 2 ) w0(α/2) 2 w0(α/2) 2 + w0(α) + α w0(α) w0(α/2) 2 + w0(α) ) P (dα) = ∫ R+ (−α 2 ) w0(α/2) 4 + α * w0(α) 2 dα = 0 (9.22) Therefore, we see that the swapping is even, i.e., no advantage and no disadvantage. 234 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) 9.6 Monty Hall problem (The Bayesian approach) 9.6.1 The review of Problem5.14 ( Monty Hall problem in pure measurement) Problem 9.15. [Monty Hall problem (The answer to Fisher's maximum likelihood method) ] You are on a game show and you are given the choice of three doors. Behind one door is a car, and behind the other two are goats. You choose, say, door 1, and the host, who knows where the car is, opens another door, behind which is a goat. For example, the host says that ([) the door 3 has a goat. And further, He now gives you the choice of sticking with door 1 or switching to door 2? What should you do? ? ? ? door door door No. 1 No. 2 No. 3 Figure 9.6: Monty Hall problem Answer: Put Ω = {ω1, ω2, ω3} with the discrete topology dD and the counting measure ν. Thus consider the classical basic structure: [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Assume that each state δωm(∈ Sp(C0(Ω)∗)) means δωm ⇔ the state that the car is behind the door 1 (m = 1, 2, 3) Define the observable O1 ≡ ({1, 2, 3}, 2{1,2,3}, F1) in L∞(Ω) such that [F1({1})](ω1) = 0.0, [F1({2})](ω1) = 0.5, [F1({3})](ω1) = 0.5, [F1({1})](ω2) = 0.0, [F1({2})](ω2) = 0.0, [F1({3})](ω2) = 1.0, 235 Ishikawa's Homepage 9.6 Monty Hall problem (The Bayesian approach) [F1({1})](ω3) = 0.0, [F1({2})](ω3) = 1.0, [F1({3})](ω3) = 0.0, (9.23) where it is also possible to assume that F1({2})(ω1) = α, F1({3})(ω1) = 1 − α (0 < α < 1). The fact that you say "the door 1" means that we have a measurement ML∞(Ω)(O1, S[∗]). Here, we assume that a) "a measured value 1 is obtained by the measurement ML∞(Ω)(O1, S[∗])" ⇔ The host says "Door 1 has a goat" b) "measured value 2 is obtained by the measurement ML∞(Ω)(O1, S[∗]) " ⇔ The host says "Door 2 has a goat" c) "measured value 3 is obtained by the measurement ML∞(Ω)(O1, S[∗]) " ⇔ The host says "Door 3 has a goat" Since the host said "Door 3 has a goat", this implies that you get the measured value "3" by the measurement ML∞(Ω)(O1, S[∗]). Therefore, Theorem 5.6 (Fisher's maximum likelihood method) says that you should pick door number 2. That is because we see that max{[F1({3})](ω1), [F1({3})](ω2), [F1({3})](ω3)} = max{0.5, 1.0, 0.0} = 1.0 = [F1({3})](ω2) and thus, there is a reason to infer that [∗] = δω2 . Thus, you should switch to door 2. This is the first answer to Monty-Hall problem. 9.6.2 Monty Hall problem in mixed measurement Next, let us study Monty Hall problem in mixed measurement theory (particularly, Bayesian statistics). Problem 9.16. [Monty Hall problem(The answer by Bayes' method) ] Suppose you are on a game show, and you are given the choice of three doors (i.e., "number 1", "number 2", "number 3"). Behind one door is a car, behind the others, goats. You pick a door, say number 1. Then, the host, who set a car behind a certain door, says (]1) the car was set behind the door decided by the cast of the distorted dice. That is, the host set the car behind the k-th door (i.e., "number k") with probability pk (or, weight such that p1 + p2 + p3 = 1, 0 ≤ p1, p2, p3 ≤ 1 ). And further, the host says, for example, 236 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) ([) the door 3 has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors? Answer: In the same way as we did in Problem 9.15 (Monty Hall problem:the answer by Fisher's maximum likelihood method), consider the state space Ω = {ω1, ω2, ω3} with the discrete metric dD and the observable O1. Under the hypothesis (]1), define the mixed state ν0 ( ∈M+1(Ω)) such that ν0 = p1δω1 + p2δω2 + p3δω3 namely, ν0({ω1}) = p1, ν0({ω2}) = p2, ν0({ω3}) = p3 Thus we have a mixed measurement ML∞(Ω)(O1, S[∗](ν0)). Note that a) "measured value 1 is obtained by the mixed measurement ML∞(Ω)(O1, S[∗](ν0))" ⇔ the host says "Door 1 has a goat" b) "measured value 2 is obtained by the mixed measurement ML∞(Ω)(O1, S[∗](ν0))" ⇔ the host says "Door 2 has a goat" c) "measured value 3 is obtained by the mixed measurement ML∞(Ω)(O1, S[∗](ν0))" ⇔ the host says "Door 3 has a goat" Here, assume that, by the mixed measurement ML∞(Ω)(O1, S[∗](ν0)), you obtain a measured value 3, which corresponds to the fact that the host said "Door 3 has a goat". Then, Theorem 9.11 (Bayes' theorem) says that the posterior state νpost ( ∈M+1(Ω)) is given by νpost = F1({3})× ν0〈 ν0, F1({3}) 〉 . That is, νpost({ω1}) = p1 2 p1 2 + p2 , νpost({ω2}) = p2 p1 2 + p2 , νpost({ω3}) = 0. Particularly, we see that (]2) if p1 = p2 = p3 = 1/3, then it holds that νpost({ω1}) = 1/3, νpost({ω2}) = 2/3, νpost({ω3}) = 0, and thus, you should pick Door 2. 237 Ishikawa's Homepage 9.6 Monty Hall problem (The Bayesian approach) ♠Note 9.3. It is not natural to assume the rule (]1) in Problem 9.16. That is because the host may intentionally set the car behind a certain door. Thus we think that Problem 9.16 is temporary. For our formal assertion, see Problem 9.17 latter. 238 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) 9.7 Monty Hall problem (The principle of equal weight) 9.7.1 The principle of equal weight- The most famous unsolved problem Let us reconsider Monty Hall problem (Problem 9.14, Problem9.15) in what follows. We think that the following is one of the most reasonable answers (also, see Problem 19.5). Problem 9.17. [Monty Hall problem (The principle of equal weight) ] Suppose you are on a game show, and you are given the choice of three doors (i.e., "number 1", "number 2", "number 3"). Behind one door is a car, behind the others, goats. (]2) You choose a door by the cast of the fair dice, i.e., with probability 1/3. According to the rule (]2), you pick a door, say number 1, and the host, who knows where the car is, opens another door, behind which is a goat. For example, the host says that ([) the door 3 has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors? Answer: By the same way of Problem9.15 and Problem9.16 (Monty Hall problem), define the state space Ω = {ω1, ω2, ω3} and the observable O = (X,F, F ). And the observable O = (X,F, F ) is defined by the formula (9.23). The map φ : Ω→ Ω is defined by φ(ω1) = ω2, φ(ω2) = ω3, φ(ω3) = ω1 we get a causal operator Φ : L∞(Ω)→ L∞(Ω) by [Φ(f)](ω) = f(φ(ω)) (∀f ∈ L∞(Ω), ∀ω ∈ Ω). Assume that a car is behind the door k (k = 1, 2, 3). Then, we say that (a) By the dice-throwing, you get  1, 23, 4 5, 6  , then, take a measurement  ML∞(Ω)(O, S[ωk])ML∞(Ω)(ΦO, S[ωk]) ML∞(Ω)(Φ 2O, S[ωk])  We, by the argument in Chapter 11 (cf. the formula (11.7))2, see the following identifications: ML∞(Ω)(ΦO, S[ωk]) = ML∞(Ω)(O, S[φ(ωk)]), ML∞(Ω)(Φ 2O, S[ωk]) = ML∞(Ω)(O, S[φ2(ωk)]). Thus, the above (a) is equal to 2Thus, from the pure theoretical point of view, this problem should be discussed after Chapter 11 239 Ishikawa's Homepage 9.7 Monty Hall problem (The principle of equal weight) (b) By the dice-throwing, you get  1, 23, 4 5, 6  then, take a measurement  ML∞(Ω)(O, S[ωk])ML∞(Ω)(O, S[φ(ωk)]) ML∞(Ω)(O, S[φ2(ωk)])  Here, note that 1 3 (δωk + δφ(ωk) + δφ2(ωk)) = 1 3 (δω1 + δω2 + δω3) (∀k = 1, 2, 3). Thus, this (b) is identified with the mixed measurement ML∞(Ω)(O, S[∗](νe)) , where νe = 1 3 (δω1 + δω2 + δω3) Therefore, Problem 9.17 is the same as Problem 9.16. Hence, you should choose the door 2. ♠Note 9.4. The above argument is easy. That is, since you have no information, we choose the door by a fair dice throwing. In this sense, the principle of equal weight - unless we have sufficient reason to regard one possible case as more probable than another, we treat them as equally probable - is clear in measurement theory. However, it should be noted that the above argument is based on dualism. From the above argument, we have the following theorem. Theorem 9.18. [The principle of equal weight] Consider a finite state space Ω, that is, Ω = {ω1, ω2, . . . , ωn}. Let O = (X,F, F ) be an observable in L∞(Ω, ν), where ν is the counting measure. Consider a measurement ML∞(Ω)(O, S[∗]). If the observer has no information for the state [∗], there is a reason to that this measurement is identified with the mixed measurement ML∞(Ω)(O, S[∗](we)) ( or, ML∞(Ω)(O, S[∗](νe)) ) , where we(ωk) = 1/n (∀k = 1, 2, ..., n) or νe = 1 n n∑ k=1 δωk Proof. The proof is a easy consequence of the above Monty Hall problem (or, see [30, 33]). ♠Note 9.5. Concerning the principle of equal weight, we deal the following three kinds: (]1) the principle of equal weight in Remark 5.19 (]2) the principle of equal weight in Theorem 9.18 (]3) the principle of equal weight in Proclaim 19.4 240 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) 9.8 Averaging information ( Entropy ) As one of applications (of Bayes theorem), we now study the "entropy (cf. [82])" of the measurement. This section is due to the following refs. (]) Ref. [27]: S. Ishikawa, A Quantum Mechanical Approach to Fuzzy Theory, Fuzzy Sets and Systems, Vol. 90, No. 3, 277-306, 1997, doi: 10.1016/S0165-0114(96)00114-5 (]) Ref. [30]: S. Ishikawa, "Mathematical Foundations of Measurement Theory," Keio University Press Inc. 2006. Let us begin with the following definition. Definition 9.19. [Entropy (cf. [27, 30]) ] Assume Classical basic structure [C0(Ω) ⊆ L∞(Ω, ν) ⊆ B(L2(Ω, ν))] Consider a mixed measurement ML∞(Ω,ν) (O = (X, 2 X , F ), S[∗](w0)) with a countable measured value space X = {x1, x2, . . .}. The probability P ({xn}) that a measured value xn is obtained by the mixed measurement ML∞(Ω)(O, S[∗](w0)) is given by P ({xn}) = ∫ Ω [F ({xn})](ω)w0(ω)ν(dω) (9.24) Further, when a measured value xn is obtained, the information I({xn}) is, from Bayes' theorem 9.11, is calculated as follows. I({xn}) = ∫ Ω [F ({xn})](ω)∫ Ω [F ({xn})](ω)w0(ω)ν(dω) log [F ({xn})](ω)∫ Ω [F ({xn})](ω)w0(ω)ν(dω) w0(ω)ν(dω) Therefore, the averaging informationH ( ML∞(Ω)(O, S[∗](w0)) ) of the mixed measurement ML∞(Ω) (O, S[∗](w0)) is naturally defined by H ( ML∞(Ω)(O, S[∗](w0)) ) = ∞∑ n=1 P ({xn}) * I({xn}) (9.25) Also, the following is clear: H ( ML∞(Ω)(O, S[∗](w0)) ) = ∞∑ n=1 ∫ Ω [F ({xn})](ω) log[F ({xn})](ω)w0(ω)ν(dω) − ∞∑ n=1 P ({xn}) logP ({xn}) (9.26) 241 Ishikawa's Homepage 9.8 Averaging information ( Entropy ) Example 9.20. [The offender is man or female? fast or slow?] Assume that (a) There are 100 suspected persons such as {s1, s2, . . . , s100}, in which there is one criminal. Define the state space Ω = {ω1, ω2, . . . , ω100} such that stateωn * * * the state such that suspect sn is a criminal (n = 1, 2, ..., 100) Assume the counting measure ν such that ν({ωk}) = 1(∀k = 1, 2, * * * , 100) Define a maleobservable Om = (X = {ym, nm}, 2X ,M) in L∞(Ω) by [M({ym})](ωn) = mym(ωn) = { 0 (n is odd) 1 (n is even) [M({nm})](ωn) = mnm(ωn) = 1− [M({ym})](ωn) For example, Taking a measurement ML∞(Ω)(Om, S[ω17]) - the sex of the criminal s17 -, we get the measured value nm(=female). Also, define the fast-observable Of = (Y = {yf , nf}, 2Y , F ) in L∞(Ω) by [F ({yf})](ωn) = fyf (ωn) = n− 1 99 , [F ({nf})](ωn) = fnf (ωn) = 1− [F ({yf})](ωn) 0 1 Ω 100 f{yf} f{nf} According to the principle of equal weight (=Theorem 9.18 ), there is a reason to consider that a mixed state w0 (∈ L1+1(Ω)) is equal to the state we such that w0(ωn) = we(ωn) = 1/100 (∀n). Thus, consider two mixed measurement ML∞(Ω)(Om, S[∗](we)) and ML∞(Ω)(Of , S[∗](we)). Then, we see: H ( ML∞(Ω)(Om, S[∗](we)) ) = ∫ Ω mym(ω)we(ω)ν(dω) * log ∫ Ω mym(ω)we(ω)ν(dω) − ∫ Ω m{nm}(ω)we(ω)ν(dω) * log ∫ Ω mnm(ω)we(ω)ν(dω) 242 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) = −1 2 log 1 2 − 1 2 log 1 2 = log2 2 = 1 (bit) 3. Also, H ( ML∞(Ω)(Of , S[∗](we)) ) = ∫ Ω fyf (ω) log fyf (ω)we(ω)ν(dω) + ∫ Ω fnf (ω) log fnf (ω)we(ω)ν(dω)− ∫ Ω fyf (ω)we(ω)ν(dω) * log ∫ Ω fyf (ω)we(ω)ν(dω) − ∫ Ω fnf (ω)we(dω) * log ∫ Ω fnf (ω)we(ω)ν(dω) +2 ∫ 1 0 λ log2 λdλ+ 1 = − 1 2 loge 2 + 1 = 0.278 * * * (bit) Therefore, as eyewitness information, "male or female" has more valuable than "fast or slow". 243 Ishikawa's Homepage 9.9 Fisher statistics:Monty Hall problem [three prisoners problem] 9.9 Fisher statistics:Monty Hall problem [three prisoners problem] This section is extracted from the following: Ref. [46]: S. Ishikawa; The Final Solutions of Monty Hall Problem and Three Prisoners Problem ( arXiv:1408.0963v1 [stat.OT] 2014 ) It is usually said that Monty Hall problem and three prisoners problem are so-called isomorphism problem But, we think that the meaning of "isomorphism problem" is not clarified, or, it is not able to be clarified without measurement (or, the dualism). Therefore, in order to understand "isomorphism", we simultaneously discuss the two • { Monty Hall problem three prisoners problem 9.9.1 Fisher statistics: Monty Hall problem [resp. three prisoners problem] Problem 9.21. (=Problem9.15: [Monty Hall problem]). Suppose you are on a game show, and you are given the choice of three doors (i.e., "Door A1", "Door A2", "Door A3"). Behind one door is a car, behind the others, goats. You do not know what's behind the doors However, you pick a door, say "Door A1", and the host, who knows what's behind the doors, opens another door, say "Door A3", which has a goat. He says to you, "Do you want to pick Door A2?" Is it to your advantage to switch your choice of doors? ? ? ? Door A1 Door A2 Door A3 244 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) Problem 9.22. [three prisoners problem]. Three prisoners, A1, A2, and A3 were in jail. They knew that one of them was to be set free and the other two were to be executed. They did not know who was the one to be spared, but the emperor did know. A1 said to the emperor, "I already know that at least one the other two prisoners will be executed, so if you tell me the name of one who will be executed, you won't have given me any information about my own execution". After some thinking, the emperor said, "A3 will be executed." Thereupon A1 felt happier because his chance had increased from 1 3(=Num{A1,A2,A3}]) to 1 2(=Num{A1,A2}]) . This prisoner A1's happiness may or may not be reasonable? E A1 A2 A3 - - " A3 will be executed" (Emperor) 9.9.2 The answer in Fisher statistics: Monty Hall problem [resp. three prisoners problem] Let rewrite the spirit of dualism (Descartes figure) as follows. • observer (I(=mind)) system (matter)  - [observable] [measured value] a©interfere b©perceive a reaction [state] Descartes Figure 9.7: The image of "measurement(= a©+ b©)" in dualism 245 Ishikawa's Homepage 9.9 Fisher statistics:Monty Hall problem [three prisoners problem] In the dualism, we have the confrontation "observer←→system" as follows. Table 9.1: Correspondence: observer * system Problems dualism Mind(=I=Observer) Matter(=System) Monty Hall problem you Three doors Three prisoners problem Prisoner A1 Emperor's mind In what follows, we present the first answer to [ Problem 9.21 (Monty-Hall problem) Problem 9.22 (Three prisoners problem) ] in classical pure measurement theory. The two will be simultaneously solved as follows. The spirit of dualism (in Figure 9.7) urges us to declare that (A) [ "observer ≈ you" and "system ≈ three doors" in Problem 9.21 "observer ≈ prisoner A1" and "system ≈ emperor's mind" in Problem 9.22 ] Put Ω = {ω1, ω2, ω3} with the discrete topology. Assume that each state δωm(∈ Sp(C(Ω)∗)) means [ δωm ⇔ the state that the car is behind the door Am δωm ⇔ the state that the prisoner Am is will be executed ] (m = 1, 2, 3) (9.27) Define the observable O1 ≡ ({1, 2, 3}, 2{1,2,3}, F1) in L∞(Ω) such that [F1({1})](ω1) = 0.0, [F1({2})](ω1) = 0.5, [F1({3})](ω1) = 0.5, [F1({1})](ω2) = 0.0, [F1({2})](ω2) = 0.0, [F1({3})](ω2) = 1.0, [F1({1})](ω3) = 0.0, [F1({2})](ω3) = 1.0, [F1({3})](ω3) = 0.0, (9.28) where it is also possible to assume that F1({2})(ω1) = α, F1({3})(ω1) = 1 − α (0 < α < 1). Thus we have a measurement ML∞(Ω)(O1, S[∗]), which should be regarded as the measurement theoretical representation of the measurement that [ you say "Door A1" "Prisoner A1" asks to the emperor ] . Here, we assume that a) "measured value 1 is obtained by the measurement ML∞(Ω)(O1, S[∗])" ⇔ [ the host says "Door A1 has a goat" the emperor says "Prisoner A1 will be executed" ] b) "measured value 2 is obtained by the measurement ML∞(Ω)(O1, S[∗]) " ⇔ [ the host says "Door A2 has a goat" the emperor says "Prisoner A2 will be executed" ] 246 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) c) "measured value 3 is obtained by the measurement ML∞(Ω)(O1, S[∗]) " ⇔ [ the host says "Door A3 has a goat" the emperor says "Prisoner A3 will be executed" ] Recall that [ the host said "Door 3 has a goat" the emperor said "Prisoner A3 will be executed" ] . This implies that [ you Prisoner A1 ] get the measured value "3" by the measurement ML∞(Ω)(O1, S[∗]). Note that [F1({3})](ω2) = 1.0 = max{0.5, 1.0, 0.0} = max{[F1({3})](ω1), [F1({3})](ω2), [F1({3})](ω3)}, (9.29) Therefore, Theorem 5.6 (Fisher's maximum likelihood method) says that (B1) In Problem 9.21 (Monty-Hall problem), there is a reason to infer that [∗] = δω2 . Thus, you should switch to Door A2. (B2) In Problem 9.22 (Three prisoners problem), there is a reason to infer that [∗] = δω2 . However, there is no reasonable answer for the question: whether Prisoner A1's happiness increases. That is, Problem 9.22 is not within Fisher's maximum likelihood method. 247 Ishikawa's Homepage 9.10 Bayesian statistics: Monty Hall problem [three prisoners problem] 9.10 Bayesian statistics: Monty Hall problem [three prisoners problem] This section is extracted from the following: Ref. [46]: S. Ishikawa; The Final Solutions of Monty Hall Problem and Three Prisoners Problem ( arXiv:1408.0963v1 [stat.OT] 2014 ) 9.10.1 Bayesian statistics: Monty Hall problem [resp. three prisoners problem] Problem 9.23. [(=Problem9.16)Monty Hall problem (the case that the host throws the dice)]. Suppose you are on a game show, and you are given the choice of three doors (i.e., "Door A1", "Door A2", "Door A3"). Behind one door is a car, behind the others, goats. You do not know what's behind the doors. However, you pick a door, say "Door A1", and the host, who knows what's behind the doors, opens another door, say "Door A3", which has a goat. And he adds that (]1) the car was set behind the door decided by the cast of the (distorted) dice. That is, the host set the car behind Door Am with probability pm (where p1 + p2 + p3 = 1, 0 ≤ p1, p2, p3 ≤ 1 ). He says to you, "Do you want to pick Door A2?" Is it to your advantage to switch your choice of doors? ? ? ? Door A1 Door A2 Door A3 Problem 9.24. [three prisoners problem]. Three prisoners, A1, A2, and A3 were in jail. They knew that one of them was to be set free and the other two were to be executed. They did not know who was the one to be 248 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) spared, but they know that (]2) the one to be spared was decided by the cast of the (distorted) dice. That is, Prisoner Am is to be spared with probability pm (where p1 + p2 + p3 = 1, 0 ≤ p1, p2, p3 ≤ 1 ). but the emperor did know the one to be spared. A1 said to the emperor, "I already know that at least one the other two prisoners will be executed, so if you tell me the name of one who will be executed, you won't have given me any information about my own execution". After some thinking, the emperor said, "A3 will be executed." Thereupon A1 felt happier because his chance had increased from 1 3(=Num[{A1,A2,A3}]) to 1 2(=Num[{A1,A2}]) . This prisoner A1's happiness may or may not be reasonable? E A1 A2 A3 - - "A3 will be executed" (Emperor) 9.10.2 The answer in Bayesian statistics: Monty Hall problem [resp. three prisoners problem] In the dualism, we have the confrontation "observer←→system" as follows. Table 9.2: Correspondence: observer * system Problems dualism Mind(=I=Observer) Matter(=System) Monty Hall problem you Three doors Three prisoners problem Prisoner A Emperor's mind In what follows we study these problems. Let Ω and O1 be as in Section 9.8. Under the hypothesis { (]1) (]2) } , define the mixed state ν0 ( ∈Mm+1(Ω)) such that: ν0({ω1}) = p1, ν0({ω2}) = p2, ν0({ω3}) = p3 (9.30) 249 Ishikawa's Homepage 9.10 Bayesian statistics: Monty Hall problem [three prisoners problem] Thus we have a mixed measurement ML∞(Ω)(O1, S[∗](ν0)). Note that a) "measured value 1 is obtained by the measurement ML∞(Ω)(O1, S[∗])" ⇔ [ the host says "Door A1 has a goat" the emperor says "Prisoner A1 will be executed" ] b) "measured value 2 is obtained by the measurement ML∞(Ω)(O1, S[∗]) " ⇔ [ the host says "Door A2 has a goat" the emperor says "Prisoner A2 will be executed" ] c) "measured value 3 is obtained by the measurement ML∞(Ω)(O1, S[∗]) " ⇔ [ the host says "Door A3 has a goat" the emperor says "Prisoner A3 will be executed" ] Here, assume that, by the statistical measurement ML∞(Ω)(O1, S[∗](ν0)), you obtain a measured value 3, which corresponds to the fact that [ the host said "Door A3 has a goat" the emperor said "Prisoner A3 is to be executed" ] Then, Bayes' theorem 9.11 says that the posterior state νpost ( ∈Mm+1(Ω)) is given by νpost = F1({3})× ν0〈 ν0, F1({3}) 〉 . (9.31) That is, νpost({ω1}) = p1 2 p1 2 + p2 , νpost({ω2}) = p2 p1 2 + p2 , νpost({ω3}) = 0. (9.32) Then, (I1) In Problem 9.23, if νpost({ω1}) < νpost({ω2}) (i.e., p1 < 2p2), you should pick Door A2 if νpost({ω1}) = νpost({ω2}) (i.e., p1 = 2p2), you may pick Doors A1 or A2 if νpost({ω1}) > νpost({ω2}) (i.e., p1 > 2p2), you should not pick Door A2 (I2) In Problem 9.24, if ν0({ω1}) < νpost({ω1}) (i.e., p1 < 1− 2p2), the prisoner A1's happiness increases if ν0({ω1}) = νpost({ω1}) (i.e., p1 = 1− 2p2), the prisoner A1's happiness is invariant if ν0({ω1}) > νpost({ω1}) (i.e., p1 > 1− 2p2), the prisoner A1's happiness decreases 250 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) 9.11 Equal probability: Monty Hall problem [three prisoners problem] This section is extracted from the following: ref. [46]: S. Ishikawa; The Final Solutions of Monty Hall Problem and Three Prisoners Problem ( arXiv:1408.0963v1 [stat.OT] 2014 ) Problem 9.25. [(=Problem9.16)Monty Hall problem (the case that you throws the dice)]. Suppose you are on a game show, and you are given the choice of three doors (i.e., "Door A1", "Door A2", "Door A3"). Behind one door is a car, behind the others, goats. You do not know what's behind the doors. Thus, (]1) you select Door A1 by the cast of the fair dice. That is, you say "Door A1" with probability 1/3. The host, who knows what's behind the doors, opens another door, say "Door A3", which has a goat. He says to you, "Do you want to pick Door A2?" Is it to your advantage to switch your choice of doors? ? ? ? Door A1 Door A2 Door A3 Problem 9.26. [three prisoners problem( the case that the prisoner throws the dice)]. Three prisoners, A1, A2, and A3 were in jail. They knew that one of them was to be set free and the other two were to be executed. They did not know who was the one to be spared, but the emperor did know. Since three prisoners wanted to ask the emperor, (]2) the questioner was decided by the fair die throw. And Prisoner A1 was selected with probability 1/3 Then, A1 said to the emperor, "I already know that at least one the other two prisoners 251 Ishikawa's Homepage 9.11 Equal probability: Monty Hall problem [three prisoners problem] will be executed, so if you tell me the name of one who will be executed, you won't have given me any information about my own execution". After some thinking, the emperor said, "A3 will be executed." Thereupon A1 felt happier because his chance had increased from 1 3(=Num[{A1,A2,A3}]) to 1 2(=Num[{A1,A2}]) . This prisoner A1's happiness may or may not be reasonable? E A1 A2 A3 - - "A3 will be executed" (Emperor) Answer By Theorem 9.18(The principle of equal weight), the above Problems 9.25 and 9.26 is respectively the same as Problems 9.23 and 9.24 in the case that p1 = p2 = p3 = 1/3. Then, the formulas (9.30) and (9.32) say that (A1) In Problem9.25, since νpost({ω1}) = 1/3 < 2/3 = νpost({ω2}), you should pick Door A2. (A2) In Problem9.26, since ν0({ω1}) = 1/3 = νpost({ω1}), the prisoner A1's happiness is invariant. Therefore, (B1) Problem9.25 [Monty Hall problem ( the case that you throw a fair dice)] νpost({ω1}) < νpost({ω2}) (i.e., p1 = 1/3 < 2/3 = 2p2), thus, you should choose the door A2 (B2) Problem9.26 [three prisoners problem ( the case that the emperor throws a fair dice)], ν0({ω1}) = νpost({ω1}) (i.e., p1 = 1/3 = 1− 2p2), Thus, the happiness of the prisoner A1 is invariant 252 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) ♠Note 9.6. These problems (i.e., Monty Hall problem and the three prisoners problem) continued attracting the philosopher's interest. This is not due to that these are easy to make a mistake for high school students, but these problems include the essence of "dualism". 253 Ishikawa's Homepage 9.12 Bertrand's paradox( "randomness" depends on how you look at) 9.12 Bertrand's paradox( "randomness" depends on how you look at) Theorem9.18(the principle of equal weight) implies that • the "randomness" may be related to the invariant probability measure. However, this is due to the finiteness of the state space. In the case of infinite state space, "randomness" depends on how you look at This is explained in this section. 9.12.1 Bertrand's paradox("randomness" depends on how you look at) Let us explain Bertrand's paradox as follows. Consider classical basic structure: [C0(Ω) ⊆ L∞(Ω,m) ⊆ B(L2(Ω,m))] We can define the exact observable OE = (Ω,BΩ, FE) in L ∞(Ω,m) such that [FE(Ξ)](ω) = χΞ(ω) = { 1 (ω ∈ Ξ) 0 (ω /∈ Ξ) (∀ω ∈ Ω, Ξ ∈ BΩ) Here, we have the following problem: (A) Can the measurement ML∞(Ω,m)(OE, S[∗](ρ)) that represents "at random" be determined uniquely? This question is of course denied by so-called Bertrand paradox. Here, let us review the argument about the Bertrand paradox (cf. [22, 30, 44]). Consider the following problem: Problem 9.27. (Bertrand paradox) Given a circle with the radius 1. Suppose a chord of the circle is chosen at random. What is the probability that the chord is shorter than √ 3? 254 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) -x1 1 6 x2 l Figure 9.8: Bertrand' paradox Define the rotation map T θrot : R 2 → R2 (0 ≤ θ < 2π) and the reverse map Trev : R2 → R2 such that T θrotx = [ cos θ − sin θ sin θ cos θ ] * [ x1 x2 ] , Trevx = [ 0 1 1 0 ] * [ x1 x2 ] Problem 9.28. (Bertrand paradox and its answer) Given a circle with the radius 1. -x1 1 6 x2 l Figure 9.9: Bertrand' paradox Put Ω = {l | l is a chord}, that is, the set of all chords. (B) Can we uniquely define an invariant probability measure on Ω? Here, "invariant" means "invariant concerning the rotation map T θrot and reverse map Trev". In what follows, we show that the above invariant measure exists but it is not determined uniquely. 255 Ishikawa's Homepage 9.12 Bertrand's paradox( "randomness" depends on how you look at) α β (Pic.2)(Pic.1) (x, y)• 0 10 1 l(α,β) l(x,y) Figure 9.10: Two cases in Bertrand' paradox [The first answer (Pic.1(in Figure 9.10))]. In Pic.1, we see that the chord l is represented by a point (α, β) in the rectangle Ω1 ≡ {(α, β) | 0 < α ≤ 2π, 0 < β ≤ π/2(radian)}. That is, we have the following identification: Ω(= the set of all chords) 3 l(α,β) ←→ identification (α, β) ∈ Ω1(⊂ R2). Note that we have the natural probability measure nu1 on Ω1 such that ν1(A) = Meas[A] Meas[Ω1] = Meas[A] π2 (∀A ∈ BΩ1), where " Meas" = " Lebesgue measure". Transferring the probability measure ν1 on Ω1 to Ω, we get ρ1 on Ω. That is, M+1(Ω) 3 ρ1 ←→ identification ν1 ∈M+1(Ω1) (]) It is clear that the measure ρ1 is invariant concerning the rotation map T θ rot and reverse map Trev. Therefore, we have a natural measurement ML∞(Ω,m)(OE ≡ (Ω,BΩ, FE), S[∗](ρ1)). Consider the identification: Ω ⊇ Ξ√3 ←→ identification {(α, β) ∈ Ω1 : "the length of l(α,β)" < √ 3} ⊆ Ω1 Then, Axiom(m) 1 says that the probability that a measured value belongs to Ξ√3 is given by∫ Ω [FE(Ξ√3)](ω) ρ1(dω) = ∫ Ξ√3 1 ρ1(dω) =m1({l(α,β) ≈ (α, β) ∈ Ω1 | "the length of l(α,β)" ≤ √ 3}) = Meas[{(α, β) | 0 ≤ α ≤ 2π, π/6 ≤ β ≤ π/2}] Meas[{(α, β) | 0 ≤ α ≤ 2π, 0 ≤ β ≤ π/2}] 256 Ishikawa's Homepage Chap. 9 Mixed measurement theory (⊃Bayesian statistics) = 2π × (π/3) π2 = 2 3 . [The second answer (Pic.2(in Figure 9.10))]. In Pic.2, we see that the chord l is represented by a point (x, y) in the circle Ω2 ≡ {(x, y) | x2 + y2 < 1}. That is, we have the following identification: Ω(= the set of all chords) 3 l(x,y) ←→ identification (x, y) ∈ Ω2(⊂ R2). We have the natural probability measure ν2 on Ω2 such that ν2(A) = Meas[A] Meas[Ω2] = Meas[A] π (∀A ∈ BΩ2). Transferring the probability measure ν2 on Ω2 to Ω, we get ρ2 on Ω. That is, M+1(Ω) 3 ρ2 ←→ identification ν2 ∈M+1(Ω2) (]) It is clear that the measure ρ2 is invariant concerning the rotation map T θ rot and reverse map Trev. Therefore, we have a natural measurement ML∞(Ω,m)(OE ≡ (Ω,BΩ, FE), S[∗](ρ2)). Consider the identification: Ω ⊇ Ξ√3 ←→ identification {(x, y) ∈ Ω2 : "the length of l(α,β)" < √ 3} ⊆ Ω1 Then, Axiom(m) 1 says that the probability that a measured value belongs to Ξ√3 is given by ∫ Ω [FE(Ξ√3)](ω) ρ2(dω) = ∫ Ξ√3 1 ρ2(dω) =ν2({l(x,y) ≈ (x, y) ∈ Ω2 | "the length of l(x,y)" ≤ √ 3}) = Meas[{(x, y) | 1/4 ≤ x2 + y2 ≤ 1}] π = 3 4 . Conclusion 9.29. Thus, even if there is a custom to regard a natural probability measure (i.e., an invariant measure concerning natural maps) as "random", the first answer and the second answer say that (]) the uniqueness in (B) of Problem 9.28 is denied. 257 Ishikawa's Homepage

Chapter 10 Axiom 2-causality Measurement theory has the following classification: (A) measurement theory (=quantum language)  pure type (A1) { classical system : Fisher statistics quantum system : usual quantum mechanics mixed type (A2) { classical system : including Bayesian statistics, Kalman filter quantum system : quantum decoherence This is formulated as follows. (B)  (B1): pure measurement theory (=quantum language) := [(pure)Axiom 1] pure measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } the manual to use spells (B2): mixed measurement theory (=quantum language) := [(mixed)Axiom(m) 1] mixed measurement (cf. §9.1) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } the manual to use spells In this chapter, we devote ourselves to the last theme [Axiom 2] Causality (cf. §10.3) , which is common to both (B1) and (B2). 259 10.1 The most important unsolved problem-what is causality? 10.1 The most important unsolved problem-what is causality? The importance of "measurement" and "causality" should be reconfirmed in the following famous maxims: (C1) There is no science without measurement. (C2) Science is the knowledge about causal relationship. They should be also regarded as one of the linguistic interpretation in a wider sense. 10.1.1 Modern science started from the discovery of "causality." When a certain thing happens, the cause always exists. This is called causality. You should just remember the proverb of "smoke is not located on the place which does not have fire." It is not so simple although you may think that it is natural. For example, if you consider This morning I feel good. Is it because that I slept sound yesterday? or is it because I go to favorite golf from now on? you may be able to understand the difficulty of how to use the word "causality". In daily conversation, it is used in many cases, mixing up "a cause (past)", "a reason (connotation)", and "the purpose and a motive (future)." It may be supposed that the pioneers of research of movement and change are Heraclitus(BC.540 -BC.480): "Everything changes." Parmenides (born around BC. 515): "Movement does not exist." (Zeno's teacher) though their assertions are not clear. However, these two pioneers (i.e., Heraclitus and Parmenides ) noticed first that "movement and change" were the primary importance keywords in science(= "world description") , i.e., it is [The beginning of World description ] =[The discovery of movement and change ] =  Heraclitus(BC.540 -BC.480) Parmenides(born around BC. 515) However, Aristotle(BC384–BC322) further investigated about the essence of movement and change, and he thought that 260 Ishikawa's Homepage Chap. 10 Axiom 2-causality all the movements had the "purpose." For example, supposing a stone falls, that is because the stone has the purpose that the stone tries to go downward. Supposing smoke rises, that is because smoke has the purpose that smoke rises upwards. Under the influence of Aristotle, "Purpose" continued remaining as a mainstream idea of "Movement" for a long time of 1500 years or more. Although "the further investigation" of Aristotle was what should be praised, it was not able to be said that "the purpose was to the point." In order to free ourselves from Purpose and for human beings to discover that the essence of movement and change is "causal relationship", we had to wait for the appearance of Galileo, Bacon, Descartes, Newton, etc. Revolution to "Causality" from "Purpose" is the greatest history-of-science top paradigm shift. It is not an overstatement even if we call it "birth of modern science". the birth of world description Movement (Heraclitus, Parmenides, Zeno) "purpose"−−−−−−−−−−−−−−−−−−−→ Aristotle :( About 1500 years) the birth of modern science Causality ( Galileo, Bacon, Descartes, Newton) ♠Note 10.1. I cannot emphasize too much the importance of the discovery of the term: "causality". That is, (]) Science is the discipline about phenomena can be represented by the term "causality". (i.e., "No smoke without fire" ) Thus, I consider that the discovery of "causality" is equal to that of science. 10.1.2 Four answers to "what is causality?" As mentioned above, about "what is an essence of movement and change?", it was once settled with the word "causality." However, not all were solved now. We do not yet understand "causality" fully. In fact, Problem 10.1. Problem: "What is causality?" is the most important outstanding problems in modern science. 