One world is (probably) just as good as many

Abstract

One of our most sophisticated accounts of objective chance in quantum mechanics involves the Deutsch–Wallace theorem, which uses state-space symmetries to justify agents’ use of the Born rule when the quantum state is known. But Wallace argues that this theorem requires an Everettian approach to measurement. I find that this argument is unsound. I demonstrate a counter-example by applying the Deutsch–Wallace theorem to the de Broglie–Bohm pilot-wave theory.

Proof of the symmetry theorem

Before proving the symmetry theorem, we prove as a lemma an important condition that follows from the structural links assumption.

Branching link. When $\{{\mathcal {S}}^i\}$ is strictly $\Psi$-branching and ${\mathcal {T}}(\alpha _i, \beta _j) \ne 0$ for $t_i < t_j$,

$$\begin{aligned} ch _{\left( \Psi ,\alpha _i\right) } \left( \beta _j \right) = \frac{ ch _\Psi \left( \beta _j\right) }{ ch _\Psi \left( \alpha _i \right) }. \end{aligned}$$

(31)

Proof of the branching link

First, we show that chances respect branching, i.e., $\mathbf {ch}_\Psi (\gamma ) = 0$ whenever $\gamma$ contains $\alpha _i, \beta _j$ such that ${\mathcal {T}}(\alpha _i,\beta _j) = 0$. Explicitly,

$$\begin{aligned} \mathbf {ch}_\Psi ( \gamma ) \le \mathbf {ch}_{\Psi }\! \left( \bigvee \{ \gamma \,|\, \gamma _j = \beta _j \wedge \gamma _i = \alpha _i \} \right)&= \mathbf {ch}_{\Psi } \!\left( \bigvee \{ \gamma \,|\, \gamma _{j} = \beta _j \} \,\bigg | \, \bigvee \{ \gamma \,|\, \gamma _i = \alpha _i \} \right) \end{aligned}$$

(32)

$$\begin{aligned}&= \mathbf {ch}_{(\Psi ,\alpha _i)} \!\left( \bigvee \{ \gamma \,|\, \gamma _{j} = \beta _j \} \right) \end{aligned}$$

(33)

$$\begin{aligned}&= ch _{(\Psi ,\alpha _i)} \left( \beta _j \right) \end{aligned}$$

(34)

$$\begin{aligned}&= 0, \end{aligned}$$

(35)

where (32) follows from additivity and the definition of conditional probability, (33) follows from temporal link, i.e., (29), (34) follows from the definition (28) in the probability assumption, and (35) follows from normalization link, the rules of probability, and the fact that ${\mathcal {T}}(\alpha _i,\beta _j) = 0$. The rules of probability then imply that $\mathbf {ch}_\Psi ( \gamma ) =0$.

Now suppose that $\{{\mathcal {S}}^i\}$ is strictly $\Psi$-branching and ${\mathcal {T}}(\alpha _i, \beta _j) \ne 0$ for $t_i < t_j$. Note that

$$\begin{aligned} ch _{(\Psi ,\alpha _i)} \left( \beta _j \right)&= \mathbf {ch}_{(\Psi ,\alpha _i)} \!\left( \bigvee \{ \gamma \,|\, \gamma _{j} = \beta _j \} \right) \end{aligned}$$

(36)

$$\begin{aligned}&= \mathbf {ch}_{\Psi } \!\left( \bigvee \{ \gamma \,|\, \gamma _{j} = \beta _j \} \,\bigg | \, \bigvee \{ \gamma \,|\, \gamma _i = \alpha _i \} \right) \end{aligned}$$

(37)

$$\begin{aligned}&= \frac{\mathbf {ch}_{\Psi }\! \left( \bigvee \{ \gamma \,|\, \gamma _j = \beta _j \wedge \gamma _i = \alpha _i \} \right) }{\mathbf {ch}_\Psi \! \left( \bigvee \{ \gamma \,|\, \gamma _i = \alpha _i \} \right) } \end{aligned}$$

(38)

$$\begin{aligned}&= \frac{\mathbf {ch}_\Psi \left( \bigvee \{ \gamma \,|\, \gamma _{j} = \beta _j \} \right) }{\mathbf {ch}_\Psi \! \left( \bigvee \{ \gamma \,|\, \gamma _i = \alpha _i \} \right) } \end{aligned}$$

(39)

$$\begin{aligned}&= \frac{ ch _\Psi \left( \beta _j \right) }{ ch _\Psi (\alpha _i )}, \end{aligned}$$

(40)

where (36) follows from probability, (37) follows from temporal link, (38) follows from the definition of conditional probability (and a bit of algebra), (39) follows from $\mathbf {ch}_\Psi$ respecting branching and additivity (and our supposition), and (40) follows from (28). $\square$

With branching link in hand, we can quickly prove the symmetry theorem.

