Quantum metametaphysics

Abstract

Say that metaphysical indeterminacy occurs just when there is a fact such that neither it nor its negation obtains. The aim of this work is to shed light on the issue of whether orthodox quantum mechanics provides any evidence of metaphysical indeterminacy by discussing the logical, semantic, and broadly methodological presuppositions of the debate. I argue that the dispute amounts to a verbal disagreement between classical and quantum logicians, given Eli Hirsch’s account of substantivity; but that it need not be so if Ted Sider’s naturalness-based account of substantivity is adopted instead. Given the latter approach, can anything be said in order to tip the balance of the dispute either way? Some prima facie reasonable constraints on naturalness entail that the classicist is right, and the quantum world is therefore determinate. Nevertheless, there are reasons for weakening those constraints, to the effect that the dispute remains very much open. Finally, I discuss alternative accounts of metaphysical indeterminacy, and argue that they are unsuitable for framing the quantum indeterminacy debate.

Appendix

1.1 The paraphrase $\mathbf { f}$

It needs to be proven that the function f from Sect. 4, which maps categorical $\mathcal {L}_{Q}$-sentences to $\mathcal {L}_{C}$ sentences, preserves coarse-grained content. The proof is by induction on the syntactical complexity of categorical $\mathcal {L}_{Q}$-sentences. (I will explicitly state the justification of a step only when the step requires appealing to a new principle.)

(At$_{Q}$):

We need to show that for any atomic categorical $\mathcal {L}_{Q}$-sentence p, and any choice of $|\psi _{@}\rangle$:

p is $\mathcal {L}_{Q}$-true iff $\mathbf {Pr}([p]_{C},|\psi _{@}\rangle )=1$ is $\mathcal {L}_{C}$-true.

Proof: p is $\mathcal {L}_{Q}$-true iff $\mathrm {T}_{Q}([p]_{Q},|\psi _{@}\rangle )$ (by definition of $\mathcal {L}_{Q}$-truth), iff $\mathbf {Pr}([p]_{Q},|\psi _{@}\rangle )=1$ (by cr together with its converse, which is trivial), iff $\mathbf {Pr}([p]_{C},|\psi _{@}\rangle )=1$ (since $[p]_{C}=[p]_{Q}$ for atomic p), iff $\mathbf {Pr}([p]_{C},|\psi _{@}\rangle )=1$ is $\mathcal {L}_{C}$-true (since C assigns the correct probabilities to classical facts).

($\lnot _{Q}$):

We need to show that, for any categorical $\mathcal {L}_{Q}$-sentence P, and any choice of $|\psi _{@}\rangle$:

$\lnot P$ is $\mathcal {L}_{Q}$-true iff $\mathbf {Pr}([\mathbf { f}(P)]_{C},|\psi _{@}\rangle )=0$ is $\mathcal {L}_{C}$-true.

Proof: $\lnot P$ is $\mathcal {L}_{Q}$-true iff $\mathrm {T}_{Q}([\lnot P]_{Q},|\psi _{@}\rangle )$, iff $\mathrm {T}_{Q}(-_{Q}[P]_{Q},|\psi _{@}\rangle )$, iff $\mathbf {Pr}(-_{Q}[P]_{Q},|\psi _{@}\rangle )=1$, iff $\mathbf {Pr}(-_{Q}[\mathbf {f}(P)]_{C},|\psi _{@}\rangle )=1$ (by inductive hypothesis), iff $\mathbf {Pr}([\mathbf {f}(P)]_{C},|\psi _{@}\rangle )=0$, iff $\mathbf {Pr}([\mathbf {f}(P)]_{C},|\psi _{@}\rangle )=0$ is $\mathcal {L}_{C}$-true.

($\wedge _{Q}$):

We need to show that, for any categorical $\mathcal {L}_{Q}$-sentences S, P and any choice of $|\psi _{@}\rangle$:

$S\wedge P$ is $\mathcal {L}_{Q}$-true iff both $\mathbf {Pr}([\mathbf { f}(S)]_{C},|\psi _{@}\rangle )=1$ and $\mathbf {Pr}([\mathbf { f}(P)]_{C},|\psi _{@}\rangle )=1$ are $\mathcal {L}_{C}$-true.

Proof: $S\wedge P$ is $\mathcal {L}_{Q}$-true iff $\mathrm {T}_{Q}([S\wedge P]_{Q},|\psi _{@}\rangle )$, iff $\mathrm {T}_{Q}([S]_{Q}\sqcap _{Q} [P]_{Q},|\psi _{@}\rangle )$, iff $\mathbf {Pr}([S]_{Q}\sqcap _{Q} [P]_{Q},|\psi _{@}\rangle )=1$, iff $\mathbf {Pr}([\mathbf { f}(S)]_{C}\sqcap _{Q} [\mathbf { f}(P)]_{C},|\psi _{@}\rangle )=1$ (by inductive hypothesis), iff $\mathbf {Pr}([\mathbf { f}(S)]_{C}\sqcap _{C} [\mathbf { f}(P)]_{C},|\psi _{@}\rangle )=1$ (because $\sqcap _{C}=\sqcap _{Q}$), iff both $\mathbf {Pr}([\mathbf { f}(S)]_{C},|\psi _{@}\rangle )=1$ and $\mathbf {Pr}([\mathbf { f}(P)]_{C},|\psi _{@}\rangle )=1$, iff both $\mathbf {Pr}([\mathbf { f}(S)]_{C},|\psi _{@}\rangle )=1$ and $\mathbf {Pr}([\mathbf { f}(P)]_{C},|\psi _{@}\rangle )=1$ are $\mathcal {L}_{C}$-true.