261 Ishikawa's Homepage 10.1 The most important unsolved problem-what is causality? Answer this problem! There may be some readers who are surprised with saying like this, although it is the outstanding problems in the present. Below, I arrange the history of the answer to this problem. (a) [Realistic causality]: Newton advocated the realistic describing method of Newtonian mechanics as a final settlement of accounts of ideas, such as Galileo, Bacon, and Descartes, and he thought as follows. : "Causality" actually exists in the world. Newtonian equation described faithfully this "causality". That is, Newtonian equation is the equation of a causal chain. This realistic causality may be a very natural idea, and you may think that you cannot think in addition to this. In fact, probably, we may say that the current of the realistic causal relationship which continues like "Newtonian mechanics−→ Electricity and magnetism−→ Theory of relativity−→ * * * " is a scientific flower. However, there are also other ideas, i.e., three "non-realistic causalities" as follows. (b) [Cognitive causality]: David Hume, Immanuel Kant, etc. who are philosophers thought as follows. : We can not say that "Causality" actually exists in the world, or that it does not exist in the world. And when we think that "something" in the world is "causality", we should just believe that the it has "causality". Most readers may regard this as "a kind of rhetoric", however, several readers may be convinced in "Now that you say that, it may be so." Surely, since you are looking through the prejudice "causality", you may look such. This is Kant's famous "Copernican revolution", that is, "recognition constitutes the world." which is considered that the recognition circuit of causality is installed in the brain, and when it is stimulated by "something" and reacts, "there is causal relationship." Probably, many readers doubt about the substantial influence which this (b) had on the science after it. However, in this book, I adopted the friendly story to the utmost to Kant. 262 Ishikawa's Homepage Chap. 10 Axiom 2-causality (c) [Mathematical causality(Dynamical system theory)]: Since dynamical system theory has developed as the mathematical technique in engineering, they have not investigated "What is causality?" thoroughly. However, In dynamical system theory, we start from the state equation (i.e., simultaneous ordinary differential equation of the first order) such that dω1 dt (t) = v1(ω1(t), ω2(t), . . . , ωn(t), t) dω2 dt (t) = v2(ω1(t), ω2(t), . . . , ωn(t), t) * * * * * * dωn dt (t) = vn(ω1(t), ω2(t), . . . , ωn(t), t) (10.1) and, we think that (]) the phenomenon described by the state equation has "causality." This is the spirit of dynamical system theory (= statistics ). Although this is proposed under the confusion of mathematics and world description, it is quite useful. In this sense, I think that (c) should be evaluated more. (d) [Linguistic causal relationship (MeasurementTheory)]: The causal relationship of measurement theory is decided by the Axiom 2 (causality; §10.3) of this chapter. If I say in detail,: Although measurement theory consists of the two Axioms 1 and 2, it is the Axiom 2 that is concerned with causal relationship. When describing a certain phenomenon in quantum language (i.e., a language called measurement theory) and using Axiom 2 (causality; §10.3) , we think that the phenomenon has causality. Summary 10.2. The above is summarized as follows. (a) World is first (b) Recognition is first (c) Mathematics(buried into ordinary language) is first (d) Language (= quantum language) is first Now, in measurement theory, we assert the next as said repeatedly: Quantum language is a basic language which describes various sciences. Supposing this is recognized, we can assert the next. Namely, 263 Ishikawa's Homepage 10.1 The most important unsolved problem-what is causality? In science, causality is just as mentioned in the above (d). This is my answer to "What is causality ?". I explain this in detail in the following. ♠Note 10.2. Consider the following problems: (]1) What is time (space, causality, probability, etc.) ? There are two ways to answer. (]2) The answer of "What is XX ?"  (a): To show the definition of XX (b): To show how to use the term "XX" In this note, the answer to the question (]1) is presented from the linguistic point of view (b). 264 Ishikawa's Homepage Chap. 10 Axiom 2-causality 10.2 Causality-Mathematical preparation 10.2.1 The Heisenberg picture and the Schrödinger picture First, let us review the general basic structure (cf. §2.1.3 ) as follows. (A): General basic structure and State spaces Sp(A∗) C∗-pure state ⊂ Sm(A∗) C∗-mixed state ⊂ A∗xdual A ⊆−−−−−−−−−−−−−−→ subalgebra*weak-closure A ⊆−−−−−−→ subalgebra B(H)y pre-dual (10.2) S m (A∗) W ∗-mixed state ⊂ A∗ Remark 10.3. [A∗ ⊆ A∗] : Consider the basic structure [A ⊆ A]B(H). For each ρ ∈ A∗, F ∈ A(⊆ A ⊆ B(H)), we see that ∣∣∣ A∗ ( ρ, F ) A ∣∣∣ ≤ C‖F‖B(H) = C‖F‖A (10.3) Thus, we can consider that ρ ∈ A∗. That is, in the sense of (10.3), we consider that A∗ ⊆ A∗ When ρ(∈ A∗) is regarded as the element of A∗, it is sometimes denoted by ρ. Therefore, A∗ ( ρ, F ) A = A∗ ( ρ, F ) A (∀F ∈ A(⊆ A)) (10.4) Definition 10.4. [Causal operator (= Markov causal operator)] Consider two basic structures: [A1 ⊆ A1 ⊆ B(H1)] and [A2 ⊆ A2 ⊆ B(H2)] A continuous linear operator Φ1,2 : A2 → A1 is called a causal operator(or, Markov causal operator , the Heisenberg picture of "causality"), if it satisfies the following (i)-(iv): (i) F2 ∈ A2 F2 = 0 =⇒ Φ12F2 = 0 (ii) Φ12IA2 = IA1 (where, IA1(∈ A1) is the identity) (iii) there exists the continuous linear operator (Φ1,2)∗ : (A1)∗ → (A2)∗ such that (a) (A1)∗ ( ρ1,Φ1,2F2 ) A1 = (A2)∗ ( (Φ1,2)∗ρ1, F2 ) A2 (∀ρ1 ∈ (A1)∗, ∀F2 ∈ A2) (10.5) 265 Ishikawa's Homepage 10.2 Causality-Mathematical preparation (b) (Φ1,2)∗(S m ((A1)∗)) ⊆ S m ((A2)∗) (10.6) This (Φ1,2)∗ is called the pre-dual causal operator of Φ1,2. (iv) there exists the continuous linear operator Φ∗1,2 : A ∗ 1 → A∗2 such that (a) (A1)∗ ( ρ1,Φ1,2F2 ) A1 = A ∗ 2 ( Φ∗1,2ρ1, F2 ) A2 (∀ρ1 = ρ1 ∈ (A1)∗(⊆ A∗1), ∀F2 ∈ A2) (10.7) (b) (Φ1,2) ∗(Sp(A∗1)) ⊆ Sm(A∗2) (10.8) This Φ∗1,2 is called the dual operator of Φ1,2. In addition, the causal operator Φ1,2 is called a deterministic causal operator , if it satisfies that (Φ1,2) ∗(Sp(A∗1)) ⊆ Sp(A∗2) (10.9) ♠Note 10.3. [ Causal operator in Classical systems] Consider the two basic structures: [C0(Ω1) ⊆ L∞(Ω1, ν1)]B(H1) and [C0(Ω2) ⊆ L ∞(Ω2, ν2)]B(H2) A continuous linear operator Φ1,2 : L ∞(Ω2)→ L∞(Ω1) called a causal operator, if it satisfies the following (i)-(iii): (i) f2 ∈ L∞(Ω2), f2 = 0 =⇒ Φ12f2 = 0 (ii) Φ1212 = 11 where, 1k(ωk) = 1 (∀ωk ∈ Ωk, k = 1, 2) (iii) There exists a continuous linear operator (Φ1,2)∗ : L 1(Ω1) → L1(Ω2) (and (Φ1,2)∗ : L1+1(Ω1)→ L1+1(Ω2) ) such that∫ Ω1 [Φ1,2f2](ω1) ρ1(ω1)ν1(dω1) = ∫ Ω2 f2(ω2) [(Φ1,2)∗ρ1](ω2)ν2(dω2) (∀ρ1 ∈ L1(Ω1),∀f2 ∈ L∞(Ω2)) This (Φ1,2)∗ is called a pre-dual causal operator of Φ1,2. (iv) There exists a continuous linear operator Φ∗1,2 : M(Ω1) → M(Ω2) (and Φ∗1,2 : M+1(Ω1) → M+1(Ω2) ) such that L1(Ω1) ( ρ1,Φ1,2F2 ) L∞(Ω1) = M(Ω2) ( Φ∗1,2ρ1, F2 ) C0(Ω2) (∀ρ1 = ρ1 ∈M(Ω1),∀F2 ∈ C0(Ω2)) where, ρ1(D) = ∫ D ρ1(ω1)ν1(dω1) (∀D ∈ BΩ1). This (Φ1,2) ∗ is called a dual causal operator of Φ1,2. In addition, a causal operator Φ1,2 is called a deterministic causal operator, if there exists a continuous map φ1,2 : Ω1 → Ω2 such that [Φ1,2f2](ω1) = f2(φ1,2(ω1)) (∀f2 ∈ C(Ω2),∀ω1 ∈ Ω1) (10.10) This φ1,2 : Ω1 → Ω2 is called a deterministic causal map. Here, it is clear that Ω1 ≈ Sp(C0(Ω1)∗) 3 δω1 −−→ Φ∗12 δφ12(ω1) ∈ S p(C0(Ω2) ∗) ≈ Ω2 266 Ishikawa's Homepage Chap. 10 Axiom 2-causality ω1 φ1,2(ω1) Ω2Ω1 f2Φ1,2f2 Figure 10.1: Deterministic causal map φ1,2 and deterministic causal operator Φ1,2 Theorem 10.5. [Continuous map and deterministic causal map] Let (Ω1,BΩ1 , ν1) and (Ω2,BΩ2 , ν2) be measure spaces. Assume that a continuous map φ1,2 : Ω1 → Ω2 satisfies: D2 ∈ BΩ2 , ν2(D2) = 0 =⇒ ν1(φ−11,2(D2)) = 0. Then, the continuous map φ1,2 : Ω1 → Ω2 is deterministic, that is, the operator Φ1,2 : L∞(Ω2, ν2) → L∞(Ω1, ν1) defined by (10.10) is a deterministic causal operator. Proof. For each ρ1 ∈ L1(Ω1, ν1), define a measure μ2 on (Ω2,BΩ2) such that μ2(D2) = ∫ φ−11,2(D2) ρ1(ω1) ν1(dω1) (∀D2 ∈ BΩ2) Then, it suffices to consider the Radon-Nikodym derivative (cf. [87]) [Φ1,2]∗(ρ1) = dμ2/dν2. That is because D2 ∈ BΩ2 , ν2(D2) = 0 =⇒ ν1(φ−11,2(D2)) = 0 =⇒ μ2(D2) = 0 (10.11) Thus, by the Radon-Nikodym theorem, we get a continuous linear operator [Φ1,2]∗ : L 1(Ω1, ν1) → L1(Ω2, ν2). Theorem 10.6. Let Φ1,2 : L ∞(Ω2) → L∞(Ω1) be a deterministic causal operator. Then, it holds that Φ1,2(f2 * g2) = Φ1,2(f2) * Φ1,2(g2) (∀f2, ∀g2 ∈ L∞(Ω2)) Proof. Let f2, g2 be in L ∞(Ω2). Let φ1,2 : Ω1 → Ω2 be the deterministic causal map of the deterministic causal operator Φ1,2. Then, we see [Φ1,2(f2 * g2)](ω1) = (f2 * g2)(φ1,2(ω1)) = f2(φ1,2(ω1)) * g2(φ1,2(ω1)) =[Φ1,2(f2)](ω1) * [Φ1,2(g2)](ω1) = [Φ1,2(f2) * Φ1,2(g2)](ω1) (∀ω1 ∈ Ω1) This completes the theorem. 267 Ishikawa's Homepage 10.2 Causality-Mathematical preparation 10.2.2 Simple example-Finite causal operator is represented by matrix Example 10.7. [Deterministic causal operator, deterministic dual causal operator, deterministic causal map ] Define the two states space Ω1 and Ω2 such that Ω1 = Ω2 = R with the Lebesgue measure ν. Thus we have the classical basic structures: [C0(Ωk) ⊆ L∞(Ωk, ν) ⊆ B(L2(Ωk, ν))] (k = 1, 2) Define the deterministic causal map φ1,2 : Ω1 → Ω2 such that ω2 = φ1,2(ω1) = 3(ω1) 2 + 2 (∀ω1 ∈ Ω1 = R) Then, by (10.10), we get the deterministic dual causal operator Φ∗1,2 : M(Ω1)→M(Ω2) such that Φ∗1,2δω1 = δ3(ω1)2+2 (∀ω1 ∈ Ω1) where δ(*) is the point measure. Also, the deterministic causal operatorΦ1,2 : L ∞(Ω2) → L∞(Ω1) is defined by [Φ1,2(f2)](ω1) = f2(3(ω1) 2 + 2) (∀f2 ∈ C0(Ω2), ∀ω1 ∈ Ω1) Example 10.8. [Dual causal operator, causal operator] Recall Remark 2.13, that is, if Ω (= {1, 2, ..., n}) is finite set ( with the discrete metric dD and the counting measure ν,), we can consider that C0(Ω) = L ∞(Ω, ν) = Cn, M(Ω) = L1(Ω, ν) = Cn, M+1(Ω) = L1+1(Ω, ν) For example, put Ω1 = {ω11, ω21, ω31} and Ω2 = {ω12, ω22}. And define ρ1(∈M+1(Ω1)) such that ρ1 = a1δω11 + a2δω21 + a3δω31 (0 5 a1, a2, a3 5 1, a1 + a2 + a3 = 1) Then, the dual causal operator Φ∗1,2 : M+1(Ω1)→M+1(Ω2) is represented by Φ∗1,2(ρ1) =(c11a1 + c12a2 + c13a3)δω12 + (c21a1 + c22a2 + c23a3)δω22 (0 5 cij 5 1, 2∑ i=1 cij = 1) and, consider the identification:M(Ω1) ≈ C3, M(Ω2) ≈ C2, That is, M(Ω1) 3 α1δω11 + α2δω21 + α3δω31 ←→(identification) α1α2 α3  ∈ C3 268 Ishikawa's Homepage Chap. 10 Axiom 2-causality M(Ω2) 3 β1δω12 + β2δω22 ←→(identification) [ β1 β2 ] ∈ C2 Then, putting Φ∗1,2(ρ1) = β1δω12 + β2δω12 = [ β1 β2 ] , ρ1 = α1δω11 + α2δω21 + α3δω31 = α1α2 α3  write, by matrix representation, as follows. Φ∗1,2(ρ1) = [ β1 β2 ] = [ c11 c12 c13 c21 c22 c23 ]α1α2 α3  Next, from this dual causal operator Φ∗1,2 : M(Ω1) → M(Ω2), we shall construct a causal operator Φ1,2 : C0(Ω2)→ C0(Ω1). Consider the identification:C0(Ω1) ≈ C3, C0(Ω2) ≈ C2, that is, C0(Ω1) 3 f1 ←→ (identification) f1(ω11)f1(ω21) f1(ω 3 1)  ∈ C3, C0(Ω2) 3 f2 ←→ (identification) [ f2(ω 1 2) f2(ω 2 2) ] ∈ C2 Let f2 ∈ C0(Ω2), f1 = Φ1,2f2. Then, we seef1(ω11)f1(ω21) f1(ω 3 1)  = f1 = Φ1,2(f2) = c11 c21c12 c22 c13 c23 [f2(ω12) f2(ω 2 2) ] Therefore, the relation between the dual causal operatorΦ∗1,2 and causal operatorΦ1,2 is represented as the the transposed matrix. Example 10.9. [ Deterministic dual causal operator, deterministic causal map, deterministic causal operator ] Consider the case that dual causal operator Φ∗1,2 : M(Ω1)(≈C3) → M(Ω2)(≈C2) ha s the matrix representation such that Φ∗1,2(ρ1) = [ b1 b2 ] = [ 0 1 1 1 0 0 ]a1a2 a3  In this case, it is the deterministic dual causal operator. This deterministic causal operator Φ1,2 : C0(Ω2)→ C0(Ω1) is represented byf1(ω11)f1(ω21) f1(ω 3 1)  = f1 = Φ1,2(f2) = 0 11 0 1 0 [f2(ω12) f2(ω 2 2) ] with the deterministic causal map φ1,2 : Ω1 → Ω2 such that φ1,2(ω 1 1) = ω 2 2, φ1,2(ω 2 1) = ω 1 2, φ1,2(ω 3 1) = ω 1 2 269 Ishikawa's Homepage 10.2 Causality-Mathematical preparation 10.2.3 Sequential causal operator - A chain of causalities Let (T,≤) be a finite tree1, i.e., a tree like semi-ordered finite set such that "t1 ≤ t3 and t2 ≤ t3" implies "t1 ≤ t2 or t2 ≤ t1". Assume that there exists an element t0 ∈ T , called the root of T , such that t0 ≤ t (∀t ∈ T ) holds. Put T 2≤ = {(t1, t2) ∈ T 2 : t1 ≤ t2}. An element t0 ∈ T is called a root if t0 ≤ t (∀t ∈ T ) holds. Since we usually consider the subtree Tt0 ( ⊆ T ) with the root t0, we assume that the tree has a root. In this chapter, assume, for simplicity, that T is finite (though it is sometimes infinite in applications). For simplicity, assume that T is finite, or a finite subtree of a whole tree. Let T ( = {0, 1, ..., N}) be a tree with the root 0. Define the parent map π : T \{0} → T such that π(t) = max{s ∈ T : s < t}. It is clear that the tree (T ≡ {0, 1, ..., N},≤ ) can be identified with the pair (T ≡ {0, 1, ..., N}, π : T \ {0} → T ). Also, note that, for any t ∈ T \ {0}, there uniquely exists a natural number h(t) (called the height of t ) such that πh(t)(t) = 0. Here, π2(t) = π(π(t)), π3(t) = π(π2(t)), etc. Also, put {0, 1, ..., N}2 ≤ = {(m,n) | 0 ≤ m ≤ n ≤ N}. In Fig. 10.2, see the root t0, the parent map: π(t3) = π(t4) = t2, π(t2) = π(t5) = t1, π(t1) = π(t6) = π(t7) = t0 t0 t1 t2 t3 t4 t5 t6 t7 ) i k + k ) k π π π π π π π Figure 10.2: Tree: (T = {t0, t1, ..., t7}, π : T \ {t0} → T ) Definition 10.10. [Sequential causal operator; Heisenberg picture of causality] The family {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 ( or, { At2 Φt1,t2→ At1}(t1,t2)∈T 25 ) is called a sequential causal operator, if it satisfies that (i) For each t (∈ T ), a basic structure [At ⊆ At ⊆ B(Ht)] is determined. (ii) For each (t1, t2) ∈ T 25, a causal operator Φt1,t2 : At2 → At1 is defined such as Φt1,t2Φt2,t3 = Φt1,t3 (∀(t1, t2), ∀(t2, t3) ∈ T 25). Here, Φt,t : At → At is the identity operator. 1In Chapter 14, we discuss the infinite case 270 Ishikawa's Homepage Chap. 10 Axiom 2-causality A0 A1 A2 A3 A4 A5A6 A7 ) i k + k ) k Φ0,6 Φ0,1 Φ0,7 Φ1,2 Φ1,5 Φ2,3 Φ2,4 Figure 10.3: Heisenberg picture( sequential causal operator) Definition 10.11. (i): [pre-dual sequential causal operator : Schrödinger picture of causality ] The sequence {(Φt1,t2)∗ : (At1)∗ → (At1)∗}(t1,t2)∈T 25 is called a pre-dual sequential causal operator of {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 (ii): [Dual sequential causal operator : Schrödinger picture of causality ] A sequence {Φ∗t1,t2 : A ∗ t1 → A∗t1}(t1,t2)∈T 25 is called a dual sequential causal operator of {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 . (A0)∗ (A1)∗ (A2)∗ (A3)∗ (A4)∗ (A5)∗(A6)∗ (A7)∗ 1 z s 3 s : z (Φ0,6)∗ (Φ0,1)∗ (Φ0,7)∗ (Φ1,2)∗ (Φ1,5)∗ (Φ2,3)∗ (Φ2,4)∗ (i):pre-dual sequential causal operator A∗0 A∗1 A∗2 A∗3 A∗4 A∗5A∗6 A∗7 1 z s 3 s : z Φ∗0,6 Φ∗0,1 Φ∗0,7 Φ∗1,2 Φ∗1,5 Φ∗2,3 Φ∗2,4 (ii):dual sequential causal operator Figure 10.4: Schrödinger picture ( dual sequential causal operator) Remark 10.12. [The Heisenberg picture is formal; the Schrödinger picture is makeshift ] The Schrödinger picture is intuitive and handy. Consider the Schrödinger picture{Φ∗t1,t2 : A ∗ t1 → A∗t1}(t1,t2)∈T 25 . For C ∗-mixed state ρt1(∈ Sm(A∗t1) (i.e., a state at time t1), • C∗-mixed state ρt2(∈ Sm(A∗t2)) (at time t2(≥ t1)) is defined by ρt2 = Φ ∗ t1,t2ρt1 However, the linguistic interpretation says "state does not move", and thus, we consider that •  the Heisenberg picture is formal the Schrödinger picture is makeshift 271 Ishikawa's Homepage 10.3 Axiom 2 -Smoke is not located on the place which does not have fire 10.3 Axiom 2 -Smoke is not located on the place which does not have fire 10.3.1 Axiom 2 (A chain of causal relations) Now we can propose Axiom 2 (i.e., causality), which is the measurement theoretical representation of the maxim (Smoke is not located on the place which does not have fire ): (C): Axiom 2 (A chain of causalities) (Under the preparation to this section, we can read this) For each t(∈ T="tree")), consider the basic structure: [At ⊆ At ⊆ B(Ht)] Then, the chain of causalities is represented by a sequential causal operator {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 . ♠Note 10.4. Axiom 2 (causality) as well as Axiom 1 (measurement) are a kind of spells. There are several spells concerning "motion". For example, (]1) [Aristotle]: final cause (]2) [Darwin]: evolution theory (survival of the fittest) (]3) [Hegel]: dialectic (Thesis, antithesis, synthesis) (]4) law of entropy increase (]1)–(]3) are non-quantitative, but (]4) is quantitative. Everybody agrees that these ((]1)–(]4)) move the world. 10.3.2 Sequential causal operator-State equation, etc. In what follows, we shall exercise the chain of causality in terms of quantum language. Example 10.13. [State equation] Let T = R be a tree which represents the time axis. (Don't mind the infinity of T . Cf. Chapter 14.) For each t(∈ T ), consider the state space Ωt = Rn (n-dimensional real space). And consider simultaneous ordinary differential equation of the first order  dω1 dt (t) = v1(ω1(t), ω2(t), . . . , ωn(t), t) dω2 dt (t) = v2(ω1(t), ω2(t), . . . , ωn(t), t) * * * * * * dωn dt (t) = vn(ω1(t), ω2(t), . . . , ωn(t), t) (10.12) 272 Ishikawa's Homepage Chap. 10 Axiom 2-causality which is called a state equation . Let φt1,t2 : Ωt1 → Ωt2 , (t1 5 t2) be a deterministic causal map induced by the state equation (10.12). It is clear that φt2,t3(φt1,t2(ωt1)) = φt1,t3(ωt1) (ωt1 ∈ Ωt1 , t1 5 t2 5 t3). Therefore, we have the deterministic sequential causal operator {Φt1,t2 : L∞(Ωt2)→ L∞(Ωt1)}(t1,t2)∈T 25 . Example 10.14. [Difference equation of the second order] Consider the discrete time T = {0, 1, 2, . . .} with the parent map π : T \ {0} → T such that π(t) = t − 1 (∀t = 1, 2, ...). For each t(∈ T ), consider a state space Ωt such that Ωt = R ( with the Lebesgue measure). For example, consider the following difference equation, that is, φ : Ωt × Ωt+1 → Ωt+2 satisfies as follows. ωt+2 = φ(ωt, ωt+1) = ωt + ωt+1 + 2 (∀t ∈ T ) Here, note that the state ωt+2 depends on both ωt+1 and ωt (i.e., multiple markov property). This must be modified as follows. For each t(∈ T ) consider a new state space Ωt = Ωt×Ωt+1 = R× R. And define the deterministic causal map φt,t+1 : Ωt → Ωt+1 as follows. (ωt+1, ωt+2) = φt,t+1(ωt, ωt+1) = (ωt+1, ωt + ωt+1 + 2) (∀(ωt, ωt+1) ∈ Ωt, ∀t ∈ T ) Therefore, by Theorem 10.5, the deterministic causal operator Φt,t+1 : L ∞(Ωt+1)→ L∞(Ωt) is defined by [Φt,t+1ft](ωt, ωt+1) = ft(ωt+1, ωt + ωt+1 + 2) (∀(ωt, ωt+1) ∈ Ωt,∀ft ∈ L∞(Ωt+1),∀t ∈ T \ {0})) Thus, we get the deterministic sequential causal operator {Φt,t+1 : L∞(Ωt+1)→ L∞(Ωt)}t∈T\{0}. ♠Note 10.5. In order to analyze multiple markov process and time-lag process, such ideas in Example 10.14 are needed. 273 Ishikawa's Homepage 10.4 Kinetic equation (in classical mechanics and quantum mechanics) 10.4 Kinetic equation (in classical mechanics and quantum mechanics) 10.4.1 Hamiltonian ( Time-invariant system) In this section, we consider the simplest kinetic equation in classical system and quantum system. Consider the state space Ω such that Ω = R2, that is, R2 = Rq × Rp={(q, p) = (position , momentum ) | q, p ∈ R} (10.13) Hamiltonian H(q, p) is defined by the total energy, for example, as the typical case (m: particle mass), we consider that [Hamiltonian (= H(q, p))] =[kinetic energy(= p2 2m )] + [potential energy(= V (q))] (10.14) 10.4.2 Newtonian equation(=Hamilton's canonical equation) Concerning Hamiltonian H(q, p), Hamilton's canonical equation is defined by Hamilton's canonical equation =  dp dt = −H(q,p) ∂q dq dt = H(q,p) ∂p (10.15) And thus, in the case of (10.14), we get Hamilton's canonical equation =  dp dt = −H(q,p) ∂q = −∂V (q,p) ∂q dq dt = ∂H(q,p) ∂p = p m (10.16) which is the same as Newtonian equation. That is, m d2q dt2 = [Mass]× [Acceleration] = −∂V (q, p) ∂q (= Force) Now, let us describe the above (10.16) in terms of quantum language. For each t ∈ T = R, define the state space Ωt by Ωt = Ω = R2 = Rq × Rp={(q, p) = (position , momentum ) | q, p ∈ R} (10.17) 274 Ishikawa's Homepage Chap. 10 Axiom 2-causality and assume Lebesgue measure ν. Then, we have the classical basic structure: [C0(Ωt) ⊆ L∞(Ωt) ⊆ B(L2(Ωt))] (∀t ∈ T = R) The solution of the canonical equation (10.16) is defined by Ωt1 3 ωt1 7→ φt1,t2(ωt1) = ωt2 ∈ Ωt2 (10.18) Since (10.18) determines the deterministic causal map, we have the deterministic sequential causal operator {Φt1,t2 : L∞(Ωt2)→ L∞(Ωt1) }(t1,t2)∈T 2≤ such that [Φt1,t2(ft2)](ωt1) = ft2(φt1,t2(ωt1)) (∀ft2 ∈ L∞(Ω2),∀ωt1 ∈ Ωt1 , t1 ≤ t2) (10.19) 10.4.3 Schrödinger equation (quantizing Hamiltonian) The quantization is the following procedure: quantization2  total energyE −−−−−−−−→ quantumization ~ √ −1∂ ∂t momentum p −−−−−−−−→ quantumization ~∂√ −1∂q position q −−−−−−−−→ quantumization q (10.20) Substituting the quantumization (10.20) to the classical Hamiltonian: E = H(q, p) = p2 2m + V (q) we get ~ √ −1 ∂ ∂t = H(q, ~√ −1 ∂ ∂q ) = − ~ 2 2m ∂2 ∂q2 + V (q) (10.21) And therefore, we get the Schrödinger equation: ~ √ −1∂u(t, q) ∂t = H(q, ~√ −1 ∂ ∂q )u(t, q) = − ~ 2 2m ∂2 ∂q2 u(t, q) + V (q)u(t, q) (10.22) Putting u(t, *) = ut ∈ L2(R) (∀t ∈ T = R) we denote the Schrödinger equation (10.22) by ut = 1 ~ √ −1 Hut 2 Learning the (10.20) by rote, we can derive Schrödinger equation (10.22). However, the meaning of "quantumization" is not clear. 275 Ishikawa's Homepage 10.4 Kinetic equation (in classical mechanics and quantum mechanics) Solving this formally, we see ut = e H ~ √ −1 tu0 (Thus, the state representation is |ut〉〈ut| = |e H ~ √ −1 tu0〉〈e H ~ √ −1 tu0| ) (10.23) where, u0 ∈ L2(R) is an initial condition. Now, put Hilbert spaceHt = L 2(R) (∀t ∈ T = R), and consider the quantum basic structure: [C(L2(R)) ⊆ B(L2(R)) ⊆ B(L2(R))] The dual sequential causal operator {Φ∗t1,t2 : Tr(Ht1)→ Tr(Ht2)}(t1,t2)∈T 2≤ is defined by Φ∗t1,t2(ρ) = e H ~ √ −1 (t2−t1)ρe −H ~ √ −1 (t2−t1) (∀ρ ∈ Tr(Ht1) = (B(Ht1))∗ = C(Ht1)∗) (10.24) And therefore, the sequential causal operator {Φt1,t2 : B(Ht2)→ B(Ht1)}(t1,t2)∈T 2≤ is defined by Φt1,t2(A) = e −H ~ √ −1 (t2−t1)Ae H ~ √ −1 (t2−t1) (∀A ∈ B(Ht2)) (10.25) Also, since Φ∗t1,t2(S p(C(Ht1) ∗) ⊆ Sp(C(Ht2)∗), the sequential causal operator {Φt1,t2 : B(Ht2) → B(Ht1)}(t1,t2)∈T 2≤ is deterministic. Since we deal with the time-invariant system, putting t = t2 − t1, we see that (10.25) is equal to At = Φt(A0) = e −H ~ √ −1 tA0e H ~ √ −1 t (10.26) And thus, we get the differential equation: dAt dt = −H ~ √ −1 e −H ~ √ −1 tA0e H ~ √ −1 t + −H ~ √ −1 e −H ~ √ −1 tA0e H ~ √ −1 t H ~ √ −1 = −H ~ √ −1 At + At H ~ √ −1 = 1 ~ √ −1 ( AtH −HAt ) (10.27) which is just Heisenberg's kinetic equation. In quantum language, we say that • Heisenberg's kinetic equation is formal, and Schrödinger equation is makeshift, though the two are usually said to be equivalent. 276 Ishikawa's Homepage Chap. 10 Axiom 2-causality 10.5 Exercise:Solve Schrödinger equation by variable separation method Consider a particle with the mass m in the box (i.e., the closed interval [0, 2]) in the one dimensional space R. The motion of this particle (i.e., the wave function of the particle) is represented by the following Schrödinger equation i~ ∂ ∂t ψ(q, t) = − ~ 2∂2 2m∂q2 ψ(q, t) + V0(q)ψ(q, t) ( in H = L 2(R)) where V0(q) = { 0 (0 ≤ q ≤ 2) ∞ ( otherwise ) q R ψ(q, t) V0(q) ∞ 0 2 Figure 10.5: Particle in a box Put φ(q, t) = T (t)X(q) (0 ≤ q ≤ 2). And consider the following equation: i~ ∂ ∂t φ(q, t) = − ~ 2∂2 2m∂q2 φ(q, t). Then, we see iT ′(t) T (t) = − X ′′(q) 2mX(q) = K(= constant ). Then, φ(q, t) = T (t)X(q) = C3 exp(iKt) ( C1 exp(i √ 2mK/~ q) + C2 exp(− i √ 2mK/~ q). ) 277 Ishikawa's Homepage 10.5 Exercise:Solve Schrödinger equation by variable separation method Since X(0) = X(2) = 0 (perfectly elastic collision), putting K = n 2π2~ 8m , we see φ(q, t) = T (t)X(q) = C3 exp( in2π2~t 8m ) sin(nπq/2) (n = 1, 2, ...). Assume the initial condition: ψ(q, 0) = c1 sin(πq/2) + c2 sin(2πq/2) + c3 sin(3πq/2) + * * * . where ∫ R |ψ(q, 0)| 2dq = 1. Then we see ψ(q, t) =c1 exp( iπ2~t 8m ) sin(πq/2) + c2 exp( i4π2~t 8m ) sin(2πq/2) + c3 exp( i9π2~t 8m ) sin(3πq/2) + * * * . And thus, we have the time evolution of the state by ρt = |ψ(*, t)〉〈ψ(*, t)| (∈ Sp(Tr(H)) ⊆ B(H)) (∀t ≥ 0) 278 Ishikawa's Homepage Chap. 10 Axiom 2-causality 10.6 Random walk and quantum decoherence 10.6.1 Diffusion process Example 10.15. [Random walk] Let the state space Ω be Z = {0,±1,±2, . . .} with the counting measure ν. Define the dual causal operator Φ∗ : M+1(Z)→M+1(Z) such that Φ∗(δi) = δi−1 + δi+1 2 (i ∈ Z) where δ(*)(∈ M+1(Z)) is a point measure. Therefore, the causal operator Φ : L∞(Z)→ L∞(Z) is defined by [Φ(F )](i) = F (i− 1) + F (i+ 1) 2 (∀F ∈ L∞(Z),∀i ∈ Z) and the pre-dual causal operator Φ∗ : L 1(Z)→ L1(Z) is defined by [Φ∗(f)](i) = f(i− 1) + F (i+ 1) 2 (∀f ∈ L1(Z), ∀i ∈ Z) Now, consider the discrete time T = {0, 1, 2, . . . , N}, where the parent map π : T \ {0} → T is defined by π(t) = t− 1 (t = 1, 2, ...). For each t(∈ T ), a state space Ωt is define by Ωt = Z. Then, we have the sequential causal operator {Φπ(t),t(= Φ) : L∞(Ωt)→ L∞(Ωπ(t))}t∈T\{0} . 10.6.2 Quantum decoherence: non-deterministic causal operator Consider the quantum basic structure: [C(H) ⊆ B(H) ⊆ B(H)] Let P = {Pn}∞n=1 be the spectrum decomposition in B(H), that is, Pn is a projection (i.e., Pn = (Pn) 2 ), and, ∞∑ n=1 Pn = I Define the operator (ΨP)∗ : Tr(H)→ Tr(H) such that (ΨP)∗(|u〉〈u|) = ∞∑ n=1 |Pnu〉〈Pnu| (∀u ∈ H) Clearly we see 〈v, (ΨP)∗(|u〉〈u|)v〉 = 〈v, ( ∞∑ n=1 |Pnu〉〈Pnu|)v〉 = ∞∑ n=1 |〈v, |Pnu〉|2 ≥ 0 (∀u, v ∈ H) 279 Ishikawa's Homepage 10.6 Random walk and quantum decoherence and, Tr((ΨP)∗(|u〉〈u|)) =Tr( ∞∑ n=1 |Pnu〉〈Pnu|) = ∞∑ n=1 ∞∑ k=1 |〈ek, Pnu〉|2 = ∞∑ n=1 ‖Pnu‖2 = ‖u‖2 (∀u ∈ H) where {ek}∞k=1 is CONS in H. And so, (ΨP)∗(Tr p +1(H)) ⊆ Tr+1(H) Therefore, ΨP(= ((ΨP)∗) ∗) : B(H) → B(H) is a causal operator, but it is not deterministic. In this note, a non-deterministic (sequential) causal operator is called a quantum decoherence. Remark 10.16. [Quantum decoherence] For the relation between quantum decoherence and quantum Zeno effect, see § 11.4. Also, for the relation between quantum decoherence and Schrödinger's cat, see § 11.5. In tis note, we assume that the don-deterministic causal operator belongs to the mixed measurement theory. Thus, we consider that quantum language (= measurement theory ) is classified as follows. (A) measurement theory (=quantum language)  pure type (A1) { classical system : Fisher statistics quantum system : usual quantum mechanics mixed type (A2) { classical system : including Bayesian statistics, Kalman filter quantum system : quantum decoherence 280 Ishikawa's Homepage Chap. 10 Axiom 2-causality 10.7 Leibniz-Clarke Correspondence: What is space-time? This section is published in the following: • ref. [57]: S. Ishikawa; Leibniz-Clarke correspondence, Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. are clarified in quantum language Open Journal of philosophy, Vol. 8, No.5 , 466-480, 2018, DOI: 10.4236/ojpp.2018.85032 (https://www.scirp.org/Journal/PaperInformation.aspx?PaperID=87862) • ref. [58]; S. Ishikawa; Leibniz-Clarke correspondence, Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. are clarified in quantum language; [Revised version] ; Keio Reseach report; 2018; KSTS/RR-18/001, 1-15 (https://philpapers.org/rec/ISHLCB) (http://www.math.keio.ac.jp/academic/research_pdf/report/2018/18001.pdf) The problems ("What is space?" and "What is time?") are the most important in modern science as well as the traditional philosophies. In this section, we give the quantum linguistic answer to these problems. As seen later, our answer is similar to Leibniz's relationalism concerning space-time. In this sense, we consider that Leibniz is one of the discoverers of the linguistic Copenhagen interpretation 10.7.1 "What is space?" and "What is time?") 10.7.1.1 Space in quantum language ( How to describe "space" in quantum language) In what follows, let us explain "space" in measurement theory (= quantum language ). For example, consider the simplest case, that is, (A) "space"=Rq( one dimensional space) Since classical system and quantum system must be considered, we see (B)  (B1): a classical particle in the one dimensional space Rq (B2): a quantum particle in the one dimensional space Rq In the classical case, we start from the following state: (q, p) = ("position", "momentum") ∈ Rq × Rp Thus, we have the classical basic structure: 281 Ishikawa's Homepage 10.7 Leibniz-Clarke Correspondence: What is space-time? (C1) [C0(Rq × Rp) ⊆ L∞(Rq × Rp) ⊆ B(L2(Rq × Rp)] Also, concerning quantum system, we have the quantum basic structure: (C2) [C(L 2(Rq) ⊆ B(L2(Rq) ⊆ B(L2(Rq)] Summing up, we have the basic structure (C) [A ⊆ A ⊆ B(H)]  (C1): classical [C0(Rq × Rp) ⊆ L∞(Rq × Rp) ⊆ B(L2(Rq × Rp)] (C2): quantum [C(L 2(Rq) ⊆ B(L2(Rq) ⊆ B(L2(Rq)] Since we always start from a basic structure in quantum language, we consider that How to describe "space" in quantum language ⇔ How to describe [(A):space] by [(C):basic structure] (10.28) This is done in the following steps. Assertion 10.17. [The linguistic Copenhagen interpretation concerning "space"] How to describe "space" in quantum language (D1) Begin with the basic structure: [A ⊆ A ⊆ B(H)] (D2) Next, consider a certain commutative C ∗-algebra A0(= C0(Ω)) such that A0 ⊆ A (D3) Lastly, the spectrum Ω (≈ Sp(A∗)) is used to represent "space". For example, (E1) in the classical case (C1): [C0(Rq × Rp) ⊆ L∞(Rq × Rp) ⊆ B(L2(Rq × Rp))] we have the commutative C0(Rq) such that C0(Rq) ⊆ L∞(Rq × Rp) And thus, we get the space Rq as mentioned in (A) 282 Ishikawa's Homepage Chap. 10 Axiom 2-causality (E2) in the quantum case (C2): [C(L2(Rq) ⊆ B(L2(Rq)) ⊆ B(L2(Rq))] we have the commutative C0(Rq) such that C0(Rq) ⊆ B(L2(Rq)) And thus, we get the space Rq as mentioned in (A) 10.7.1.2 Time in quantum language ( How to describe "time" in quantum language) In what follows, let us explain "time" in measurement theory (= quantum language ). This is easily done in the following steps. Assertion 10.18. [The linguistic Copenhagen interpretation concerning "time"] How to describe "time" in quantum language (F1) Let T be a tree. (Don't mind the finiteness or infinity of T . Cf. Chapter 14.) For each t ∈ T , consider the basic structure: [At ⊆ At ⊆ B(Ht)] (F2) Next, consider a certain linear subtree T ′(⊆ T ), which can be used to represent "time". 10.7.2 Leibniz-Clarke Correspondence The above argument urges us to recall Leibniz-Clarke Correspondence (1715–1716: cf. [1]), which is important to know both Leibniz's and Clarke's (=Newton's) ideas concerning space and time. (G) [The realistic space-time] Newton's absolutism says that the space-time should be regarded as a receptacle of a "thing." Therefore, even if "thing" does not exits, the space-time exists. On the other hand, (H) [The metaphysical space-time] Leibniz's relationalism says that 283 Ishikawa's Homepage 10.7 Leibniz-Clarke Correspondence: What is space-time? (H1) Space is a kind of state of "thing". (H2) Time is an order of occurring in succession which changes one after another. Therefore, I regard this correspondence as Newton (≈ Clarke) (realistic view) ←→ v.s. Leibniz (linguistic view) which should be compared to Einstein (realistic view) ←→ v.s. Bohr (linguistic view) (also, recall Note 4.3). Again, we emphasize that Leibniz's relationalism in Leibniz-Clarke correspondence is clarified in quantum language, and it should be regarded as one of the most important parts of the linguistic Copenhagen interpretation of quantum mechanics. ♠Note 10.6. Many scientists may think that Newton's assertion is understandable, in fact, his idea was inherited by Einstein. On the other, Leibniz's assertion is incomprehensible and literary. Thus, his idea is not related to science. However, recall the classification of the world-description (Figure 1.1): 1© : Newton, Clarke (realistic world view) * * * (space-time in physics) realistic space-time "What is space-time?" (successors: Einstein, etc.) 2© : Leibniz (linguistic world view) * * * (space-time in measurement theory) linguistic space-time "How should space-time be represented?" (i.e., spectrum, tree) in which Newton and Leibniz respectively devotes himself to 1© and 2©. Although Leibniz's assertion is not clear, we believe that • Leibniz found the importance of "linguistic space and time" in science, Also, it should be noted that (]1) Newton proposed the scientific language called Newtonian mechanics, on the other hand, Leibniz could not propose a scientific language After all, we conclude that (]2) the cause of philosophers' failure is not to propose a language. Talking cynically, we say that 284 Ishikawa's Homepage Chap. 10 Axiom 2-causality (]3) Philosophers continued investigating "linguistic interpretation" (="how to use Axioms 1 and 2") without language (i.e., Axiom 1(measurement:§2.7) and Axiom 2(causality:§10.3)). ♠Note 10.7. I want to believe that "realistic" vs. "linguistic" is always hidden behind the great disputes in the history of the world view (cf. ref. [49]). That is, realistic world view ←→ v.s. linguistic world view (idealistic) For example, Table 10.1 : The realistic world view vs the linguistic world view Dispute R vs. L R:= the realistic world view L:= the linguistic world view Greek philosophy Aristotle Plato Problem of universals Nominalisme(William of Ockham) Realismus(Anselmus) Space*times Clarke( Newton) Leibniz Quantum mechanics Einstein (cf. [14]) Bohr (cf. [5]) It is usally said that the Problem of universals is not easy to understand. The reason is that the two problems ( i,e., "Trialism in Table 3.1" and "realistic view or linguistic view" in Table 10.1) were simultaneously discussed and confused in the history. 285 Ishikawa's Homepage 10.8 Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. 10.8 Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. This section is published in the following: (A1) ref. [57]: S. Ishikawa; Leibniz-Clarke correspondence, Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. are clarified in quantum language Open Journal of philosophy, Vol. 8, No.5 , 466-480, 2018, DOI: 10.4236/ojpp.2018.85032 (https://www.scirp.org/Journal/PaperInformation.aspx?PaperID=87862) (A2) ref. [58]; S. Ishikawa; Leibniz-Clarke correspondence, Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. are clarified in quantum language; [Revised version] ; Keio Reseach report; 2018; KSTS/RR-18/001, 1-15 (https://philpapers.org/rec/ISHLCB) (http://www.math.keio.ac.jp/academic/research_pdf/report/2018/18001.pdf) Before we mention the main section 10.8.3, we review Section 10.8.1 ( the linguistic Copenhagen interpretation (mentioned in Chapter 4)) and Section 10.8.2 ( Review; main assertions of quantum language). 10.8.1 Review; The linguistic Copenhagen interpretation (= the manual to use Axioms 1 and 2) Assume that Axiom 1 [ Measurement ] (in Section 2.7) and Axiom [ Causality ] ( in Section 10.3 ) are known. Now I will review a little the linguistic Copenhagen interpretation (mentioned in Chapter 4). Since so-called Copenhagen interpretation is not firm (cf. ref. [21] ), we propose the linguistic Copenhagen interpretation in what follows. In the above, Axioms 1 and 2 are kinds of spells, (i.e., incantation, magic words, metaphysical statements), and thus, it is nonsense to verify them experimentally. Therefore, what we should do is not "to understand" but "to use". After learning Axioms 1 and 2 by rote, we have to improve how to use them through trial and error. We can do well even if we do not know the linguistic Copenhagen interpretation (= the manual to use Axioms 1 and 2). However, it is better to know the linguistic Copenhagen interpretation , if we would like to make progress quantum language early. I believe that the linguistic Copenhagen interpretation is the true Copenhagen interpretation (cf. ref. [21]). In Figure 10.1 (mentioned later), I remark: (B1) x©: it suffices to understand that "interfere" is, for example, "apply light". y©: perceive the reaction. 