Proof of the symmetry theorem

Following (Wallace 2012, Ch. 4), we aim to prove that $ch _\Psi (\alpha _i) = \langle \Psi (t_i), \alpha _i \Psi (t_i) \rangle$ in four steps. First (i), we generalize the intuitive re-labeling argument to show that projections with equal Born weights must have equal chances—i.e., we show that $ch _\Psi$ must be a function of Born weights. Second (ii), we invoke some of our environmental degrees of freedom to show that this function is increasing. Third (iii), we invoke N environmental degrees of freedom to show that function must equal the Born weight when it is rational. Fourth (iv), we use a simple limiting argument (and the second and third steps) to obtain agreement for arbitrary Born weights.

• (i) Suppose $\langle \Psi (t_i) , \alpha _i \Psi (t_i) \rangle = \langle \Psi (t_i), \beta _i \Psi (t_i) \rangle$.

  First, suppose that both sides equal zero. Then $\Psi (t_i)$ lies in the range of both $\lnot \alpha _i$ and $\lnot \beta _i$, and so by normalization link, $ch _\Psi (\lnot \alpha _i ) = ch _\Psi (\lnot \beta _i ) = 1$. Thus, by probability, $ch _\Psi ( \alpha _i) = ch _\Psi ( \beta _i ) = 0$.

  Now suppose otherwise. By the above reasoning, we get that $ch _\Psi ( \alpha _i ) \ne 0$ and $ch _\Psi ( \beta _i ) \ne 0$. Next, we run the analog of the intuitive re-labelling argument. By decoherence availability, we can consider two different strictly branching history algebras for the next step of decoherence: one that gives a projection $\alpha _{i+1}$ weight from $\alpha _i$ and one that gives it weight from $\beta _i$. To distinguish these, let $\Psi$ evolve with the first dynamics, and let $\Phi$ evolve with the second dynamics. Now define $\Phi (t_j) =\Psi (t_j)$ for $j \le i$ and let $\Psi (t_{i+1}) = X \Psi (t_i)$ and $\Phi (t_{i+1}) = Y \Psi (t_i)$ for the unitary operators

  $$\begin{aligned} X := V^\alpha \alpha _i + W^\alpha (1 - \alpha _i) \qquad Y := V^\beta \beta _i + W^\beta (1 - \beta _i) \end{aligned}$$

  (41)

  where $V^\alpha$ and $V^\beta$ share the range of $\alpha _{i+1}$ and $W^\alpha$ and $W^\beta$ share the range of some mutually orthogonal projection $\beta _{i+1}$. By construction, $\Psi (t_{i+1}) = \Phi (t_{i+1})$.

  Note, too, that $X \alpha _i \Psi (t_i)$ and $Y \beta _i \Psi (t_i)$ both lie in the range of $\alpha _{i+1}$. So by branching link, we have

  $$\begin{aligned} ch _{(\Psi ,\alpha _i)} \left( \alpha _{i+1} \right) = \frac{ ch _\Psi (\alpha _{i+1} )}{ ch _\Psi (\alpha _i )} = 1, \end{aligned}$$

  (42)

  and similarly

  $$\begin{aligned} ch _{(\Phi ,\beta _i)} \left( \alpha _{i+1} \right) = \frac{ ch _\Phi (\alpha _{i+1} )}{ ch _{\Phi } (\beta _i )} = 1. \end{aligned}$$

  (43)

  By state supervenience, $ch _\Phi (\alpha _{i+1} ) = ch _\Psi (\alpha _{i+1} )$, and so $ch _\Psi (\alpha _i \ ) = ch _\Phi (\beta _i )$. Applying state supervenience once more, we get

  $$\begin{aligned} ch _\Psi (\alpha _i ) = ch _\Psi (\beta _i ). \end{aligned}$$

  (44)

  (Note that, again by state supervenience, this last equality holds even if our original history algebra was not strictly branching.)