1.2 The paraphrase $\mathbf { g}$

It needs to be proven that the function g from Sect. 4, which maps categorical $\mathcal {L}_{C}$-sentences to $\mathcal {L}_{Q}$ sentences, preserves coarse-grained content. The proof is by induction on the syntactical complexity of categorical $\mathcal {L}_{C}$-sentences. (Once again, I will explicitly state the justification of a step only when a new principle is appealed to.)

(At$_{C}$):

We need to show that for any atomic categorical $\mathcal {L}_{C}$-sentence p, and any choice of $|\psi _{@}\rangle$:

p is $\mathcal {L}_{C}$-true iff $\mathcal {T}(p)$ is $\mathcal {L}_{Q}$-true.

Proof: p is $\mathcal {L}_{C}$-true iff $\mathrm {T}_{C}([p]_{C},|\psi _{@}\rangle )$ (by definition of $\mathcal {L}_{C}$-truth), iff $\mathrm {T}_{C}([p]_{Q},|\psi _{@}\rangle )$ (since $[p]_{C}=[p]_{Q}$ for atomic p), iff $\mathcal {T}(p)$ is $\mathcal {L}_{Q}$-true (by definition of $\mathcal {T}$).

($\lnot _{C}$):

We need to show that, for any categorical $\mathcal {L}_{C}$-sentence P, and any choice of $|\psi _{@}\rangle$:

$\lnot P$ is $\mathcal {L}_{C}$-true iff $\sim \mathbf {g}(P)$ is $\mathcal {L}_{Q}$-true.

Proof: $\lnot P$ is $\mathcal {L}_{C}$-true iff $\mathrm {T}_{C}([\lnot P]_{C},|\psi _{@}\rangle )$, iff $\mathrm {T}_{C}(-_{C}[P]_{C},|\psi _{@}\rangle )$, iff not $\mathrm {T}_{C}([P]_{C},|\psi _{@}\rangle )$, iff not $\mathrm {T}_{C}([\mathbf { g}(P)]_{Q},|\psi _{@}\rangle )$ (by inductive hypothesis), iff $\mathbf {g}(P)$ is not $\mathcal {L}_{Q}$-true, iff $\sim \mathbf {g}(P)$ is $\mathcal {L}_{Q}$-true (by definition of $\sim$).

($\wedge _{C}$):

We need to show that, for any categorical $\mathcal {L}_{C}$-sentences S, P and any choice of $|\psi _{@}\rangle$:

$S\wedge P$ is $\mathcal {L}_{C}$-true iff $\mathbf { g}(S) \& \mathbf { g}(P)$ is $\mathcal {L}_{Q}$-true.

Proof: analogous to the case of ($\lnot _{C}$).

1.3 Verbal disagreement about MI

We want to show that the dispute between classicists ($\mathcal {L}_{C}$-speakers) and quantum logicians ($\mathcal {L}_{Q}$-speakers) about whether quantum mechanics leads to MI is verbal in Hirsch’s sense (Sect. 4). A rigorous proof that the dispute is verbal will require (i) introducing an indeterminacy operator in both object languages; (ii) extending the paraphrase schemes $\mathbf { f}$, $\mathbf { g}$ in order to account for statements of indeterminacy; (iii) showing that, for some categorical sentence p, ‘it is metaphysically indeterminate whether p’ is true-in-$\mathcal {L}_{C}$ iff it is not true-in-$\mathcal {L}_{Q}$.

Let us start by expanding both $\mathcal {L}_{C}$ and $\mathcal {L}_{Q}$ with a sentential operator $\nabla$ with the intended meaning ‘it is metaphysically indeterminate whether’ as defined by ind. We then extend $\mathbf { f}$ to a mapping $\mathbf { f}^{*}$ from categorical $\mathcal {L}_{Q}$-sentences to $\mathcal {L}_{C}$ sentences such that:

($\nabla _{Q}$):

If P is a categorical sentence of $\mathcal {L}_{Q}$, then $\mathbf { f}^{*}(\nabla P)$ is $\lnot \mathbf {Pr}([\mathbf { f}^{*}(P)]_{C},|\psi _{@}\rangle )$ $=1\wedge \lnot \mathbf {Pr}([\mathbf { f}^{*}(P)]_{C},|\psi _{@}\rangle )=0$

whereas $\mathbf { f}^{*}(P)=\mathbf { f}(P)$ if P is atomic, negated, or conjunctive. We need to show that for any categorical $\mathcal {L}_{Q}$-sentence P, and any choice of $|\psi _{@}\rangle$:

$\nabla P$ is $\mathcal {L}_{Q}$-true iff $\lnot \mathbf {Pr}([\mathbf { f}^{*}(P)]_{C},|\psi _{@}\rangle )$ $=1\wedge \lnot \mathbf {Pr}([\mathbf { f}^{*}(P)]_{C},|\psi _{@}\rangle )=0$ is $\mathcal {L}_{C}$-true.