286 Ishikawa's Homepage Chap. 10 Axiom 2-causality That is, "measurement" is characterized as the interaction between "observer" and "measuring object (= matter)". However, (B2) in measurement theory (=quantum language), "interaction" must not be emphasized. Therefore, in order to avoid confusion, it might better to omit the interaction " x© and y©" in Figure 10.1. After all, we think that: (B3) it is clear that there is no measured value without observer (i.e., "I", "mind"). Thus, we consider that measurement theory is composed of three key-words: measured value (I, observer, mind) , observable (= measuring instrument ) (body(= sensory organ), eye, ear, compass (e.g., polar star) ) , state (matter) , (10.29) The essence of the manual is as follows: • observer (I(=mind)) system (matter, measuring object)  - [observable] [(=measuring instrument)] (body) [measured value] x©interfere y©perceive reaction [state] Figure 10.1: (=Figure 3.1)[Descartes Figure]: Image of "measurement(= x©+ y©)" in mindmatter dualism The linguistic Copenhagen interpretation says that (C1) Only one measurement is permitted. Thus, Axiom 1 can be used only once. And therefore, the state after a measurement is meaningless since it can not be measured any longer. Thus, the collapse of the wavefunction is prohibited (cf. ref. [48]; projection postulate ). We are not concerned with anything after measurement. Strictly speaking, the phrase "after the measurement" should not be used. Also, the causality should be assumed only in the side of system, however, a state never moves. Thus, the Heisenberg picture should be adopted, and thus, the Schrödinger picture should be prohibited. 287 Ishikawa's Homepage 10.8 Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. (C2) "Observer"(="I") and "system" are completely separated. Hence, the measurement MA(O :=(X,F, F ), S[ρ]) does not depend on the choice of observers. That is, any proposition (except Axiom 1) in quantum language is not related to "observer"(="I"), therefore, there is no "observer's space and time" in quantum language. And thus, it does not have tense (i.e., past, present, future). (C3) there is no probability without measurements (Bertrand's paradox in Section 9.12) ) (C4) Leibniz's relationalism concerning space-time. See Section 10.7. and so on. We consider that the above (C1) is closely related to Parmenides' saying (born around BC. 515 in ancient Greek)[ There are no "plurality",but only "one"] and Kolmogorov's extension theorem (cf. [63]). For details, see ref. [53]. Remark 10.19. ["Who measured?" is not essential] "Who measured?" is not essential. An observer may be satisfactory for anyone. For example consider the following cases: (]1) Jack measures Tom's body temperature. (]2) A doctor measures Tom's body temperature. (]3) Tom's body temperature is measured. (]4) An observer measures Tom's body temperature (]5) I measure Tom's body temperature. (]6) Tom measures Tom's body temperature. (]7) I measure my body temperature ( when I am Tom) The above are all the same. See the above (]6) and (]7), which may be misleading, since (C2) says that "observer" and "system" are completely (or, almost completely) separated. The meaning of "separation" will be clarified in Section 10.8.3; Brain in a Vat. Also, identification of "observer" and "I" in (C2) may be misleading. Thus, we may say that ([) any statement in quantum language should be expressed without using "I" if it is possible. In this sense, quantum language is quite different from Descartes philosophy. 288 Ishikawa's Homepage Chap. 10 Axiom 2-causality Remark 10.20. [Experiment verification] Experiment verification must be possible also for any statement in quantum language. For example, "Apple falls down a tree" can carry out experiment verification. Thus, this is a statement in quantum language. On the other hand, the statement "Now I am here" can not carry out experiment verification. Thus, this is not a statement in quantum language. 10.8.2 Review; Main assertions of quantum language 10.8.2.1 The history of world description Figure 10.21. (=Figure 1.1) [The location of quantum language in the history of worlddescription (cf. refs.[32, 53]) ] Parmenides Socrates 0©:Greek philosophy Plato Aristotle Schola-−−−−→ sticism 1© −−→ (monism) Newton (realism) 2© → relativity theory −−−−−−→ 3© → quantum mechanics −−−−−−→ 4© −→ (dualism) Descartes Locke,... Kant (idealism) 6©−→ (linguistic view) linguistic philosophy language−−−−−→ 8© language−−−−−−→ 7©  5©−→ (unsolved) theory of everything (quantum phys.)  10©−→ (=MT) quantum language (language) Figure 1.1: The history of the world-view statistics system theory language−−−−−→ 9© (Descartes, Locke may belong to substance dualism) the linguistic world view ( dualism, idealism ) the realistic world view (monism, realism) In refs. [53, 49], I asserted that the following four are equivalent: (D0) to propose quantum language (cf. 10© in Figure 10.21) (D1) to clarify so-called Copenhagen interpretation of quantum mechanics (cf. 7© in Figure 10.21) (D2) to find the final goal of the dualistic idealism (cf. 8© in Figure 10.21) (D3) to reconstruct statistics in the dualistic idealism (cf. 9© in Figure 10.21) 289 Ishikawa's Homepage 10.8 Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. 10.8.2.2 The Copernican Revolution In Figure 10.21, " language; 7©−−−−−−−→ (D1) " should be called "the linguistic CR(=Copernican Revolution)" in the sense below: (substance dualism) Descartes(dualism) (the world is previous, recognition is later) idealism−−−−−−−−−−−−→ recognitive CR a priori + a posteriori Kant (dualism) (recognition is previous, the world is later)ylinguistic turn (realism) quantum mechanics(dualism) (the world is previous, language is later) idealism(≈language) −−−−−−−−−−−−−−−−→ linguistic CR Axioms+Copenhagen interpretation quantum language (dualism) (language is previous, the world is later) Kant's Copernican revolution (i.e., the above recognitive CR (cf. ref. [62])) should be praised as the discovery of "idealism", though the true discovery may be due to the above linguistic CR. 10.8.2.3 Philosophy made progress In the above Figure 10.21, let us focus on the history of the dualistic idealism in the linguistic world view such as Plato −−−−−→ Descartes −−−−−→ Kant −−−−−→ Wittgenstein (10.30) Note that physics obviously made progress in Figure 10.21, on the other hand, the (10.30)'s progress is not clear. In ref. [49], we asserted that, if "(philosophical) progress" is defined by "approaching quantum language", then (E) the (10.30) does not only imply time series but also progress, that is, Plato starting point −−−−−→ progress dualism Descartes −−−−−→ progress dualism Kant idealism −−−−−→ progress Wittgenstein language −−−−−→ progress dualism Quantum language idealism(≈language) (if "progress" is defined by "approaching quantum language") (10.31) Here, • Plato: the founder • Descartes: the discoverer of dualism (though the true scientific discovery is due to N. Bohr (cf. [5])). Also, Berkeley's saying: "To be is to be perceived" is essential to idealism (cf. ref. [49]). 290 Ishikawa's Homepage Chap. 10 Axiom 2-causality • Kant: the discoverer of idealism (in the sense of the above Section 10.8.3) • Wittgenstein: he emphasized the importance of language. This is natural since we assume [(D2); quantum language is the final goal of the dualistic idealism]. That is, we consider that the (10.31) is the history which gropes after the language in which science is written. Also, for the linguistic approach to the mind-body problem, see ref. [51], i.e., • Ishikawa,S., A Final solution to mind-body problem by quantum language, Journal of quantum information science, Vol. 7, No.2 , 48-56, 2017, DOI: 10.4236/jqis.2017.72005 (http://www.scirp.org/Journal/PaperInformation.aspx?PaperID=76391) Remark 10.22. [Brain science?] As mentioned in ref. [49], we do not agree with the following "progress"; Plato starting point −−−−−→ progress dualism Descartes −−−−−→ progress dualism Kant idealism −−−−−→ progress Husserl −−−−−→ progress brain science That is because we think that • philosophy should be metaphysics, and thus it isn't in the immature state of the science. 10.8.2.4 Quantum language is the language to describe science Also, since the (D) says that "statistics" ( 9© in Figure 10.21) ∪ "quantum information theory" ( 7© in Figure 10.21) ∪ "dualistic idealism" ( 8© in Figure 10.21) ⊂"quantum language" it is natural to assume that (F) quantum language is the language to describe science, that is, proposition in quantum language ⇐⇒scientific proposition (=experiment verifiable proposition) which is the most important assertion of quantum language. Also, we assume that this (i.e., to make the language to describe science) is the true purpose of the philosophy of science. 291 Ishikawa's Homepage 10.8 Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. Remark 10.23. [The theory of relativity] Note that the theory of relativity cannot be described by quantum language. However, we want to assert the (F). We think that the theory of relativity ( and more, the theory of everything ) is too special, an exception. 10.8.3 What we cannot speak about in quantum language In this section we clarifies the following well-known philosophical statements: (G) "brain in a vat problem", "the Cogito proposition", "five-minute hypothesis", "only the present exists", "McTaggart's paradox" and so on. which are "what we cannot speak about in quantum language", that is, non-scientific propositions. 10.8.3.1 Brain in a vat argument Suppose (cf. ref. [75]); (H1) a mad scientist has removed your brain, and placed it into a vat of liquid to keep it alive and active. The scientist has also connected your brain to a powerful computer, which sends neurological signals to the brain in the way the brain normally receives them. Thus, the computer is able to send your brain data to fool you into believing that you are still walking around in your body. Then, you may say; (H2) "Am I a brain in a vat?" Or, "Can I check whether I am a brain vat or not?" Note that the question (H2) is related to "I". Or, precisely, "observer"="I", "system (=measuring object)"="I", thus, "observer" and "system" are not separated. Thus, the linguistic Copenhagen interpretation (C2) says that this (H2) is not a statement in quantum language. Thus, the (H2) is not scientific, that is, there is no experiment to verify the statement (H2). 292 Ishikawa's Homepage Chap. 10 Axiom 2-causality Remark 10.24. [Experiment verification] Since we receive several questions for the above argument in ref. [57], we add the following. If you are Tom, (H2) is the same as (H3) "Can Tom himself check whether Tom is a brain vat or not?" Here, "observer"="Tom", "system (=measuring object)"="Tom", thus, "observer" and "system" are not separated. Thus, this is not the statement in quantum language. This is obvious compared to the following. (H4) "Can Jack check whether Tom is a brain vat or not?" which is the statement in quantum language. 10.8.3.2 The Cogito proposition It is well known that Descartes proposed the Cogito proposition "I think, therefore I am", as the first principle of philosophy since he believed that this proposition has no room for doubt. That is, Descartes think that (I1) I confirm "I think, therefore I am" However, this is doubtful. Note that the proposition (I1) is related to "I". Or, precisely, "observer"="I", "system (=measuring object)"="I", thus, "observer" and "system" are not separated. Thus, the linguistic Copenhagen interpretation (C2) says that this (I) is not a statement in quantum language. Thus, the (I1) is not scientific, that is, there is no experiment to verify the statement (I1). Remark 10.25. Since we receive several questions for the above argument in ref. [57], we add the following. As brain death determination, (I2) A doctor confirms "Tom thinks, therefore Tom is alive" In this case, we see that "observer"="doctor", "system (=measuring object)"="Tom". Hence "Tom thinks, therefore Tom is alive" is the proposition in quantum language. For the more precise argument, see Section 8.4 [Cogito – I think, therefore I am]. 293 Ishikawa's Homepage 10.8 Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. 10.8.3.3 What is "I"? Descartes proclaimed that he discovered "I". Then, we have the natural question: What is "I(discovered by Descartes)"? If (E) is true (i.e., Descartes −−−−→ progress Quantum language ), this question can be answered as follows. In quantum language, several words ("I"(="observer"), "observable", "matter", "measurement", etc.) are undefined such as point, line, plane etc. in Hilbert's geometry (i.e., The Foundations of Geometry (1899)). D. Hilbert said that • The elements, such as point, line, plane, and others, could be substituted by tables, chairs, glasses of beer and other such objects. For example, the readers should note that the term "measurement" is used trickily in the quantum linguistic answer of Monty-Hall problem (cf. ref. [34]). 10.8.3.4 Five-minute hypothesis The five-minute hypothesis, proposed by B. Russell (cf. ref. [79]), is as follows. (J1) The universe was created five minutes ago. Or equivalently, the universe was created ten years ago. Now we show that this (J1) is not the statement in quantum language as follows (i.e., The first answer (i) and the second answer (ii)) The first answer (i): Note that this hypothesis (J1) is related to" tense". Thus, the linguistic Copenhagen interpretation (C2) says that this (J1) is not a statement in quantum language. Thus, the (J1) is not scientific, that is, there is no experiment to verify the statement (J1). The second answer (ii): There may be another understanding as follows. If we consider that ["observer"∈"the universe"], the proposition (J1) cannot be described in quantum language. That is because the linguistic Copenhagen interpretation (C2) says that "observer" (="I" ) and "measuring object"(="the universe") have to be completely separated. ( Also, see Remark 10.26 (b) later.) Some may want to relate this hypothesis to skepticism (cf. ref. [79]), However we do not think that this direction is productive. Remark 10.26. (a): Also, the above (J1) should be compared to the following (J2) (J2) The universe was created in A.D. 2008. ( Or equivalently, now is A.D. 2018, and the universe was created ten years ago.) 294 Ishikawa's Homepage Chap. 10 Axiom 2-causality This (J2) can be denied by experiment, that is, it is different from the fact. Thus, this is a proposition in quantum language. (b): If the (J2) is a proposition in quantum language, the hypothesis ["observer"∈"the universe"] in (J1) may be doubtful. We may not understand the meaning of ["observer"∈"the universe"] completely. Thus, the second answer (ii): may be doubtful. 10.8.3.5 Only the present exists It is well known that St. Augustinus (AD.354-AD.430) said that • the past does not exist because of its being already gone, that the future does not exist because of its not coming yet, and that the present really exists. Here, consider (K) "Only the present exists" Note that this proposition (K) is related to "tense". Thus, the linguistic Copenhagen interpretation (C2) says that this (K) is not a statement in quantum language. Thus, the (K) is not scientific, that is, there is no experiment to verify the (K). 10.8.3.6 McTaggart's paradox In ref. [67], McTaggart asserted "the Unreality of Time" as follows. The sketch of McTaggart's proof (L1) Assume that there are two kinds of times. i.e., "observer's time ( A-series)" and "objective time (B-series)". (Note that this assumption is against the linguistic Copenhagen interpretation (C2).) (L2) * * * * * * (L3) After all, the contradiction is obtained Therefore, by the reduction to the absurd, we get; (L4) A-series does not exist (in science). About this proof, there are various opinions also among philosophers. Although I can not understand the above part (L2) ( since the properties of A-series are not clear), I agree to him if his assertion is (L4) (cf. ref. [32]). That is, I agree that McTaggart noticed first that observer's time is not scientific. 295 Ishikawa's Homepage 10.8 Brain in a vat, Five-minute hypothesis, McTaggart's paradox, etc. 10.8.3.7 Is "What we cannot speak about we must pass over in silence" true? It should be noted that "what we cannot speak about" depends on language. As mentioned in the above, the Cogito proposition "I think, therefore I am" is "what we cannot speak about in quantum language". However, thanks to Descartes said "I think, therefore I am", dualism was developed. This fact may imply that (M) "What we cannot speak about we must pass over in silence" is not true. However, we think that Descartes' success is accidental luck. Or, we may consider that the true discoverer of dualism is N. Bohr, the leader of the Copenhagen school (cf. [5]). Since Wittgenstein (cf. ref. [86]) said "The limits of my language mean the limits of my world.", he had should propose "my language". We are sure that it will fall into a play on words by the argument without "my language". 10.8.4 Conclusion Dr. Hawking said in his best seller book [ ref.[18]; A Brief History of Time: From the Big Bang to Black Holes, Bantam, Boston, 1990]: • Philosophers reduced the scope of their inquiries so much that Wittgenstein the most famous philosopher this century, said "The sole remaining task for philosophy is the analysis of language." What a comedown from the great tradition of philosophy from Aristotle to Kant! We think that this is not only his opinion but also most scientists'opinion. And moreover, we mostly agree with him. However, we believe that, if "the analysis of language" was rewritten to "the creation of language", then Dr. Hawking would not have been critical to philosophy. That is because the task of phycisists is just the creation of language, i.e., the language called Newtonian mechanics, the language called the theory of relativity, etc. Also, since Wittgenstein (cf. ref. [86]) said "The limits of my language mean the limits of my world.", he had should propose "my language". We are sure that the argument without "my language" will fall into a play on words. In this paper, we introduced quantum language, and in the framework of quantum language, we discussed the followings: (N) "brain in a vat argument", "the Cogito proposition", "five-minute hypothesis", "only the present exists", "McTaggart's paradox", and so on. 296 Ishikawa's Homepage Chap. 10 Axiom 2-causality And we showed that the above propositions in (N) are not in quantum language, that is, these are not scientific. Or equivalently, we have no experiment to verify the above propositions in (N). 297 Ishikawa's Homepage

Chapter 11 Simple measurement and causality Until the previous chapter, we studied all of quantum language, that is, (])  (]1): pure measurement theory (=quantum language) := [(pure)Axiom 1] pure measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } the manual to use spells (]2): mixed measurement theory (=quantum language) := [(mixed)Axiom(m) 1] mixed measurement (cf. §9.1) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } the manual to use spells However, what is important is • to exercise the relationship of measurement and causality Since measurement theory is a language, we have to note the following wise sayings: • experience is the best teacher, or custom makes all things 11.1 The Heisenberg picture and the Schrödinger picture 11.1.0.1 State does not move- the Heisenberg picture - We consider that "only one measurement" =⇒"state does not move" 299 11.1 The Heisenberg picture and the Schrödinger picture That is because (a) In order to see the state movement, we have to take measurement at least more than twice. However, the "plural measurement" is prohibited. Thus, we conclude "state does not move" We want to believe that this is associated with Parmenides' words: There is no movement which is related to the Heisenberg picture. This will be explained in what follows. Theorem 11.1. [Causal operator and observable] Consider the basic structure: [Ak ⊆ Ak ⊆ B(Hk)] (k = 1, 2) Let Φ1,2 : A2 → A1 be a causal operator, and let O2 = (X,F, F2) be an observable in A2. Then, Φ1,2O2 = (X,F,Φ1,2F2) is an observable in A2. Proof. Let Ξ (∈ F). And consider the countable decomposition {Ξ1,Ξ2, . . . ,Ξn, . . .} of Ξ( i.e., Ξ = ∞∪ n=1 Ξn, Ξn ∈ F, (n = 1, 2, . . .), Ξm ∩ Ξn = ∅ (m 6= n) ) . Then we see, for any ρ1(∈ (A1)∗), (A1)∗ ( ρ1,Φ1,2F2( ∞∪ n=1 Ξn) ) A1 = (A1)∗ ( (Φ1,2)∗ρ1, F2( ∞∪ n=1 Ξn) ) A2 = ∞∑ n=1 (A1)∗ ( (Φ1,2)∗ρ1, F2(Ξn) ) A2 = ∞∑ n=1 (A1)∗ ( ρ1,Φ1,2F2(Ξn) ) A2 Thus,Φ1,2O2 = (X,F,Φ1,2F2) is an observable in A1. Let us begin from the simplest case. Consider a tree T = {0, 1}. For each t ∈ T , consider the basic structure: [At ⊆ At ⊆ B(Ht)] (t = 0, 1) And consider the causal operator Φ0,1 : A1 → A0. That is, A0 Φ0,1←−− A1 (11.1) Therefore, we have the pre-dual operator (Φ0,1)∗ and the dual operator Φ ∗ 0,1: (A0)∗ −−−−→ (Φ0,1)∗ (A1)∗ A ∗ 0 −−→ Φ∗0,1 A∗1 (11.2) 300 Ishikawa's Homepage Chap. 11 Simple measurement and causality If Φ0,1 : A1 → A0 is deterministic, we see that A∗0 ⊃ Sp(A∗0) 3 ρ −−→ Φ∗0,1 Φ∗0,1ρ ∈ Sp(A∗1) ⊂ A∗1 (11.3) Under the above preparation, we shall explain the Heisenberg picture and the Schrödinger picture in what follows. Assume that (A1) Consider a deterministic causal operator Φ0,1 : A1 → A0. (A2) a state ρ0 ∈ Sp(A∗0) : pure state (A3) Let O1 = (X1,F1, F1) be an observable in A1. Explanation 11.2. [the Heisenberg picture]. The Heisenberg picture is just the following (a): (a1) To identify an observable O1 in A1 with an Φ0,1O1 in A0 . That is, Φ0,1O1 ( in A0) Φ0,1←−−−−−−−− identification O1 ( in A1) Therefore, (a2) a measurement of an observable O1 (at time t = 1) for a pure state ρ0 (at time t = 0) ∈ Sp(A∗0) is represented by MA0(Φ0,1O1, S[ρ0]) Thus, Axiom 1 ( measurement: §2.7) says that (a3) the probability that a measured value belongs to Ξ(∈ F) is given by A∗0 ( ρ0,Φ0,1(F1(Ξ) ) A0 (11.4) Explanation 11.3. [the Schrödinger picture]. The Schrödinger picture is just the following (b): (b1) To identify a pure state Φ∗0,1ρ0(∈ Sp(A∗1)) with ρ0(∈ Sp(A∗0)), That is, A∗0 ⊃ Sp(A∗0) 3 ρ0 Φ∗0,1−−−−−−−−→ identification Φ∗0,1ρ0 ∈ Sp(A∗1) ⊂ A∗1 Therefore, Axiom 1 ( measurement: §2.7) says that 301 Ishikawa's Homepage 11.1 The Heisenberg picture and the Schrödinger picture (b2) a measurement of an observable O1 (at time t = 1) for a pure state ρ0 (at time t = 0) ∈ Sp(A∗1) is represented by MA1(O1, S[Φ∗0,1ρ0]) Thus, (a3) the probability that a measured value belongs to Ξ(∈ F) is given by A∗1 ( Φ∗0,1ρ0, F1(Ξ) ) A1 (11.5) which is equal to A∗0 ( ρ0,Φ0,1(F1(Ξ)) ) A0 (11.6) In the above sense (i.e., (11.5) and (11.6) ), we conclude that, under the condition (A1), the Heisenberg picture and the Schrödinger picture are equivalent That is, MA0(Φ0,1O1, S[ρ0]) (Heisenberg picture) ←→ (identification) MA1(O1, S[Φ∗0,1ρ0]) (Schrödenger picture) (11.7) Remark 11.4. In the above, the conditions (A1) is indispensable, that is, (A1) Consider a deterministic causal operator Φ0,1 : A1 → A0. Without the deterministic conditions (A1), the Schrödinger picture can not be formulated completely. That is because Φ∗0,1ρ0 is not necessarily a pure state. In this sense, we consider that •  the Heisenberg picture is formal the Schrödinger picture is makeshift 302 Ishikawa's Homepage Chap. 11 Simple measurement and causality 11.2 Wave function collapse ( i.e., the projection postulate ) does not occur, but we look at somthing just like this. The lingistic interpretation says that the post measurement state is meaningless. However, considering a tricky measurement, we can realize the wave function collapse. In this section, we shall explain this idea in the following paper: • [48] S. Ishikawa, Linguistic interpretation of quantum mechanics; Projection Postulate, Journal of quantum information science, Vol. 5, No.4 , 150-155, 2015, DOI: 10.4236/jqis.2015.54017 (http://www.scirp.org/Journal/PaperInformation.aspx?PaperID=62464) Or see the following preprint; (http://www.math.keio.ac.jp/academic/research_pdf/report/2015/15009.pdf) 11.2.1 Problem: The von Neumann-Lüders projection postulate Let [C(H), B(H)]B(H) be a quantum basic structure. Let Λ be a countable set. Consider the projection valued observable OP = (Λ, 2 λ, P ) in B(H). Put Pλ = P ({λ}) (∀λ ∈ Λ) (11.8) Axiom 1 (measurement; §2.7) says: (A1) The probability that a measured value λ0 (∈ Λ) is obtained by the measurement MB(H)(OP :=(Λ, 2λ, P ), S[ρ]) is given by Tr H (ρPλ0)(= 〈u, Pλ0u〉 = ‖Pλ0u‖2), ( where ρ = |u〉〈u|) (11.9) Also, the von Neumann-Lüders projection postulate ( in the Copenhagen interpretation, cf. [83, 66]) says: (A2) When a measured value λ0 (∈ Λ) is obtained by the measurement MB(H)(OP :=(Λ, 2λ, P ), S[ρ]), the post-measurement state ρpost is given by ρpost = Pλ0 |u〉〈u|Pλ0 ‖Pλ0u‖2 (11.10) 303 Ishikawa's Homepage 11.2 Wave function collapse ( i.e., the projection postulate ) does not occur, but we look at somthing just like this. And therefore, when a next measurement MB(H)(OF :=(X,F, F ), S[ρpost]) is taken (where OF is arbitrary observable in B(H)), the probability that a measured value belongs to Ξ(∈ F) is given by Tr H (ρpostF (Ξ)) ( = 〈 Pλ0u ‖Pλ0u‖ , F (Ξ) Pλ0u ‖Pλ0u‖ 〉 ) (11.11) Problem 11.5. In the linguistic interpretation, the phrase:"post-measurement state" in the (A2) is meaningless. Also, the above (=(A1)+(A2)) is equivalent to the simultaneous measurement MB(H)(OF ×OP , S[ρ]), which does not exist in the case that OP and OF do not commute. Hence the (A2) is meaningless in general. Therefore, we have the following problem: (B) Instead of the OF × OP in MB(H)(OF × OP , S[ρ]), what observable should be chosen? In the following section, I answer this problem within the framework of the linguistic interpretation. 11.2.2 The derivation of von Neumann-Lüders projection postulate in the linguistic interpretation Consider two basic structure [C(H), B(H)]B(H) and [C(H⊗K), B(H⊗K)]B(H⊗K). Let {Pλ | λ ∈ Λ} be as in Section 11.2.1, and let {eλ}λ∈Λ be a complete orthonormal system in a Hilbert space K. Define the predual Markov operator Ψ∗ : Tr(H)→ Tr(H ⊗K) by, for any u ∈ H, Ψ∗(|u〉〈u|) = | ∑ λ∈Λ (Pλu⊗ eλ)〉〈 ∑ λ∈Λ (Pλu⊗ eλ)| (11.12) or Ψ∗(|u〉〈u|) = ∑ λ∈Λ |Pλu⊗ eλ〉〈Pλu⊗ eλ| (11.13) Thus the Markov operator Ψ : B(H ⊗K)→ B(H) ( in Axiom 2) is defined by Ψ = (Ψ∗)∗. Define the observable OG = (Λ, 2 Λ, G) in B(K) such that G({λ}) = |eλ〉〈eλ| (λ ∈ Λ) Let OF = (X,F, F ) be arbitrary observable in B(H). Thus, we have the tensor observable OF ⊗ OG = (X × Λ,F  2Λ, F ⊗G) in B(H ⊗K), where F  2Λ is the product σ-field. Fix a pure state ρ = |u〉〈u| (u ∈ H, ‖u‖H = 1). Consider the measurement MB(H)(Ψ(OF ⊗ OG), S[ρ]). Then, we see that 304 Ishikawa's Homepage Chap. 11 Simple measurement and causality (C) the probability that a measured value (x, λ) obtained by the measurement MB(H)(Ψ(OF ⊗ OG), S[ρ]) belongs to Ξ× {λ0} is given by Tr H [(|u〉〈u|)Ψ(F (Ξ)⊗G({λ0}))] = Tr(H) ( |u〉〈u|,Ψ(F (Ξ)⊗G({λ0})) ) B(H) = Tr(H⊗K) ( Ψ∗(|u〉〈u|), F (Ξ)⊗G({λ0}) ) B(H⊗K) = TrH⊗K [(Ψ∗(|u〉〈u|))(F (Ξ)⊗G({λ0}))] =Tr H⊗K [(| ∑ λ∈Λ (Pλu⊗ eλ)〉〈 ∑ λ∈Λ (Pλu⊗ eλ)|)(F (Ξ)⊗ |eλ0〉〈eλ0 |)] =〈Pλ0u, F (Ξ)Pλ0u〉 (∀Ξ ∈ F) ( In a similar way, the same result is easily obtained in the case of (7)). Thus, we see the following. (D1) if Ξ = X, then Tr H [(|u〉〈u|)Ψ(F (X)⊗G({λ0}))] = 〈Pλ0u, Pλ0u〉 = ‖Pλ0u‖2 (11.14) (D2) in case that a measured value (x, λ) belongs to X×{λ0}, the conditional probability such that x ∈ Ξ is given by 〈Pλ0u, F (Ξ)Pλ0u〉 ‖Pλ0u‖2 ( = 〈 Pλ0u ‖Pλ0u‖ , F (Ξ) Pλ0u ‖Pλ0u‖ 〉 ) (∀Ξ ∈ F) (11.15) where it should be recalled that OF is arbitrary. Also note that the above (i.e., the projection postulate (D)) is a consequence of Axioms 1 and 2. Considering the correspondence: (A)⇔ (D), that is, MB(H)(OP , S[ρ]) ( or, meaningless MB(H)(OF × OP , S[ρ]) ) ⇔ MB(H)(Ψ(OF ⊗ OG), S[ρ]), namely, (11.9)⇔ (11.14), (11.11)⇔ (11.15) there is a reason to assume that the true meaning of the (A) is just the (D). Also, note the taboo phrase "post-measurement state" is not used in (D2) but in (A2). Hence, we obtain the answer of Problem 1 (i.e., Ψ(OF ⊗ OG) ). Postulate 11.6. [Projection postulate] In the sense of the (D2), the statement (A2) is often used. That is, we often say: 305 Ishikawa's Homepage 11.2 Wave function collapse ( i.e., the projection postulate ) does not occur, but we look at somthing just like this. (E) When a measured value λ0 (∈ Λ) is obtained by the measurement MB(H)(OP :=(Λ, 2λ, P ), S[ρ]), the post-measurement state ρpost is given by ρpost = Pλ0 |u〉〈u|Pλ0 ‖Pλ0u‖2 (11.16) Remark 11.7. So called Copenhagen interpretation may admit the post-measurement state (cf. [21]). Thus, in this case, readers may think that the post-measurement state is equal to Pλ0 |u〉〈u|Pλ0 ‖Pλ0u‖ 2 , which is obtained by the (D2) ( since OF is arbitrary). However, this idea would not be generally approaved. That is because, if the post-measurement state is admitted, a series of problems occur, that is, "When is a measurement taken?", "When does the wave function collapse happen?", or "How fast is the wave function collapse?", which is beyond Axioms 1 and 2. Hence, the projection postulate is usually regarded as "postulate". On the other hand, in the linguistic interpretation, the projection postulate is completely clarified, and therefore, it should be regarded as a theorem. Recall the Wittgenstein's words: "The limits of my language mean the limits of my world", or "What we cannot speak about we must pass over in silence. " 306 Ishikawa's Homepage Chap. 11 Simple measurement and causality 11.3 de Broglie's paradox(non-locality=faster-than-light) In this section, we explain de Broglie's paradox in B(L2(R)) (cf. §2.10:de Broglie's paradox in B(C2) ). Putting q = (q1, q2, q3) ∈ R3, and ∇2 = ∂ 2 ∂q21 + ∂2 ∂q22 + ∂2 ∂q23 consider Schrödinger equation (concerning one particle): i~ ∂ ∂t ψ(q, t) = [−~2 2m ∇2 + V (q, t) ] ψ(q, t) (11.17) where, m is the mass of the particle, V is a potential energy. In order to demonstrate in the picture, regard R3 as R. Therefore, consider the Hilbert space H = L2(R, dq). Putting Ht = H (t ∈ R), consider the quantum basic structure: [C(H) ⊆ B(H) ⊆ B(H)] Equation 11.8. [Schrödinger equation]. There is a particle P (with mass m) in the box (that is, the closed interval [0, 2](⊆ R)). Let ρt0 = |ψt0〉〈ψt0 | ∈ Sp(C(H)∗) be an initial state (at time t0) of the particle P . Let ρt = |ψt〉〈ψt| (t0 ≤ t ≤ t1) be a state at time t, where ψt = ψ(*, t) ∈ H = L2(R, dq) satisfies the following Schrödinger equation: initial state:ψ(*, t0) = ψt0 i~ ∂ ∂t ψ(q, t) = [ −~2 2m ∂2 ∂q2 + V (q, t) ] ψ(q, t) (11.18) Consider the same situation in §10.5, i.e., a particle with the mass m in the box (i.e., the closed interval [0, 2]) in the one dimensional space R. R ψ(q, t) V0(q) ∞ 0 2 Figure 11.1(1) 307 Ishikawa's Homepage 11.3 de Broglie's paradox(non-locality=faster-than-light) Now let us partition the box [0, 2]] into [0, 1]] and [1, 2]. That is, we change V0(q) to V1(q), where V1(q) =  0 (0 ≤ q < 1) ∞ (q = 1) 0 (1 < q ≤ 2) ∞ ( otherwise ) (11.19) ψ1(q, t)0 1 ψ2(q, t) V1(q) ∞ 2 Figure 11.1(2) Next, we carry the box [0, 1] [ resp. the box [1, 2] ] to New York (or, the earth) [ resp. Tokyo (or, the polar star) ] . New York 0 1 ψ1(q, t1) ψ2(q, t1) Tokyo a+1 a+2 Figure 11.1(3) Here, 1 a. Solving the Schrödinger equation (11.18), we see that ψ1(*, t1) + ψ2(*, t1) = Ut0,t1ψt0 where Ut0,t1 : L 2(Rt1) → L2(Rt0) is the unitary operator. Define the causal operator Φt0,t1 : B(L2(Rt2))→ B(L2(Rt1)) by Φt0,t1(A) = U ∗ t0,t1 AUt0,t1 (∀A ∈ B(L2(Rt2))) 308 Ishikawa's Homepage Chap. 11 Simple measurement and causality Put T = {t0, t1}. And consider the observable O = (X = {N, T.E}, 2X , F ) in B(L2(Rt1)) (where "N"=New York, "T"=Tokyo, "E"=elsewhere ) such that [F ({N})](q) = { 1 0 ≤ q < 1 0 elsewhere , [F ({T})](q) = { 1 a+ 1 ≤ q < a+ 2 0 elsewhere , [F ({E})](q) = 1− [F ({N})](q)− [F ({T})](q). Hence we have the measurement MB(L2(Rt0 )) ( Φt0,t1O, S[|ψt0 〉〈ψt0 |] ) . Conclusion 11.9. In Heisenberg picture, we see, by Axiom 1 ( measurement: §2.7), that (A1) the probability that a measured value  NT E  is obtained by the measurement MB(L2(Rt0 )) ( Φt0,t1O, S[|ψt0 〉〈ψt0 |] ) is given by  〈ut0 ,Φt0.t1F ({N})ut0〉 = ∫ 10 |ψ1(q, t1)|2dq〈ut0 ,Φt0.t1F ({T})ut0〉 = ∫ a+2a+1 |ψ2(q, t1)|2dq 〈ut0 ,Φt0.t1F ({E})ut0〉 = 0  . Also, In Schrödinger picture, we see Axiom 1 ( measurement: §2.7), that (A2) the probability that a measured value  NT E  is obtained by the measurement MB(L2(Rt0 )) ( O, S[Φ∗t0,t1 (|ψt0 〉〈ψt0 |)] ) is given by Tr ( Φ∗t0,t1(|ψt0〉〈ψt0 |) * F ({N}) ) = 〈Ut0,t1ψt0 , F ({N})Ut0,t1ψt0〉 = ∫ 1 0 |ψ1(q, t1)|2dq Tr ( Φ∗t0,t1(|ψt0〉〈ψt0 |) * F ({T}) ) = 〈Ut0,t1ψt0 , F ({T})Ut0,t1ψt0〉 = ∫ a+2 a+1 |ψ2(q, t1)|2dq Tr ( Φ∗t0,t1(|ψt0〉〈ψt0 |) * F ({E}) ) = 〈Ut0,t1ψt0 , F ({E})Ut0,t1ψt0〉 = 0  Note that the probability that we find the particle in the box [0, 1] [ resp. the box [a + 1, a+ 2] ] is given by ∫ R |ψ1(q, t1)| 2dq [ resp. ∫ R |ψ2(q, t1)| 2dq ] . That is, (A1)=(A2) Remark 11.10. In the above, assume that we get a measured value "N", that is, we open the box [0, 1] at New York. And assume that we find the particle in the box [0, 1]. Then, in the sense of Postulate 11.6, we say that at the moment the wave function ψ2 vanishes. That is, 309 Ishikawa's Homepage 11.3 de Broglie's paradox(non-locality=faster-than-light) New York 0 1 ψ′1(q, t1) "Vanish" Tokyo a+1 a+2 Figure 11.1(4) (The wave function after measurement) where ψ′1(q, t1) = ψ1(q, t1) ‖ψ′1(*, t1)‖ . Thus, we may consider "the collapse of wave function" such as ψ1(*, t1) + ψ2(*, t1) −−−−−−−−−−−−−−−−→ the collapse of wave function ψ′1(*, t1) (11.20) Also, note that New York [ resp. Tokyo ] may be the earth [ resp. the polar star ] . Thus, • the above argument (in both cases (A1) and (A2)) implies that there is something faster than light. This is called "the de Broglie paradox"(cf. [13, 81]). This is a true paradox, which is not clarified even in quantum language. 310 Ishikawa's Homepage Chap. 11 Simple measurement and causality 11.4 Quantum Zeno effect; watched pot effect This section is extracted from • Ref. [40]: S. Ishikawa; Heisenberg uncertainty principle and quantum Zeno effects in the linguistic interpretation of quantum mechanics ( arXiv:1308.5469 [quant-ph] 2014 ) 11.4.1 Quantum decoherence: non-deterministic sequential causal operator Let us start from the review of Section 10.6.2 (quantum decoherence). Consider the quantum basic structure: [C(H) ⊆ B(H) ⊆ B(H)] Let P = [Pn]∞n=1 be the spectrum decomposition in B(H), that is, Pn is a projection, and, ∞∑ n=1 Pn = I Define the operator (ΨP)∗ : Tr(H)→ Tr(H) such that (ΨP)∗(|u〉〈u|) = ∞∑ n=1 |Pnu〉〈Pnu| (∀u ∈ H) Clearly we see 〈v, (ΨP)∗(|u〉〈u|)v〉 = 〈v, ( ∞∑ n=1 |Pnu〉〈Pnu|)v〉 = ∞∑ n=1 |〈v, |Pnu〉|2 ≥ 0 (∀u, v ∈ H) and, Tr((ΨP)∗(|u〉〈u|)) =Tr( ∞∑ n=1 |Pnu〉〈Pnu|) = ∞∑ n=1 ∞∑ k=1 |〈ek, Pnu〉|2 = ∞∑ n=1 ‖Pnu‖2 = ‖u‖2 (∀u ∈ H) And so, (ΨP)∗(Tr p +1(H)) ⊆ Tr+1(H) Therefore, (]) ΨP(= ((ΨP)∗) ∗) : B(H)→ B(H) is a causal operator, but it is not deterministic. 