• (ii) Suppose $\langle \Psi (t_i) , \alpha _i \Psi (t_i) \rangle > \langle \Psi (t_i), \beta _i \Psi (t_i) \rangle$.

  By decoherence availability, we may assume that the $i+1$ step of decoherence is given by a unitary Y such that, for $\gamma _{i+1}$ and $\omega _{i+1}$ two mutually orthogonal projections, we have

  $$\begin{aligned} \begin{aligned} \langle Y^\dagger \Psi (t_i) , (\gamma _{i+1} \vee \omega _{i+1}) Y\Psi (t_i) \rangle&= \langle \Psi ( t_i), \alpha _i \Psi (t_i)\rangle \\ \langle Y^\dagger \Psi (t_i), \gamma _{i+1} Y \Psi (t_i) \rangle&= \langle \Psi (t_i), \beta _i\Psi (t_i) \rangle \end{aligned} \end{aligned}$$

  (45)

  where Y maps vectors in the range of $\alpha _i$ to the range of $\gamma _{i+1} \vee \omega _{i+1}$. Then, by step (i),

  $$\begin{aligned} ch _\Psi (\gamma _{i+1} \vee \omega _{i+1} ) = ch _\Psi ( \alpha _i ),\qquad ch _\Psi ( \gamma _{i+1}) = ch _\Psi ( \beta _{i} ). \end{aligned}$$

  (46)

  By probability,

  $$\begin{aligned} ch _\Psi (\gamma _{i+1} \vee \omega _{i+1} ) \ge ch _\Psi (\gamma _{i+1} ) \end{aligned}$$

  (47)

  and so

  $$\begin{aligned} ch _\Psi (\alpha _i ) \ge ch _\Psi (\beta _i ). \end{aligned}$$

  (48)

• (iii) Suppose $\langle \Psi (t_i), \alpha _i \Psi (t_i) \rangle$ is rational, i.e. equal to $\frac{M}{N}$ for some positive integers M, N.

  By decoherence availability, we may pick some ${\mathcal {H}}_{SE}$-spanning sequence of orthogonal projections $\gamma ^1, \ldots , \gamma ^m, \ldots , \gamma ^N$ that generate a $\sigma$-algebra containing $\alpha _i$ such that, for some $\Phi$ such that $\Phi (t_i) = \Psi (t_i)$,

  $$\begin{aligned} \langle \Phi (t_i), \gamma ^m \Phi (t_i) \rangle = \frac{1}{N} \end{aligned}$$

  (49)

  for all m. By (i), $ch _\Phi (\gamma ^m )$ must be independent of m. Thus, by probability, $ch _\Phi (\gamma ^m ) = \frac{1}{N}$.

  Now let $\omega := \bigvee _{i\le M} \gamma ^i$. We have that $\langle \Psi (t_i), \alpha _i \Psi (t_i) \rangle = \langle \Phi (t_i), \omega \Phi (t_i) \rangle = \langle \Psi (t_i), \omega \Psi (t_i) \rangle$, so by step (i), $ch _\Psi (\alpha _i) = ch _\Psi (\omega )$. By probability, $ch _\Psi (\omega ) = \sum _{m=1}^M ch _\Psi (\gamma ^m) = \frac{M}{N}$, and so we get that

  $$\begin{aligned} ch _\Psi ( \alpha _i) = \frac{M}{N}. \end{aligned}$$

  (50)

• (iv) Suppose $\langle \Psi (t_i), \alpha _i \Psi (t_i) \rangle = r \in [0,1]$, where r may not be rational.

  By (i), $ch$ is a function of Born-rule weights, i.e.

  $$\begin{aligned} ch _\Psi (\alpha _i ) = f(\langle \Psi (t_i), \alpha _i \Psi (t_i) \rangle ) \end{aligned}$$

  (51)

  for some $f:[0,1] \rightarrow [0,1]$. By step (ii), f is increasing; by step (iii), $f(M/N)=M/N$.

  So let $\{a_i\}$ and $\{b_i\}$ be, respectively, increasing and decreasing sequences of rational numbers in [0, 1] converging to r. Since $f(b_i) = b_i$ for all i, $f(r)\le r$. And since $f(a_i) = a_i$ for all i, $f(r)\ge r$—thus, $f(r) = r$, and so

  $$\begin{aligned} ch _\Psi ( \alpha _i ) = \langle \Psi (t_i), \alpha _i \Psi (t_i) \rangle . \end{aligned}$$

  (52)

$\square$