Proof: $\nabla P$ is $\mathcal {L}_{Q}$-true iff not $\mathrm {T}_{Q}([P]_{Q},|\psi _{@}\rangle )$ and not $\mathrm {T}_{Q}([\lnot P]_{Q},|\psi _{@}\rangle )$, iff not $\mathrm {T}_{Q}([P]_{Q},|\psi _{@}\rangle )$ and not $\mathrm {T}_{Q}(-_{Q}[P]_{Q},|\psi _{@}\rangle )$, iff not $\mathbf {Pr}([P]_{Q},|\psi _{@}\rangle )=1$ and not $\mathbf {Pr}(-_{Q}[P]_{Q},|\psi _{@}\rangle )=1$, iff not $\mathbf {Pr}([\mathbf { f}^{*}(P)]_{C},|\psi _{@}\rangle )=1$ and not $\mathbf {Pr}(-_{Q}[\mathbf { f}^{*}(P)]_{C},|\psi _{@}\rangle )=1$ (by inductive hypothesis), iff not $\mathbf {Pr}([\mathbf { f}^{*}(P)]_{C},|\psi _{@}\rangle )$ $=1$ and not $\mathbf {Pr}([\mathbf { f}^{*}(P)]_{C},|\psi _{@}\rangle )=0$, iff $\lnot \mathbf {Pr}([\mathbf { f}^{*}(P)]_{C},|\psi _{@}\rangle )$ $=1\wedge \lnot \mathbf {Pr}([\mathbf { f}^{*}(P)]_{C},|\psi _{@}\rangle )=0$ is $\mathcal {L}_{C}$-true.

We now extend $\mathbf { g}$ to a mapping $\mathbf { g}^{*}$ from categorical $\mathcal {L}_{C}$-sentences to $\mathcal {L}_{Q}$ sentences such that

($\nabla _{C}$):

If P is a categorical sentence of $\mathcal {L}_{C}$, then $\mathbf { g}^{*}(\nabla P)$ is $0=1$

whereas $\mathbf { g}^{*}(P)=\mathbf { g}(P)$ if P is atomic, negated, or conjunctive. We need to show that for any categorical $\mathcal {L}_{C}$-sentence P, and any choice of $|\psi _{@}\rangle$:

$\nabla P$ is $\mathcal {L}_{C}$-true iff 0=1 is $\mathcal {L}_{Q}$-true.

Since $\mathcal {L}_{C}$ is bivalent, the proof is trivial. This concludes the proof that there is a two-way paraphrase between expanded $\mathcal {L}_{Q}$ and expanded $\mathcal {L}_{C}$.

The final step is to find a categorical sentence p in the common fragment of $\mathcal {L}_{C}$ and $\mathcal {L}_{Q}$ such that $\nabla p$ is true-in-$\mathcal {L}_{Q}$ and not true-in-$\mathcal {L}_{C}$. Relative to a single-electron system in a superposition of z-spin up and z-spin down, one such sentence is ‘e is z-spin up.’ This suffices to show that the disagreement between C and Q concerning quantum MI is verbal in Hirsch’s sense.

1.4 Collective completeness of $\mathcal {Q}$ relative to $\mathcal {C}$

Let $\mathcal {C},\mathcal {Q}$ be classical and quantum logical spaces, respectively, on a set of quantum states W. We need to show that $\mathcal {Q}$ is collectively complete relative to $\mathcal {C}$, which is to say, that the collection of all classical facts obtaining at $|\psi _{@}\rangle$ is collectively empirically equivalent to the singleton of the quantum fact which corresponds to the closed set having as members all vectors indistinguishable from $|\psi _{@}\rangle$. The proof is as follows.

In quantum mechanics, states differing by a scalar verify the same experimental facts. Therefore, quantum states are indistinguishable just in case they differ by a scalar. Let $\mathbf {P}$ be the set of facts obtaining at $|\psi _{@}\rangle$ in $\mathcal {C}$. Consider now the set of facts $\mathbf {Q}=\{Q\}$ in $\mathcal {Q}$, where Q is the set of unitary vectors indistinguishable from $|\psi _{@}\rangle$. If Q obtains at some state indistinguishable from a state $|\psi \rangle$, then $|\psi _{@}\rangle$ and $|\psi \rangle$ are indistinguishable. Therefore, every $P\in \mathbf {P}$ obtains at some state, namely $|\psi _{@}\rangle$, which is indistinguishable from $|\psi \rangle$. Conversely, let every $P\in \mathbf {P}$ obtain at some state indistinguishable from a state $|\psi \rangle$. Since $Q\in \mathbf {P}$, $|\psi _{@}\rangle$ and $|\psi \rangle$ are indistinguishable. Therefore, Q obtains at some state, namely $|\psi _{@}\rangle$, which is indistinguishable from $|\psi \rangle$. QED.