311 Ishikawa's Homepage 11.4 Quantum Zeno effect; watched pot effect In this note, a non-deterministic (sequential) causal operator is called a quantum decoherence. Example 11.11. [Quantum decoherence in quantum Zeno effect cf. [37]]. Further consider a causal operator (Ψ∆tS )∗ : Tr(H)→ Tr(H) such that (Ψ∆tS )∗(|u〉〈u|) = |e− iH∆t ~ u〉〈e− iH∆t ~ u| (∀u ∈ H) where the Hamiltonian H (cf. (10.22) ) is, for example, defined by H = [−~2 2m ∂2 ∂q2 + V (q, t) ] Let P = [Pn]∞n=1 be the spectrum decomposition in B(H), that is, for each n, Pn ∈ B(H) is a projection such that ∞∑ n=1 Pn = I Define the (ΨP)∗ : Tr(H)→ Tr(H) such that (ΨP)∗(|u〉〈u|) = ∞∑ n=1 |Pnu〉〈Pnu| (∀u ∈ H) Also, we define the Schrödinger time evolution (Ψ∆tS )∗ : Tr(H)→ Tr(H) such that (Ψ∆tS )∗(|u〉〈u|) = |e− iH∆t ~ u〉〈e− iH∆t ~ u| (∀u ∈ H) where H is the Hamiltonian (10.21). Consider t = 0, 1. Putting ∆t = 1 N , H = H0 = H1, we can define the (Φ (N) 0,1 )∗ : Tr(H0)→ Tr(H1) such that (Φ (N) 0,1 )∗ = ((Ψ 1/N S )∗(ΨP)∗) N which induces the Markov operator Φ (N) 0,1 : B(H1) → B(H0) as the dual operator Φ (N) 0,1 = ((Φ (N) 0,1 )∗) ∗. Let ρ = |ψ〉〈ψ| be a state at time 0. Let O1 :=(X,F, F ) be an observable in B(H1). Then, we see ρ=|ψ〉〈ψ| B(H0) ←−−− Φ (N) 0,1 B(H1) O1 :=(X,F,F ) Thus, we have a measurement: MB(H0)(Φ (N) 0,1 O1, S[ρ])( or more precisely, MB(H0)(Φ (N) 0,1 O :=(X,F,Φ (N) 0,1 F ), S[|ψ〉〈ψ|]) ) . Here, Axiom 1 ( §2.7) says that 312 Ishikawa's Homepage Chap. 11 Simple measurement and causality (A) the probability that the measured value obtained by the measurement belongs to Ξ(∈ F) is given by Tr(|ψ〉〈ψ| * Φ(N)0,1 F (Ξ)) (11.21) Now we shall explain "quantum Zeno effect" in the following example. Example 11.12. [Quantum Zeno effect] Let ψ ∈ H such that ‖ψ‖ = 1. Define the spectrum decomposition P = [P1(= |ψ〉〈ψ|), P2(= I − P1)] (11.22) And define the observable O1 :=(X,F, F ) in B(H1) such that X = {x1, x2}, F = 2X and F ({x1}) = |ψ〉〈ψ|(= P1), F ({x2}) = I − |ψ〉〈ψ|(= P2), Now we can calculate (11.21)(i.e., the probability that a measured value x1 is obtained) as follows. (11.21) = 〈ψ, ((Ψ1/NS )∗(ΨP)∗) N(|ψ〉〈ψ|)ψ〉 ≥ |〈ψ, e− iH ~N ψ〉〈ψ, e iH ~N ψ〉|N ≈ ( 1− 1 N2 ( ||(H ~ )ψ||2 − |〈ψ, (H ~ )ψ〉|2 ))N → 1 (N →∞) (11.23) Thus, if N is sufficiently large, we see that MB(H0)(Φ (N) 0,1 O1, S[|ψ〉〈ψ|]) ≈ MB(H0)(ΦIO1, S[|ψ〉〈ψ|]) (where ΦI : B(H1)→ B(H0) is the identity map) = MB(H0)(O1, S[|ψ〉〈ψ|]) Hence, we say, roughly speaking in terms of the Schrödinger picture, that the state |ψ〉〈ψ| does not move. 313 Ishikawa's Homepage 11.4 Quantum Zeno effect; watched pot effect Remark 11.13. The above argument is motivated by B. Misra and E.C.G. Sudarshan [69]. However, the title of their paper: "The Zeno's paradox in quantum theory" is not proper. That is because (B) the spectrum decomposition P should not be regarded as an observable (or moreover, measurement). The effect in Example 11.12 should be called "brake effect" and not "watched pot effect". 314 Ishikawa's Homepage Chap. 11 Simple measurement and causality 11.5 Schrödinger's cat, Wigner's friend and Laplace's demon 11.5.1 Schrödinger's cat and Wigner's friend Let us explain Schrödinger's cat paradox in the Schrödinger picture. Problem 11.14. [Schrödinger's cat] (a) Suppose we put a cat in a cage with a radioactive atom, a Geiger counter, and a poison gas bottle; further suppose that the atom in the cage has a half-life of one hour, a fiftyfifty chance of decaying within the hour. If the atom decays, the Geiger counter will tick; the triggering of the counter will get the lid off the poison gas bottle, which will kill the cat. If the atom does not decay, none of the above things happen, and the cat will be alive. Geiger counter radioactive atom * * * cat poison gas Figure 11.2: Schrödinger's cat Here, we have the following question: (b) Is the cat dead or alive after 1 hour (= 6060 seconds ) ? Of course, we say that it is half-and-half whether the cat is alive. However, our problem is Clarify the meaning of "half-and-half" ♠Note 11.1. [Wigner's friend]: Instead of the above (b), we consider as follows. 315 Ishikawa's Homepage 11.5 Schrödinger's cat, Wigner's friend and Laplace's demon (b′) after one hour, Wigner's friend look at the inside of the box, and thus, he knows whether the cat is dead or alive after one hour. And further, after two hours, Wigner's friend informs you of the fact. How is the cat ? This problem is not difficult. That is because the linguistic interpretation says that "the moment you measured" is out of quantum language. Recall the spirit of the linguistic world-view (i.e., Wittgenstein's words) such as The limits of my language mean the limits of my world and What we cannot speak about we must pass over in silence. 11.5.2 The usual answer Answer 11.15. [The first answer to Problem11.14(i.e., the pure state, projection postulate )]. Put q = (q11, q12, q13, q21, q22, q23, . . . , qn1, qn2, qn3) ∈ R3n. And put ∇2i = ∂2 ∂q2i1 + ∂2 ∂q2i2 + ∂2 ∂q2i3 Consider the quantum system basic structure: [C(H) ⊆ B(H) ⊆ B(H)] ( where, H = L2(R3n, dq) ) And consider the Schrödinger equation (concerning n-particles system): i~ ∂ ∂t ψ(q, t) = [∑n i=1 −~2 2mi ∇2i + V (q, t) ] ψ(q, t) ψ0(q) = ψ(q, 0) : initial condition (11.24) where, mi is the mass of a particle Pi, V is a potential energy. If we believe in quantum mechanics, it suffices to solve this Schrödinger equation (11.24). That is, (A1) Assume that the wave function ψ(*, 602) = U0,602ψ0 after one hour (i.e., 602 seconds) is calculated. Then, the state ρ602 (∈ Trp+1(H)) after 602 seconds is represented by ρ602 = |ψ602〉〈ψ602 | (11.25) (where, ψ602 = ψ(*, 602)). Now, define the observable O = (X = {life, death}, 2X , F ) in B(H) as follows. 316 Ishikawa's Homepage Chap. 11 Simple measurement and causality (A2) that is, putting Vlife(⊆ H) = { u ∈ H | " the state |u〉〈u| ‖u‖2 "⇔ "cat is alive" } Vdeath(⊆ H) = the orthogonal complement space of Vlife = {u ∈ H | 〈u, v〉 = 0 (∀v ∈ Vlife)} define F ({life})(∈ B(H)) is the projection of the closed subspace Vlife and F ({death}) = I − F ({life}), Here, (A3) Consider the measurement MB(H)(O = (X, 2 X , F ), S[ρ602]). The probability that a measured value [ life death ] is obtained is given by Tr(H)(ρ602 , F ({life}))B(H) = 〈ψ602 , F ({life})ψ602〉 = 0.5 Tr(H) ( ρ602 , F ({death}) ) B(H) = 〈ψ602 , F ({death})ψ602〉 = 0.5  Therefore, we can assure that ψ602 = 1√ 2 (ψlife + ψdeath) (11.26) (where, ψlife ∈ Vlife, ‖ψlife‖ = 1 ψdeath ∈ Vdeath, ‖ψdeath‖ = 1) Hence. we can conclude that (A4) the state (or, wave function) of the cat (after one hour ) is represented by (11.26), that is, "Fig.(]1)"+"Fig.(]2)"√ 2 Fig. (]1) ≈ ψlife Geiger counter radioactive atom * * * click! 6 Geiger counter radioactive atom Fig. (]2)≈ ψdeath cat poison gas cat poison gas Figure 11.3: Schrödinger's cat(half and half) 317 Ishikawa's Homepage 11.5 Schrödinger's cat, Wigner's friend and Laplace's demon And, (A5) After one hour (i.e, to the moment of opening a window), It is decided "the cat is dead" or "the cat is vigorously alive." That is, "half-dead" ( = 1 2 (|ψlife + ψdeath〉〈ψlife + ψdeath|) ) in the sense of Postulate 11.6 ( precisely speaking, by the misunderstanding of Postulate 11.6), to the moment of opening a window−−−−−−−−−−−−−−−−−−−−−−−−→ the collapse of wave function  "alive"(= |ψlife〉〈ψlife|) "dead"(= |ψdeath〉〈ψdeath|) 11.5.3 The answer by quantum decoherence Answer 11.16. [The second answer to Problem11.14 (i.e., decoherence)]. In quantum language, the quantum decoherence is permitted. That is, we can assume that (B1) the state ρ ′ 602 after one hour is represented by the following mixed state ρ′602 = 1 2 ( |ψlife〉〈ψlife|+ |ψdeath〉〈ψdeath| ) That is, we can assume the decoherent causal operator Φ0,602 : B(H)→ B(H) such that (Φ0,602)∗(ρ0) = ρ ′ 602 Here, consider the measurement MB(H)(O = (X, 2 X , F ), S[ρ ′ 602 ]), or, its Heisenberg picture MB(H)(Φ0,602O = (X, 2 X ,Φ0,602F ), S[ρ ′ 0]). Of course we see: (B2) The probability that a measured value [ life death ] is obtained by the measurement MB(H)(Φ0,602O = (X, 2 X ,Φ0,602F ), S[ρ ′ 0]) is given by Tr(H)(ρ0,Φ0,602F ({life}))B(H) = 〈ψ′602 , F ({life})ψ602〉 = 0.5 Tr(H) ( ρ0,Φ0,602F ({death}) ) B(H) = 〈ψ′602 , F ({death})ψ602〉 = 0.5  Also, "the moment of measuring" and "the collapse of wave function" are prohibited in the linguistic interpretation, but the statement (B2) is within quantum language. 318 Ishikawa's Homepage Chap. 11 Simple measurement and causality Summary 11.17. [Schrödinger's cat in quantum language] Here, let us examine Answer11.15 :(A5) v.s. Answer11.16 :(B2) (C1) the answer (A5) may be unnatural, but it is an argument which cannot be confuted, On the other hand, (C2) the answer (B2) is natural. but the non-deterministic time evolution is used. Since the non-deterministic causal operator (i.e., quantum decoherence) is permitted in quantum language, we conclude that (C3) Answer11.16:(B2) is superior to Answer11.15:(A1) For the reason that the non-deterministic causal operator (i.e., quantum decoherence) is permitted in quantum language, we add the following. • If Newtonian mechanics is applied to the whole universe, Laplace's demon appears. Also, if Newtonian mechanics is applied to the microworld, chaos appears. This kind of supremacy of physics is not natural, and thus, we consider that these are out of "the limit of Newtonian mechanics" And, • when we want to apply Newton mechanics to phenomena out of "the limit of Newtonian mechanics", we often use the stochastic differential equation (and Brownian motion). This approach is called "dynamical system theory", which is not physics but metaphysics. Newtonian mechanics physics out of the limits−−−−−−−−−−−−→ linguistic turn dynamical system theory; statistics metaphysics In the same sense, we consider that quantum mechanics has "the limit". That is, • Schrödinger's cat is out of quantum mechanics. And thus, • When we want to apply quantum mechanics to phenomena out of "the limit of quantum mechanics", we often use the quantum decoherence. Although this approach is not physics but metaphysics, it is quite powerful. quantum mechanics physics out of the limits−−−−−−−−−−−−→ linguistic turn quantum language metaphysics 319 Ishikawa's Homepage 11.5 Schrödinger's cat, Wigner's friend and Laplace's demon ♠Note 11.2. If we know the present state of the universe and the kinetic equation (=the theory of everything), and if we calculate it, we can know everything (from past to future). There may be a reason to believe this idea. This intellect is often referred to as Laplace's demon. Laplace's demon is sometimes discussed as the realistic-view over which the degree passed. Thus, we consider the following correspondence: Laplace's Demon Newtonian mechanics ←→ correspondence Schrödinger's cat in Answer 11.15 quantum mechanics 320 Ishikawa's Homepage Chap. 11 Simple measurement and causality 11.6 Wheeler's Delayed choice experiment: "Particle or wave?" is a foolish question This section is extracted from (]) [45] S. Ishikawa, The double-slit quantum eraser experiments and Hardy's paradox in the quantum linguistic interpretation, arxiv:1407.5143[quantum-ph],( 2014) 11.6.1 "Particle or wave?" is a foolish question In the conventional quantum mechanics, the question: "particle or wave?" may frequently appear. However, this is a foolish question. On the other hand, the argument about the "particle vs. wave" is clear in quantum language. As seen in the following table, this argument is traditional: Table 11.1: Particle vs. Wave in several world-views (cf. Table 2.1, Table 3.1) World-views \ P or W Particle(=symbol) Wave(= mathematical representation ) Aristotle hyle eidos Newton mechanics point mass state (=(position, momentum)) Statistics population parameter Quantum mechanics particle state (≈ wave function) Quantum language system (=measuring object) state In the table 11.1, Newtonian mechanics (i.e., mass point↔ state) may be easiest to understand. Thus, "particle" and "wave" are not confrontation concepts. Concerning "particle or wave", we have the following statements: (A1) "Particle or wave" is a foolish question. (A2) Wheeler's delayed choice experiment is related to the question "particle or wave" If so, it may be interesting to answer the following: (A3) How is Wheeler's delayed choice experiment described in terms of quantum mechanics? This is the purpose of this section. And we answer it in the conclusion (H). 321 Ishikawa's Homepage 11.6 Wheeler's Delayed choice experiment: "Particle or wave?" is a foolish question 11.6.2 Preparation Let us start from the review of Section 2.10 (de Broglie paradox in B(C2)) Let H be a two dimensional Hilbert space, i.e., H = C2. Consider the basic structure [B(C2) ⊆ B(C2) ⊆ B(C2)] Let f1, f2 ∈ H such that f1 = [ 1 0 ] , f2 = [ 0 1 ] Put u = f1 + f2√ 2 Thus, we have the state ρ = |u〉〈u| (∈ Sp(B(C2))). Let U(∈ B(C2)) be an unitary operator such that U = [ 1 0 0 eiπ/2 ] and let Φ : B(C2)→ B(C2) be the homomorphism such that Φ(F ) = U∗FU (∀F ∈ B(C2)) Consider two observable Of = ({1, 2}, 2{1,2}, F ) and Og = ({1, 2}, 2{1,2}, G) in B(C2) such that F ({1}) = |f1〉〈f1|, F ({2}) = |f2〉〈f2| and G({1}) = |g1〉〈g1|, G({2}) = |g2〉〈g2| where g1 = f1 + f2√ 2 , g2 = f1 − f2√ 2 322 Ishikawa's Homepage Chap. 11 Simple measurement and causality 11.6.3 de Broglie's paradox in B(C2) (No interference) D1(= (|f1〉〈f1|)) (photon detector) D2(= (|f2〉〈f2|)) (photon detector) u= 1√ 2 (f1+f2) −−−−−−−−→ 1√ 2 f1 ? √ −1√ 2 f2 ? 1√ 2 f1 1√ 2 f1 - √ −1√ 2 f2 √ −1√ 2 f2 half mirror 1 Figure 11.4(1). [D1 +D2]=ObservableOf mirror 2 mirror 1course 1 course 2 Photon P Now we shall explain, by the Schrödinger picture, Figure 11.4(1) as follows. The photon P with the state u = 1√ 2 (f1 + f2) ( precisely, ρ = |u〉〈u| ) rushed into the half-mirror 1, (B1) the f1 part in u = 1√ 2 (f1 +f2) passes through the half-mirror 1, and goes along the course 1. And it is reflected in the mirror 1, and goes to the photon detector D1. (B2) the f2 part in u = 1√ 2 (f1 + f2) rebounds on the half-mirror 1 (and strictly saying, the f2 changes to √ −1f2, we are not concerned with it ), and goes along the course 2. And it is reflected in the mirror 2, and goes to the photon detector D2. This is, by the Heisenberg picture, represented by the following measurement: MB(C2)(ΦOf , S[ρ]) (11.27) Then, we see: (C) the probability that [ a measured value 1 a measured value 2 ] is obtained by MB(C2)(ΦOf , S[ρ]) is given by [ 〈Uu, F ({1})Uu〉 〈Uu, F ({2})Uu〉 ] = [ |〈Uu, f1〉|2 |〈Uu, f2〉|2 ] = [ 1 2 1 2 ] (11.28) 323 Ishikawa's Homepage 11.6 Wheeler's Delayed choice experiment: "Particle or wave?" is a foolish question Remark 11.18. [Projection postulate] By the analogy of Section 11.2 ( The projection postulate ), Figure 11.4(1) is also described as follows. That is, putting e1 = [ 1 0 ] and e2 = [ 0 1 ] (∈ C2), we have the observable OE = ({1, 2}, 2{1,2}, E) in B(C2) such that E({1}) = |e1〉〈e1 and E({1}) = |e1〉〈e1. Hence, D1(= (Of ⊗ |e1〉〈e1|)) (photon detector) D2(= (Of ⊗ |e2〉〈e2|)) (photon detector) u= 1√ 2 (f1+f2) −−−−−−−−→ 1√ 2 f1⊗e1 ? √ −1√ 2 f2⊗e2 ? 1√ 2 f1⊗e1 1√ 2 f1⊗e1 - √ −1√ 2 f2⊗e2 √ −1√ 2 f2⊗e2 half mirror 1 Figure 11.4(1′). [D1 +D2]=Of ⊗ OE mirror 2 mirror 1course 1 course 2 Photon P Thus, using the Schrödinger picture, in the above figure we see: u = 1√ 2 (f1 + f2) −−−−−−−−−−−→ time evolution 1√ 2 f1⊗e1 + √ −1√ 2 f2⊗e2 which may imply that spacetime and quantum entanglement are related. 11.6.4 Mach-Zehnder interferometer (Interference) Next, consider the following figure: 324 Ishikawa's Homepage Chap. 11 Simple measurement and causality D1(= (|g2〉〈g2|)) (photon detector) D2(= (|g1〉〈g1|)) (photon detector) u= 1√ 2 (f1+f2) −−−−−−−−→ 1√ 2 f1 ? √ −1√ 2 f2 ? 1√ 2 f1 1√ 2 f1 − 1√2f2 - √ −1√ 2 f2 0 half mirror 1 half mirror 2 Figure 11.4(2). [D1 +D2]=ObservableOg mirror 1 mirror 2course 1 course 2 Photon P Now we shall explain, by the Schrödinger picture, Figure 11.4(2) as follows. The photon P with the state u = 1√ 2 (f1 + f2) ( precisely, ρ = |u〉〈u| ) rushed into the half-mirror 1, (D1) the f1 part in u = 1√ 2 (f1 +f2) passes through the half-mirror 1, and goes along the course 1. And it is reflected in the mirror 1, and passes through the half-mirror 2, and goes to the photon detector D1. (D2) the f2 part in u = 1√ 2 (f1 + f2) rebounds on the half-mirror 1 (and strictly saying, the f2 changes to √ −1f2, we are not concerned with it ), and goes along the course 2. And it is reflected in the mirror 2, and further reflected in the half-mirror 2, and goes to the photon detector D2. This is, by the Heisenberg picture, represented by the following measurement: MB(C2)(Φ 2Og, S[ρ]) (11.29) Then, we see: (E) the probability that [ a measured value 1 a measured value 2 ] is obtained by MB(C2)(Φ 2Og, S[ρ]) is given by [ 〈u,Φ2G({1})u〉 〈u,Φ2G({2})u〉 ] = [ |〈u, UUg1〉|2 |〈u, UUg2〉|2 ] = [ 0 1 ] 325 Ishikawa's Homepage 11.6 Wheeler's Delayed choice experiment: "Particle or wave?" is a foolish question 11.6.5 Another case Consider the following Figure 11.4(3). D2(= (|f2〉〈f2|)) (photon detector) D1(= (|f1〉〈f1|)) (photon detector) u= 1√ 2 (f1+f2) −−−−−−−−→ 1√ 2 f1 ? √ −1√ 2 f2 ? −1√ 2 f2 - √ −1√ 2 f2 half mirror 1 half mirror 2mirror Figure 11.4(3). [D2 +D1] =ObservableOf 1 mirror 2course 1 course 2 Photon P Now we shall explain, by the Schrödinger picture, Figure 11.4(3) as follows. The photon P with the state u = 1√ 2 (f1 + f2) ( precisely, ρ = |u〉〈u| ) rushed into the half-mirror 1, (F1) the f1 part in u = 1√ 2 (f1 +f2) passes through the half-mirror 1, and goes along the course 1. And it reaches to the photon detector D1. (F2) the f2 part in u = 1√ 2 (f1 + f2) rebounds on the half-mirror 1 (and strictly saying, the f2 changes to √ −1f2, we are not concerned with it ), and goes along the course 2. And it is again reflected in the mirror 1, and further reflected in the half-mirror 2, and goes to the photon detector D2. This is, by the Heisenberg picture, represented by the following measurement: MB(C2)(Φ 2Of , S[ρ]) (11.30) Therefore, we see the following: (G) The probability that [ measured value 1 measured value 2 ] is obtained by the measurement MB(C2)(Φ 2Of , S[ρ]) is given by[ Tr(ρ * Φ2F ({1})) Tr(ρ * Φ2F ({2})) ] = [ 〈UUu, F ({1})UUu〉 〈UUu, F ({2})UUu〉 ] = [ |〈UUu, f1〉|2 |〈UUu, f2〉|2 ] = [ 1 2 1 2 ] 326 Ishikawa's Homepage Chap. 11 Simple measurement and causality Therefore, if the photon detector D1 does not react, it is expected that the photon detector D2 reacts. 11.6.6 Conclusion The above argument is just Wheeler's delayed choice experiment. It should be noted that the difference among Examples in §11.5.3 (Figure 11.4(1))– §11.5 (Figure 11.4(3)) is that of the observables (= measuring instrument ). That is, §11.5.3 (Figure 11.4(1)) −−−−−−−−−−→ Heisenberg picture ΦOf §11.5.4 (Figure 11.4(2)) −−−−−−−−−−→ Heisenberg picture Φ2Og §11.5.5 (Figure 11.4(3)) −−−−−−−−−−→ Heisenberg picture Φ2Of Hence, it should be noted that (H) Wheeler's delayed choice experiment -"after the photon P passes through the halfmirror 1, one of Figure 11.4(1), Figure 11.4(2) and Figure 11.4(3) is chosen" - can not be described paradoxically in quantum language. However, it should be noted that the non-locality paradox (i.e., "there is some thing faster than light") is not solved even in quantum language. ♠Note 11.3. What we want to assert in this book may be the following: (]) everything (except "there is some thing faster than light") can not be described paradoxically in terms of quantum language 327 Ishikawa's Homepage 11.7 Hardy's paradox: total probabilty is less than 1 11.7 Hardy's paradox: total probabilty is less than 1 In this section, we shall introduce the Hardy's paradox (cf. ref.[17]) in terms of quantum language1. Let H be a two dimensional Hilbert space, i.e., H = C2. Let f1, f2, g1, g2 ∈ H such that f1 = f ′ 1 = [ 1 0 ] , f2 = f ′ 2 = [ 0 1 ] , g1 = g ′ 1 = f1 + f2√ 2 , g2 = g ′ 2 = f1 − f2√ 2 Put u = f1 + f2√ 2 ( = g1 ) Consider the tensor Hilbert space H ⊗H = C2 ⊗ C2 and define the state ρ such that û = u⊗ u′ = f1 + f2√ 2 ⊗ f ′ 1 + f ′ 2√ 2 , ρ = |u⊗ u′〉〈u⊗ u′| As shown in the next section (e.g., annihilation (i.e., f1 ⊗ f1 7→ 0), etc.), define the operator P : C2 ⊗ C2 → C2 ⊗ C2 such that P (α11f1 ⊗ f1 + α12f1 ⊗ f2 + α21f2 ⊗ f1 + α22f2 ⊗ f2) = −α12f1 ⊗ f2 − α21f2 ⊗ f1 + α22f2 ⊗ f2 Here, it is clear that P 2(α11f1 ⊗ f1 + α12f1 ⊗ f2 + α21f2 ⊗ f1 + α22f2 ⊗ f2) = α12f1 ⊗ f2 + α21f2 ⊗ f1 + α22f2 ⊗ f2 hence, we see that P 2 : C2 ⊗ C2 → C2 ⊗ C2 is a projection. Also, define the causal operator Ψ : B(C2 ⊗ C2)→ B(C2 ⊗ C2) by Ψ(Â) = PÂP (Â ∈ B(C2 ⊗ C2)) Here, it is easy to see that Ψ : B(C2 ⊗ C2)→ B(C2 ⊗ C2) satisfies (A1) Ψ(Â ∗Â) ≥ 0 (∀Â ∈ B(C2 ⊗ C2)) (A2) Ψ(I) = P 2 Since it is not always assured that Ψ(I) = I, strictly speaking, the Ψ : B(C2⊗C2)→ B(C2⊗C2) is a causal operator in the wide sense. 1This section is extracted from (]) [45] S. Ishikawa, The double-slit quantum eraser experiments and Hardy's paradox in the quantum linguistic interpretation, arxiv:1407.5143[quantum-ph],( 2014) 328 Ishikawa's Homepage Chap. 11 Simple measurement and causality 11.7.1 Observable Og ⊗ Og Consider the following figure D′1(= (|g′2〉〈g′2|)) (Detector) D′2(= (|g′1〉〈g′1|)) (Detector) ? 1√ 2 (f ′1 + f ′ 2) √ −1√ 2 f ′2 ? 1√ 2 f ′1 ? √ −1√ 2 f ′2 if no annihilation, 1√ 2 f ′1 half mirror 2′ half mirror 1′ mirror 2′ mirror 1′ course 2′ course 1′ Positron P′ D1(= (|g2〉〈g2|)) (Detector) D2(= (|g1〉〈g1|)) (Detector) 1√ 2 (f1+f2) −−−−−−→ 1√ 2 f1 ? √ −1√ 2 f2 ? if no annihilation, 1√ 2 f1 - √ −1√ 2 f2 half mirror 1 half mirror 2 Figure 11.5(1). Electron P and Positron P′ are annihilated at • mirror 1 mirror 2course 1 course 2 Electron P In the above, Electron P and Positron P ′ rush into the half-mirror 1 and the half-mirror 1′ respectively. Here, "half-mirror" has the following property:[ 1 0 ] (= f1 = f ′ 1) −−−−−−−−−−−−−−−−−−−→ pass through half-mirror [ 1 0 ] (= f1 = f ′ 1)[ 0 1 ] (= f2 = f ′ 2) −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→ be reflected in half-mirror, and × √ −1 √ −1 [ 0 1 ] (= f2 = f ′ 2) Assume that the initial state of Electron P [resp. Positron P ′] is β1f1 +β2f2 [resp. β ′ 1f ′ 1 +β ′ 2f ′ 2]. Then, we see, by the Schrödinger picture, that (β1f1 + β2f2)⊗ (β′1f ′1 + β′2f ′2) = β1β′1f1 ⊗ f ′1 + β1β′2f1 ⊗ f ′2 + β2β′1f2 ⊗ f ′1 + β2β′2f2 ⊗ f ′2 −−−−−−−−→ (half-mirror) 329 Ishikawa's Homepage 11.7 Hardy's paradox: total probabilty is less than 1 β1β ′ 1f1 ⊗ f ′1 + √ −1β1β′2f1 ⊗ f ′2 + √ −1β2β′1f2 ⊗ f ′1 − β2β′2f2 ⊗ f ′2 −−−−−−−−−−−−−−−−−−−−→ (annihilation(i.e., f1 ⊗ f ′1 = 0))√ −1β1β′2f1 ⊗ f ′2 + √ −1β2β′1f2 ⊗ f ′1 − β2β′2f2 ⊗ f ′2 −−−−−−−−−−−−−→ (second half-mirror) − β1β′2f1 ⊗ f ′2 − β2β′1f2 ⊗ f ′1 + β2β′2f2 ⊗ f ′2 The above is written by the Schrödinger picture Ψ∗ : Tr(C2 ⊗ C2) → Tr(C2 ⊗ C2). Thus, we have the Heisenberg picture (i.e., the causal operator ) Ψ : B(C2 ⊗ C2) → B(C2 ⊗ C2) by Ψ = (Ψ∗) ∗. Define the observable Ôgg = ({1, 2} × {1, 2}, 2{1,2}×{1,2}, Ĥgg) in B(C2 ⊗ C2) by the tensor observable Og ⊗ Og, that is, Ĥgg({(1, 1)}) = |g1 ⊗ g1〉〈g1 ⊗ g1|, Ĥgg({(1, 2)}) = |g1 ⊗ g2〉〈g1 ⊗ g2|, Ĥgg({(2, 1)}) = |g2 ⊗ g1〉〈g2 ⊗ g1|, Ĥgg({(2, 2)}) = |g2 ⊗ g2〉〈g2 ⊗ g2| Consider the measurement: MB(C2⊗C2)(ΨÔgg, S[ρ]) (11.31) Then, the probability that a measured value (2, 2) is obtained by MB(C2⊗C2)(ΨÔ, S[ρ]) is given by 〈u⊗ u, PĤgg({(2, 2)})P (u⊗ u)〉 = |〈(f1 − f2)⊗ (f1 − f2), f1 ⊗ f2 + f2 ⊗ f1 + f2 ⊗ f2〉|2 16 = |〈f1 ⊗ f1 − f1 ⊗ f2 − f2 ⊗ f1 + f2 ⊗ f2, f1 ⊗ f2 + f2 ⊗ f1 + f2 ⊗ f2〉|2 16 = 1 16 Also, the probability that a measured value (1, 1) is obtained by MB(C2⊗C2)(ΨÔgg, S[ρ]) is given by 〈u⊗ u, PĤgg({(1, 1)})P (u⊗ u)〉 = |〈(f1 + f2)⊗ (f1 + f2), f1 ⊗ f2 + f2 ⊗ f1 + f2 ⊗ f2〉|2 16 = |〈f1 ⊗ f1 + f1 ⊗ f2 + f2 ⊗ f1 + f2 ⊗ f2, f1 ⊗ f2 + f2 ⊗ f1 + f2 ⊗ f2〉|2 16 = 9 16 Further, the probability that a measured value (1, 2) is obtained by MB(C2⊗C2)(ΨÔgg, S[ρ]) is given by 〈u⊗ u, PĤgg({(1, 2)})P (u⊗ u)〉 330 Ishikawa's Homepage Chap. 11 Simple measurement and causality = |〈(f1 + f2)⊗ (f1 − f2), f1 ⊗ f2 + f2 ⊗ f1 + f2 ⊗ f2〉|2 16 = |〈f1 ⊗ f1 − f1 ⊗ f2 + f2 ⊗ f1 − f2 ⊗ f2, f1 ⊗ f2 + f2 ⊗ f1 + f2 ⊗ f2〉|2 16 = 1 16 Similarly, 〈u⊗ u, PĤgg({(2, 1)})P (u⊗ u)〉 = 1 16 Remark 11.19. Note that 1 16 + 9 16 + 1 16 + 1 16 = 3 4 < 1 which is due to the annihilation. Thus, the probability that no measured value is obtained by the measurement MB(C2⊗C2)(ΨÔ, S[ρ]) is equal to 1 4 . 11.7.2 The case that there is no half-mirror 2′ Consider the case that there is no half-mirror 2′, the case described in the following figure: D′1(= (|f ′2〉〈f ′2|)) (Detector) D′2(= (|f ′1〉〈f ′1|)) (Detector) ? 1√ 2 (f ′1 + f ′ 2) √ −1√ 2 f ′2 ? 1√ 2 f ′1 ? √ −1√ 2 f ′2 if no annihilation, 1√ 2 f ′1 half mirror 1′ mirror 2′ mirror 1′ course 2′ course 1′ Positron P′ D1(= (|g2〉〈g2|)) (Detector) D2(= (|g1〉〈g1|)) (Detector) 1√ 2 (f1+f2) −−−−−−→ 1√ 2 f1 ? √ −1√ 2 f2 ? if no annihilation, 1√ 2 f1 - √ −1√ 2 f2 half mirror 1 half mirror 2 Figure 11.5(2). Electron P and Positron P′ are annihilated at • mirror 1 mirror 2course 1 course 2 Electron P 331 Ishikawa's Homepage 11.7 Hardy's paradox: total probabilty is less than 1 Define the observable Ôgf = ({1, 2} × {1, 2}, 2{1,2}×{1,2}, Ĥgf ) in B(C2 ⊗ C2) by the tensor observable Og ⊗ Of , that is, Ĥgf ({(1, 1)}) = |g1 ⊗ f1〉〈g1 ⊗ f1|, Ĥgf ({(1, 2)}) = |g1 ⊗ f2〉〈g1 ⊗ f2|, Ĥgf ({(2, 1)}) = |g2 ⊗ f1〉〈g2 ⊗ f1|, Ĥgf ({(2, 2)}) = |g2 ⊗ f2〉〈g2 ⊗ f2| Since the causal operator Ψ : B(C2⊗C2)→ B(C2⊗C2) is the same, we get the measurement: MB(C2⊗C2)(ΨÔgf , S[ρ]) (11.32) Then, the probability that a measured value (2, 2) is obtained by MB(C2⊗C2)(ΨÔgf , S[ρ]) is given by 〈u⊗ u, PĤgf ({(2, 2)})P (u⊗ u)〉 = |〈(f1 − f2)⊗ f2, f1 ⊗ f2 + f2 ⊗ f1 + f2 ⊗ f2〉|2 8 = 0 Also, the probability that a measured value (1, 1) is obtained by MB(C2⊗C2)(ΨÔgf , S[ρ]) is given by 〈u⊗ u, PĤgf ({(1, 1)})P (u⊗ u)〉 = |〈(f1 + f2)⊗ f1, f1 ⊗ f2 + f2 ⊗ f1 + f2 ⊗ f2〉|2 8 = 1 8 Further, the probability that a measured value (1, 2) is obtained by MB(C2⊗C2)(ΨÔgf , S[ρ]) is given by 〈u⊗ u, PĤgf ({(1, 2)})P (u⊗ u)〉 = |〈(f1 + f2)⊗ f2, f1 ⊗ f2 + f2 ⊗ f1 + f2 ⊗ f2〉|2 16 = 4 8 Similarly, 〈u⊗ u, PĤgf ({(2, 1)})P (u⊗ u)〉 = |〈(f1 − f2)⊗ f1, f1 ⊗ f2 + f2 ⊗ f1 + f2 ⊗ f2〉|2 8 = 1 8 Remark 11.20. It is usual to consider that "Which way pass problem" is nonsense. It should be noted that, in the Heisenberg picture, the observable (= measuring instrument ) does not only include detectors but also mirrors. 332 Ishikawa's Homepage Chap. 11 Simple measurement and causality 11.8 quantum eraser experiment Let us explain quantum eraser experiment(cf. [84]). This section is extracted from (]) [45] S. Ishikawa, The double-slit quantum eraser experiments and Hardy's paradox in the quantum linguistic interpretation, arxiv:1407.5143[quantum-ph],( 2014) 11.8.1 Tensor Hilbert space Let C2 be the two dimensional Hilbert space, i,e., C2 = {[z1 z2 ] | z1, z2 ∈ C } . And put e1 = [ 1 0 ] , e2 = [ 0 1 ] Here, define the observable Ox = ({−1, 1}, 2{−1,1}, Fx) in B(C2) such that Fx({1}) = 1 2 [ 1 1 1 1 ] , Fx({−1}) = 1 2 [ 1 −1 −1 1 ] , Here, note that Fx({1})e1 = 1 2 (e1 + e2), Fx({1})e2 = 1 2 (e1 + e2) Fx({−1})e1 = 1 2 (e1 − e2), Fx({−1})e2 = 1 2 (−e1 + e2) Let H be a Hilbert space such that L2(R). And let O = (X,F, F ) be an observable in B(H). For example, consider the position observable, that is, X = R, F = BR, and [F (Ξ)](q) = { 1 (q ∈ Ξ ∈ F) 0 (q /∈ Ξ ∈ F) Let u1 and u2 (∈ H) be orthonormal elements, i.e., ‖u1‖H = ‖u2‖H = 1 and 〈u1, u2〉 = 0. Put u = α1u1 + α2u2 where αi ∈ C such that |α1|2 + |α2|2 = 1. Further, define ψ ∈ C2 ⊗H ( the tensor Hilbert space of C2 and H) such that ψ = α1e1 ⊗ u1 + α2e2 ⊗ u2 where αi ∈ C such that |α1|2 + |α2|2 = 1. 333 Ishikawa's Homepage 11.8 quantum eraser experiment 11.8.2 Interference Consider the measurement: MB(C2⊗H)(Ox ⊗ O, S[|ψ〉〈ψ|]) (11.33) Then, we see: (A1) the probability that a measured value (1, x)(∈ {−1, 1} ×X) belongs to {1} × Ξ is given by 〈ψ, (Fx({1})⊗ F (Ξ))ψ〉 =〈α1e1 ⊗ u1 + α2e2 ⊗ u2, (Fx({1} ⊗ F (Ξ)))(α1e1 ⊗ u1 + α2e2 ⊗ u2)〉 = 1 2 〈α1e1 ⊗ u1 + α2e2 ⊗ u2, α1(e1 + e2)⊗ F (Ξ)u1 + α2(e1 + e2)⊗ F (Ξ)u2〉 = 1 2 ( |α1|2〈u1, F (Ξ)u1〉+ |α2|2〈u2, F (Ξ)u2〉+ α1α2〈u1, F (Ξ)u2〉+ α1α2〈u2, F (Ξ)u1〉 ) = 1 2 ( |α1|2〈u1, F (Ξ)u1〉+ |α2|2〈u2, F (Ξ)u2〉+ 2[Real part](α1α2〈u1, F (Ξ)u2〉) ) where the interference term (i.e., the third term) appears. Define the probability density function p1 by∫ Ξ p1(q)dq = 〈ψ, (Fx({1})⊗ F (Ξ))ψ〉 〈ψ, (Fx({1})⊗ I)ψ〉 (∀Ξ ∈ F) Then, by the interference term (i.e., 2[Real part](α1α2〈u1, F (Ξ)u2〉) ), we get the following graph. q p1 Figure 11.6(1): The graph of p1 Also, we see: (A2) the probability that a measured value (−1, x)(∈ {−1, 1} × X) belongs to {−1} × Ξ is given by 〈ψ, (Fx({−1})⊗ F (Ξ))ψ〉 =〈α1e1 ⊗ u1 + α2e2 ⊗ u2, (Fx({−1} ⊗ F (Ξ)))(α1e1 ⊗ u1 + α2e2 ⊗ u2)〉 334 Ishikawa's Homepage Chap. 11 Simple measurement and causality = 1 2 〈α1e1 ⊗ u1 + α2e2 ⊗ u2, α1(e1 − e2)⊗ F (Ξ)u1 + α2(−e1 + e2)⊗ F (Ξ)u2〉 = 1 2 ( |α1|2〈u1, F (Ξ)u1〉+ |α2|2〈u2, F (Ξ)u2〉 − α1α2〈u1, F (Ξ)u2〉 − α1α2〈u2, F (Ξ)u1〉 ) = 1 2 ( |α1|2〈u1, F (Ξ)u1〉+ |α2|2〈u2, F (Ξ)u2〉 − 2[Real part](α1α2〈u1, F (Ξ)u2〉) ) where the interference term (i.e., the third term) appears. Define the probability density function p2 by∫ Ξ p2(q)dq = 〈ψ, (Fx({−1})⊗ F (Ξ))ψ〉 〈ψ, (Fx({−1})⊗ I)ψ〉 (∀Ξ ∈ F) Then, by the interference term (i.e., −2[Real part](α1α2〈u1, F (Ξ)u2〉) ), we get the following graph. q p2 Figure 11.6(2): The graph of p2 11.8.3 No interference Consider the measurement: MB(C2⊗H)(Ox ⊗ O, S[|ψ〉〈ψ|]) (11.34) Then, we see (A3) the probability that a measured value (u, x)(∈ {1,−1} × X) belongs to {1,−1} × Ξ is given by 〈ψ, (I ⊗ F (Ξ))ψ〉 =〈α1e1 ⊗ u1 + α2e2 ⊗ u2, (I ⊗ F (Ξ))(α1e1 ⊗ u1 + α2e2 ⊗ u2)〉 =〈α1e1 ⊗ u1 + α2e2 ⊗ u2, α1e1 ⊗ F (Ξ)u1 + α2e2 ⊗ F (Ξ)u2〉 =|α1|2〈u1, F (Ξ)u1〉+ |α2|2〈u2, F (Ξ)u2〉 where the interference term disappears. Define the probability density function p3 by∫ Ξ p3(q)dq = 〈ψ, (I ⊗ F (Ξ))ψ〉 (∀Ξ ∈ F) 335 Ishikawa's Homepage 11.8 quantum eraser experiment Since there is no interference term, we get the following graph. q p1 p2 p3 = p1 + p2 Figure 11.6(3): The graph of p3 = p1 + p2 Remark 11.21. Note that (A3) no interference = (A1)+(A2) interferences are canceled This was experimentally examined in [84]. 336 Ishikawa's Homepage Chapter 12 Realized causal observable in general theory Until the previous chapter, we studied all of quantum language, that is, (])  (]1): pure measurement theory (=quantum language) := [(pure)Axiom 1] pure measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } the manual to use spells (]2): mixed measurement theory (=quantum language) := [(mixed)Axiom(m) 1] mixed measurement (cf. §9.1) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } the manual to use spells As mentioned in the previous chapter, what is important is • to exercise the relationship of measurement and causality In this chapter, we discuss the relationship more systematically. 12.1 Finite realized causal observable In dualism (i.e., quantum language), Axiom 2 (Causality) is not used independently, but is always used with Axiom 1 (measurement), just as George Berkeley (A.D. 1685A.D.1753) said : (A1) To be is to be perceived. 337 12.1 Finite realized causal observable ♠Note 12.1. Note that Berkeley's words is opposite to Einstein's words: (]3) The moon is there whether one looks at it or not. in Einstein and Tagore's conversation. In this chapter, we devote ourselves to finite realized causal observable. ( For the infinite realized causal observable, see Chapter 14.) The readers should understand: • "realized causal observable" is a direct consequence of the linguistic interpretation, that is, Only one measurement is permitted. Now we shall review the following theorem: Theorem 12.1. [=Theorem 11.1:Causal operator and observable] Consider the basic structure: [Ak ⊆ Ak ⊆ B(Hk)] (k = 1, 2) Let Φ1,2 : A2 → A1 be a causal operator, and let O2 = (X,F, F2) be an observable in A2. Then, Φ1,2O2 = (X,F,Φ1,2F2) is an observable in A1. Proof. See the proof of Theorem 11.1 In this section, we consider the case that the tree ordered set T (t0) is finite. Thus, putting T (t0) = {t0, t1, . . . , tN}, consider the finite tree (T (t0), 5 ) with the root t0, which is represented by (T={t0, t1, . . . , tN}, π : T \ {t0} → T ) with the the parent map π. . Definition 12.2. [(finite)sequential causal observable] Consider the basic structure: [Ak ⊆ Ak ⊆ B(Hk)] (t ∈ T (t0) = {t0, t1, * * * , tn}) in which, we have a sequential causal operator {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 (cf. Definition 10.10 ) such that (i) for each (t1, t2) ∈ T 25, a causal operator Φt1,t2 : At2 → At1 satisfies that Φt1,t2Φt2,t3 = Φt1,t3 (∀(t1, t2), ∀(t2, t3) ∈ T 25). Here, Φt,t : At → At is the identity. 338 Ishikawa's Homepage Chap. 12 Realized causal observable in general theory [A0 : O0] [A1 : O1] [A2 : O2] [A3 : O3] [A4 : O4] [A5 : O5][A6 : O6] [A7 : O7] ) i k + k ) k Φ0,6 Φ0,1 Φ0,7 Φ1,2 Φ1,5 Φ2,3 Φ2,4 Figure 12.1 : Simple example of sequential causal observable For each t ∈ T , consider an observable Ot=(Xt,Ft, Ft) in At. The pair [{Ot}t∈T , {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 ] is called a sequential causal observable, denoted by [OT ] or [OT (t0)]. That is, [OT ] = [{Ot}t∈T , {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 ]. Using the parent map π : T \{t0} → T , [OT ] is also denoted by [OT ] = [{Ot}t∈T , {At Φπ(t),t−−−→ Aπ(t)}t∈T\{t0})]. Now we can show our present problem. Problem 12.3. We want to formulate the measurement of a sequential causal observable[OT ] = [{Ot}t∈T , {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 ] for a system S with an initial state ρt0(∈ S p(A∗t0)). How do we formulate this measurement? Now let us solve this problem as follows. Note that the linguistic interpretation says that only one measurement (and thus, only one observable) is permitted Thus, we have to combine many observables in a sequential causal observable[OT ] = [{Ot}t∈T , {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 ]. This is realized as follows. Definition 12.4. [Realized causal observable] Let T (t0) = {t0, t1, . . . , tN} be a finite tree. Let [OT (t0)] = [{Ot}t∈T , {Φπ(t),t : At Φπ(t),t−−−→ Aπ(t)}t∈T\{t0} ] be a sequential causal observable. For each s (∈ T ), put Ts = {t ∈ T | t = s}. Define the observable Ôs=(×t∈Ts Xt,  t∈TsFt, Fs) in As such that 339 Ishikawa's Homepage 12.1 Finite realized causal observable Ôs =  Os ( if s ∈ T \ π(T ) ) Os×(×t∈π−1({s}) Φπ(t),tÔt) ( if s ∈ π(T ) ) (12.1) (In quantum case, the existence of Ôs is not always guaranteed). And further, iteratively, we get the observable Ôt0 = (×t∈T Xt,  t∈TFt, Ft0) in At0 . Put Ôt0 = ÔT (t0). The observable ÔT (t0) = (×t∈T Xt,  t∈TFt, Ft0) is called the (finite) realized causal observable of the sequential causal observable[OT (t0)] = [{Ot}t∈T , {Φπ(t),t : At → Aπ(t)}t∈T\{t0} ]. Summing up the above arguments, we have the following theorem: In the classical case, the realized causal observable ÔT (t0) = (×t∈T Xt,  t∈TFt, Ft0) always exists. ♠Note 12.2. In the above (12.1), the product "×" may be generalized as the quasi-product " qp ×". However, in this note we are not concerned with such generalization. Example 12.5. [A simple classical example ] Suppose that a tree (T ≡ {0, 1, ..., 6, 7}, π) has an ordered structure such that π(1) = π(6) = π(7) = 0, π(2) = π(5) = 1, π(3) = π(4) = 2. [L∞(Ω0) : O0] [L∞(Ω1) : O1] [L∞(Ω2) : O2] [L∞(Ω3) : O3] [L∞(Ω4) : O4] [L∞(Ω5) : O5][L∞(Ω6) : O6] [L∞(Ω7) : O7] ) i k + k ) k Φ0,6 Φ0,1 Φ0,7 Φ1,2 Φ1,5 Φ2,3 Φ2,4 Figure 12.2 : Simple classical example of sequential causal observable Consider a sequential causal observable [OT ] = [{Ot}t∈T , {L∞(Ωt)Φπ(t),t→ L ∞(Ωπ(t))}t∈T\{0})]. Now, we shall construct its realized causal observable ÔT (t0) = (×t∈T Xt,  t∈TFt, Ft0) in what follows. Put Ôt = Ot and thus Ft = Ft (t = 3, 4, 5, 6, 7). First we construct the product observable Ô2 in L ∞(Ω2) such as Ô2 = (X2 ×X3 ×X4,F2  F3  F4, F2) where F2 = F2×( × t=3,4 Φ2,tFt), 340 Ishikawa's Homepage Chap. 12 Realized causal observable in general theory Iteratively, we construct the following: L∞(Ω0) Φ0,1←−−− L∞(Ω1)P Φ1,2←−−− L∞(Ω2) F0×Φ0,6F6×Φ0,7F7 F1×Φ1,5F5y y F0 (F0×Φ0,6F6×Φ0,7F7×Φ0,1F1) Φ0,1←−−− F1 (F1×Φ1,5F5×Φ1,2F2) Φ1,2←−−− F2 (F2×Φ2,3F3×Φ2,4F4) . That is, we get the product observable Ô1 ≡ (×5t=1Xt,  5t=1Ft, F1) of O1, Φ1,2Ô2 and Φ1,5Ô5, and finally, the product observable Ô0 ≡ (×7t=0Xt,  7t=0Ft, F0(= F0 × ( × t=1,6,7 Φ0,tFt)) of O0, Φ0,1Ô1, Φ0,6Ô6 and Φ0,7Ô7. Then, we get the realization of a sequential causal observable [{Ot}t∈T , {L∞(Ωt) Φπ(t),t→ L∞(Ωπ(t))}t∈T\{0}]. For completeness, F0 is represented by F0(Ξ0 × Ξ1 × Ξ2 × Ξ3 × Ξ4 × Ξ5 × Ξ6 × Ξ7)] =F0(Ξ0)× Φ0,1 ( F1(Ξ1)× Φ1,5F5(Ξ5)× Φ1,2 ( F2(Ξ2)× Φ2,3F3(Ξ3)× Φ2,4F4(Ξ4) )) × Φ0,6(F6(Ξ6))× Φ0,7(F7(Ξ7)) (12.2) (In quantum case, the existence of Ô0 in not guaranteed). Remark 12.6. In the above example, consider the case that Ot (t = 2, 6, 7) is not determined. In this case,it suffices to define Ot by the existence observable O (exi) t =(Xt, {∅, Xt}, F (exi) t ). Then, we see that F0(Ξ0 × Ξ1 ×X2 × Ξ3 × Ξ4 × Ξ5 ×X6 ×X7) =F0(Ξ0)× Φ0,1 ( F1(Ξ1)× Φ1,5F5(Ξ5)× Φ1,2 ( Φ2,3F3(Ξ3)× Φ2,4F4(Ξ4) )) (12.3) This is true. However, the following is not wrong. Putting T ′ = {0, 1, 3, 4, 5}, consider the [OT ′ ] = [{Ot}t∈T ′ , {Φt1,t2 : L∞(Ωt2) → L∞(Ωt1)}(t1,t2)∈(T ′)25 ]. Then, the realized causal observable ÔT ′(0) = (×t∈T ′ Xt,  t∈T ′Ft, F ′0) is defined by F ′0(Ξ0 × Ξ1 × Ξ3 × Ξ4 × Ξ5) = F0(Ξ0) × Φ0,1 ( F1(Ξ1)× Φ1,5F5(Ξ5)× Φ1,4F4(Ξ4)× Φ1,3F3(Ξ3)× Φ1,4F4(Ξ4) ) (12.4) which is different from the true (12.2). We may sometimes omit "existence observable". However, if we do so, we omit it on the basis of careful cautions. 341 Ishikawa's Homepage 12.1 Finite realized causal observable Thus, we can answer Problem 12.3 as follows. Problem 12.7. [=Problem 12.3] (written again) We want to formulate the measurement of a sequential causal observable[OT ] = [{Ot}t∈T , {Φt1,t2 : At2 → At1}(t1,t2)∈T 25 ] for a system S with an initial state ρt0(∈ S p(A∗t0)). How do we formulate the measurement ? Answer: If the realized causal observable Ôt0 exists, the measurement is formulated by measurement MAt0 (Ôt0 , S[ρt0 ]) Thus, according to Axiom 1 ( measurement: §2.7), we see that (A) The probability that a measured value (xt)t∈T obtained by the measurement MAt0 (ÔT , S[ρt0 ]) belongs to Ξ(∈  t∈TFt) is given by A∗0 ( ρt0 , Ft0(Ξ) ) At0 (12.5) The following theorem, which holds in classical systems, is frequently used. Theorem 12.8. [The realized causal observable of deterministic sequential causal observable in classical systems ] Let (T (t0), 5 ) be a finite tree. For each t ∈ T (t0), consider the classical basic structure [C0(Ωt) ⊆ L∞(Ωt, νt) ⊆ B(L2(Ωt, νt))] Let [OT ] = [{Ot}t∈T , {Φt1,t2 : L∞(Ωt2)→ L∞(Ωt1)}(t1,t2)∈T 25 ] be deterministic causal observable. Then, the realization Ôt0 ≡ (×t∈TXt,  t∈TFt, Ft0) is represented by Ôt0 = × t∈T Φt0,tOt That is, it holds that [Ft0(× t∈T Ξt )](ωt0) = × t∈T [Φt0,tFt(Ξt)](ωt0) = × t∈T [Ft(Ξt)](φt0,tωt0) (∀ωt0 ∈ Ωt0 ,∀Ξt ∈ Ft) 342 Ishikawa's Homepage Chap. 12 Realized causal observable in general theory Proof. It suffices to prove the simple classical case of Example 12.5. Using Theorem 10.6 repeatedly, we see that F0 = F0 × ( × t=1,6,7 Φ0,tFt) =F0 × (Φ0,1F1 × Φ0,6F6 × Φ0,7F7) = F0 × (Φ0,1F1 × Φ0,6F6 × Φ0,7F7) = ( × t=0,6,7 Φ0,tFt ) × (Φ0,1F1) = ( × t=0,6,7 Φ0,tFt ) × Φ0,1(F1 × ( × t=2,5 Φ1,tFt)) = ( × t=0,1,6,7 Φ0,tFt ) × Φ0,1( × t=2,5 Φ1,tFt) = ( × t=0,1,6,7 Φ0,tFt ) × Φ0,1(Φ1,2F2 × Φ1,5F5) = ( × t=0,1,5,6,7 Φ0,tFt ) × Φ0,1(Φ1,2F2) = ( × t=0,1,5,6,7 Φ0,tFt ) × Φ0,1(Φ1,2(F2 × ( × t=3,4 Φ2,tFt))) = 7 × t=0 Φ0,tFt This completes the proof. 343 Ishikawa's Homepage 12.2 Double-slit experiment and projection postulate 12.2 Double-slit experiment and projection postulate 12.2.1 Interference For each t ∈ T = [0,∞), define the quantum basic structure [C(Ht) ⊆ B(Ht) ⊆ B(Ht)], where Ht = L 2(R2) (∀t ∈ T ). Let u0 ∈ H0 = L2(R2) be an initial wave-function such that (k0 > 0, small σ > 0): u0(x, y) ≈ ψx(x, 0)ψy(y, 0) = 1√ π1/2σ exp ( ik0x− x2 2σ2 ) * 1√ π1/2σ exp ( − y 2 2σ2 ) , where the average momentum (p01, p 0 2) is calculated by (p01, p 0 2) = (∫ R ψx(x, 0) * ~∂ψx(x, 0) i∂x dx, ∫ R ψy(y, 0) * ~∂ψy(y, 0) i∂y dy ) = (~k0, 0). That is, we assume that the initial state of the particle P is equal to |u0〉〈u0|. Picture 12.9. MB(H0)(Φ0,t2O2 = (R,BR,Φ0,t2F2), S[|u0〉〈u0|]) 6 6 x y y ρ1(y)P • → a b A u↑1 B u↓1 t = 0 t = t1 t = t2 Figure 12.3(1) Potential V (x, y) =∞ on the thick line, = 0 (elsewhere) Thus, we have the following Schrödinger equation: i~ ∂ ∂t ut(x, y) = Hut(x, y), H = − ~2 2m ∂2 ∂x2 − ~ 2 2m ∂2 ∂y2 + V (x, y) Let s, t be 0 < s < t < ∞. Thus, we have the causal relation: {Φs,t : B(Ht) → B(Hs)}0<s<t<∞ where Φs,tA = e H(t−s) i~ Ae− H(t−s) i~ (∀A ∈ B(Ht) = B(L2(R2))) 344 Ishikawa's Homepage Chap. 12 Realized causal observable in general theory Thus, (Φ0,t1)∗(u0) = u ↑ 1 + u ↓ 1 in Picture 12.9. Let O2 = (R,BR, F2) be the position observable in B(L2(R2) such that [F (Ξ)](x, y) = χΞ(y) =  1 (x, y) ∈ R× Ξ 0 (x, y) ∈ R× R \ Ξ Hence, we have the measurement MB(H0)(Φ0,t2O2 = (R,BR,Φ0,t2F2), S[|u0〉〈u0|]). Axiom 1 ( measurement: §2.7) says that (A) the probability that a measured value a ∈ R by MB(H0)(Φ0,t2O, S|u0〉〈u0|) belongs to (−∞, y] is given by 〈u0, (Φ0,t2F ((−∞, y]))u0〉 = ∫ y −∞ ρ1(y)dy ♠Note 12.3. Precisely speaking, we say as follows. Let ∆, ε be small positive real numbers. For each k ∈ Z = {k | k = 0,±1,±2,±3, , , , , }, define the rectangle Dk such that D0 = {(x, y) ∈ R2 | x < b}, Dk = {(x, y) ∈ R2 | b ≤ x, (k − 1)∆ < y ≤ k∆}, k = 1, 2, 3, ... Dk = {(x, y) ∈ R2 | b ≤ x, k∆ < y ≤ (k + 1)∆}, k = −1,−2,−3, ... Thus we have the projection observable O∆2 = (Z, 2Z, F∆2 ) in L2(R2) such that [F ({k})](x, y) = 1 ((x, y) ∈ Dk), = 0 ((x, y) ∈ R2 \Dk) (k ∈ Z) Then it suffices to consider • for each time tn = t2 + nε(n = 0, 1, 2, ...), the projection observable O∆2 is measured in the sense of Projection Postulate 11.6. 12.2.2 Which-way path experiment Picture 12.10. Which-way path experiment: A measured value by MB(L2(R2))(Φ0,t1(Ψ(OG⊗ Φt1,t2O2)), S[|u0〉〈u0|]) belongs to {↑} × (−∞, y] 345 Ishikawa's Homepage 12.2 Double-slit experiment and projection postulate 6 6 x y y ρ2(y)P • → a b A u↑1 B t = 0 t = t1 t = t2 Figure 12.3(2) Potential V (x, y) =∞ on the thick line, = 0 (elsewhere) Next, let us explain the above figure. Define the projection observable O1 = ({↑, ↓}, 2{↑,↓}, F1) in B(L2(R2)) such that [F1({↑})](x, y) = { 1 y ≥ 0 0 y < 0 [F1({↓})](x, y) = 1− [F1({↑})](x, y) According to Section 11.2 ( Projection postulate ), consider the CONS {e1, e2} (∈ C2). Define the predual operator Ψ∗ : Tr(L 2(R2))→ Tr(C2 ⊗ L2(R2)) such that Ψ∗(|u〉〈u|) = |(e1 ⊗ F1({↑})u) + (e2 ⊗ F1({↓})u)〉〈(e1 ⊗ F1({↑})u) + (e2 ⊗ F1({↓})u)| Then we have the causal operator Ψ : B(C2 ⊗ L2(R2))→ L2(R2) such that Ψ = (Ψ∗)∗. Define the observable OG = ({↑, ↓}, 2{↑,↓}, G) in B(C2) such that G({↑}) = |e1〉〈e1|, G({↓}) = |e2〉〈e2| Hence we have the tensor observable OG⊗Φt1,t2O2 in B(C2⊗L2(R2)), and hence, the measurement MB(L2(R2))(Φ0,t1(Ψ(OG ⊗ Φt1,t2O2)), S[|u0〉〈u0|]). Then, Axiom 1 ( measurement: §2.7) says that (B) the probability that a measured value (λ, y) ∈ {↑, ↓} × R by MB(L2(R2))(Φ0,t1(Ψ(OG ⊗ Φt1,t2O2)), S[|u0〉〈u0|]) belongs to {↑} × (−∞, y] is given by 〈u↑1, (Φt1,t2F2((−∞, y]))u ↑ l 〉 = 1 2 ∫ y −∞ ρ2(y)dy 346 Ishikawa's Homepage Chap. 12 Realized causal observable in general theory ♠Note 12.4. Precisely speaking, in the above case, it suffices to consider the following procedure (1) and (ii): (i) for time t1, the projection observable O1 is measured in the sense of Projection Postulate 11.6 (ii) for each time tn = t2 + nε(n = 0, 1, 2, ...), the projection observable O ∆ 2 is measured in the sense of Projection Postulate 11.6. 347 Ishikawa's Homepage 12.3 Wilson cloud chamber in double slit experiment 12.3 Wilson cloud chamber in double slit experiment In this section, we shall analyze a discrete trajectory of a quantum particle, which is assumed one of the models of the Wilson cloud chamber ( i.e., a particle detector used for detecting ionizing radiation). The main idea is due to. [24, 25, (1991, 1994, S. Ishikawa, et al.)]. 12.3.1 Trajectory of a particle is non-sense We shall consider a particle P in the one-dimensional real line R, whose initial state function is u(x) ∈ H = L2(R). Since our purpose is to analyze the discrete trajectory of the particle in the double-slit experiment, we choose the state u(x) as follows: u(x) =  l/ √ 2, x ∈ (−3/2,−1/2) ∪ (1/2, 3/2) 0, otherwise (12.6) 0 1/ √ 2 6 -3/2 -1/2 1/2 3/2 x Figure 12.4 The initial wave function u(x) Let A0 be a position observable in H, that is, (A0v)(x) = xv(x) (∀x ∈ R, ( for v ∈ H = L2(R) which is identified with the observable O = (R,BR, EA0) defined by the spectral representation: A0 =∫ R xEA0(dx). We treat the following Heisenberg's kinetic equation of the time evolution of the observable A, (−∞ < t <∞) in a Hilbert space H with a Hamiltonian H such that H = −(~2/2m)∂2/∂x2 (i.e., the potential V (x) = 0), that is, −i~dAt dt = HAt −AtH, −∞ < t <∞, where A0 = A (12.7) The one-parameter unitary group Ut is defined by exp(−itA). An easy calculation shows that At = U ∗ t AUt = U ∗ t xUt = x+ ~t im d dx (12.8) Put t = 1/4, ~/m = 1. And put A = A0(= x), B = A1/4(= x+ 1 4i d dx ) = U∗1/4A0U1/4 = Φ0,1/4A0 Thus, we have the sequential causal observable position observable: A0 B(H0) initial wave function:u0 ←−−−−−− Φ0,1/4 position observable: A0 B(H1/4) 348 Ishikawa's Homepage Chap. 12 Realized causal observable in general theory However, A0(= A) and Φ0,1/4A0(= B) do not commute, that is, we see: AB −BA = x(x+ 1 4i d dx )− (x+ 1 4i d dx )x = i/4 6= 0 Therefore, the realized causal observable does not exist. In this sense, the trajectory of a particle is non-sense 12.3.2 Approximate measurement of trajectories of a particle In spite of this fact, we want to consider "trajectories" as follows. That is, we consider the approximate simultaneous measurement of self-adjoint operators {A,B} for a particle P with an initial state u(x). Recall Definition 4.13, that is, Definition 12.11. (=Definition 4.13). The quartet (K, s, Â, B) is called an approximately simultaneous observable of A and B, if it satisfied that (A1) K is a Hilbert space. s ∈ K, ‖s‖K = 1, Â and B are commutative self-adjoint operators on a tensor Hilbert space H ⊗K that satisfy the average value coincidence condition, that is, 〈u⊗ s, Â(u⊗ s)〉 = 〈u,Au〉, 〈u⊗ s, B(u⊗ s)〉 = 〈u,Bu〉 (12.9) (∀u ∈ H, ‖u‖H = 1) Also, the measurement MB(H⊗K)(OÂ×OB, S[ρus]) is called the approximately simultaneous measurement of MB(H)(OA, S[ρu]) and MB(H)(OB, S[ρu]), where ρus = |u⊗ s〉〈u⊗ s| (‖s}K = 1) And we define that (A2) ∆ ρus N1 (= ‖(Â − A ⊗ I)(u ⊗ s)‖) and ∆ρus N2 (= ‖(B − B ⊗ I)(u ⊗ s)‖) are called errors of the approximate simultaneous measurement measurement MB(H⊗K)(OÂ × OB, S[ρus]) Now, let us constitute the approximately observable (K, s, Â, B) as follows. Put K = L2(Ry), s(y) == (ω1 π )1/4 exp ( − ω1|y| 2 2 ) where ω1 is assumed to be ω1 = 4, 16, 64 later. It is easy to show that ‖s‖L2(Ry) = 1 (i.e., ‖s‖K = 1 ) and 〈s,As〉 = 〈s,Bs〉 = 0 (12.10) And further, put Â = A⊗ I + 2I ⊗A B = B ⊗ I − 1 2 I ⊗B 349 Ishikawa's Homepage 12.3 Wilson cloud chamber in double slit experiment Note that the two commute (i.e., ÂB = BÂ ). Also, we see, by (12.10), 〈u⊗ s, Â(u⊗ s)〉 = 〈u⊗ s, (A⊗ I + 2I ⊗A)(u⊗ s)〉 = 〈u,Au〉 (12.11) 〈u⊗ s, Â(u⊗ s)〉 = 〈u⊗ s, (B ⊗ I − 2I ⊗A)(u⊗ s)〉 = 〈u,Bu〉 (12.12) (∀u ∈ H, i = 1, 2) Thus, we have the approximately simultaneous measurementMB(H⊗K)(OÂ×OB, S[ρus]), and the errors are calculated as follows: δ0 = ∆ ρus N1 = ‖(Â−A⊗ I)(u⊗ s)‖ = ‖2(I ⊗A)(u⊗ s)‖ = 2‖As‖ (12.13) δ1/4 = ∆ ρus N2 = ‖(B −B ⊗ I)(u⊗ s)‖ = (1/2)‖(I ⊗B)(u⊗ s)‖ = (1/2)‖Bs‖ (12.14) 350 Ishikawa's Homepage Chap. 12 Realized causal observable in general theory By the parallel measurement ⊗N k=1MB(H⊗K)(OÂ × OB, S[ρus]), assume that a measured value:( (x1, x ′ 1), (x2, x ′ 2), * * * , (xN , x′N ) ) is obtained. This is numerically calculated as follows. Figure 12.5: The lines connecting two points (i.e., xk and x ′ k) (k = 1, 2, ...) Here, note that δθ(= δ1/4) and δ0 are depend on ω1. ♠Note 12.5. For the further arguments, see the following refs. 351 Ishikawa's Homepage 12.3 Wilson cloud chamber in double slit experiment (]1) [24]: S. Ishikawa, Uncertainties and an interpretation of nonrelativistic quantum theory, International Journal of Theoretical Physics 30, 401–417 (1991) doi: 10.1007/BF00670793 (]2) [25]: Ishikawa, S., Arai, T. and Kawai, T. Numerical Analysis of Trajectories of a Quantum Particle in Two-slit Experiment, International Journal of Theoretical Physics, Vol. 33, No. 6, 1265-1274, 1994 doi: 10.1007/BF00670793 352 Ishikawa's Homepage Chap. 12 Realized causal observable in general theory 12.4 Two kinds of absurdness - idealism and dualism This section is extracted from ref. [39]. Measurement theory (= quantum language ) has two kinds of absurdness. That is, (]) Two kinds of absurdness  idealism* * *linguistic world-view The limits of my language mean the limits of my world dualism * * *Descartes=Kant philosophy The dualistic description for monistic phenomenon In what follows, we explain these. 12.4.1 The linguistic interpretation - A spectator does not go up to the stage Problem 12.12. [A spectator does not go up to the stage] Consider the elementary problem with two steps (a) and (b): (a) Consider an urn, in which 3 white balls and 2 black balls are. And consider the following trial: • Pick out one ball from the urn. If it is black, you return it in the urn If it is white, you do not return it and have it. Assume that you take three trials. . (b) Then, calculate the probability that you have 2 white ball after (a)(i.e., three trials). Answer Put N0 = {0, 1, 2, . . .} with the counting measure. Assume that there are m white balls and n black balls in the urn. This situation is represented by a state (m,n) ∈ N20. We can define the dual causal operator Φ∗ : M+1(N20) →M+1(N20) such that Φ∗(δ(m,n)) = { m m+nδ(m−1,n) + n m+nδ(m,n) (when m 6= 0 ) δ(0,n) (when m = 0 ). (12.15) where δ(*) is the point measure. Let T = {0, 1, 2, 3} be discrete time. For each t ∈ T , put Ωt = N20. Thus, we see: [Φ∗]3(δ(3,2)) = [Φ ∗]2 ( 3 5 δ(2,2) + 2 5 δ(3,2) ) =Φ∗ ( ( 3 5 ( 2 4 δ(1,2) + 2 4 δ(2,2)) + 2 5 ( 3 5 δ(2,2) + 2 5 δ(3,2)) ) =Φ∗ ( 3 10 δ(1,2) + 27 50 δ(2,2) + 4 25 δ(3,2) ) = 3 10 ( 1 3 δ(0,2) + 2 3 δ(1,2)) + 27 50 ( 2 4 δ(1,2) + 2 4 δ(2,2)) + 4 25 ( 3 5 δ(2,2) + 2 5 δ(3,2)) = 1 10 δ(0,2) + 47 100 δ(1,2) + 183 500 δ(2,2) + 8 125 δ(3,2) (12.16) Define the observable O = (N0, 2N0 , F ) in L∞(Ω3) such that [F (Ξ)](m,n) = { 1 (m,n) ∈ Ξ× N0 ⊆ Ω3 0 (m,n) /∈ Ξ× N0 ⊆ Ω3 353 Ishikawa's Homepage 12.4 Two kinds of absurdness - idealism and dualism Therefore, the probability that a measured value "2" is obtained by the measurement ML∞(N20)(Φ 3O, S[(3,2)]) is given by [Φ3(F ({2}))](3, 2) = ∫ Ω3 [F ({2})](ω)([Φ∗]3(δ(3,2)))(dω) = 183 500 (12.17) The above may be easy, but we should note that (c) the part (a) is related to causality, and the part (b) is related to measurement. Thus, the observer is not in the (a). Figuratively speaking, we say: A spectator does not go up to the stage Thus, someone in the (a) should be regard as "robot". ♠Note 12.6. The part (a) is not related to "probability". That is because The spirit of measurement theory says that there is no probability without measurements. although something like "probability" in the (a) is called "Markov probability". 12.4.2 In the beginning was the words-Fit feet to shoes Remark 12.13. [The confusion between measurement and causality ( Continued from Example2.31)] Recall Example2.31 [The measurement of "cold or hot" for water]. Consider the measurement ML∞(Ω)(Och, S[ω]) where ω = 5( ◦C). Then we say that (a) By the measurement ML∞(Ω)(Och, S[ω(=5)]), the probability that a measured value x(∈ X = {c, h}) belongs to a set  ∅(= empty set) {c} {h} {c,h}  is equal to  0 [F ({c})](5) = 1 [F ({h})](5) = 0 1  Here, we should not think: "5 ◦C" is the cause and "cold" is a result. That is, we never consider that (b) 5 ◦C (cause) −→ cold (result) That is because Axiom 2 (causality; §10.3) is not used in (a), though the (a) may be sometimes regarded as the causality (b) in ordinary language. 354 Ishikawa's Homepage Chap. 12 Realized causal observable in general theory ♠Note 12.7. However, from the different point of view, the above (b) can be justified as follows. Define the dual causal operator Φ∗ : M([0, 100])→M({c, h}) by [Φ∗δω](D) = fc(ω) * δC(D) + fh(ω) * δH(D) (∀ω ∈ [0, 100], ∀D ⊆ {c, h}) Then, the (b) can be regarded as "causality". That is, (]) "measurement or causality" depends on how to describe a phenomenon. This is the linguistic world-description method. Remark 12.14. [Mixed measurement and causality ] Reconsider Problem 9.5(urn problem:mixed measurement). That is, consider a state space Ω = {ω1, ω2}, and define the observable O = ({w, b}, 2{w,b}, F ) in L∞(Ω) in Problem 9.5. Define the mixed state by ρm = pδω1 + (1 − p)δω2 . Then the probability that a measured value x ( ∈ {w, b}) is obtained by the mixed measurement ML∞(Ω)(O, S[∗](ρ m)) is, by (9.3), given by P ({x}) = ∫ Ω [F ({x})](ω)ρm(dω) = p[F ({x})](ω1) + (1− p)[F ({x})](ω2) = { 0.8p+ 0.4(1− p) (when x = w ) 0.2p+ 0.6(1− p)) (when x = b ) (12.18) Now, define a new state space Ω0 by Ω0 = {ω0}. And define the dual (non-deterministic) causal operator Φ∗ : M+1(Ω0) →M+1(Ω) by Φ∗(δω0) = pδω1 + (1− p)δω2 . Thus, we have the (non-deterministic) causal operator Φ : L∞(Ω)→ L∞(Ω0). Here, consider a pure measurement ML∞(Ω0)(ΦO, S[ω0]). Then, the probability that a measured value x ( ∈ {w, b}) is obtained by the measurement is given by P ({x}) = [Φ(F ({x}))](ω0) = ∫ Ω [F ({x})](ω)ρm(dω) = { 0.8p+ 0.4(1− p) (when x = w ) 0.2p+ 0.6(1− p)) (when x = b ) which is equal to the (12.18). Therefore, the mixed measurement ML∞(Ω)(O, S[∗](ν0)) can be regarded as the pure measurement ML∞(Ω0)(ΦO, S[ω0]). ♠Note 12.8. In the above arguments, we see that (]) Concept depends on the description This is the linguistic world-description method. As mentioned frequently, we are not concerned with the question "what is ©©?". The reason is due to this (]). "Measurement or Causality" depends on the description. Some may recall Nietzsche's famous saying: There are no facts, only interpretations. This is just the linguistic world-description method with the spirit: "Fit feet (=world) to shoes (language)". 355 Ishikawa's Homepage 12.4 Two kinds of absurdness - idealism and dualism ♠Note 12.9. In the book "The astonishing hypothesis" ([11] by F. Click (the most noted for being a co-discoverer of the structure of the DNA molecule in 1953 with James Watson)), Dr. Click said that (a) You, your joys and your sorrows, your memories and your ambitions,your sense of personal identity and free will,are in fact no more than the behavior of a vast assembly of nerve cells and their associated molecules. It should be note that this (a) and the dualism do not contradict. That is because quantum language says: (b) Describe any monistic phenomenon by the dualistic language (= quantum language )! Also, if the above (a) is due to David Hume, he was a scientist rather than a philosopher. 356 Ishikawa's Homepage Chapter 13 Fisher statistics (II) Measurement theory (= quantum language ) is formulated as follows. • measurement theory (=quantum language) := [Axiom 1] Measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } manual to use spells In Chapter 5 (Fisher statistics (I)), we discuss "inference" in the relation of "measurement". In this chapter, we discuss "inference" in the relation of "measurement" and "causality". Thus, we devote ourselves to regression analysis. This chapter is extracted from the following: (]) Ref. [30]: S. Ishikawa, "Mathematical Foundations of Measurement Theory," Keio University Press Inc. 2006. 13.1 "Inference" = "Control" It is usually considered that{ • statistics is closely related to inference • dynamical system theory is closely related to control However, in this chapter, we show that "inference" = "control" In this sense, we conclude that statistics and dynamical system theory are essentially the same. 13.1.1 Inference problem(statistics) 357 13.1 "Inference" = "Control" Problem 13.1. [Inference problem and regression analysis] Let Ω ≡ {ω1, ω2, ..., ω100} be a set of all students of a certain high school. Define h : Ω→ [0, 200] and w : Ω→ [0, 200] such that: h(ωn) = "the height of a student ωn" (n = 1, 2, ..., 100) w(ωn) = "the weight of a student ωn" (n = 1, 2, ..., 100) (13.1) For simplicity, put, N = 5. For example, see Table 13.1. Table 13.1: Height and weight Height* Weight Student ω1 ω2 ω3 ω4 ω5 Height (h(ω) cm) 150 160 165 170 175 Weight(w(ω) kg) 65 55 75 60 65 ω h(ω) w(ω) Ω 0 100 200 0 100 200 Assume that: (a1) The principal of this high school knows the both functions h and w. That is, he knows the exact data of the height and weight concerning all students. Also, assume that: (a2) Some day, a certain student helped a drowned girl. But, he left without reporting the name. Thus, all information that the principal knows is as follows: (i) he is a student of his high school. (ii) his height [resp. weight] is about 170 cm [resp. about 80 kg]. Now we have the following question: (b) Under the above assumption (a1) and (a2), how does the principal infer who is he? This will be answered in Answer 13.5. 358 Ishikawa's Homepage Chap. 13 Fisher statistics (II) 13.1.2 Control problem(dynamical system theory) Adding the measurement equation g : R3 → R to the state equation, we have dynamical system theory(13.2). That is, dynamical system theory =  (i) : dω(t)dt = v(ω(t), t, e1(t), β) (initialω(0)=α) * * * ( state equation) (ii) : x(t) = g(ω(t), t, e2(t)) * * * ( measurement) (13.2) where α, β are parameters, e1(t) is noise, e2(t) is measurement error. The following example is the simplest problem concerning inference. Problem 13.2. [Control problem and regression analysis] We have a rectangular water tank filled with water. h(t) ? 6 Figure 13.1: Water tank Assume that the height of water at time t is given by the following function h(t): dh dt = β0, then h(t) = α0 + β0t, (13.3) where α0 and β0 are unknown fixed parameters such that α0 is the height of water filling the tank at the beginning and β0 is the increasing height of water per unit time. The measured height hm(t) of water at time t is assumed to be represented by hm(t) = α0 + β0t+ e(t), where e(t) represents a noise (or more precisely, a measurement error) with some suitable conditions. And assume that we obtained the measured data of the heights of water at t = 1, 2, 3 as follows: hm(1) = 1.9, hm(2) = 3.0, hm(3) = 4.7. (13.4) Under this setting, we consider the following problem: 359 Ishikawa's Homepage 13.1 "Inference" = "Control" (c1) [Control]: Settle the state (α0, β0) such that measured data (13.4) will be obtained. or, equivalently, (c2) [Inference]: when measured data (13.4) is obtained, infer the unknown state (α0, β0). This will be answered in Answer 13.6. Note that (c1)=(c2) from the theoretical point of view. Thus we consider that (d) Inference problem and control problem are the same problem. And these are characterized as the reverse problem of measurements. Remark 13.3. [Remark on dynamical system theory (cf. [30]) ] Again recall the formulation (13.2) of dynamical system theory, in which (]) the noise e1(t) and the measurement error e2(t) have the same mathematical structure (i.e., stochastic processes ). This is a weak point of dynamical system theory. Since the noise and the measurement error are different, I think that the mathematical formulations should be different. In fact, the confusion between the noise and the measurement error frequently occur. This weakness is clarified in quantum language, as shown in Answer 13.6. 360 Ishikawa's Homepage Chap. 13 Fisher statistics (II) 13.2 Regression analysis According to Fisher's maximum likelihood method (Theorem5.6) and the existence theorem of the realized causal observable, we have the following theorem: Theorem 13.4. [Regression analysis (cf. [30]) ] Let (T={t0, t1, . . . , tN}, π : T \ {t0} → T ) be a tree. Let ÔT =(×t∈T Xt,  t∈TFt, Ft0) be the realized causal observable of a sequential causal observable [{Ot}t∈T , {Φπ(t),t : L∞(Ωt)→ L∞(Ωπ(t))}t∈T\{t0} ]. Consider a measurement ML∞(Ωt0 )(ÔT=(×t∈T Xt,  t∈TFt, Ft0), S[∗]) Assume that a measured value obtained by the measurement belongs to Ξ (∈  t∈TFt). Then, there is a reason to infer that [ ∗ ] = ωt0 where ωt0 (∈ Ωt0) is defined by [Ft0(Ξ)](ωt0) = max ω∈Ωt0 [Ft0(Ξ)](ω) The poof is a direct consequence of Axiom 2 (causality; §10.3) and Fisher maximum likelihood method (Theorem 5.6). Thus, we omit it. It should be noted that (]) regression analysis is related to Axiom 1 (measurement; §2.7) and Axiom 2 (causality; §10.3) Now we shall answer Problem13.1 in terms of quantum language, that is, in terms of regression analysis (Theorem13.4). Answer 13.5. [(Continued from Problem13.1(Inference problem))Regression analysis] Let (T= {0, 1, 2}, π : T \ {0} → T ) be the parent map representation of a tree, where it is assumed that π(1) = π(2) = 0 Put Ω0 = {ω1, ω2, . . . , ω5}, Ω1 = interval[100, 200], Ω2 = interval[30, 110]. Here, we consider that Ω0 3 ωn * * * * * * a state such that "the girl is helped by a student ωn" (n = 1, 2, ..., 5) For each t (∈ {1, 2}), the deterministic map φ0,t : Ω0 → Ωt is defined by φ0,1 = h(height function), φ0,2 = w(weight function). Thus, for each t (∈ {1, 2}), the deterministic causal operator Φ0,t : L ∞(Ωt)→ L∞(Ω0) is defined by [Φ0,tft](ω) = ft(φ0,t(ω)) (∀ω ∈ Ω0, ∀ft ∈ L∞(Ωt)) 361 Ishikawa's Homepage 13.2 Regression analysis L∞(Ω1) L∞(Ω0) L∞(Ω2) + k Φ0,1 Φ0,2 For each t = 1, 2, let OGσt=(R,BR, Gσt) be the normal observable with a standard deviation σt > 0 in L ∞(Ωt). That is, [Gσt(Ξ)](ω) = 1√ 2πσ2t ∫ Ξ e − (x−ω) 2 2σ2t dx (∀Ξ ∈ BR,∀ω ∈ Ωt) Thus, we have a deterministic sequence observable [{OGσt}t=1,2, {Φ0,t : L ∞(Ωt)→ L∞(Ω0)}t=1,2]. Its realization ÔT = (R2,FR2 , F0) is defined by [F0(Ξ1 × Ξ2)](ω) = [Φ0,1Gσ1 ](ω) * [Φ0,2Gσ2 ](ω) = [Gσ1(Ξ1)](φ0,1(ω)) * [Gσ2(Ξ2)](φ0,2(ω)) (∀Ξ1,Ξ2 ∈ BR, ∀ω ∈ Ω0 = {ω1, ω2, . . . , ω5}) Let N be sufficiently large. Define intervals Ξ1,Ξ2 ⊂ R by Ξ1 = [ 165− 1 N , 165 + 1 N ] , Ξ2 = [ 65− 1 N , 65 + 1 N ] The measured data obtained by a measurement ML∞(Ω0)(ÔT , S[∗]) is (165, 65) (∈ R2) Thus, measured value belongs to Ξ1×Ξ2. Using regression analysis ( Theorem 13.4) is characterized as follows: (]) Find ω0 (∈ Ω0) such as [F0({Ξ1 × Ξ2)](ω0) = max ω∈Ω [F0({Ξ1 × Ξ2)](ω) Since N is sufficiently large, (]) =⇒max ω∈Ω0 1√ (2π)2σ21σ 2 2 ∫ ∫ Ξ1×Ξ2 exp [− (x1 − h(ω)) 2 2σ21 − (x2 − w(ω)) 2 2σ22 ]dx1dx2 =⇒max ω∈Ω0 exp [− (165− h(ω)) 2 2σ21 − (65− w(ω)) 2 2σ22 ] =⇒ min ω∈Ω0 [ (165− h(ω))2 2σ21 + (65− w(ω))2 2σ22 ] ( for simplicity, assume that σ1 = σ2) 362 Ishikawa's Homepage Chap. 13 Fisher statistics (II) =⇒When ω4, minimum value (165− 170)2 + (65− 60)2 2σ21 is obtained =⇒The student is ω4 Therefore, we can infer that the student who helps the girl is ω4. Now, let us answer Problem 13.2 in terms of quantum language (or, by using regression analysis (Theorem13.4)). Answer 13.6. [(Continued from Problem 13.2(Control problem))Regression analysis] In Problem 13.2, it is natural to consider that the tree T = {0, 1, 2, 3} is discrete time, that is, the linear ordered set with the parent map π : T \ {0} → T such that π(t) = t − 1 (t = 1, 2, 3). For example, put Ω0 = [0, 1]× [0, 2], Ω1 = [0, 4]× [0, 2], Ω2 = [0, , 6]× [0, 2], Ω3 = [0, 8]× [0, 2] For each t = 1, 2, 3, define the deterministic causal map φπ(t),t : Ωπ(t) → Ωt by (13.3), that is, φ0,1(ω0) = (α + β, β) (∀ω0 = (α, β) ∈ Ω0 = [0, 1]× [0, 2]) φ1,2(ω1) = (α + β, β) (∀ω1 = (α, β) ∈ Ω1 = [0, 4]× [0, 2]) φ2,3(ω2) = (α + β, β) (∀ω2 = (α, β) ∈ Ω2 = [0, 6]× [0, 2]) Thus, we get the deterministic sequence causal map {φπ(t),t : Ωπ(t) → Ωt}t∈{1,2,3}, and the deterministic sequence causal operator {Φπ(t),t : L∞(Ωt)→ L∞(Ωπ(t))}t∈{1,2,3}. That is, (Φ0,1f1)(ω0)=f1(φ0,1(ω0)) (∀f1 ∈ L∞(Ω1),∀ω0 ∈ Ω0) (Φ1,2f2)(ω1)=f2(φ1,2(ω1)) (∀f2 ∈ L∞(Ω2),∀ω1 ∈ Ω1) (Φ2,3f3)(ω2)=f3(φ2,3(ω2)) (∀f3 ∈ L∞(Ω3),∀ω1 ∈ Ω2). Illustrating by the diagram, we see L∞(Ω0) Φ0,1←−L∞(Ω1) Φ1,2←−L∞(Ω2) Φ2,3←−L∞(Ω3) And thus, φ0,2(ω0) = φ1,2(φ0,1(ω0)), φ0,3(ω0) = φ2,3(φ1,2(φ0,1(ω0))), Therefore, note that Φ0,2 = Φ0,1 * Φ1,2, Φ0,3 = Φ0,1 * Φ1,2 * Φ2,3. L∞(Ω1) L∞(Ω0) L∞(Ω2) L∞(Ω3) +  k Φ0,1 Φ0,2 Φ0,3 363 Ishikawa's Homepage 13.2 Regression analysis Let R be the set of real numbers. Fix σ > 0. For each t = 0, 1, 2, define the normal observable Ot≡(R,BR, Gσ) in L∞(Ωt) such that [Gσ(Ξ)](ωt) = 1√ 2πσ2 ∫ Ξ exp(−(x− α) 2 2σ2 )dx (∀Ξ ∈ BR,∀ωt = (α, β) ∈ Ωt=[0, 2t+ 2]× [0, 2]). Thus, we have the deterministic sequential causal observable [{Ot}t=1,2,3, {Φπ(t),t : L∞(Ωt) → L∞(Ωπ(t))}t∈{1,2,3}]. And thus, we have the realized causal observable ÔT = (R3,FR3 , F0) in L∞(Ω0) such that ( using Theorem 12.8 ) [F0(Ξ1 × Ξ2 × Ξ3)](ω0) = [ Φ0,1 ( Gσ(Ξ1)Φ1,2(Gσ(Ξ2)Φ2,3(Gσ(Ξ3))) )] (ω0) =[Φ0,1Gσ(Ξ1)](ω0) * [Φ0,2Gσ(Ξ2)](ω0) * [Φ0,3Gσ(Ξ3)](ω0) =[Gσ(Ξ1)](φ0,1(ω0)) * [Gσ(Ξ2)](φ0,2(ω0)) * [Gσ(Ξ3)](φ0,3(ω0)) (∀Ξ1,Ξ2,Ξ3 ∈ BR, ∀ω0 = (α, β) ∈ Ω0 = [0, 1]× [0, 2]) Our problem (i.e., Problem 13.2) is as follows, (]1) Determine the parameter (α, β) such that the measured value of ML∞(Ω0)( ÔT , S[∗]) is equal to (1.9, 3.0, 4.7) For a sufficiently large natural number N , put Ξ1 = [ 1.9− 1 N , 1.9 + 1 N ] ,Ξ2 = [ 3.0− 1 N , 3.0 + 1 N ] ,Ξ3 = [ 4.7− 1 N , 4.7 + 1 N ] Fisher's maximum likelihood method (Theorem 5.6)) says that the above (]1) is equivalent to the following problem (]2) Find (α, β) (= ω0 ∈ Ω0) such that [F0(Ξ1 × Ξ2 × Ξ3)](α, β) = max (α,β) [F0(Ξ1 × Ξ2 × Ξ3)] Since N is assumed to be sufficiently large, we see (]2) =⇒ max (α,β)∈Ω0 [F0(Ξ1 × Ξ2 × Ξ3)](α, β) =⇒ max (α,β)∈Ω0 1 √ 2πσ2 3 ∫ ∫ ∫ Ξ1×Ξ2×Ξ3 e[− (x1−(α+β)) 2+(x2−(α+2β)) 2+(x3−(α+3β)) 2 2σ2 ] 364 Ishikawa's Homepage Chap. 13 Fisher statistics (II) × dx1dx2dx3 =⇒ max (α,β)∈Ω0 exp(−J/(2σ2)) =⇒ min (α,β)∈Ω0 J where J = (1.9− (α + β))2 + (3.0− (α + 2β))2 + (4.7− (α + 3β))2 ( ∂ ∂α {* * * } = 0, ∂ ∂β {* * * } = 0 and thus, ) =⇒ { (1.9− (α + β)) + (3.0− (α + 2β)) + (4.7− (α + 3β)) = 0 (1.9− (α + β)) + 2(3.0− (α + 2β)) + 3(4.7− (α + 3β)) = 0 =⇒ (α, β) = (0.4, 1.4) Therefore, in order to obtain a measured value (1.9, 3.0, 4.7), it suffices to put (α, β) = (0.4, 1.4) Remark 13.7. For completeness, note that, • From the theoretical point of view, "inference" = "control" Thus, we conclude that statistics and dynamical system theory are essentially the same. 365 Ishikawa's Homepage

Chapter 14 Realized causal observable in classical systems As mentioned in the previous chapters, what is important is • to exercise the relationship of measurement and causality In this chapter, we discuss the relationship more systematically. That is, we add the further argument concerning the realized causal observable. This field is too vast, thus, we mainly concentrate our interest to classical systems, particularly, Zeno's paradox. That is, ([) to describe the flying arrow ( the best work in Zeno's paradoxes ) in terms of quantum language (cf. refs.[37, 39])1 We believe that this is the final answer to Zeno's paradox. 14.1 Infinite realized causal observable in classical systems In what follows, we shall generalize the argument ( concerning the finite realized causal observable in Chapter 12) to infinite case. In the case of infinite trees, it is impossible to discuss quantum system deeply. thus, in this chapter, we devote ourselves to classical systems 1 This chapter is extracted from [37]: S. Ishikawa, "Zeno's paradoxes in the Mechanical World View," arXiv:1205.1290v1 [physics.hist-ph], (2012) [39]: S. Ishikawa, Measurement Theory in the Philosophy of Science, arXiv:1209.3483 [physics.hist-ph] 2012, (177 pages) 367 14.1 Infinite realized causal observable in classical systems Let (T,≤) be an infinite tree, i.e., an infinite tree like semi-ordered set such that "t1 5 t3 and t2 5 t3" =⇒ "t1 5 t2 or t2 5 t1" Put T 2≤ = {(t1, t2) ∈ T 2 : t1 ≤ t2}. An element t0 ∈ T is called a root if t0 ≤ t (∀t ∈ T ) holds. If T has the root t0, we sometimes denote T by T (t0). T ′(⊆ T ) is called lower bounded if there exists an element ti(∈ T ) such that ti 5 t (∀t ∈ T ′). Therefore, if T has the root, any T ′(⊆ T ) is lower bounded. We always assume that T is complete, that is, for any T ′(⊆ T ) which is lower bounded, there exists an element InfT (T ′)(∈ T ) that satisfies the following (i) and (ii): (i) InfT (T ′) 5 t (∀t ∈ T ′) (ii) If s 5 t (∀t ∈ T ′), then it holds that s 5 InfT (T ′) /// Let (T (t0), 5 ) be an infinite tree with the root t0. For each t ∈ T , consider the classical basic structure: [C0(Ωt) ⊆ L∞(Ωt, νt) ⊆ B(L2(Ωt, νt))] Also, for each t ∈ T , define the separable complete metric space Xt, and the Borel field BXt , and further, define the observable Ot=(Xt,Ft, Ft) in L ∞(Ωt, νt). That is, we have a sequential causal observable: [OT (t0)] = [{Ot}t∈T , {Φt1,t2 : L∞(Ωt2 , νt2)→ L∞(Ωt1 , νt1)}(t1,t2)∈T 25 ] Now let us construct the realized causal observable in what follows: Here, define, P0(T ) (= P0(T (t0)) ⊆ P(T )) such that P0(T (t0)) ={T ′ ⊆ T | T ′ is finite, t0 ∈ T ′ and satisfies InfT ′S = InfTS (∀S ⊆ T ′)} Let T ′(t0) ∈ P0(T (t0)). Since (T ′(t0), 5 ) is finite, we can put (T ′={t0, t1, . . . , tN}, π : T ′ \ {t0} → T ′), where π is a parent map. Review 14.1. [The review of Definition 12.4]. Let T ′(= T ′(t0)) ∈ P0(T ). Consider the sequential causal observable [{Ot}t∈T ′ , {Φπ(t),t : L∞(Ωt, νt) → L∞(Ωπ(t), νπ(t))}t∈T ′\{t0} ]. For each s ( ∈ T ′), putting Ts = {t ∈ T ′ | t = s}, define the observable Ôs=(×t∈Ts Xt, ×t∈Ts Ft, Fs) in 368 Ishikawa's Homepage Chap. 14 Realized causal observable in classical systems L∞(Ωt, νt) such that Ôs =  Os (s ∈ T ′ \ π(T ′) and ) Os×( × t∈π−1({s}) Φπ(t),tÔt) (s ∈ π(T ′) and ) (14.1) And further, iteratively, we get Ôt0=(×t∈T ′ Xt, ×t∈T ′ Ft, Ft0), which is also denoted by ÔT ′=(×t∈T ′ Xt,×t∈T ′ Ft, FT ′).( In classical cases, the existence is guaranteed by Definition 12.4 ) For any subsets T1 ⊆ T2( ⊆ T ), define the natural map πT1,T2 :×t∈T2 Xt −→×t∈T1 Xt by × t∈T2 Xt 3 (xt)t∈T2 7→ (xt)t∈T1 ∈ × t∈T1 Xt It is clear that the observables { ÔT ′=(×t∈T ′ Xt, ×t∈T ′ Ft, FT ′) | T ′ ∈ P0(T ) } in L∞(Ωt0 , νt0) satisfy the following consistency condition, that is, • for any T1, T2 (∈ P0(T )) such that T1 ⊆ T2, it holds that FT2 ( π−1T1,T2(ΞT1) ) = FT1 ( ΞT1 ) (∀ΞT1 ∈ × t∈T1 Ft) Then, by Theorem 4.1[ Kolmogorov extension theorem in measurement theory ], there uniquely exists the observable ÔT = (×t∈T Xt,  t∈T Ft, FT ) in L∞(Ωt0 , νt0) such that: FT ( π−1T ′,T (ΞT ′) ) = FT ′ ( ΞT ′ ) (∀ΞT ′ ∈  t∈T ′ Ft, ∀T ′ ∈ P0(T )) This observable ÔT = (×t∈T Xt,  t∈T Ft, FT ) is called the realization of the sequential causal observable [OT (t0)] = [{Ot}t∈T , {Φt1,t2 : L∞(Ωt2 , νt2) → L∞(Ωt1 , νt1)}(t1,t2)∈T 25 ]. Summing up the above argument, we have the following theorem in classical systems. This is the infinite version of Definition 12.4. Theorem 14.2. [The existence theorem of an infinite realized causal observable in classical systems] Let T be an infinite tree with the root t0. For each t ∈ T , consider the basic structure: [C0(Ωt) ⊆ L∞(Ωt, νt) ⊆ B(L2(Ωt, νt))] Also, for each t ∈ T , define the separable complete metric space Xt, the Borel field (Xt,Ft) and an observable Ot=(Xt,Ft, Ft) in L ∞(Ωt, νt). And, consider the sequential causal 369 Ishikawa's Homepage 14.1 Infinite realized causal observable in classical systems observable[OT (t0)] = [{Ot}t∈T , {Φt1,t2 : L∞(Ωt2 , νt2) → L∞(Ωt1 , νt1)}(t1,t2)∈T 25 ]. Then, there uniquely exists the realized causal observable ÔT = (×t∈T Xt,  t∈TFt, FT ) in L∞(Ωt0 , νt0), that is, it satisfies that FT ( π−1T ′,T (ΞT ′) ) = FT ′ ( ΞT ′ ) (∀ΞT ′ ∈  t∈T ′Ft, ∀T ′ ∈ P0(T )) (14.2) 370 Ishikawa's Homepage Chap. 14 Realized causal observable in classical systems 14.2 Is Brownian motion a motion or a measured value? 14.2.1 Brownian motion in probability theory There is a reason to consider that (A) Brownian motion should be understood in measurement theory. That is because Brownian motion is not in Newtonian mechanics. As one of applications of Theorem 14.2, we discuss the Brown motion in quantum language. t ω0 B(t, λ) = ω( ≡ (ωt)t∈R+) R 6 Let us explain the above figure as follows. Definition 14.3. [The review of Brownian motion in probability theory [63]]. Let (Λ,FΛ, P ) be a probability space. For each λ ∈ Λ, define the real-valued continuous function B(*, λ) : T (=[0,∞))→ R such that, for any t0 = 0 < t1 < t2 < * * * < tn, P ({λ ∈ Λ | B(tk, λ) ∈ Ξk ∈ BR (k = 1, 2, . . . , n)}) = ∫ Ξ1 ( * * * ( ∫ Ξtn−1 ( ∫ Ξtn n × k=1 G√tk−tk−1(ωk − ωk−1)dωn)dωn−1) * * * ) dω1 (14.3) where, ω0 ∈ R, dωk is the Lebesgue measure on R, and G√t(q) = 1√2πtexp [ − q2 2t ] . The B(*, λ) : T (=[0,∞))→ R is called the Brownian motion. 371 Ishikawa's Homepage 14.2 Is Brownian motion a motion or a measured value? 14.2.2 Brownian motion in quantum language Now consider the diffusion equation: ∂ρt(q) ∂t = ∂2ρt(q) ∂q2 , (∀q ∈ R,∀t ∈ T=R+ = [t0 = 0,∞) ) By the solution ρt, we get predual operator {[Φt1,t2 ]∗ : L1(R, dq)→ L1(R, dq)} as follows. That is, for each ρt1 ∈ L1(R,m), define( [Φt1,t2 ]∗(ρt1) ) (q) = ρt2(q) = ∫ ∞ −∞ ρt1(y)G √ t2−t1(q − y)m(dy) (∀q ∈ R, ∀(t1, t2) ∈ T 2 5) For simplicity, we put (Ωt.BΩt , dωt) = (Ω,B, dω) = (Rq,BRq , dq). And thus, for each t ∈ T , consider the classical basic structure: [C0(Ωt) ⊆ L∞(Ωt, dωt) ⊆ B(L2(Ωt, dωt))] Putting Φt1,t2 = ([Φt1,t2 ]∗) ∗, we get the sequential causal operator {Φt1,t2 : L∞(Ωt2 , dωt2)→ L∞(Ωt1 , dωt1) | (t1, t2) ∈ T 2≤} For each t ∈ T , consider the exact observable O(exa)t = (Ω,BΩ, F (exa)) in L∞(Ω, dω). Thus, we get the sequential causal exact observable [OT ] = [{O(exa)t }t∈T ; {Φt1,t2 | (t1, t2) ∈ T 2≤}]. The existence theorem of the infinite classical realized causal observable (Theorem 14.2) says that OT has the realized causal observable Ôt0 = (ΩT ,B(ΩT ), Ft0) in L∞(Ω, dω). Assume that (B) a measured value ω (= (ωt)t∈T ∈ ΩT ) is obtained by ML∞(Ω)(Ôt0 , S[δω0 ]). Let T ′ = {t0, t1, t2, * * * , tn} be a finite subset of T , where t0 = 0 < t1 < t2 < * * * < tn. Put Ξ =×T ′t∈TΞt ( ∈ BR +) where Ξt = Ω (∀t /∈ T ′). Then, by Axiom 1 (measurement; §2.7) , we see the probability that ω( = (ωt)t∈T ) belongs to the set Ξ ≡ ×T ′t∈TΞt is given by [Ft0(×T ′ t∈TΞt)](ω0) where [Ft0(×T ′ t∈TΞt)](ω0) = ( F (Ξ0)Φ0,t1 ( F (Ξt1) * * *Φtn−2,tn−1 ( F (Ξtn−1) ( Φtn−1,tnF (Ξtn) )) * * * ) (ω0) = ∫ Ξ1 ( * * * ( ∫ Ξtn−1 ( ∫ Ξtn ×nk=1G√tk−tk−1(ωk − ωk−1)dωn)dωn−1) * * * ) dω1 (14.4) 372 Ishikawa's Homepage Chap. 14 Realized causal observable in classical systems which is equal to the (14.3). Thus, we see that probability theory( B(t, *) ) t∈T Brownian motion = quantum language( ωt ) t∈T measured value ♠Note 14.1. Thus, the following assertion has a reason in some sense: • The Brownian motion B(t, λ) is not a motion but a measured value. Some may recall Parmenides' saying: (]) There are no "plurality", but only "one". And therefore, there is no movement. which is the same as the essence of the linguistic interpretation. That is, the spirit of quantum language says that (]) Describe "plurality" as if only "one". (]) Describe moving one as if not moving. 373 Ishikawa's Homepage 14.3 The Schrödinger picture of the sequential deterministic causal operator 14.3 The Schrödinger picture of the sequential deterministic causal operator 14.3.1 The preparation of the next section (§14.4: Zeno's paradox) The linguistic interpretation (§3.1) says that a state does no move, which is called the Heisenberg picture (i.e., a state does not move, and, an observable moves). This is formal. On the other hand, we sometimes use the Schrödinger picture (i.e., a state moves, and, an observable does not move), which is handy and makeshift. In this section, we explain something about the Schrödinger picture in classical deterministic systems. This section is the preparation of the next section (Zeno's paradoxes). Let (T (t0), 5 ) be an infinite tree with the root t0. For each t ∈ T , consider the classical basic structure: [C0(Ωt) ⊆ L∞(Ωt, νt) ⊆ B(L2(Ωt, νt))] Definition 14.4. [State changes - the Schrödinger picture] Let {Φt1,t2 : L∞(Ωt2 , νt2) → L∞(Ωt1), νt1)}(t1,t2)∈T 25 be a deterministic causal relation with the deterministic causal maps φt1,t2 : Ωt1 → Ωt2 (∀(t1, t2) ∈ T 25). Let ωt0 ∈ Ωt0 be an initial state. Then, the {φt0,t(ωt0)}t∈T (or, {δφt0,t(ωt0 )}t∈T is called the Schrödinger picture representation. The following is the infinite version of Theorem12.8. Theorem 14.5. [Deterministic sequential causal operator and realized causal observable ] Let (T (t0), 5 ) be an infinite tree with the root t0. Let [OT ] = [{Ot}t∈T , {Φt1,t2 : L∞(Ωt2 , νt2) → L∞(Ωt1 , νt1)}(t1,t2)∈T 25 ] be a deterministic sequential causal observable. Then, the realization Ôt0 ≡ (×t∈TXt,  t∈TFt, Ft0) is represented by Ôt0 = × t∈T Φt0,tOt That is, it holds that [Ft0(× t∈T Ξt )](ωt0) = × t∈T [Φt0,tFt(Ξt)](ωt0) = × t∈T [Ft(Ξt)](φt0,t(ωt0)) 374 Ishikawa's Homepage Chap. 14 Realized causal observable in classical systems (∀ωt0 ∈ Ωt0 ,∀Ξt ∈ Ft) Proof. The proof is similar to that of Theorem12.8 Theorem 14.6. Let [OT (t0)] = [{O (exa) t }t∈T , {Φt1,t2 : L∞(Ωt2 , νt2) → L∞(Ωt1 , νt1)}(t1,t2)∈T 25 ] be a deterministic sequential causal exact observable, which has the deterministic causal maps φt1,t2 : Ωt1 → Ωt2 (∀(t1, t2) ∈ T 25). And let Ôt0 = (×t∈T Xt,×t∈T Ft, FT ) be its realized causal observable in L∞(Ωt0 , νt0). Assume that the measured value (xt)t∈T is obtained by ML∞(Ωt0 )(ÔT = (×t∈T Xt,×t∈T Ft, F0), S[ωt0 ]). Then, we surely believe that xt = φt0,t(ωt0) (∀t ∈ T ) Thus, we say that, as far as a deterministic sequential causal observable, (a) exact measured value (xt)t∈T = the Schrödinger picture representation (φt0,t(ωt0))t∈T Proof. Let D = {t1, t2, . . . , tn}(⊆ T ) be any finite subset of T . Put Ξ = ×Dt∈TΞt = (×t∈D Ξt) × (×t∈T\DXt), where Ξt ⊆ Xt(= Ωt) is an open set such that φt0,t(ωt0) ∈ Ξt (∀t ∈ D). Then, we see that (b) the probability that the measured value (xt)t∈T belongs to Ξ =×Dt∈TΞt is equal to 1. That is because Theorem 14.5 says that( FT (Ξ) ) (ωt0) = ( n × k=1 ( Φt0,tkF (exa)(Ξtk) )) (ωt0) = ( n × k=1 F (exa)(φ−1t0,tk(Ξtk) )) (ωt0) = n × k=1 χ Ξtk (φt0,tk(ωt0)) = 1 Thus, from the arbitrariness of Ξt, we surely believe that (c) (xt)t∈T = φt0,t(ωt0) (∀t ∈ T ) ♠Note 14.2. Note that "(b) ⇔(c)" in the above. That is, (b) is the definition of (c). Thus, we have the following corollary, which is the generalization of Theorem 3.15. 375 Ishikawa's Homepage 14.3 The Schrödinger picture of the sequential deterministic causal operator Corollary 14.7. [System quantity and exact observable]. For each t ∈ T (t0), consider the exact observable O (exa) t = (X,Ft, F (exa))(= (Ωt,Bt, χ)) in L ∞(Ωt, νt) and a system quantity gt : Ωt → R on Ωt. Let O′t = (R,BR, Gt) be the observable representation of the quantity gt in L∞(Ωt). Assuming the simultaneous observable O (exa) t ×O′t, define the sequential deterministic causal observable: [OT (t0)] = [{O (exa) t × O′t}t∈T , {Φt1,t2 : L∞(Ωt2 , νt2)→ L∞(Ωt1 , νt1)}(t1,t2)∈T 25 ] Let φt1,t2 : Ωt1 → Ωt2 (∀(t1, t2) ∈ T 25) be the deterministic causal map. Let Ôt0 = (×t∈T (Xt×R), t∈T (Ft  BR), Ft0) be the realized causal observable. Thus, we have the measurement ML∞(Ωt0 )(Ôt0 , S[ωt0 ]). Let (xt, yt)t∈T be the measured value obtained by the measurement ML∞(Ωt0 )(Ôt0 , S[ωt0 ]). Then, we can surely believe that xt = φt0,t(ωt0) and yt = gt(φt0,t(ωt0)) (∀t ∈ T ) Remark 14.8. [Why doesn't Newtonian mechanics have measurement?]. Newtonian mechanics and quantum mechanics are formulated as follows: (])  Newtoinan mechanics = Nothing + Causality (Newtonian equation) quantum mechanics = Measurement (Born's quantum measurement) + Causality (Heisenberg (and Schrödinger) equation) Thus, the following question is natural: (]2) Why doesn't Newtonian mechanics have measurement ? Some may think that the reason is due to Theorem 14.6 (or, Corollary 14.7 ), which says that we need only φt0,t(ωt0) and not xt. However, this answer is superficial. The question (]2) is significant in the light of Einstein's words: (]3) The moon is there whether one looks at it or not. in Einstein and Tagore's conversation. This should be compared with Berkley's words "To be is to be perceived". We believe that the (]3) is the same as (]4) (= (]5) ): (]4) Physics should exist without measurement (]5) The concept of "measurement" is metaphysical and not physical 376 Ishikawa's Homepage Chap. 14 Realized causal observable in classical systems 14.4 Even Zeno's paradoxes can be soloved-Flying arrow is at rest First we explain what Zeno's paradox means, one of the oldest paradoxes in science. 14.4.1 What is Zeno's paradox? Although Zeno's paradox has some types (i.e., "flying arrow", "Achilles and a tortoise", "dichotomy", "stadium", etc.), I think that these are essentially the same problem. And I think that the flying arrow expresses the essence of the problem exactly and is the first masterpiece in Zeno's paradoxes. However, since "Achilles and the tortoise" may be more famous, I will also describe this as follows. Paradox 14.9. [Zeno's paradox] [Flying arrow is at rest] • Consider a flying arrow. In any one instant of time, the arrow is not moving. Therefore, If the arrow is motionless at every instant, and time is entirely composed of instants, then motion is impossible. [Achilles and a tortoise] • I consider competition of Achilles and a tortoise. Let the start point of a tortoise (a late runner) be the front from the starting point of Achilles (a quick runner). Suppose that both started simultaneously. If Achilles tries to pass a tortoise, Achilles has to go to the place in which a tortoise is present now. However, then, the tortoise should have gone ahead more. Achilles has to go to the place in which a tortoise is present now further. Even Achilles continues this infinite, he can never catch up with a tortoise. 377 Ishikawa's Homepage 14.4 Even Zeno's paradoxes can be soloved-Flying arrow is at rest In order to explain "What is Zeno's paradox?" we have to start from the following Figure. That is, we assert that Zeno's paradox can not be understood without the following figure: Figure 14.10. [=Figure 1.1: The location of quantum language in the history of world-description (cf. ref.[32]) ] Parmenides Socrates 0©:Greek philosophy Plato Aristotle Schola-−−−−→ sticism 1© −−→ (monism) Newton (realism) 2© → relativity theory −−−−−−→ 3© → quantum mechanics −−−−−−→ 4© −→ (dualism) Descartes Locke,... Kant (idealism) 6©−→ (linguistic view) linguistic philosophy language−−−−−→ 8© language−−−−−−→ 7©  5©−→ (unsolved) theory of everything (quantum phys.)  10©−→ (=MT) quantum language (language) Figure 1.1: The history of the world-view statistics system theory language−−−−−→ 9© the linguistic view the realistic view It is clear that (A) Descartes=Kant philosophy and the philosophy of language have no power to describe Zeno's paradox 14.9. However, we have the following problems: 378 Ishikawa's Homepage Chap. 14 Realized causal observable in classical systems (B1) How do we describe Zeno's paradox 14.9 in terms of Newtonian mechanics? (B2) How do we describe Zeno's paradox 14.9 in terms of quantum mechanics? (B3) How do we describe Zeno's paradox 14.9 in terms of the theory of relativity? (B4) How do we describe Zeno's paradox 14.9 in terms of statistics (i.e., the dynamical system theory) ? (B5) How do we describe Zeno's paradox 14.9 in terms of quantum language? And, finally, we have (C) What is the most proper world description for Zeno's paradox 14.9? We assert that (D) "to solve Zeno's paradox 14.9" ⇐⇒ "to answer the above (C)" and conclude that (E) The answer of the above (C) is just quantum language Therefore, it suffices to answer the above (B5), that is, Problem 14.11. [The meaning of Zeno's paradox] Describe "flying arrow" and "Achilles an a tortoise" in (classical) quantum language! 14.4.2 The answer to (B4): the dynamical system theoretical answer to Zeno's paradox Before the answer of Problem 14.11, we give the answer to the Problem (B4), i.e., the dynamical system theoretical answer. However, in order to do it, we have to start from the formulation of dynamical system theory in what follows . 379 Ishikawa's Homepage 14.4 Even Zeno's paradoxes can be soloved-Flying arrow is at rest 14.4.2.1 The formulation of dynamical system theory Although statistics and dynamical system theory have no clear formulations, as mentioned in Chapter 13, we have the opinion that statistics and dynamical system theory are the same things. At least, the following formulation (i.e., the formulation of dynamical system theory in the narrow sense) should belong to statistics. Formulation 14.12. [The formulation of dynamical system theory in the narrow sense] Dynamical system theory is formulated as follows. Dynamical system theory = 1©:State equation + 2©:Measurement equation (14.5) 1©: State equation is as follows. Let T = R be the time axis. For each t(∈ T ), consider the state space Ωt = Rn (n-dimensional real space). The state equation (Chap. 13(13.2)) is defined by the following simultaneous ordinary differential equation of the first order State equation =  dω1 dt (t) = v1(ω1(t), ω2(t), . . . , ωn(t), ε1(t), t) dω2 dt (t) = v2(ω1(t), ω2(t), . . . , ωn(t), ε2(t), t) * * * * * * dωn dt (t) = vn(ω1(t), ω2(t), . . . , ωn(t), εn(t), t) (14.6) where εk(t) is a noise (k = 1, 2, * * * , n). 2©: Measurement equation is as follows. Consider the measured value space X = Rm (mdimensional real space). The measurement equation (Chap. 13(13.2)) is defined by Measurement equation =  x1(t) = g1(ω1(t), ω2(t), . . . , ωn(t), η1(t), t) x2(t) = g2(ω1(t), ω2(t), . . . , ηn(t), η2(t), t) * * * * * * xm(t) = gm(ω1(t), ω2(t), . . . , ηn(t), ηn(t), t) (14.7) where g(= (g1, g2, * * * , gn)) : Ω × R2 → X is the system quantity and ηk(t) is a noise (k = 1, 2, * * * ,m). Here, x(t)(= (x1(t), x2(t), * * * , xn(t))) is called a motion function. 14.4.2.2 The dynamical system theoretical answer to Zeno's paradox Answer 14.13. [The dynamical system theoretical answer to "flying arrow (in Paradox 14.9)"] Let q(t) be the position of the flying arrow at time t. That is, consider the motion function q(t). 380 Ishikawa's Homepage Chap. 14 Realized causal observable in classical systems • Note that the following logic (i.e., Zeno's logic ) is wrong: • for each time t, the position q(t) of the flying arrow is determined. =⇒ the motion function q is a constant function Thus, Zeno's logic is wrong. [The dynamical system theoretical answer to "Achilles and a tortoise (in Paradox 14.9)"] For example, assume that the velocity vq [resp. vs] of the quickest [resp. slowest] runner is equal to v(> 0) [resp. γv (0 < γ < 1)]. And further, assume that the position of the quickest [resp. slowest] runner at time t = 0 is equal to 0 [resp. a (> 0)]. Thus, we can assume that the position ξ(t) of the quickest runner and the position η(t) of the slowest runner at time t (≥ 0) is respectively represented by{ ξ(t) = vt η(t) = γvt+ a (14.8) • Calculations The formula (14.8) can be calculated as follows (i.e., (i) or (ii)): [(i): Algebraic calculation of (14.8)]: Solving ξ(s0) = η(s0), that is, vs0 = γvs0 + a we get s0 = a (1−γ)v . That is, at time s0 = a (1−γ)v , the fast runner catches up with the slow runner. [(ii): Iterative calculation of (14.8)]: Define tk (k = 0, 1, ...) such that, t0 = 0 and tk+1 = γvtk + a (k = 0, 1, 2, ...) Thus, we see that tk = (1−γk)a (1−γ)v (k = 0, 1, ...). Then, we have that ( ξ(tk), η(tk) ) = ((1− γk)a 1− γ , (1− γk+1)a 1− γ ) → ( a 1− γ , a 1− γ ) (14.9) 381 Ishikawa's Homepage 14.4 Even Zeno's paradoxes can be soloved-Flying arrow is at rest as k →∞. Therefore, the quickest runner catches up with the slowest at time s0 = a(1−γ)v . [(iii): Conclusion]: After all, by the above (i) or (ii), we can conclude that (]) the quickest runner can overtake the slowest at time s0 = a (1−γ)v . 6 t 6 q1(t) = vt ? q2(t) = γvt+ a 0 (= t0) a v (= t1) (1−γ2)a (1−γ)v (= t2) (1−γ3)a (1−γ)v (= t3) a (1−γ)v (= s0) * * * * * * * * * ... ...... a (1−γ2)a 1−γ (1−γ3)a 1−γ a 1−γ q1, q2 The graph of q1(t) = vt, q2(t) = γvt+ a 14.4.2.3 Why isn't the Answer 14.13 authorized? We believe that the Answer 14.13 is not the wrong answer of Zeno's paradox. If so, we have to answer the following question: (F) Why isn't the Answer 14.13 accepted as the final answer of Zeno's paradox? We of course believe that (G1) the reason is due to the fact that statistics (=dynamical system theory) is not accepted as the world-view in Figure 14.10. Or equivalently, (G1) the linguistic world-view is not accepted as the world-view in Figure 14.10. If so, the readers note that (H) the purpose of this note is to assert that the linguistic world view should be authorized in Figure 14.10. 382 Ishikawa's Homepage Chap. 14 Realized causal observable in classical systems 14.4.3 Quantum linguistic answer to Zeno's paradoxes Before reading Answer 14.14 ( Zeno's paradox(flying arrow) ), confirm our spirit: (I) The theory described in ordinary language should be described in a certain world description. That is because almost ambiguous problems are due to the lack of "the worlddescription method". Therefore, (J) it suffices to describe "motion function q(t) in Answer 14.13 (flying arrow)" in terms of quantum language. Here, the motion function should be a measured value, in which the causality is concealed. This will be done as follows. Answer 14.14. [The answer to Problem14.11] or [Answer to Problem 14.9: Zeno's paradox(flying arrow) (cf. ref. [37, 39])] In Corollary 14.7, putting q(t) = yt(= gt(φt0,t(ωt0))) we get the time-position function q(t). Although there may be several opinions, we consider that the followings (i.e., (K1) and (K2)) are equivalent: (K1) to accept Figure 14.10:[The history of the world-view] (K2) to believe in Answer 14.14 as the final answer of Zeno's paradox ♠Note 14.3. I think that "the flying arrow" is Zeno's best work. If readers agree to the above answer, they can easily answer the other Zeno's paradoxes. Also, it should be noted that Zeno of Elea (BC. 490-430) was a Greek philosopher (about 2500 years ago). Hence, we are not concerned with the historical aspect of Zeno's paradoxes. Therefore, we think that (]) "How did Zeno think Zeno's paradoxes?" is not important from the scientific point of view. and (]) What is important is "How do we think Zeno's paradoxes?" Also, for the quantum linguistic space-time, see §10.7 ( Leibniz-Clarke correspondence). I doubt great philosophers' opinions concerning Zeno's paradoxes. 383 Ishikawa's Homepage

Chapter 15 Least-squares method and Regression analysis Although regression analysis has a great history, we consider that it has always continued being confused. For example, the fundamental terms in regression analysis (e.g., "regression", "leastsquares method", "explanatory variable", "response variable", etc.) seem to be historically conventional, that is, these words do not express the essence of regression analysis. In this chapter, we show that the least squares method acquires a quantum linguistic story as follows. The least squares method (Section 15.1) describe by−−−−−−−−−−−→ quantum language Regression analysis (Section 15.2) natural−−−−−−−−→ generalization Generalized linear model (Section 15.4) (]) In this story, the terms "explanatory variable" and "response variable" are clarified in terms of quantum language. As the general theory of regression analysis, it suffices to devote ourselves to Theorem 13.4. However, from the practical point of view, we have to add the above story (])1. 15.1 The least squares method Let us start from the simple explanation of the least-squares method. Let {(ai, xi)}ni=1 be a sequence in the two dimensional real space R2. Let φ(β1,β2) : R → R be the simple function such that R 3 a 7→ x = φ(β1,β2)(a) = β1a+ β0 ∈ R (15.1) 1This chapter is extracted from • Ref. [43]: S. Ishikawa; Regression analysis in quantum language ( arxiv:1403.0060[math.ST],( 2014) ) 385 15.1 The least squares method where the pair (β1, β2)(∈ R2) is assumed to be unknown. Define the error σ by σ2(β1, β2) = 1 n n∑ i=1 (xi − φ(β1,β2)(ai))2 ( = 1 n n∑ i=1 (xi − (β1ai + β0))2 ) (15.2) Then, we have the following minimization problem: Problem 15.1. [The least squares method]. Let {(ai, xi)}ni=1 be a sequence in the two dimensional real space R2. Find the (β0, β1) (∈ R2) such that σ2(β0, β1) = min (β1,β2)∈R2 σ2(β1, β2) ( = min (β1,β2)∈R2 1 n n∑ i=1 (xi − (β1ai + β0))2 ) (15.3) where (β0, β1) is called "sample regression coefficients". This is easily solved as follows. Taking partial derivatives with respect to β0, β1, and equating the results to zero, gives the equations (i.e., "likelihood equations"), ∂σ2(β1, β2) ∂β0 = n∑ i=1 (xi − β0 − β1ai) = 0, (i = 1, ..., n) (15.4) ∂σ2(β1, β2) ∂β1 = n∑ i=1 (xi − β0 − β1ai)ai = 0, (i = 1, ..., n) (15.5) Solving it, we get that β1 = sax saa , β0 = x− sax saa a, σ2(= 1 n n∑ i=1 (xi − (β1ai + β0))2 ) = sxx − s2ax saa (15.6) where ā = a1 + * * *+ an n , x = x1 + * * *+ xn n , (15.7) saa = (a1 − ā)2 + * * *+ (an − ā)2 n , sxx = (x1 − x)2 + * * *+ (xn − x)2 n , (15.8) sax = (a1 − ā)(x1 − x) + * * *+ (an − ā)(xn − x) n . (15.9) Remark 15.2. [Applied mathematics]. Note that the above result is in (applied) mathematics, that is, • the above is neither in statistics nor in quantum language. The purpose of this chapter is to add a quantum linguistic story to Problem 15.1 (i.e., the least-squares method) in the framework of quantum language. 386 Ishikawa's Homepage Chap. 15 Least-squares method and Regression analysis 15.2 Regression analysis in quantum language Put T = {0, 1, 2, * * * , i, * * * , n}. And let (T, τ : T \ {0} → T ) be the parallel tree such that τ(i) = 0 (∀i = 1, 2, * * * , n) (15.10) 1 2 n 0 + ) k τ τ * * * * * * * * * * * * τ Figure 15.1: Parallel structure ♠Note 15.1. In regression analysis, we usually devote ourselves to "classical deterministic causal relation". Thus, Theorem 12.8 is important, which says that it suffices to consider only the parallel structure. For each i ∈ T , define a locally compact space Ωi such that Ω0 = R2 = { β = [ β0 β1 ] : β0, β1 ∈ R } (15.11) Ωi = R = { μi : μi ∈ R } (i = 1, 2, * * * , n) (15.12) where the Lebesgue measures mi are assumed. Assume that ai ∈ R (i = 1, 2, * * * , n), (15.13) which are called explanatory variables in the conventional statistics. Consider the deterministic causal map ψai : Ω0(= R2)→ Ωi(= R) such that Ω0 = R2 3 β = (β0, β1) 7→ ψai(β0, β1) = β0 + β1ai = μi ∈ Ωi = R (15.14) which is equivalent to the deterministic causal operator Ψai : L ∞(Ωi)→ L∞(Ω0) such that [Ψai(fi)](ω0) = fi(ψai(ω0)) (∀fi ∈ L∞(Ωi), ∀ω0 ∈ Ω0,∀i ∈ 1, 2, * * * , n) (15.15) 387 Ishikawa's Homepage 15.2 Regression analysis in quantum language L∞(Ω1(≡ R)) L∞(Ω2(≡ R)) L∞(Ωn(≡ R)) L∞(Ω0(≡ R2)) + ) k Ψa1 Ψa2 * * * * * * * * * * * * Ψan Figure 15.2: Parallel structure (Causal relation Ψai) Thus, under the identification: ai ⇔ Ψai , the term "explanatory variable" means a kind of causal relation Ψai . For each i = 1, 2, * * * , n, define the normal observable Oi≡(R,BR, Gσ) in L∞(Ωi(≡ R)) such that [Gσ(Ξ)](μ) = 1 ( √ 2πσ2) ∫ Ξ exp [ −(x− μ) 2 2σ2 ] dx (∀Ξ ∈ BR, ∀μ ∈ Ωi(≡ R)) (15.16) where σ is a positive constant. Thus, we have the observable Oai0 ≡(R,BR,ΨaiGσ) in L∞(Ω0(≡ R2)) such that [Ψai(Gσ(Ξ))](β) = [(Gσ(Ξ))](ψai(β)) = 1 ( √ 2πσ2) ∫ Ξ exp [ −(x− (β0 + aiβ1)) 2 2σ2 ] dx (15.17) (∀Ξ ∈ BR,∀β = (β0, β1) ∈ Ω0(≡ R2) Hence, we have the simultaneous observable ×ni=1Oai0 ≡(Rn,BRn ,×ni=1 ΨaiGσ) in L∞(Ω0(≡ R2)) such that [( n × i=1 ΨaiGσ)( n × i=1 Ξi)](β) = n × i=1 ( [ΨaiGσ)(Ξi)](β) ) = 1 ( √ 2πσ2)n ∫ * * * ∫ ×ni=1 Ξi exp [ − ∑n i=1(xi − (β0 + aiβ1))2 2σ2 ] dx1 * * * dxn = ∫ * * * ∫ ×ni=1 Ξi p(β0,β1,σ)(x1, x2, * * * , xn)dx1 * * * dxn (15.18) (∀ n × i=1 Ξi ∈ BRn , ∀β = (β0, β1) ∈ Ω0(≡ R2)) Assuming that σ is variable, we have the observable O = ( Rn(= X),BRn(= F), F ) in L∞(Ω0× R+) such that [F ( n × i=1 Ξi)](β, σ) = [( n × i=1 ΨaiGσ)( n × i=1 Ξi)](β) (∀Ξi ∈ BR, ∀(β, σ) ∈ R2(≡ Ω0)× R+) (15.19) 388 Ishikawa's Homepage Chap. 15 Least-squares method and Regression analysis Problem 15.3. [Regression analysis in quantum language] Assume that a measured value x =  x1 x2 ... xn  ∈ X = Rn is obtained by the measurement ML∞(Ω0×R+)(O ≡ (X,F, F ), S[(β0,β1,σ)]). (The measured value is also called a response variable.) And assume that we do not know the state (β0, β1, σ 2). Then, • from the measured value x = (x1, x2, . . . , xn) ∈ Rn, infer the β0, β1, σ! That is, represent the (β0, β1, σ) by (β0(x), β1(x), σ(x)) (i.e., the functions of x). Answer. Taking partial derivatives with respect to β0, β1, σ 2, and equating the results to zero, gives the log-likelihood equations. That is, putting L(β0, β1, σ 2, x1, x2, * * * , xn) = log ( p(β0,β1,σ)(x1, x2, * * * , xn) ) , (where "log" is not essential), we see that ∂L ∂β0 = 0 =⇒ n∑ i=1 (xi − (β0 + aiβ1)) = 0 (15.20) ∂L ∂β1 = 0 =⇒ n∑ i=1 ai(xi − (β0 + aiβ1)) = 0 (15.21) ∂L ∂σ2 = 0 =⇒ − n 2σ2 + 1 2σ4 n∑ i=1 (xi − β0 − β1ai)2 = 0 (15.22) Therefore, using the notations (15.7)-(15.9), we obtain that β0(x) = x− β1(x)a = x− sax saa a, β1(x) = sax saa (15.23) and (σ(x))2 = ∑n i=1 ( xi − (β0(x) + aiβ1(x)) )2 n = ∑n i=1 ( xi − (x− saxsaaa)− ai sax saa )2 n = ∑n i=1 ( (xi − x) + (a− ai) saxsaa )2 n =sxx − 2sax sax saa + saa( sax saa )2 = sxx − s2ax saa (15.24) 389 Ishikawa's Homepage 15.2 Regression analysis in quantum language Note that the above (15.23) and (15.24) are the same as (15.6). Therefore, Problem 15.3 (i.e., regression analysis in quantum language) is a quantum linguistic story of the least squares method (Problem 15.1). Remark 15.4. Again, note that (A) the least squares method (15.6) and the regression analysis (15.23) and (15.24) are the same. Therefore, a small mathematical technique (the least squares method) can be understood in a grand story (regression analysis in quantum language). The readers may think that (B) Why do we choose "complicated (Problem 15.3)" rather than "simple (Problem 15.1)"? Of course, such a reason is unnecessary for quantum language! That is because (C) the spirit of quantum language says that "Everything should be described by quantum language" However, this may not be a kind answer. The reason is that the grand story has a merit such that statistical methods (i.e., the confidence interval method and the statistical hypothesis testing ) can be applicable. This will be mentioned in the following section. 390 Ishikawa's Homepage Chap. 15 Least-squares method and Regression analysis 15.3 Regression analysis(distribution , confidence interval and statistical hypothesis testing) As mentioned in Problem 15.3 ( regression analysis), consider the measurement ML∞(Ω0×R+)(O ≡ (X(= Rn),F, F ), S[(β0,β1,σ)]) For each (β, σ) ∈ R2 × R+, define the sample probability space (X,F, P(β,σ)), where P(β,σ)(Ξ) = [F (Ξ)](β0, β1, σ) (∀Ξ ∈ F) Define L2(X,P(β,σ)) (or in short, L 2(X)) by L2(X) = {measurable function f : X → R | [ ∫ X |f(x)|2P(β,σ)(dx)]1/2 <∞}. (15.25) Further, for each f, g ∈ L2(X), define E(f) and V (f) such that E(f) = ∫ X f(x)P(β,σ)(dx), V (f) = ∫ X |f(x)− E(f)|2P(β,σ)(dx). (15.26) Our main assertion is to mention Problem 15.3 (i.e., regression analysis in quantum language). This section should be regarded as an easy consequence of Problem 15.3 ( regression analysis). For the detailed proof of Lemma 15.5, see standard books of statistics (e.g., ref. [8]). Lemma 15.5. Consider the measurement ML∞(Ω0×R+)(O ≡ (X,F, F ), S[(β0,β1,σ)]) in Problem 15.3 ( regression analysis). And assume the above notations. Then, we see: (A1) (1): V (β0) = σ2 n (1 + a 2 saa ), (2): V (β1) = σ2 n 1 saa , (A2) [Studentization]. Motivated by the (A1), we see: Tβ0 := √ n(β0 − β0)√ σ2(1 + a2/saa) ∼ tn−2, Tβ1 := √ n(β1 − β1)√ σ2/saa ∼ tn−2 (15.27) where tn−2 is the student's distribution with n− 2 degrees of freedom. For the proof. see ref. [8]. Let ML∞(Ω0(=R2)×R+)(O ≡ (X(= Rn),F, F ), S[(β0,β1,σ)]) be the measurement in Problem 15.3 ( regression analysis). For each k = 0, 1, define the estimator Êk : X(= Rn) → Θk(= R) and the quantity πk : Ω(= R2 × R+)→ Θk(= R) as follows. Ê0(x)(= β0(x)) = x− sax saa a, Ê1(x)(= β1(x)) = sax saa , π0(β0, β1, σ) = β0. π1(β0, β1, σ) = β1, (15.28) 391 Ishikawa's Homepage 15.3 Regression analysis(distribution , confidence interval and statistical hypothesis testing) (∀(β0, β1, σ) ∈ R2 × R+) Let α be a real number such that 0 < α 1, for example, α = 0.05. For any state ω = (β, σ)( ∈ Ω = R2 × R+), define the positive number ηαω,k ( > 0) by (6.9), (6.15), that is, ηαω,k(= δ 1−α ω,k ) = inf{η > 0 : [F ({x ∈ X : d x Θk (Êk(x), πk(ω)) ≥ η})](ω) ≤ α} (15.29) where, for each θ0k, θ 1 k(∈ Θk), the semi-distance dxΘk in Θk is defined by dxΘk(θ 0 k, θ 1 k) =  √ n|θ00−θ10 |√ σ2(1+a2/saa) (if k = 0) √ n|θ01−θ11 |√ σ2/saa (if k = 1) (15.30) Therefore, we see, by Lemma 15.5, that ηαω,k =  inf{η > 0 : [F ({x ∈ X : √ n|β0(x)−β0|√ σ2(1+a2/saa) ≥ η})](ω) ≤ α} (if k = 0) inf{η > 0 : [F ({x ∈ X : √ n|β1(x)−β1|√ σ2(x)/saa ≥ η})](ω) ≤ α} (if k = 1) (15.31) = tn−2(α/2) (15.32) Summing up the above arguments, we have the following proposition: Proposition 15.6. [confidence interval]. Assume that a measured value x ∈ X is obtained by the measurement ML∞(Ω0×R+)(O ≡ (X,F, F ), S[(β0,β1,σ)]). Here, the state (β0, β1, σ) is assumed to be unknown. Then, we have the (1−α)-confidence interval I1−αx,k in Corollary 6.6 as follows. I1−αx,k = {πk(ω)(∈ Θk) : d x Θk (Êk(x), πk(ω)) < η 1−α ω,k } =  I1−αx,0 = { β0 = π0(ω)(∈ Θ0) : |β0(x)−β0|√ σ2(x) n (1+a2/saa) ≤ tn−2(α/2) } (if k = 0) I1−αx,1 = { β1 = π1(ω)(∈ Θ1) : |β1(x)−β1|√ σ2(x) n (1/saa) ≤ tn−2(α/2) } (if k = 1) (15.33) Proposition 15.7. [Statistical hypothesis testing]. [Hypothesis test]. Consider the measurement ML∞(Ω0×R+)(O ≡ (X,F, F ), S[(β0,β1,σ)]). Here, the state (β0, β1, σ) is assumed to be unknown. Then, according to Corollary 6.6, we say: 392 Ishikawa's Homepage Chap. 15 Least-squares method and Regression analysis (B1) Assume the null hypothesis HN = {β0}(⊆ Θ0 = R). Then, the rejection region is as follows: Rα;XHN = Ê −1 0 (R α;Θ0 HN ) = ∩ ω∈Ω such that π0(ω)∈HN {x(∈ X) : dxΘ0(Ê0(x), π0(ω)) ≥ η α ω} = { x ∈ X : |β0(x)− β0|√ σ2(x) n (1 + a2/saa) ≥ tn−2(α/2) } (15.34) (B2) Assume the null hypothesis HN = {β1}(⊆ Θ1 = R). Then, the rejection region is as follows: Rα;XHN = Ê −1 1 (R α;Θ1 HN ) = ∩ ω∈Ω such that π1(ω)∈HN {x(∈ X) : dxΘ1(Ê1(x), π1(ω)) ≥ η α ω} = { x ∈ X : |β1(x)− β1|√ σ2(x) n (1/saa) ≥ tn−2(α/2) } (15.35) 393 Ishikawa's Homepage 15.4 Generalized linear model 15.4 Generalized linear model Put T = {0, 1, 2, * * * , i, * * * , n}, which is the same as the tree (15.10), that is, τ(i) = 0 (∀i = 1, 2, * * * , n) (15.36) 1 2 n 0 + ) k τ τ * * * * * * * * * * * * τ Figure 15.3: Parallel structure For each i ∈ T , define a locally compact space Ωi such that Ω0 = Rm+1 = { β =  β0 β1 ... βm  : β0, β1, * * * , βm ∈ R} (15.37) Ωi = R = { μi : μi ∈ R } (i = 1, 2, * * * , n) (15.38) Assume that aij ∈ R (i = 1, 2, * * * , n, j = 1, 2, * * * ,m, (m+ 1 ≤ n)) (15.39) which are called explanatory variables in the conventional statistics. Consider the deterministic causal map ψai• : Ω0(= Rm+1)→ Ωi(= R) such that Ω0 = Rm+1 3 β = (β0, β1, * * * , βm) 7→ ψai•(β0, β1, * * * , βm) = β0 + m∑ j=1 βjaij = μi ∈ Ωi = R (15.40) (i = 1, 2, * * * , n) Summing up, we see β =  β0 β1 β2 ... βm  7→  ψa1•(β0, β1, * * * , βm) ψa2•(β0, β1, * * * , βm) ψa3•(β0, β1, * * * , βm) ... ψan•(β0, β1, * * * , βm)  =  1 a11 a12 * * * a1m 1 a21 a22 * * * a2m 1 a31 a32 * * * a3m 1 a41 a42 * * * a4m ... ... ... ... ... 1 an1 an2 * * * anm  *  β0 β1 β2 ... βm  (15.41) 394 Ishikawa's Homepage Chap. 15 Least-squares method and Regression analysis which is equivalent to the deterministic Markov operator Ψai• : L ∞(Ωi)→ L∞(Ω0) such that [Ψai•(fi)](ω0) = fi(ψai•(ω0)) (∀fi ∈ L∞(Ωi), ∀ω0 ∈ Ω0,∀i ∈ 1, 2, * * * , n) (15.42) Thus, under the identification: aij ⇔ Ψai• , the term "explanatory variable" means a kind of causality. L∞(Ω1(≡ R)) L∞(Ω2(≡ R)) L∞(Ωn(≡ R)) L∞(Ω0(≡ Rm+1)) + ) k Ψa1• Ψa2• * * * * * * * * * * * * Ψan• Figure 15.4: Parallel structure(Causal relation Ψai•) Therefore, we have the observable Oai•0 ≡(R,BR,Ψai•Gσ) in L∞(Ω0(≡ Rm+1)) such that [Ψai•(Gσ(Ξ))](β) = [(Gσ(Ξ))](ψai•(β)) = 1 ( √ 2πσ2) ∫ Ξ exp [ − (x− (β0 + ∑m j=1 aijβj)) 2 2σ2 ] dx (15.43) (∀Ξ ∈ BR,∀β = (β0, β1, * * * , βm) ∈ Ω0(≡ Rm+1)) Hence, we have the simultaneous observable ×ni=1Oai•0 ≡(Rn,BRn ,×ni=1 Ψai•Gσ) in L∞(Ω0(≡ Rm+1)) such that [( n × i=1 Ψai•Gσ)( n × i=1 Ξi)](β) = n × i=1 ( [Ψai•Gσ)(Ξi)](β) ) = 1 ( √ 2πσ2)n ∫ * * * ∫ ×ni=1 Ξi exp [ − ∑n i=1(xi − (β0 + ∑m j=1 aijβj)) 2 2σ2 ] dx1 * * * dxn (15.44) (∀ n × i=1 Ξi ∈ BRn ,∀β = (β0, β1, * * * , βm) ∈ Ω0(≡ Rm+1)) Assuming that σ is variable, we have the observable O = ( Rn(= X),BRn(= F), F ) in L∞(Ω0× R+) such that [F ( n × i=1 Ξi)](β, σ) = [( n × i=1 Ψai•Gσ)( n × i=1 Ξi)](β) (∀ n × i=1 Ξi ∈ BRn , ∀(β, σ) ∈ Rm+1(≡ Ω0)× R+) (15.45) 395 Ishikawa's Homepage 15.4 Generalized linear model Thus, we have the following problem. Problem 15.8. [Generalized linear model in quantum language] Assume that a measured value x =  x1 x2 ... xn  ∈ X = Rn is obtained by the measurement ML∞(Ω0×R+)(O ≡ (X,F, F ), S[(β0,β1,*** ,βm,σ)]). (The measured value is also called a response variable.) And assume that we do not know the state (β0, β1, * * * , βm, σ2). Then, • from the measured value x = (x1, x2, . . . , xn) ∈ Rn, infer the β0, β1, * * * , βm, σ! That is, represent the (β0, β1, * * * , βm, σ) by (β0(x), β1(x), * * * , βm(x), σ(x)) (i.e., the functions of x). The answer is easy, since it is a slight generalization of Problem 15.3. Also, it suffices to follow ref. [8]. However, note that the purpose of this chapter is to propose Problem 15.8 (i.e, the quantum linguistic formulation of the generalized linear model) and not to give the answer to Problem 15.8. Remark 15.9. As a generalization of regression analysis, we also see measurement error model (cf. §5.5 (117 page) in ref. [30]), That is, we have two different generalizations such as Regression analysis −−−−−−−→ generalization  1© : generalized linear model 2© : measurement error model (15.46) However, we believe that the 1© is the main street. 396 Ishikawa's Homepage Chapter 16 Kalman filter (calculation) The Kalman filter [61, 65] is located as in the following (]): (]) : Statistics  Fisher's maximum likelihood method + causality−−−−−−−−−−−−→ usually deterministic regression analysis Bayes' method + causality−−−−−−−−−−→ non-deterministic Kalman filter Thus, I can not emphasize too much the importance of the Kalman filter. Though Kalman filter belongs to Bayes' statistics, this fact may not be a common sense. This present state is due to the confusion between Fisher's statistics and Bayes' statistics. I hope that such confusion should be clarified by the above (]) (based on quantum language). This chapter is extracted from the following paper: • S. Ishikawa, K. Kikuchi: Kalman filter in quantum language, arXiv:1404.2664 [math.ST] 2014. 16.1 Bayes=Kalman method (in L∞(Ω,m)) Recall Theorem 9.11(Bayes' theorem), particularly, the Bayes operator (9.5). This will be generalized as Bayes=Kalman operator as follows. Let t0 be the root of a tree T . For each t ∈ T , consider the classical basic structure: [C0(Ωt) ⊆ L∞(Ωt,mt) ⊆ B(L2(Ωt,mt))] Let [OT ] = [{Ot( ≡ (Xt, Ft, Ft))}t∈T , {Φt1,t2 : L∞(Ωt2) → L∞(Ωt1)}(t1,t2)∈T 2≤ ] be a sequential causal observable with the realization Ôt0 ≡ (×t∈T Xt,  t∈TGt, Ft0) in L∞(Ωt0). For example, 397 16.1 Bayes=Kalman method (in L∞(Ω,m)) [L∞(Ω0) : O0] [L∞(Ω1) : O1] [L∞(Ω2) : O2] [L∞(Ω3) : O3] [L∞(Ω4) : O4] [L∞(Ω5) : O5][L∞(Ω6) : O6] [L∞(Ω7) : O7] ) i k + k ) k Φ0,6 Φ0,1 Φ0,7 Φ1,2 Φ1,5 Φ2,3 Φ2,4 Figure 16.1 : Simple classical example of sequential causal observable For each t ∈ T , consider another observable O′t = (Yt,Gt, Gt) in L∞(Ωt,mt), and the simultaneous observable O × O′t = (Xt × Yt,Ft  Gt, Ft × Gt) in L∞(Ωt,mt). And let [O×T ] = [{O×t ( ≡ (Xt×Yt, FtGt, Ft×Gt))}t∈T , {Φt1,t2 : L∞(Ωt2)→ L∞(Ωt1)}(t1,t2)∈T 2≤ ] be a sequential causal observable with the realization Ô×t0 ≡ (×t∈T (Xt× Yt),  t∈T (FtGt), Ĥt0) in L∞(Ωt0). For example, [L∞(Ω0) : O × 0 ] [L∞(Ω1) : O × 1 ] [L∞(Ω2) : O × 2 ] [L∞(Ω3) : O × 3 ] [L∞(Ω4) : O × 4 ] [L∞(Ω5) : O × 5 ][L∞(Ω6) : O × 6 ] [L∞(Ω7) : O × 7 ] ) i k + k ) k Φ0,6 Φ0,1 Φ0,7 Φ1,2 Φ1,5 Φ2,3 Φ2,4 Figure 16.2 : Simple classical example of sequential causal observable Thus we have the mixed measurement ML∞(Ωt0 )(Ô × t0 , S[∗](z0)), where z0 ∈ L1+1(Ωt0). Assume that we know that the measured value (x, y) (= ((xt)t∈T , (yt)t∈T , ) ∈ (×t∈T Xt)×(×t∈T Yt)) obtained by the measurement ML∞(Ωt0)(Ô × t0 , S[∗](z0)) belongs to (×t∈T Ξt)× (×t∈T Yt) (∈ (t∈TFt)  (t∈TGt)). Then, by Axiom(m) 1(§9.1), we can infer that (A) the probability P×t∈TΞt((Gt(Γt))t∈T ) that y belongs to×t∈T Γt(∈ t∈TGt) is given by P×t∈TΞt((Gt(Γt))t∈T ) = ∫ Ω0 [Ĥt0((×t∈T Ξt)×(×t∈T Γt))](ω0) z0(ω0) m0(dω0)∫ Ω0 [Ĥt0(×t∈T Ξt)×(×t∈T Yt)](ω0) z0(ω0) m0(dω0) (16.1) (∀Γt ∈ Gt, t ∈ T ). 398 Ishikawa's Homepage Chap. 16 Kalman filter (calculation) Let s ∈ T be fixed. Assume that Γt = Yt (∀t ∈ T such that t 6= s) Thus, putting P×t∈TΞt(Gs(Γs)) = P×t∈TΞt((Gt(Γt))t∈T ), we see that P×t∈TΞt ∈ L1+1(Ωs,ms). That is, there uniquely exists zas ∈ L1+1(Ωs,ms) such that P×t∈TΞt((Gs(Γs)) = L1(Ωs)〈z a s , Gs(Γs)〉L∞(Ωs) = ∫ Ωs [Gs(Γs)](ωs)z a s (ωs)ms(dωs) for any observable (Ys,Gs, Gs) in L ∞(Ωs). That is because the linear functional P×t∈TΞt : L∞(Ωs)→ C (complex numbers) is weak∗ continuous. After all, (B) we can define the Bayes-Kalman operator [Bs Ôt0 (×t∈T Ξt)] : L1+1(Ωt0)→ L1+1(Ωs) such that (pretest state) z0 (∈L1+1(Ωt0 )) [Bs Ôt0 (×t∈T Ξt)] −−−−−−−−−−−−−−−−−−−→ Bayes-Kalman operator (posttest state) zas (∈L1+1(Ωs)) (16.2) which is the generalization of the Bayes operator (9.5). Remark 16.1. We have frequently discussed the Bayes=Kalman filter, for example, in [30, 33]. However, these arguments are too theoretical. In this chapter, we devote ourselves to the numerical aspect of the Kalman filter. 399 Ishikawa's Homepage 16.2 Problem establishment (concrete calculation) 16.2 Problem establishment (concrete calculation) In the previous section, we study the general theory of Kalman filter. In this section, we devote ourselves to the calculation of Kalman filter in the case of a linear ordered tree T = {0, 1, 2, * * * , n} such that the parent map π : T \ {0} → T is defined by π(k) = k − 1: 0 π←−−−− 1 π←−−−− 2 π←−−−− * * * π←−−−− n− 1 π←−−−− n Figure 16.3: Linear ordered tree For each k ∈ T , consider the classical basic structure: [C0(Ωk) ⊆ L∞(Ωk,mk) ⊆ B(L∞(Ωk,mk))] ( = [C0(R) ⊆ L∞(R, dω) ⊆ B(L2(R, dω))] ) where dω is the Lebesgue measure on R. Consider the sequential causal observable [OT ] = [{Ot}t∈T , {Φt−1,t : L∞(Ωt) → L∞(Ωt−1)}T=1,2,*** ,n ], and assume the initial state z0 ∈ L1+1(Ω0,m0). Thus, we have the following situation: initial state z0 L∞(Ω0,m0) O0=(X0,F0F0) Φ0,1←−− L∞(Ω1,m1) O1=(X1,F1F1) Φ1,2←−− * * * Φ s−1,s ←−−−− L∞(Ωs,ms) Os=(Xs,FsFs) Φs,s+1←−−−− * * * Φ n−1,n ←−−−− L∞(Ωn,mn) On=(Xn,FnFn) or, equivalently, initial state z0 L1(Ω0,m0) O0=(X0,F0,F0) Φ0,1∗−−→ L1(Ω1,m1) O1=(X1,F1,F1) Φ1,2∗−−→ * * * Φ s−1,s ∗−−−−→ L1(Ωs,ms) Os=(Xs,Fs,Fs) Φs,s+1∗−−−−→ * * * Φ n−1,n ∗−−−−→ L1(Ωn,mn) On=(Xn,Fn,Fn) In the above, the initial state z0(∈ L1+1(Ω0,m0)) is defined by z0(ω0) = 1√ 2πσ0 exp[−(ω0 − μ0) 2 2σ20 ] (∀ω0 ∈ Ω0) (16.3) where it is assumed that μ0 and σ0 are known. Also, for each t ∈ T = {0, 1, * * * , n}, consider the observable Ot = (Xt,Ft, Ft) = (R,BR, Ft) in L∞(Ωt,mt) such that [Ft(Ξt)](ωt) = ∫ Ξt 1√ 2πqt exp[−(xt − ctωt − dt) 2 2q2t ]dxt ≡ ∫ Ξt fxt(ωt)dxt (∀Ξt ∈ Ft, ∀ωt ∈ Ωt) (16.4) where it is assumed that ct, dt and qt are known (t ∈ T ). And further, the causal operator Φt−1.t : L∞(Ωt)→ L∞(Ωt−1) is defined by [Φt−1,tfxt ](ωt−1) = ∫ ∞ −∞ 1√ 2πrt exp[−(ωt − atωt−1 − bt) 2 2r2t ]fxt)dωt ≡ ft−1(ωt−1) (16.5) 400 Ishikawa's Homepage Chap. 16 Kalman filter (calculation) (∀fxt ∈ L∞(Ωt,mt), ∀ωt−1 ∈ Ωt−1) where it is assumed that at, bt and rt are known (t ∈ T ). Or, equivalently, the pre-dual causal operator Φt−1.t∗ : L 1 +1(Ωt−1)→ L1+1(Ωt) is defined by [Φt−1,t∗ zt−1](ωt) = ∫ ∞ −∞ 1√ 2πrt exp[−(ωt − atωt−1 − bt) 2 2r2t ]zt−1(ωt−1)dωt−1 (16.6) (∀zt−1 ∈ L1+1(Ωt−1,mt−1), ∀ωt ∈ Ωt) Now we have the sequential causal observable [OT ] = [{Ot}t∈T , {Φt−1,t : L∞(Ωt)→ L∞(Ωt−1)}T=1,2,*** ,n Let Ô0 (×nt=0Xt,nt=0Ft, F ) be its realization. Then we have the following problem: Problem 16.2. [Kalman filter; calculation] Assume that a measured value (x0, x2, * * * , xn) (∈×nt=0Xt) is obtained by the measurement ML∞(Ω0) (Ô0, S[∗](z0)). Let s(∈ T ) be fixed. Then, calculate the Bayes-Kalman operator [Bs Ô0 (×t∈T{xt})](z0) in (16.2), where [Bs Ô0 (× t∈T {xt})](z0) = zas = lim Ξt→xt (t∈T ) [Bs Ô0 (× t∈T Ξt)](z0) That is, L1+1(Ω0) 3 z0 measured value:(x0,x1,...,xn)−−−−−−−−−−−−−−−−→ Bs Ô0 (×t∈T {xt}) zas ∈ L1+1(Ωs) 401 Ishikawa's Homepage 16.3 Bayes=Kalman operator Bs Ô0 (×t∈T{xt}) 16.3 Bayes=Kalman operator Bs Ô0 (×t∈T{xt}) In what follows, we solve Problem 16.2. For this, it suffices to find the zs ∈ L1+1(Ωs) such that lim Ξt→xt (t∈T ) ∫ Ω0 [F0((×nt=0 Ξt)× Γs)](ω0) z0(ω0)dω0∫ Ω0 [F0(×nt=0 Ξt)](ω0) z0(ω0)dω0 = ∫ Ωs [Gs(Γs)](ωs) zs(ωs)dωs (∀Γs ∈ Fs) Let us calculate zs = [B s Ô0 (×t∈T{xt})](z0) as follows. ∫ Ω0 [F0(( n × t=0 Ξt)× Γs)](ω0) z0(ω0)dω0 = L1(Ω0) 〈z0, F0(( n × t=0 Ξt)× Γs)〉L∞(Ω0) = L1(Ω1) 〈Φ0,1∗ (F0(Ξ0)z0), F1(( n × t=1 Ξt)× Γs)〉L∞(Ω1) (16.7) (A) and, putting z0 = F0(Ξ0)z0 (or, exactly, its normalization, i.e., z0 = limΞ0→x0 F0(Ξ0)z0∫ Ω0 F0(Ξ0)z0dω0 ) , z1 = F1(Ξ1)Φ 0,1 ∗ (z0), z2 = F2(Ξ2)Φ 1,2 ∗ (z1), * * * , zs−1 = Fs−1(Ξs−1)Φs−2,s−1∗ (zs−2), we see that (16.7) = L1(Ω1) 〈Φ0,1∗ (z0), F1(( n × t=1 Ξt)× Γs)〉L∞(Ω1) = L1(Ω2) 〈Φ1,2∗ (z1), F2(( n × t=2 Ξt)× Γs)〉L∞(Ω2) * * * * * * = L1(Ωs+1) 〈Φs,s+1∗ (zs), Fs+1(( n × t=s+1 Ξt)× Γs)〉L∞(Ωs+1) = L1(Ωs) 〈Φs−1,s∗ (zs−1), Fs(( n × t=s Ξt)× Γs)〉L∞(Ωs) = L1(Ωs) 〈Φs−1,s∗ (zs−1), Fs(Ξs)Gs(Γs)Φs,s+1Fs+1( n × t=s+1 Ξt)〉L∞(Ωs) = L1(Ωs) 〈 ( Fs(Ξs)Φ s,s+1Fs+1( n × t=s+1 Ξt) )( Φs−1,s∗ (zs−1) ) , Gs(Γs)〉L∞(Ωs) (16.8) Thus, we see [Bs Ô0 (× t∈T {xt})](z0) = lim Ξt→xt (t∈T ) ( Fs(Ξs)Φ s,s+1Fs+1(×nt=s+1 Ξt) ) × ( Φs−1,s∗ zs−1) ) ∫ Ω0 [F0(×nt=0 Ξt)](ω0) z0(ω0)dω0 (16.9) 402 Ishikawa's Homepage Chap. 16 Kalman filter (calculation) 16.4 Calculation: prediction part 16.4.1 Calculation: zs = Φ s−1,s ∗ (zs−1) in (16.9) We prepare the following lemma. Lemma 16.3. It holds that (B1) ∫∞ −∞ 1√ 2πA exp[− (x−By) 2 2A2 ] 1√ 2πC exp[− (y−D) 2 2C2 ]dy = 1√ 2π √ A2+B2C2 exp[− (x−BD) 2 2(A2+B2C2) ] (B2) exp[− (Aω−B) 2 2E2 ] exp[− (Cω−D) 2 2F 2 ] ≈ exp[−1 2 (A 2F 2+C2E2 E2F 2 ) ( ω − (ABF 2+CDE2) (A2F 2+C2E2) )2 ] where the notation "≈" means as follows: "f(ω) ≈ g(ω)"⇐⇒ "there exists a positive K such that f(ω) = Kg(ω) (∀ω ∈ Ω)" Proof. It is easy, thus we omit the proof. We see, by (16.3) and (A), that z0(ω0) = lim Ξ0→x0 F (Ξ0)z0∫ R F (Ξ0)z0dω0 ≈ 1√ 2πq0 exp[−(x0 − c0ω0 − d0) 2 2q20 ] 1√ 2πσ0 exp[−(ω0 − μ0) 2 2σ20 ] ≈ 1√ 2πσ0 exp[−(ω0 − μ0) 2 2σ20 ] (16.10) where σ20 = q20σ 2 0 q20 + c 2 0σ 2 0 , μ0 = μ0 + σ 2 0( c0 q20 )(x0 − d0 − c0μ0) (16.11) Further, the (B1) in Lemma 16.3 and (16.6) imply that z1(ω1) = [Φ 0,1 ∗ z0](ω1) = ∫ ∞ −∞ 1√ 2πr1 exp[−(ω1 − a1ω0 − b1) 2 2r21 ] 1√ 2πσ0 exp[−(ω0 − μ0) 2 2σ20 ]dω0 = 1√ 2πσ1 exp[−(ω1 − μ1) 2 2σ12 ] (16.12) where σ21 = a 2 1σ 2 0 + r 2 1, μ1 = a1μ0 + b1 (16.13) Thus, we see, by (B2) in Lemma 16.3, that zt−1(ωt−1) = lim Ξt−1→xt−1 F (Ξt−1)zt−1∫ R F (Ξt−1)zt−1dωt−1 403 Ishikawa's Homepage 16.4 Calculation: prediction part ≈ 1√ 2πqt−1 exp[−(xt−1 − ct−1ωt−1 − dt−1) 2 2q2t−1 ] 1√ 2πσt−1 exp[−(ωt−1 − μt−1) 2 2σ2t−1 ] ≈ 1√ 2πσt−1 exp[−(ωt−1 − μt−1) 2 2σ2t−1 ] (16.14) where σ2t−1 = q2t−1σ 2 t−1 q2t−1 + c 2 t−1σ 2 t−1 = σ2t−1 q2t−1 + c 2 t−1σ 2 t−1 + q 2 t−1 − q2t−1 − c2t−1σ2t−1 q2t−1 + c 2 t−1σ 2 t−1 = σ2t−1(1− c2t−1σ 2 t−1 q2t−1 + c 2 t−1σ 2 t−1 ) μt−1 = μt−1 + σ 2 t−1( ct−1 q2t−1 )(xt−1 − ct−1μt−1) (16.15) Further, we see, by (B1) in Lemma 16.3, that zt(ωt) = [Φ t−1,t ∗ zt−1](ωt) ≈ ∫ ∞ −∞ 1√ 2πrt exp[−(ωt − atωt−1 − bt) 2 2r2t ] 1√ 2πσt−1 exp[−(ωt−1 − μt−1) 2 2σ2t−1 ]dωt−1 ≈ 1√ 2πσt exp[−(ωt − μt) 2 2σt2 ] (16.16) where σ2t = a 2 t σ 2 t−1 + r 2 t , μt = atμt−1 + bt (16.17) Summing up the above (16.10)–(16.17), we see: z0 μ0,σ0 x0−−−−−→ (16.11) z0 μ0,σ0 Φ0,1∗−−−−−→ (16.13) z1 μ1,σ1 x1−−→ * * * Φt−2,t−1∗−−−−−−−→ zt−1 μt−1,σt−1 xt−1−−−−−→ (16.15) zt−1 μt−1,σt−1 Φt−1,t∗−−−−−→ (16.17) zt μt,σt xt+1−−−−→ * * * Φs−1,s∗−−−−−→ zs μs,σs And thus, we get zs = Φ s−1,s ∗ (zs−1) (16.18) in (16.9). 404 Ishikawa's Homepage Chap. 16 Kalman filter (calculation) 16.5 Calculation: Smoothing part 16.5.1 Calculation: ( Fs(Ξs)Φ s,s+1Fs+1(×nt=s+1 Ξt) ) in (16.9) Put fxn(ωn) = 1√ 2πqn exp[−(xn − cnωn − dn) 2 2q2n ] ≈ exp[−(cnωn − (xn − dn)) 2 2q2n ] ≡ exp[−1 2 ( ũnωn − ṽn )2 ] (16.19) where it is assumed that cn, dn and qn are known (t ∈ T ). And thus, put ũn = cn qn , ṽn = xn − dn qn (16.20) And further, Lemma 16.3 implies that the causal operator Φt−1.t : L∞(Ωt) → L∞(Ωt−1) is defined by ft−1(ωt−1) = [Φ t−1,tfxt ](ωt−1) ≈ ∫ ∞ −∞ 1√ 2πrt exp[−(ωt − atωt−1 − bt) 2 2r2t ] exp[−(ũtωt − ṽt) 2 2 ]dωt ≈ exp[−1 2 ( ṽt√ 1 + r2t ũ 2 t − ũt(atωt−1 + bt)√ 1 + r2t ũ 2 t )2 ] ≈ exp[−1 2 ( ut−1ωt−1 − vt−1 )2 ] (16.21) where ut−1 = − atũt√ 1 + r2t ũ 2 t , vt−1 = btũt − ṽt√ 1 + r2t ũ 2 t (16.22) And also, Lemma 16.3 implies that fxt−1(ωt−1) = exp[− (ct−1ωt−1 + dt−1 − xt−1)2 2q2t−1 ] exp[−(ut−1ωt−1 − vt−1) 2 2 ] ≈ exp[−1 2 ( c2t−1 + u 2 t−1q 2 t−1 q2t−1 ) ( ωt−1 − ct−1(dt−1 − tt−1) + ut−1vt−1q2t−1 c2t−1 + u 2 t−1q 2 t−1 )2 ] ≈ exp[−1 2 ( ũt−1ωt−1 − ṽt−1 )2 ] (16.23) where ũt−1 = √ c2t−1 + u 2 t−1q 2 t−1 qt−1 , ṽt−1 = ct−1(dt−1 − tt−1) + ut−1vt−1q2t−1 qt−1 √ c2t−1 + u 2 t−1q 2 t−1 (16.24) Summing up the above (16.19)-(16.24), we see: ũs,ṽs fxs ws xs←−− * * * Φ t−2,t−1 ←−−−−−−− ũt−1,ṽt−1 fxt−1 wt−1 xt−1←−−−−− (16.24) ut−1,vt−1 ft−1 wt−1 Φt−1,t←−−−−− (16.22) ũt,ṽt fxt wt xt←−− * * * xn−1←−−−− un−1,vn−1 fn−1 wn−1 Φn−1,n←−−−−− ũnṽn fxn=(16.19) wn 405 Ishikawa's Homepage 16.5 Calculation: Smoothing part And thus, we get fxs ≈ lim Ξt→xt (t∈{s.s+1,*** ,n}) ( Fs(Ξs)Φ s,s+1Fs+1(×nt=s+1 Ξt) ) ‖Fs(Ξs)Φs,s+1Fs+1(×nt=s+1 Ξt) ) ‖L∞(Ωs) (16.25) in (16.9) After all, we solve Problem16.2(Kalman Filter), that is, Answer 16.4. [The answer to Problem16.2(Kalman Filter)] (A) Assume that a measured value (x0, x2, * * * , xn) (∈ ×nt=0Xt) is obtained by the measurement ML∞(Ω0) (Ôt0 , S[∗](z0)). Let s(∈ T ) be fixed. Then, we get the Bayes-Kalman operator [Bs Ôt0 (×t∈T{xt})](z0), that is, ( [Bs Ôt0 (× t∈T {xt})]z0 ) (ωs) = fxs(ωs) * zs(ωs)∫∞ −∞ fxs(ωs) * zs(ωs)dωs = zas (ωs) (∀ωs ∈ Ωs) where zs in (16.18) and fxs in (16.25) can be iteratively calculated as mentioned in this section. Remark 16.5. The following classification is usual (B1) Smoothing: in the case that 0 ≤ s < n (B2) Filter: in the case that s = n (B3) Prediction: in the case that s = n and, for any m such that n0 ≤ m < n, the existence observable (Xm,Fm, Fm) = ({1}, {∅, {1}}, Fm) is defined by Fm(∅) ≡ 0, Fm({1}) ≡ 1, 406 Ishikawa's Homepage Chapter 17 Equilibrium statistical mechanics and Ergodic Hypothesis In this chapter, we study and answer the following fundamental problems concerning classical equilibrium statistical mechanics: (A) Is the principle of equal a priori probabilities indispensable for equilibrium statistical mechanics? (B) Is the ergodic hypothesis related to equilibrium statistical mechanics? (C) Why and where does the concept of "probability" appear in equilibrium statistical mechanics? Note that there are several opinions for the formulation of equilibrium statistical mechanics. In this sense, the above problems are not yet answered. Thus we propose the measurement theoretical foundation of equilibrium statistical mechanics, and clarify the confusion between two aspects (i.e., probabilistic and kinetic aspects in equilibrium statistical mechanics), that is, we discuss{ the kinetic aspect (i.e, causality) * * * in Section 17.1 the probabilistic aspect (i.e., measurement) * * * in Section 17.2 And we answer the above (A) and (B), that is, we conclude that (A) is "No", but, (B) is "Yes". and further, we can understand the problem (C). This chapter is extracted from the following: [35] S. Ishikawa, " and Equilibrium Statistical Mechanics in the Quantum Mechanical World View," World Journal of Mechanics, Vol. 2, No. 2, 2012, pp. 125-130. doi: 10.4236/wim.2012.22014. 17.1 Equilibrium statistical mechanical phenomena concerning Axiom 2 (causality) 407 17.1 Equilibrium statistical mechanical phenomena concerning Axiom 2 (causality) 17.1.1 Equilibrium statistical mechanical phenomena Hypothesis 17.1. [ Equilibrium statistical mechanical hypothesis ]. Assume that about N(≈1024 ≈ 6.02 × 1023 ≈ "the Avogadro constant") particles (for example, hydrogen molecules) move in a box with about 20 liters. It is natural to assume the following phenomena 1© – 4©: 1© Every particle obeys Newtonian mechanics. 2© Every particle moves uniformly in the box. For example, a particle does not halt in a corner. 3© Every particle moves with the same statistical behavior concerning time. 4© The motions of particles are (approximately) independent of each other. U 7  W -i z U 9 M R  7 W y U ) : K R z K Y - 3  q  qy  K 9 -  * - U o O W U i 9 U  z  K * R K w W  i z K U R 9 N   s   j) 9 U I 9 N K * (17.1) In what follows we shall devote ourselves to the problem: (D) how to describe the above equilibrium statistical mechanical phenomena 1© – 4© in terms of quantum language ( =measurement theory). 17.1.2 About 1© in Hypothesis 17.1 In Newtonian mechanics, any state of a system composed of N( ≈ 1024) particles is represented by a point (q, p) ( ≡ (position, momentum) = (q1n, q2n, q3n, p1n, p2n, p3n)Nn=1 ) in a phase (or state) space R6N . Let H : R6N → R be a Hamiltonian such that H ( (q1n, q2n, q3n, p1n, p2n, p3n) N n=1 ) = momentum energy + potential energy 408 Ishikawa's Homepage Chap. 17 Equilibrium statistical mechanics and Ergodic Hypothesis =[ N∑ n=1 ∑ k=1,2,3 (pkn) 2 2× particle's mass ]+U((q1n, q2n, q3n) N n=1). (17.2) Fix a positive E > 0. And define the measure ν E on the energy surface Ω E (≡ {(q, p) ∈ R6N | H(q, p) = E}) such that ν E (B) = ∫ B |∇H(q, p)|−1dm6N−1 (∀B ∈ BΩ E , the Borel field of Ω E ) where |∇H(q, p)| = [ N∑ n=1 ∑ k=1,2,3 {( ∂H ∂pkn )2 + ( ∂H ∂qkn )2}]1/2 and dm6N−1 is the usual surface Lebesgue measure on ΩE . Let {ψEt }−∞<t<∞ be the flow on the energy surface Ω E induced by the Newton equation with the Hamiltonian H, or equivalently, Hamilton's canonical equation: dqkn dt = ∂H ∂pkn , dpkn dt = − ∂H ∂qkn , (17.3) (k = 1, 2, 3, n = 1, 2, . . . , N). Liouville's theorem (cf.[64]) says that the measure ν E is invariant concerning the flow {ψEt }−∞<t<∞. Defining the normalized measure νE such that νE = ν E ν E (Ω E ) , we have the normalized measure space (Ω E ,BΩ E , ν E ). Putting A = C0(ΩE) = C(ΩE) (from the compactness of ΩE), we have the classical basic structure: [C(Ω E ) ⊆ L∞(Ω E , ν E ) ⊆ B(L2(Ω E , ν E ))] Thus, putting T = R, and solving the (17.4), we get ωt = (q(t), p(t)), φt1.t2 = ψEt2−t1 , Φ∗t1.t2δωt1 = δφt1.t2 (ωt1 ) (∀ωt1 ∈ ΩE), and further we define the sequential deterministic causal operator {Φt1,t2 : L∞(ΩE)→ L∞(ΩE)}(t1.t2)∈T 2≤ (cf. Definition 10.4). 17.1.3 About 2© in Hypothesis 17.1 Now let us begin with the well-known ergodic theorem (cf. [64]). For example, consider one particle P1. Put SP1 = {ω ∈ ΩE | a state ω such that the particle P1 stays around a corner of the box } Clearly, it holds that SP1 ( ΩE . Also, if ψEt (SP1) ⊆ SP1 (0 5 ∀t < ∞), then the particle P1 must always stay a corner. This contradicts 2©. Therefore, 2© means the following: 409 Ishikawa's Homepage 17.1 Equilibrium statistical mechanical phenomena concerning Axiom 2 (causality) 2©′ [Ergodic property]: If a compact set S(⊆ Ω E , S 6= ∅) satisfies ψEt (S) ⊆ S (0 5 ∀t < ∞), then it holds that S = Ω E . The ergodic theorem (cf. [64]) says that the above 2©′ is equivalent to the following equality:∫ Ω E f(ω)ν E (dω) ((state) space average) = lim T→∞ 1 T ∫ α+T α f(ψEt (ω0))dt (time average) (17.4) (∀α ∈ R,∀f ∈ C(Ω E ), ∀ω0 ∈ ΩE) After all, the ergodic property 2©′ (⇔ (17.4) ) says that if T is sufficiently large, it holds that∫ Ω E f(ω)ν E (dω)≈ 1 T ∫ α+T α f(ψEt (ω0))dt. (17.5) Putm T (dt) = dt T . The probability space ([α, α+T ],B[α,α+T ],mT ) (or equivalently, ([0, T ],B[0,T ], m T ) ) is called a (normalized) first staying time space, also, the probability space (Ω E ,BΩ E , ν E ) is called a (normalized)second staying time space. Note that these mathematical probability spaces are not related to "probability" (Recall the linguistic interpretation (§3.1) :there is no probability without measurement). 17.1.4 About 3© and 4© in Hypothesis 17.1 Put KN = {1, 2, . . . , N(≈1024)}. For each k ( ∈ KN), define the coordinate map πk : ΩE( ⊂ R6N)→ R6 such that πk(ω) = πk(q, p) =πk((q1n, q2n, q3n, p1n, p2n, p3n) N n=1) =(q1k, q2k, q3k, p1k, p2k, p3k) (17.6) for all ω = (q, p) = (q1n, q2n, q3n, p1n, p2n, p3n) N n=1 ∈ ΩE( ⊂ R6N). Also, for any subset K ( ⊆ KN= {1, 2, . . . , N (≈1024)}), define the distribution map D(*)K : Ω E ( ⊂ R6N) →Mm+1(R6) such that D (q,p) K = 1 ][K] ∑ k∈K δπk(q,p) (∀(q, p) ∈ ΩE( ⊂ R 6N)) where ][K] is the number of the elements of the set K. Let ω0(∈ ΩE) be a state. For each n (∈ KN), we define the map Xω0n : [0, T ] → R6 such that Xω0n (t) = πn(ψ E t (ω0)) (∀t ∈ [0, T ]). (17.7) 410 Ishikawa's Homepage Chap. 17 Equilibrium statistical mechanics and Ergodic Hypothesis And, we regard {Xω0n }Nn=1 as random variables (i.e., measurable functions ) on the probability space ([0, T ],B[0,T ],mT ). Then, 3© and 4© respectively means 3©′ {Xω0n }Nn=1 is a sequence with the approximately identical distribution concerning time. In other words, there exists a normalized measure ρ E on R6 (i.e., ρ E ∈Mm+1(R6)) such that: m T ({t ∈ [0, T ] : Xω0n (t) ∈ Ξ})≈ ρE(Ξ) (17.8) (∀Ξ ∈ BR6 , n = 1, 2, . . . , N) 4©′ {Xω0n }Nn=1 is approximately independent, in the sense that, for any K0 ⊂ {1, 2, . . . , N(≈1024)} such that 1 5 ][K0] N ( that is, ][K0]N ≈0 ), it holds that m T ({t ∈ [0, T ] : Xω0k (t) ∈ Ξk(∈ BR6), k ∈ K0}) ≈ × k∈K0 m T ({t ∈ [0, T ] : Xω0k (t) ∈ Ξk(∈ BR6)}). Here, we can assert the advantage of our method in comparison with Ruelle's method (cf.[78]) as follows. Remark 17.2. [About the time interval [0, T ]]. For example, as one of typical cases, consider the motion of 1024 particles in a cubic box (whose long side is 0.3m). It is usual to consider that "averaging velocity"=5× 102m/s, "mean free path"=10−7m. And therefore, the collisions rarely happen among ][K0] particles in the time interval [0, T ], and therefore, the motion is "almost independent". For example, putting ][K0] = 10 10, we can calculate the number of times a certain particle collides with K0-particles in [0,T] as (10 −7 × 1024 1010 )−1 × (5× 102) × T ≈ 5 × 10−5 × T . Hence, in order to expect that 3©′ and 4©′ hold, it suffices to consider that T ≈ 5 seconds. /// Also, we see, by (17.7) and (17.5), that, for K0(⊆ KN) such that 1 ≤ ][K0] N , m T ({t ∈ [0, T ] : Xω0k (t) ∈ Ξk(∈ BR6), k ∈ K0}) =m T ({t ∈ [0, T ] : πk(ψEt (ω0) ∈ Ξk(∈ BR6), k ∈ K0}) =m T ({t ∈ [0, T ] : ψEt (ω0) ∈ ((πk)k∈K0)−1(× k∈K0 Ξk)}) ≈ ν E ( ((πk)k∈K0) −1( × k∈K0 Ξk) ) ≡ ( ν E ◦ ((πk)k∈K0)−1 ) ( × k∈K0 Ξk). (17.9) 411 Ishikawa's Homepage 17.1 Equilibrium statistical mechanical phenomena concerning Axiom 2 (causality) Particularly, putting K0 = {k}, we see: m T ({t ∈ [0, T ] : Xω0k (t) ∈ Ξ})≈ (νE ◦ π −1 k )(Ξ) (∀Ξ ∈ BR6). (17.10) Hence, we can describe the 3© and 4© in terms of {πk} in what follows. Hypothesis 17.3. [ 3© and 4© ]. Put KN = {1, 2, . . . , N(≈1024)}. Let H, E, νE , νE , πk : Ω E → R6 be as in the above. Then, summing up 3© and 4©, by (17.9) we have: (E) {πk : ΩE → R6}Nk=1 is approximately independent random variables with the identical distribution in the sense that there exists ρ E (∈Mm+1(R6)) such that⊗ k∈K0 ρ E (= "product measure")≈ ν E ◦ ((πk)k∈K0)−1. (17.11) for all K0 ⊂ KN and 1 5 ][K0] N . Also, a state (q, p)(∈ Ω E ) is called an equilibrium state if it satisfies D (q,p) KN ≈ρ E . 17.1.5 Ergodic Hypothesis Now, we have the following theorem (cf.[35]): Theorem 17.4. [Ergodic hypothesis]. Assume Hypothesis 17.3 ( or equivalently, 3© and 4© ). Then, for any ω0 = (q(0), p(0)) ∈ ΩE , it holds that [D (q(t),p(t)) KN ](Ξ)≈ m T ({t ∈ [0, T ] : Xω0k (t) ∈ Ξ}) (∀Ξ ∈ BR6 , k = 1, 2, . . . , N(≈1024)) (17.12) for almost all t. That is, 0 5 m T ({t ∈ [0, T ] : (17.12) does not hold}) 1. Proof. Let K0 ⊂ KN such that 1 ][K0] ≡ N0 N (that is, 1][K0]≈0≈ ][K0] N ). Then, from Hypothesis A, the law of large numbers (cf. [63]) says that D (q(t),p(t)) K0 ≈ ν E ◦ π−1k ( ≈ ρE ) (17.13) for almost all time t. Consider the decomposition KN = {K(1), K(2), . . . , K(L)}. (i.e., KN =∪L l=1K(l), K(l) ∩K(l′) = ∅ (l 6= l′) ), where ][K(l)]≈N0 (l = 1, 2, . . . , L). From (7.13), it holds that, for each k ( = 1, 2, . . . , N (≈1024)), D (q(t),p(t)) KN = 1 N L∑ l=1 [][K(l)]×D(q(t),p(t))K(l) ] 412 Ishikawa's Homepage Chap. 17 Equilibrium statistical mechanics and Ergodic Hypothesis ≈ 1 N L∑ l=1 [][K(l)]× ρE ]≈ νE ◦ π−1k ( ≈ ρE ), (17.14) for almost all time t. Thus, by (17.10), we get (17.12). Hence, the proof is completed. We believe that Theorem 17.4 is just what should be represented by the "ergodic hypothesis" such that "population average of N particles at each t" ="time average of one particle". Thus, we can assert that the ergodic hypothesis is related to equilibrium statistical mechanics (cf. the (B) in the abstract). Here, the ergodic property 2©′ (or equivalently, equality (17.5)) and the above ergodic hypothesis should not be confused. Also, it should be noted that the ergodic hypothesis does not hold if the box ( containing particles ) is too large. Remark 17.5. [The law of increasing entropy]. The entropy H(q, p) of a state (q, p)(∈ Ω E ) is defined by H(q, p) = k log[ν E ({(q′, p′) ∈ Ω E : D (q,p) KN ≈ D(q ′,p′) KN )})] where k = [Boltzmann constant]/([Plank constant]3NN !) Since almost every state in Ω E is equilibrium, the entropy of almost every state is equal k log ν E (Ω E ). Therefore, it is natural to assume that the law of increasing entropy holds. 413 Ishikawa's Homepage 17.2 Equilibrium statistical mechanical phenomena concerning Axiom 1 ( Measurement) 17.2 Equilibrium statistical mechanical phenomena concerning Axiom 1 ( Measurement) In this section we shall study the probabilistic aspects of equilibrium statistical mechanics. For completeness, note that (F) the argument in the previous section is not related to "probability" since Axiom 1 (measurement; §2.7) does not appear in Section 17.1. Also, Recall the linguistic interpretation (§3.1) : there is no probability without measurement. Note that the (17.12) implies that the equilibrium statistical mechanical system at almost all time t can be regarded as: (G) a box including about 1024 particles such as the number of the particles whose states belong to Ξ ( ∈ BR6) is given by ρE(Ξ)× 1024. Thus, it is natural to assume as follows. (H) if we, at random, choose a particle from 1024 particles in the box at time t, then the probability that the state (q1, q2, q3, p1, p2, p3) (∈ R6) of the particle belongs to Ξ ( ∈ BR6) is given by ρ E (Ξ). In what follows, we shall represent this (H) in terms of measurements. Define the observable O0 = (R6,BR6 , F0) in L∞(ΩE) such that [F0(Ξ)](q, p) = [D (q,p) KN ](Ξ) ( ≡ ][{k | πk(q, p) ∈ Ξ}] ][KN ] ) (∀Ξ ∈ BR6 ,∀(q, p) ∈ ΩE( ⊂ R6N)). (17.15) Thus, we have the measurement ML∞(ΩE)(O0 := (R6,BR6 , F0), S[δψt(q0 ,p0 )]). Then we say, by Axiom 1 (measurement; §2.7) , that (I) the probability that the measured value obtained by the measurement ML∞(ΩE)(O0 := (R6,BR6 , F0), S[δψt(q0 ,p0 )]) belongs to Ξ(∈ BR6) is given by ρE(Ξ). That is because Theorem A says that [F0(Ξ)](ψt(q0 , p0)) ≈ ρE(Ξ) (almost every time t). Also, let ΨEt : L ∞(Ω E ) → L∞(Ω E ) be a deterministic Markov operator determined by the continuous map ψEt : ΩE → ΩE (cf. Section 17.1.2). Then, it clearly holds ΨEt O0 = O0. And, we must take a ML∞(Ω E )(O0, S[(q(tk),p(tk))]) for each time t1, t2, . . . , tk, . . . , tn. However, the linguistic interpretation (§3.1) :( there is no probability without measurement) says that it suffices to take the simultaneous measurement MC(Ω E )(×nk=1O0, S[δ(q(0),p(0))]). 414 Ishikawa's Homepage Chap. 17 Equilibrium statistical mechanics and Ergodic Hypothesis Remark 17.6. [The principle of equal a priori probabilities ]. The (H) (or equivalently, (I)) says "choose a particle from N particles in box", and not "choose a state from the state space Ω E ". Thus, as mentioned in the abstract of this chapter, the principle of equal (a priori) probability is not related to our method. If we try to describe Ruele's method [78] in terms of measurement theory, we must use mixed measurement theory (cf. Chapter 9). However, this trial will end in failure. 17.3 Conclusions Our concern in this chapter may be regarded as the problem: "What is the classical mechanical world view?" Concretely speaking, we are concerned with the problem: "our method" vs. "Ruele's method [78] ( which has been authorized for a long time )" And, we assert the superiority of our method to Ruele's method in Remarks 17.2, 17.5, 17.6. 415 Ishikawa's Homepage

Chapter 18 Reliability in psychological tests In this chapter, we shall introduce a measurement theoretical approach to a problem of analyzing scores of tests for students. The obtained score is assumed to be a sum of a true value and a measurement error. It is also subject to a systematic error (=noise) depending on his/her health or psychological condition at the test. In such cases, statistical measurements are convenient since these two errors (i.e., measurement error and systematic error) in measurement theory can be characterized in different mathematical structures. As a result, we show that "reliability coefficient" = "correlation coefficient" in a clear formulation. This chapter is extracted from the following. [59] K. Kikuchi, S. Ishikawa, "Psychological tests in Measurement Theory," Far east journal of theoretical statistics, 32(1) 81-99, (2010) ISSN: 0972-0863 18.1 Reliability in psychological tests 18.1.1 Preparation In this section, let us consider reliability of psychological tests for a group of students. We discuss examples from measurement theoretical characterization of tests to measure mathematical ability of students. Let Θ := {θ1, θ2, . . . , θn} be a set of students, say, there are n students θ1, θ2, . . . , θn. Define the counting measure νc on Θ such that νc({θi}) = 1 (i = 1, 2, . . . , n). The Θ will be regarded as a state. For each θi (∈ Θ), we define 1θi (∈ L1+1(Θ, νc)) by 1θi(θ) = 1 (if θ = θi), = 0 (if θ 6= θi). Recall that Θ can be identified with the {1θi | θi ∈ Θ} under the identification: Θ 3 θi ↔ 1θi ∈ {1θ | θ ∈ Θ}. 417 18.1 Reliability in psychological tests For simplicity, we shall begin with the test for one student θi (∈ Θ). Let (ΩR,FΩR , dω) be the Lebesgue measure space where ΩR = R. Example 18.1. (test in mathematics for a student θi) Let Θ := {θ1, θ2, . . . , θn} be a state space which is identified with the set of the students. The mathematical ability of the student θi (∈ Θ) is assumed to be represented by a statistical state Φ∗(1θi) (∈ L1+1(ΩR, dω)) (i = 1, 2, . . . , n) where Φ∗ : L 1(Θ, νc) → L1(ΩR, dω) is a pre-dual Markov causal operator of Φ : L∞(ΩR, dω)→ L∞(Θ, νc). θ1 θ2 θn Φ∗(1θ1 ) Φ∗(1θ2 )Φ∗(1θn ) Θ = {1θ | θ ∈ Θ} ΩR Φ∗ =⇒ Let O := (XR,FXR , F ) be an observable in L ∞(ΩR, dω). Axiom (m) 1 (§9.1) asserts that (A) the probability that the score (measured value) of the student θi (∈ Θ) obtained by the statistical measurement ML∞(ΩR,dω)(O, S[∗](Φ∗(1θi))) belongs to a set Ξ (∈ FXR) is given by L1(ΩR,dω) 〈Φ∗(1θi), F (Ξ)〉L∞(ΩR,dω) ( = ∫ ΩR [F (Ξ)](ω) [Φ∗(1θi)](ω) dω ) . Remark 18.2. In the above, readers may have a question (B) What is the unknown pure state [∗] in S[∗] ? Imaging the deterministic causal map ψ : Θ→ ΩR, we may consider that [∗] = ψ(θi) = ∫ ΩR ω[Φ∗(1θi)](ω) dω. Also, note that the [∗] does not play an important role in this chapter since Bayes' theorem 9.11 is not used. 418 Ishikawa's Homepage Chap. 18 Reliability in psychological tests Remark 18.3. It should be kept in mind that the variance σ2i of the ability of θi (∈ Θ) (i = 1, 2, . . . , n) is not constant, that is to say, we do not assume that σ2i = σ 2 j (∀i, ∀j): σ2i := ∫ ΩR (ω − μi)2 [Φ∗(1θi)](ω) dω (i = 1, 2, . . . , n), (18.1) where μi is an expectation of Φ∗(1θi): μi := ∫ ΩR ω [Φ∗(1θi)](ω) dω (i = 1, 2, . . . , n). (18.2) 18.1.2 Group measurement (= parallel measurement) The above example is the test for a student θi (∈ Θ). Keeping this in mind, we will next consider the test for a group of n students. Let ΩnR = Rn, and let (ΩnR,FΩnR , dω n) be a ndimensional Lebesgue measure space. Furthermore, let O := (XR,FXR , F ) and ML∞(ΩR,dω)(O, S[∗](Φ∗(1θi))) (i = 1, 2, . . . , n) be as in above example. Here, we consider a parallel measurement ML∞(ΩnR ,dωn)(Ô, S[∗](ρ)) where Ô := (X n R,FXnR , F ) is an observable in L ∞(ΩnR, dω n). If [F (Ξ1 × Ξ2 × * * * × Ξn)](ω1, ω2, . . . , ωn) = [F (Ξ1)](ω1) * [F (Ξ2)](ω2) * * * [F (Ξn)](ωn), and ρ(ω1, ω2, . . . , ωn) = [Φ∗(1θ1)](ω1) * [Φ∗(1θ2)](ω2) * * * [Φ∗(1θn)](ωn), then, the parallel measurement ML∞(ΩnR ,dωn)(Ô, S[∗](ρ)) is denoted by ⊗θi∈ΘML∞(ΩR,dω)(O, S[∗](Φ∗(1θi))). In addition, we introduce the following notations concerning tensor product: ⊗nk=1L∞(ΩR, dω) = L∞(ΩnR, dωn) and ⊗nk=1 L1(ΩR, dω) = L1(ΩnR, dωn). By the way, we introduce the test observable. Definition 18.4. [Test observable] The Oτ = (XR,FXR , Fτ ) is called a test observable in L∞(ΩR, dω), if Fτ satisfies the following no-bias condition:∫ XR x [Fτ (dx)](ω) = ω (∀ω ∈ ΩR). (18.3) 419 Ishikawa's Homepage 18.1 Reliability in psychological tests Recall that the normal observable (cf. Example 2.24 ) and the exact observable (cf. Example 2.25 ). For each θi (∈ Θ), we use the notation M(i)Oτ to the test for θi (∈ Θ) (the measurement of the test observable Oτ for the statistical state Φ∗(1θi)): M (i) Oτ := ML∞(ΩR,dω)(Oτ , S[∗](Φ∗(1θi))). (18.4) Now we are ready to consider the test for a set of the n students in our measurement theory. Definition 18.5. [Test, Group test] Let Θ := {θ1, θ2, . . . , θn}, XR = ΩR = R and Φ∗ : L1+1(Θ, νc)→ L1+1(ΩR, dω) be as in Example 18.1. Let Oτ := (XR,FXR , Fτ ) be a test observable in L∞(ΩR, dω). The measurement ML∞(ΩR,dω)(Oτ , S[∗](Φ∗(1θi))) is called a test for a student θi (∈ Θ) and symbolized by M(i)Oτ for short. And the measurement ⊗θi∈ΘML∞(ΩR,dω)(Oτ , S[∗](Φ∗(1θi))) (or in short, ⊗θi∈ΘM (i) Oτ ), (18.5) is called a group test and symbolized by M⊗Oτ for short. Axiom(m) 1 (§9.1) says that (C) the probability that the score (x1, x2, . . . , xn) (∈ XnR) obtained by the group test ⊗θi∈ΘML∞(ΩR,dω) (Oτ , S[∗](Φ∗(1θi))) (or in short, M ⊗ Oτ ) belongs to the set×ni=1 Ξi (∈ FXnR ) is given by × θi∈Θ L1(ΩR,dω) 〈Φ∗(1θi), Fτ (Ξi)〉L∞(ΩR,dω) ( =: P1( n × i=1 Ξi) = n × i=1 Pi(Ξi) ) . (18.6) Here, (XR,FXR , Pi) is a sample probability space of M (i) Oτ . Let W : XnR → R be a statistics (i.e., measurable function). Then, EM⊗Oτ [W ], the expectation of W , is defined by EM⊗Oτ [W ] = ∫ XR * * * ∫ XR W (x1, x2, . . . , xn) P1(dx1 dx2 * * * dxn). Definition 18.6. Let Oτ := (XR,FXR , Fτ ) be a test observable in L ∞(ΩR, dω). (i: Score of θi) Let ML∞(ΩR,dω)(Oτ , S[∗](Φ∗(1θi))) (or in short, M (i) Oτ ) be a test for a student θi (∈ Θ). Here, we consider the expectation of xi (∈ XR) and its variance. 420 Ishikawa's Homepage Chap. 18 Reliability in psychological tests 1. Av[M (i) Oτ ] := E M (i) Oτ [xi], 2. Var[M (i) Oτ ] := E M (i) Oτ [ (xi − Av[M(i)Oτ ]) 2 ] . (ii: Scores of n students) Let ⊗θi∈ΘML∞(ΩR,dω)(Oτ , S[∗](Φ∗(1θi))) (or in short, M ⊗ Oτ ) be a group test. Here, we consider the expectation of 1 n (x1 + x2 + * * *+ xn) and its variance. 1. Av[M⊗Oτ ] := EM⊗Oτ [1 n (x1 + x2 + * * *+ xn) ] , 2. Var[M⊗Oτ ] := EM⊗Oτ [ 1 n n∑ k=1 (xk − Av[M⊗Oτ ]) 2 ] . From the no-bias condition (18.3), we get Av[M (i) Oτ ] = Av[M (i) OE ] = ∫ ΩR ω [Φ∗(1θi)](ω) dω = μi, (18.7) Av[M⊗Oτ ] = 1 n n∑ i=1 Av[M (i) Oτ ] = Av[M⊗OE ] = 1 n n∑ i=1 Av[M (i) OE ] = 1 n n∑ i=1 μi =: μ, (18.8) where OE := (XR,FXR , E) is an exact observable in L ∞(ΩR, dω). 18.1.3 Reliability coefficient When we suppose the group test, we can consider the reliability coefficient which can be represented by a proportion of variance of mathematical abilities to obtained variance. Definition 18.7. [Reliability coefficient] Let Oτ := (XR,FXR , Fτ ) [resp. OE := (XR,FXR , E)] be a test observable [resp. an exact observable] in L∞(ΩR, dω). And, let M⊗Oτ := ⊗θi∈ΘML∞(ΩR,dω)(Oτ , S[∗](Φ∗(1θi))) be a group test. The reliability coefficient RC[M⊗Oτ ] of the group test M ⊗ Oτ is defined by RC[M⊗Oτ ] = Var[M⊗OE ] Var[M⊗Oτ ] . Now let us consider the measurement error. First, when the ability (true value) is ω (∈ Ω), the measurement error ∆ω is as follows: ∆ω := (∫ XR (x− ω)2 [Fτ (dx)](ω) )1/2 (∀ω ∈ Ω). (18.9) 421 Ishikawa's Homepage 18.1 Reliability in psychological tests Note that the error ∆ω (∀ω ∈ Ω) depends on ω (∈ Ω) in general, that is, we do not assume that ∆ω = ∆ω′ (∀ω, ∀ω′ ∈ Ω). Next, for each θi (∈ Θ), the error ∆i for the student θi (∈ Θ) is as follows: ∆i := (∫ XR ∆ω [Φ∗(1θi)](ω) dω )1/2 = (∫ ΩR (∫ XR (x− ω)2 [Fτ (dx)](ω) ) [Φ∗(1θi)](ω) dω )1/2 (i = 1, 2, . . . , n). (18.10) Finally, the group average of the student θi's error ∆i (i = 1, 2, . . . , n) is as follows: ∆g := ( 1 n n∑ i=1 ∆2i )1/2 . (18.11) From what we have seen, we can get the following theorem. Theorem 18.8. (i: The variance Var[M (i) Oτ ]) Let M (i) Oτ := ML∞(ΩR,dω)(Oτ , S[∗](Φ∗(1θi))) be the measurement of test observable Oτ for the statistical state Φ∗(1θi). Then, we see Var[M (i) Oτ ] = Var[M (i) OE ] + ∆2i . (18.12) (ii: The variance Var[M⊗Oτ ]) We consider the group test M ⊗ Oτ := ⊗θi∈ΘM (i) Oτ = ⊗θi∈Θ ML∞(ΩR,dω)(Oτ , S[∗](Φ∗(1θi))). And, we obtain the following: Var[M⊗Oτ ] = Var[M ⊗ OE ] + ∆2g. (18.13) Proof. Let μi be an expectation of Φ∗(1θi). Then, we see Var[M (i) Oτ ] = ∫ ΩR (∫ XR (x− μi)2 [Fτ (dx)](ω) ) [Φ∗(1θi)](ω) dω = ∫ ΩR (ω − μi)2 [Φ∗(1θi)](ω) dω + ∫ ΩR (∫ XR (x− ω)2 [Fτ (dx)](ω) ) [Φ∗(1θi)](ω) dω + ∫ ΩR (∫ XR 2(x− ω)(ω − μi) [Fτ (dx)](ω) ) [Φ∗(1θi)](ω) dω = Var[M (i) OE ] + ∆2i . From the above formula, it follows that the group average of Var[M (i) Oτ ] becomes Var[M⊗Oτ ] = ∫ ΩR * * * ∫ ΩR (∫ XR * * * ∫ XR 1 n n∑ i=1 (xi − μ)2 n × i=1 [Fτ (dxi)](ωi) ) n × i=1 [Φ∗(1θi)](ωi) dωi = 1 n n∑ i=1 ∫ ΩR (∫ XR (ω − μ+ x− ω)2 [Fτ (dx)](ω) ) [Φ∗(1θi)](ω) dω 422 Ishikawa's Homepage Chap. 18 Reliability in psychological tests = 1 n n∑ i=1 ∫ ΩR (ω − μ)2 [Φ∗(1θi)](ω) dω + 1 n n∑ i=1 ∫ ΩR (∫ XR (x− ω)2 [Fτ (dx)](ω) ) [Φ∗(1θi)](ω) dω + 1 n n∑ i=1 ∫ ΩR (∫ XR 2(x− ω)(ω − μ) [Fτ (dx)](ω) ) [Φ∗(1θi)](ω) dω = ∫ ΩR * * * ∫ ΩR 1 n n∑ i=1 (ωi − μ)2 n × i=1 [Φ∗(1θi)](ωi) dωi + 1 n n∑ i=1 ∆2i = Var[M⊗OE ] + ∆ 2 g. 18.2 Correlation coefficient: How to calculate the reliability coefficient In the previous section, we define the reliability coefficient RC[M⊗Oτ ] := Var[M⊗OE ] Var[M⊗Oτ ] . However, from the measured data (x1, x2, . . . , xn) (∈ XnR), we can not get the variance of mathematical abilities of n students Var[M⊗OE ] directly (though we can calculate the Var[M ⊗ Oτ ]). Thus, we focus on the problem how to estimate the reliability coefficient. Here we consider one typical method, say the split-half method. Split-half method: This method is appropriate where the testing procedure may in some fashion be divided into two halves and two scores obtained. These may be correlated. With psychological tests, a common procedure is to obtain scores on the odd and even items. Now we introduce the measurement theoretical characterizations of the split-half method. Definition 18.9. [Group simultaneous test] Let Θ := {θ1, θ2, . . . , θn}, XR = ΩR = R and Φ∗ : L 1 +1(Θ, νc) → L1+1(ΩR, dω) be as in Example 18.1. Let Oτ1 := (XR,FXR , Fτ1) and Oτ2 := (XR,FXR , Fτ2) be test observables in L ∞(ΩR, dω). The measurement ⊗θi∈ΘML∞(ΩR,dω)(Oτ1 × Oτ2 , S[∗](Φ∗(1θi))), is called a group simultaneous test of Oτ1 and Oτ2 and it is symbolized by M ⊗ Oτ1×Oτ2 for short. Axiom(m) 1 (§9.1) says that 423 Ishikawa's Homepage 18.2 Correlation coefficient: How to calculate the reliability coefficient (A) the probability that the score ((x11, x 2 1), (x 1 2.x 2 2), . . . , (x 1 n, x 2 n)) (∈ X2nR ) obtained by the group simultaneous test ⊗θi∈ΘML∞(ΩR,dω)(Oτ1 ×Oτ2 , S[∗](Φ∗(1θi))) (or in short, M ⊗ Oτ1×Oτ2 ) belongs to the set×ni=1(Ξ1i × Ξ2i ) (∈ FX2nR ) is given by × θi∈Θ L1(ΩR,dω) 〈Φ∗(1θi), (Fτ1 × Fτ2)(Ξ1i × Ξ2i )〉L∞(ΩR,dω) ( =: P2( n × i=1 (Ξ1i × Ξ2i )) ) . (18.14) Here note that (X2nR ,FX2nR , P2) is a sample probability space. Let W2 : X 2n R → R be a statistics (i.e., measurable function). Then, EM⊗Oτ1×Oτ2 [W2], the expectation of W2, is defined by EM⊗Oτ1×Oτ2 [W2] = ∫ XnR W (x11, x 2 1, x 1 2, x 2 2, . . . , x 1 n, x 2 n) P2(dx 1 1 dx 2 1 dx 1 2 dx 2 2 * * * dx1n dx2n). We use the following notations: (i) Av(k)[M⊗Oτ1×Oτ2 ] := EM⊗Oτ1×Oτ2 [ 1 n n∑ i=1 xki ] (k = 1, 2), (ii) Var(k)[M⊗Oτ1×Oτ2 ] := EM⊗Oτ1×Oτ2 [ 1 n n∑ i=1 (xki − Av(k)[M⊗Oτ1×Oτ2 ]) 2 ] (k = 1, 2), (iii) Cov[M⊗Oτ1×Oτ2 ] := EM⊗Oτ1×Oτ2 [1 n n∑ i=1 (x1i − Av(1)[M⊗Oτ1×Oτ2 ]) × (x2i − Av(2)[M⊗Oτ1×Oτ2 ]) ] . It is clear that Av(k)[M⊗Oτ1×Oτ2 ] = Av[M⊗Oτk ] = Av[M⊗OE ] (k = 1, 2). Definition 18.10. [Equivalency of test observables] We call that test observables Oτ1 := (XR,FXR , Fτ1) and Oτ2 := (XR,FXR , Fτ2) in L ∞(ΩR, dω) are equivalent if it holds ∆(1)ω = ∆ (2) ω (∀ω ∈ ΩR), (18.15) where ∆ (k) ω := ( ∫ XR (x− ω)2 [Fτk(dx)](ω))1/2 (see (18.9)). In case that test observables Oτ1 := (XR,FXR , Fτ1) and Oτ2 := (XR,FXR , Fτ2) in L ∞(ΩR, dω) are equivalent and Oτ1 × Oτ2 is a product test observable in L∞(ΩR, dω), it holds that Var[M⊗Oτ1 ] = Var(1)[M⊗Oτ1×Oτ2 ] = Var(2)[M⊗Oτ1×Oτ2 ] = Var[M⊗Oτ2 ]. (18.16) In consequence of these properties, we introduce the correlation coefficient of the measured values (x11, x 1 2, . . . , x 1 n) (∈ XnR) and (x21, x22, . . . , x2n) (∈ XnR) which are obtained by the group simultaneous test M⊗Oτ1×Oτ2 . 424 Ishikawa's Homepage Chap. 18 Reliability in psychological tests Theorem 18.11. [The reliability coefficient and the correlation coefficient in group simultaneous tests] Let Oτ1 and Oτ2 be equivalent test observables in L ∞(ΩR, dω). And let Oτ1 × Oτ2 be a product test observable in L∞(ΩR, dω). Let M ⊗ Oτk := ⊗θi∈ΘML∞(ΩR,dω)(Oτk , S[∗](Φ∗(1θi))) (k = 1, 2) and M ⊗ Oτ1×Oτ2 := ⊗θi∈ΘM(Oτ1 × Oτ2 , S[∗](Φ∗(1θi))) be group tests as above notations. Then we see that RC[M⊗Oτ1 ] = RC[M⊗Oτ2 ] = Cov[M⊗Oτ1×Oτ2 ]√ Var[M⊗Oτ1 ] * √ Var[M⊗Oτ2 ] . (18.17) Proof. From the (18.3), we get the following: Cov[M⊗Oτ1×Oτ2 ] := EM⊗Oτ1×Oτ2 [ 1 n n∑ i=1 (x1i − Av(1)[M⊗Oτ1×Oτ2 ])(x 2 i − Av(2)[M⊗Oτ1×Oτ2 ]) ] = ∫ ΩR * * * ∫ ΩR (∫ XR * * * ∫ XR 1 n n∑ i=1 (x1i − Av(1)[M⊗Oτ1×Oτ2 ])(x 2 i − Av(2)[M⊗Oτ1×Oτ2 ]) × n × i=1 [Fτ1(dx 1 i )Fτ2(dx 2 i )](ωi) ) n × i=1 [Φ∗(1θi)](ωi) dωi = 1 n n∑ i=1 (∫ ΩR (∫ XR ∫ XR (x1i − Av[M⊗OE ])(x 2 i − Av[M⊗OE ]) × [Fτ1(dx1i )](ω) [Fτ2(dx2i )](ω) ) [Φ∗(1θi)](ω) dω ) = 1 n n∑ i=1 (∫ ΩR (∫ XR (x1i − Av[M⊗OE ]) [Fτ1(dx 1 i )](ω) × ∫ XR (x2i − Av[M⊗OE ]) [Fτ2(dx 2 i )](ω) ) [Φ∗(1θi)](ω) dω ) = 1 n n∑ i=1 ∫ ΩR (ω − Av[M⊗OE ]) 2 [Φ∗(1θi)](ω) dω = Var[M ⊗ OE ]. (18.18) Then, we see that Cov[M⊗Oτ1×Oτ2 ]√ Var[M⊗Oτ1 ] * √ Var[M⊗Oτ2 ] = Var[M⊗OE ] Var(1)[M⊗Oτ1×Oτ2 ] = Var[M⊗OE ] Var(2)[M⊗Oτ1×Oτ2 ] . (18.19) 18.3 Conclusions In this chapter, we introduce the measurement theoretical understanding of psychological test and the split-half method which estimate reliability. Measurement theoretical approach show 425 Ishikawa's Homepage 18.3 Conclusions the following correspondences: split-half method ←→ group simultaneous test. M⊗Oτ1×Oτ2 := ⊗θi∈ΘML∞(ΩR,dω)(Oτ1 × Oτ2 , S[∗](Φ∗(1θi))) And further, we show the well-known theorem: "reliability coefficient" = "correlation coefficient" in Theorem 18.11. 426 Ishikawa's Homepage Chapter 19 How to describe "belief" Recall the spirit of quantum language (i.e., the spirit of the quantum mechanical world view), that is, (]) every phenomenon should be described in quantum language. Thus, we consider that even "belief" should be described in quantum language. For this, it suffices to consider the identification: "belief" = "odds by bookmaker" This approach has a great merit such that the principle of equal weight holds. This chapter is extracted from Chapter 8 in Ref. [30]: S. Ishikawa, "Mathematical Foundations of Measurement Theory," Keio University Press Inc. 2006. 19.1 Belief, probability and odds For instance, we want to formulate the following "probability": (A) the "probability" that Japan will win the victory in the next FIFA World Cup. This is possible (cf. [30]), if "parimutuel betting (or, odds in bookmaker)" is formulated by Axiom(m) 1 ( mixed measurement ). The purpose of this chapter is to show it, and further, to propose the principle of equal weight, that is, (B) the principle that, in the absence of any reason to expect one event rather than another, all the possible events should be assigned the same probability. 427 19.1 Belief, probability and odds whose validity has not been proven yet. It is one of the most important unsolved problems in statistics. In Chapter 9, we studied the mixed measurement: that is, mixed measurement theory (=quantum language) := [(mixed)Axiom(m) 1] mixed measurement (cf. §9.1 ) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spells (a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } manual to use spells (19.1) The purpose of this chapter is to characterize "belief" as a kind of mixed measurement. 19.1.1 A simple example; how to describe "belief" in quantum language We begin with a simplest example (cf. Problem 9.5 ) as follows. Problem 19.1. [= Problem 9.5; Bayes' method] Assume the following situation: (C) You do not know which the urn behind the curtain is, U1 or U2, but the "probability": p and 1− p. Here, consider the following problem: p 1-p [∗] Assume that you pick up a ball from the urn behind the curtain. (i): What is the probability that the picked ball is a white ball ? U1 U2 (ii): If the picked ball is white, what is the probability that the urn behind the curtain is U1 ? Figure 19.1:( Mixed measurement) Answer 19.2. (=Answer 9.13) Put Ω = {ω1, ω2} with the discrete metric and the counting measure νc, thus, note that 428 Ishikawa's Homepage Chap. 19 How to describe "belief" C0(Ω) = C(Ω) = L ∞(Ω, ν). Thus, in this chapter, we devote ourselves to the C∗-algebraic formulation: Define the observables O = ({W,B}, 2{W,B}, F ) and OU = ({U1,U2}, 2{U1,U2}, GU) in C(Ω) by F ({W})(ω1) = 0.8, F ({B})(ω1) = 0.2, F ({W})(ω2) = 0.4, F ({B})(ω2) = 0.6 GU({U1})(ω1) = 1, GU({U2})(ω1) = 0, GU({U1})(ω2) = 0, GU({U2})(ω2) = 1 Here "W" and "B" means "white" and "black" respectively. Under the identification: U1 ≈ ω1 and U2 ≈ ω2, the above situation is represented by the mixed state ρ(p)prior(∈M+1(Ω)) such that ρ (p) prior = pδω1 + (1− p)δω2 , where δω is the point measure at ω. Thus, we have the mixed measurement: MC(Ω)(O× OU := ({W,B} × {U1, U2}, 2{W,B}×{U1,U2}, F ×GU), S[∗](ρ(p)prior)). (19.2) Axiom(m) 1 gives the answer to the (i) in Problem 19.1 as follows. (D) the probability that a measured value (x, y) obtained by the mixed measurement MC(Ω)(O× OU , S[∗](ρ(p)prior)) belongs to {W} × {U1, U2} is given by M(Ω)(ρ (p) prior, F ({W}))C(Ω) = 0.8p+ 0.4(1− p). Since a white ball is obtained, Answer 9.13 (=Bayes' theorem ) says that a new mixed state ρ (p) post(∈M+1(Ω)) is given by ρ (p) post = F ({W})ρ(p)prior∫ Ω [F ({W})](ω)ρ(p)prior(dω) = 0.8p 0.8p+ 0.4(1− p) δω1 + 0.4(1− p) 0.8p+ 0.4(1− p) δω2 (19.3) Hence, the answer of the (ii) is given by M(Ω)(ρ (p) post, GU({U1}))C(Ω) = 0.8p 0.8p+ 0.4(1− p). By an analogy of the above Problem 19.1 ( for simplicity, we put: p = 1/4), we consider as follows. Assume that there are 100 people. And moreover assume the following situation (E) such that, for some reasons, (E) { 25 people believe ( or vote) that [∗] = U1 (i.e., U1 is behind the curtain) 75 people believe ( or vote) that [∗] = U2 (i.e., U2 is behind the curtain) That is, we have the following picture instead of Figure 19.1: 429 Ishikawa's Homepage 19.1 Belief, probability and odds 25 people believe that [∗] = U1, 75 people believe that [∗] = U2. - [∗] Figure 19.2: Belief ( or voting ) U1(≈ ω1) U2(≈ ω2) Now, we have the following problem: Problem 19.3. Consider Situation (E) and Situation (C) ( p = 1/4, 1− p = 3/4 ). Then, (F1) Can Situation (E) be understood like Situation (C) ? or, in the same sense, (F2) Can Situation (E) be formulated in mixed measurement (i.e., Axiom (m) 1)? That is, can Situation (E) be described in quantum language ? 19.1.2 The affirmative answer to Problem 19.3 Since 100 people know the situation of the urn (i.e., Figure 19.2, the assumption (E) ) implies (G)(=Figure 19.3), that is, (G)  25 people (in 100 people) believe that [∗] = U1 =⇒ { (G1): 20 people guess (or bet) that a white ball will be picked (G2): 5 people guess (or bet) that a black ball will be picked 75 people (in 100 people) believe that [∗] = U2 =⇒ { (G3): 30 people guess (or bet) that a white ball will be picked (G4): 45 people guess (or bet) that a black ball will be picked 25 people believe that [∗] = U1. (G1): 20 people guess that a white ball will be picked. (G2): 5 people guess that a black ball will be picked. 75 people believe that [∗] = U2. (G3): 30 people guess that a white ball will be picked. (G4): 45 people guess that a black ball will be picked. - [∗] Figure 19.3: The odds in bookmaker U1(≈ ω1) U2(≈ ω2) Assume that a white ball is picked in the above figure. Then, the above (G2) and (G4) are vanished as follows. 430 Ishikawa's Homepage Chap. 19 How to describe "belief" 25 people believe that [∗] = U1. (G1): 20 people guess that a white ball will be picked. (G2): 5 people guess that a black ball will be picked. 75 people believe that [∗] = U2. (G3): 30 people guess that a white ball will be picked. (G4): 45 people guess that a black ball will be picked. - [∗] Figure 19.4: A white ball is picked U1(≈ ω1) U2(≈ ω2) After all, we get the following figure: 40 % people believe that [∗] = U1, 60 % people believe that [∗] = U2. - [∗] Figure 19.5: After all, we get the new odds U1(≈ ω1) U2(≈ ω2) Thus we see that (prior state) Fig. 19.3 1 4 δω1+ 3 4 δω2 −−−−−−−→ (a white ball is picked) Fig. 19.4 −−−−−−−→ (post state) Fig. 19.5 2 5 δω1+ 3 5 δω2 (19.4) Considering the mixed measurement (i.e., the (19.2) in the case that p = 1/4): MC(Ω)(O× OU = ({W,B} × {U1, U2}, 2{W,B}×{U1,U2}, F ×GU), S[∗](ρ(1/4)prior )) (19.5) we see that the above (19.4) is the same as the Bayesian result (19.3). Note that the measurement (19.5) is interpreted as (H) choose one person from the 100 people at random, and ask him/her "Do you guess that a white ball (or, a black ball) will be picked from the urn behind the curtain, and its urn is U1 or U2 ?" In what follows, let us explain it. Consider the product observable Ô×ÔU of Ô = ({W,B}, 2{W,B}, F ) and ÔU = ({U1, U2}, 2{U1,U2}, ĜU) in C(Θ) (where Θ = {θ1, θ2, ..., θ100}) such that [F ({W})](θk) = 4/5, [F ({B})](θk) = 1/5, (k = 1, 2, ..., 25) [F ({W})](θk) = 2/5, [F ({B})](θk) = 3/5, (k = 26, 27, ..., 100) (19.6) 431 Ishikawa's Homepage 19.2 The principle of equal odds weight [ĜU({U1})](θk) = 1, [ĜU({U2})](θk) = 0, (k = 1, 2, ..., 25) [ĜU({U1})](θk) = 0, [ĜU({U2})](θk) = 1, (k = 26, 27, ..., 100) (19.7) And put ν0 = (1/100) ∑100 k=1 δθk(∈ M+1(Θ)). Then, the above measurement (H) is formulated by MC(Θ)(Ô× ÔU = ({W,B} × {U1, U2}, 2{W,B}×{U1,U2}, F × ĜU), S[∗](ν0)) (19.8) which is identified with the measurement (19.5) under the deterministic causal operator Φ : C(Ω)→ C(Θ) such that Φ∗(δθk) = δω1 (k = 1, 2, ..., 25), = δω2 (k = 26, 27, ..., 100). That is, we see, symbolically, (H)=(19.8): the Heisenberg picture Φ←−−−−−−− identification (19.5): the Schrödinger picture Thus, as a particular case of the above arguments, we can answer Problem 19.3 such that (I1) Situation (E) can be understood like Situation (C). That is, (I2) Situation (E) can be formulated in mixed measurement (i.e., Axiom (m) 1). In the same sense, Situation (E) can be described in quantum language. 19.2 The principle of equal odds weight From the above arguments, we see that Proclaim 19.4. [The principle of equal weight] Consider a finite state space Ω with the discrete metric, that is, Ω = {ω1, ω2, . . . , ωn}. Let O = (X,F, F ) be an observable in C(Ω). Consider a measurement MC(Ω)(O, S[∗]). If the observer has no information for the unknown state [∗], there is a reason to assume that this measurement is also represented by the mixed measurement MC(Ω)(O, S[∗](ρprior)), where ρprior = 1 n n∑ k=1 δωk . (19.9) Explanation. In betting, it is certain that everybody wants to choose an unpopular ωk. Thus, I believe that everybody agrees with Proclaim 19.4. Also, it should be noted that (J) the term "probability" can be freely used within the rule of Axiom 1 or Axiom(m) 1. The reason that the justice of the (B: the principle of equal weight) is not assured yet is due to the lack of the understanding of the (J). ♠Note 19.1. In this book, we dealt with the following three kinds: 432 Ishikawa's Homepage Chap. 19 How to describe "belief" (]1) the principle of equal weight in Remark 5.19 (]2) the principle of equal weight in Theorem 9.18 (]3) the principle of equal weight in Proclaim 19.4 which are essentially the same. In order to promote the readers' understanding of the difference between Theorem 9.18 and Proclaim 19.4, we show the following example, which should be compared with Problem 5.14 and Problem 9.17 Problem 19.5. [Monty Hall problem (=Problem 5.14; The principle of equal weight) ] You are on a game show and you are given a choice of three doors. Behind one door is a car, and behind the other two are goats. You choose, say, door 1, and the host, who knows where the car is, opens another door, behind which is a goat. For example, the host says that ([) the door 3 has a goat. And further, he now gives you a choice of sticking to door 1 or switching to door 2 ? What should you do ? ? ? ? door door door No. 1 No. 2 No. 3 Figure 19.6: Monty Hall problem Proof. It should be noted that the above is completely the same as Problem 5.14. However, the proof is different. That is, it suffices to use Proclaim 19.4 and Bayes theorem (B2). That is, the proof is similar to Problem 9.16 . 433 Ishikawa's Homepage

Chapter 20 Postscript: Linguistic Copenhagen interpretation 20.1 Two kinds of (realistic and linguistic) world-views In this lecture note, we assert the following figure: Figure 20.1. [=Figure 1.1: The location of quantum language in the history of world-description (cf. refs.[32, 53]) ] Parmenides Socrates 0©:Greek philosophy Plato Aristotle Schola-−−−−→ sticism 1© −−→ (monism) Newton (realism) 2© → relativity theory −−−−−−→ 3© → quantum mechanics −−−−−−→ 4© −→ (dualism) Descartes Locke,... Kant (idealism) 6©−→ (linguistic view) linguistic philosophy language−−−−−→ 8© language−−−−−−→ 7©  5©−→ (unsolved) theory of everything (quantum phys.)  10©−→ (=MT) quantum language (language) Figure 20.1(= Figure 1.1): The history of the world-view statistics system theory language−−−−−→ 9© (Descartes, Locke may belong to substance dualism) the linguistic world view ( dualism, idealism ) the realistic world view (monism, realism) Most physicists feel that (A1) quantum mechanics has both realistic aspect and metaphysical aspect. And they want to unify the two aspects. However, quantum language asserts that (A2) Two aspects are separated, and they develop in the respectively different directions 5© and 10© in Figure 20.1. 435 20.2 The summary of quantum language 20.2 The summary of quantum language 20.2.1 The big-picture view of quantum language The big-picture view of quantum language Measurement theory (= quantum language ) is classified as follows. (B) measurement theory (=quantum language)  pure type (B1) { classical system : Fisher statistics quantum system : usual quantum mechanics mixed type (B2) { classical system : including Bayesian statistics, Kalman filter quantum system : quantum decoherence And the structure is as follows. (C)  (C1): pure measurement theory (=quantum language) := [(pure)Axiom 1] pure measurement (cf. §2.7) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } the manual to use spells (C2): mixed measurement theory (=quantum language) := [(mixed)Axiom(m) 1] mixed measurement (cf. §9.1) + [Axiom 2] Causality (cf. §10.3)} {{ } a kind of spell(a priori judgment) + [quantum linguistic interpretation] Linguistic interpretation (cf. §3.1)} {{ } the manual to use spells In the above, (D1) Axioms 1 and 2 (i.e., kinds of spells) are essential On the other hand, the linguistic interpretation (i.e., the manual to use Axioms 1 and 2) may not be indispensable. However, (D2) if we would like to make speed of acquisition of a quantum language as quick as possible, we may want the good manual to use the axioms. In this sense, this note is a manual book (=cookbook). Although all written in this note can be regarded as a part of the linguistic interpretation, the most important statement is Only one measurement is permitted 436 Ishikawa's Homepage Chap. 20 Postscript: Linguistic Copenhagen interpretation Also, since we assert that quantum language is the final goal of dualistic idealism (= Descartes=Kant philosophy) in Figure20.1, we think that (E) Many philosophers' maxims and thoughts constitute a part of the linguistic interpretation 20.2.2 The characteristic of quantum language Also, we see: The characteristic of quantum language (F1) Non-reality (metaphysics ): Quantum language is metaphysics (= language), which asserts the linguistic world-view. (F2) The collapse of wave function does not occur: According to the linguistic interpretation (i.e., only one measurement is permitted), we can not get information after the measurement. That is, the collapse of wave function can not be found. However, the projection postulate holds in the sense of Postulate 11.6. (F3) Non-deterministic: Since we usually consider non-deterministic processes in classical system, it is natural to assume non-deterministic processes (i.e., quantum decoherence) in quantum language. (F4) Dualism: The two concepts: "measurement" and "dualism" are non-separable. Thus, quantum language says (]) describe any monistic phenomenon in the dualistic language ! (F5) Non-locality, faster-than-light: Quantum language accepts "non-locality". This is the only one paradox in quantum language. (F6) Many paradoxes and unsolved problems are clarified: (a) Paradoxes and unsolved problems due to a lack of quantum language: What is probability (causality, space-time) ? Zeno's paradox, the principle of equal probability, classical syllogizm, classical Bell's inequlity (b) Paradoxes and unsolved problems solved by descriptive power of quantum language: Schrödinger's cat 437 Ishikawa's Homepage 20.3 Quantum language (≈ dualistic idealism ) is located at the center of science (c) What we cannot speak about we must pass over in silence: Heisenberg's uncertainty principle (due to the thought experiment by γ-ray microscope), Cogito proposition, Wigner's friend, delayed choice experiment, Barin in a vat, Five-minute hypothesis, Only the present exists (d) Everything should be spoken by quantum language: Several problems in statistics (Fisher's maximum likelihood method, Bayes method, semi-distance (confidence interval, statistical hypothesis, ANOVA), regression analysis, Kalman filter) 20.3 Quantum language (≈ dualistic idealism ) is located at the center of science Dr. Hawking said in his best seller book [ ref.[18]; A Brief History of Time: From the Big Bang to Black Holes, Bantam, Boston, 1990]: (G) Philosophers reduced the scope of their inquiries so much that Wittgenstein the most famous philosopher this century, said "The sole remaining task for philosophy is the analysis of language." What a comedown from the great tradition of philosophy from Aristotle to Kant! I think that this is not only his opinion but also most scientists' opinion. And moreover, I mostly agree with him. However, I believe that it is worth reconsidering the series in the linguistic world view ( 1©– 6©– 8©–10© in Figure 20.1). As mentioned in Section 10.8.4, I believe that, if "the analysis of language" was rewritten to "the creation of language", then Dr. Hawking would not have been critical to philosophy. That is because the task of phycisists is just the creation of language, i.e., the language called Newtonian mechanics, the language called the theory of relativity, etc. It is a matter of course that quantum language is different from pure mathematics. Hence, in spite of Lord Kelvin's saying: Mathematics is the only good metaphysics , I assert that (H) quantum language is located at the center of science or 438 Ishikawa's Homepage (I) quantum language is the language for science That is, I believe, from the pure theoretical point of view, that quantum language will replace statistics ( since, in statistics, the concept of measurement is not exposed). Recall two famous maxims concerning science: (J1) There is no science without measurement (J2) Sience is knowledge about causality. Here, we see the following correspondences: (J1) ⇔ Axiom 1 ( measurement ), (J2) ⇔ Axiom 2 ( causality ) which may imply the (I). Since quantum language is not physics but language (= metaphysics), quantum language (= the linguistic Copenhagen interpretation of quantum mechanics) is completely different from other interpretations. Therefore, even if someone discovers the "final" interpretation of quantum mechanics in the realistic view (i.e., 5© in Figure 20.1 ), quantum language is not affected by it. Although I don't know whether quantum language is final in the linguistic view, I believe that it is the greatest purpose of philosophy of science to pursue powerful scientific language than quantum language. I hope that my proposal will be examined from various view-points. Shiro ISHIKAWA November in 2018

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Functional analysis, Springer-Verlag (Sixth Edition) 1980 445 Index a priori synthetic judgment, 7, 8, 62 ANOVA(one-way), 181 ANOVA(two-way), 185 ANOVA(zero-way), 177 Aristotle(BC384-BC322), 66, 145, 197 Augustinus(354-430), 207, 295 averaging entropy, 241 Axiom 1[measurement], 7, 47, 62 Axiom 1[classical measurement], 148 Axiom 2[causality], 7, 272 Axiom(m) 1[mixed measurement (= statistical measurement )], 221 Bacon(1561-1626), 262 basic structure, 16 Bayes(1702-1761), 229 Bayes' method, 229 Bell's inequality, 105, 210 Bergson, Henri-Louis(1859-1941), 207 Berkeley, George (1685-1753), 37, 47 Bernoulli, J.(1654-1705), 90 blood type system, 53 Bohr(1885-1962), 104, 285 Borel field, 25, 40 Born(1882-1970), 47, 126 brain in a vat, 286 Brownian motion, 371 causal operator , 265, 266 chi-square distribution, 149 Click (The astonishing hypothesis), 356 cogito proposition, 94, 205, 293 collapse of wave function , 4 combined observable , 106, 208 compact operator, 20 conditional probability, 201 confidence interval, 147, 150 CONS, 20 consistency condition, 86, 369 contraposition, 204 control problem, 360 cookbook, 9, 436 Copenhagen interpretation, 71 Copernican revolution, 145, 262 correlation coefficient, 423 counting measure, 28, 52 Critique of Pure Reason, 8 C∗-algebra, 16 Darwin(1809–1831), 272 de Broglie(1892-1987), 58 definition functionχΞ , 40, 50 Descartes(1596-1650), 62, 205 Descartes figure, 62, 205 Descartes: I think, therefore I am, 205 deterministic causal operator , 266 dialectic(Hegel), 272 Dirac notation, 20 discrete metric, 24 double-slit experiment, 344 dual causal operator , 266 dualism, 32 dynamical system theory, 359, 380 edios(Aristotle), 31, 66 F -distribution , 179 Einstein(1879-1955), 104, 285 energy observable, 42 entangled state, 102 EPR-experiment, 99 equal weight(the principle of equal weight), 144, 432 equal weight, 240 ergodic hypothesis, 413 ergodic property, 82, 83, 409 error function, 39, 119 essentially continuous, 33 estimator, 150 evolution theory(Darwin), 272 exact observable , 40 exact measurement, 50 existence observable, 37 falsifiability, 3 446 Feynman(1918-1988), 1 final cause(Aristotle), 272 Fisher(1890-1962), 126 Fisher's maximum likelihood method, 123, 124 five-minute hypothesis, 286 flow, 409 Galileo(1564-1642), 90, 262 Gauss integral, 194 Gelfand theorem, 27 generalized linear model, 394 geocentric model, 145 group test, 420 Hamilton(1805-1865), 274 Hamilton's canonical equation, 274 Hamiltonian, 408, 409 Hamilton's canonical equation, 274, 409 Hawking(1942–), 438 Hegel(1770–1831), 272 Heidegger(1889-1976), 207 Heisenberg(1901-1976), 93, 276 Heisenberg picture, 265, 266 Heisenberg's kinetic equation, 276 Heisenberg's uncertainty relation, 93, 98 heliocentrism, 145 Heraclitus(BC.540 -BC.480), 260 Hermitian matrix, 43 hidden variable, 114, 211 Hilbert space, 15 Hume, David(1711-1776), 356 hyle(Aristotle), 31, 66 idea(Plato), 31, 66 image observable, 149, 198 increasing entropy, 413 inference problem, 360 Kalman(1930-), 397 Kalman filter, 397 Kant(1724-1804), 7, 8, 62, 262 Kelvin(1824-1907), 438 Kolmogorov(1903-1987), 10, 85 Kolmogorov extension theorem, 85, 369 law of entropy increase, 272 law of large numbers, 89 least squares method, 385 Leibniz(1646-1716), 281 Leibniz-Clarke Correspondence, 281 likelihood equation, 131, 386, 389 likelihood function, 124 Locke, John(1632-1704), 32 lower bounded, 368 Mach-Zehnder interferometer, 324 marginal observable , 198 Markov causal operator, 265 McTaggart's paradox, 286 measurable space, 34 measurable space, 34 measured value, 34, 46 measured value space, 34 measurement equation, 359, 380 measurement error model, 396 measuring instrument, 34 metaphysics, 8 mixed measurement (= statistical measurement), 221 moment method, 132 momentum observable , 42, 92 monistic phenomenon, 353, 356 Monty Hall problem, 137, 235, 236, 239, 433 Monty Hall problem ; Bayesian approach, 235 Monty Hall problem: moment method, 139 Monty Hall problrem:The principle of equal weight, 239 Monty Hall problrm: Fisher's maximamum likelihoood, 138 MT (= measurement theory=quantum language ), 3 multiple markov property, 273 natural map, 86 Newton(1643-1727), 262, 284 Newtonian equation, 274 Nietzsche(1844–1900), 355 No smoke, no fire, 265, 272 normal observable, 39, 119, 128 observable: definition, 34 Only the present exists, 295 ONS, 20 Ozawa's inequality, 101 paradox Bertrand's paradox, 53 de Broglie's paradox, 307 EPR paradox, 102 Hardy's's paradox, 328 447 Schrödinger's cat, 315 Zeno's paradox, 377 parallel measurement, 80 parallel observable, 80 parent map, 270, 368 Parmenides(born around BC. 515), 66, 260, 373 particle or wave ?, 321 Plank constant, 93 Plato(BC427-BC347), 66 point measure, 28 population, 31, 66 position observable , 42, 92 power set, 37 pre-dual sequential causal observable, 271 primary quality, secondary quality, 30–32, 66 principle of equal a priori probabilities, 415 problem of universals, 285 product measurable space, 72 product state space, 80 projection, 279 projective observable, 35 quantity, 42 quantum decoherence, 279, 311 quantum eraser experiment, 333 quantum Zeno effect, 313 quasi-product observable , 78, 197 Radon-Nikodym theorem, 267 random, 53 random walk, 279 realized causal observable , 339 regression analysis, 361, 387 reliability coefficient, 421 resolution of the identity, 37 Robertson's uncertainty relation, 91 root, 270, 368 rounding observable , 40 sample probability space, 34 state space(mixed state space, pure state space), 17 scholasticism, 66 Schrödinger(1887-1961), 275 Schrödinger equation, 275 Schrödinger picture, 266 sequential causal observable, 270, 338, 369 sequential causal operator, 270 σ-field, 34 σ-finite, 25 simultaneous measurement, 73 simultaneous observable , 72 spectrum, 27, 282 spectrum decomposition, 44 spin observable, 56 split-half method, 423 St. Petersburg two envelope problem, 227 state equation, 263, 272, 359, 380 state space(mixed state space, pure state space), 68, 69 statistical hypothesis testing deference of population means, 169 population mean, 154 student t-distribution, 173 population variance, 162 staying time space, 410 Stern=Gerlach experiment, 56 student t-distribution , 120, 173, 177 syllogism, 212 syllogism does not hold in quantum system, 216 system(=measuring object), 46 system quantity, 42 Tagore, 37, 338, 376 tensor basic structure, 70 test, 420 test observable, 419 Thomas Aquinas (1225-1274), 63 time-lag process, 273 trace, 21, 23, 44 tree (tree-like semi-ordered set), 270 tree (infinite tree-like semi-ordered set), 368 trialism, 63 triangle observable, 39 two envelope problem, 140, 227, 232 Unsolved problem What is causality?, 262 What is space-time?, 281 Monty Hall problem, equal weight, 238, 433 Zeno's paradox, 377 urn problem, 51, 119, 121, 125, 127, 133 von Neumann(1903-1957), 15 wave function collapse, 303 weak convergence, 16 Wheeler's Delayed choice experiment, 321 Wilson cloud chamber, 348 Wittgenstein(1889-1951), 207 448 W ∗-algebra, 16 Zeno(BC490-BC430), 377 Zeno's paradox, 377 Notation BalldΩ(ω; η) :Ball, 155 BallCdΩ(ω; η) :complement of Ball, 155 B(H): bounded operators space, 15 χΞ :definition function, 50 C(= the set of all complex numbers), 15 C(H): compact operators class, 20 Ξc: complement of Ξ, 26 Cn : n-dimensional complex space, 21 C0(Ω): continuous functions space, 25 δω: point measure at ω, 28 ess.sup : essential sup, 25 Φ1,2: causal operator , 265 Φ∗1,2:dual causal operator , 266 (Φ1,2)∗:pre-dual causal operator , 266 ~: Plank constant, 93 Lr(Ω, ν): r-th integrable functions space, 25 MA ( O, S[ρ] ) :pure measurement, 47 MA ( O, S[∗](w) ) :mixed measurement, 221 M(Ω): the space of measures, 26 MA ( O, S[∗] ) :inference, 122 N(= the set of all natural numbers), 16⊗n k=1Ok: parallel observable , 80 nk=1Fk:product σ-field, 72 2X(= P(X)):power set of X, 34 P0(X):power finite set of X, 86 Rn(= n-dimensional Euclidean space), 24 R(= the set of all real numbers), 13 Sp(A∗): pure state space, 17 Sm(A∗): C∗-mixed state space, 17 S m (A∗): W ∗-mixed state space, 17 Tr(H): trace class, 21 Tr: trace, 22 Trp+1(H): quantum pure state space, 22 (T, 5 ), (T (t0), 5 ):tree, 368