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In the Pusey-Barrett-Rudolph (PBR) paper “The quantum state cannot be interpreted statistically” [arXiv:1111.3328v1, later published as Nat. Phys. 8, 475 (2012)] the following claim is made:

If a full specification of a system’s physical properties (λ) uniquely determines the quantum state (|φ⟩) then the quantum state is NOT ‘statistical’.

I am having trouble understanding why this is true. I am also having trouble understanding the ‘classical analogy’ the paper uses to help explain why this is true. Here is the relevant passage from the first version of the preprint:

If the quantum state is statistical in nature (the second view), then a full specification of λ need not determine the quantum state uniquely. Some values of λ may be compatible with the quantum state being either |φ0⟩ or |φ1⟩. This can be understood via a classical analogy. Suppose there are two different methods of flipping a coin, each of which is biased. Method 1 gives heads with probability p0 > 0 and method 2 with probability 0 < p1 ≠ p0. If the coin is flipped only once, there is no way to determine by observing only the coin which method was used. The outcome heads is compatible with both. The statistical view says something similar about the quantum system after preparation. The preparation method determines either |φ0⟩ or |φ1⟩ just as the flipping method determines probabilities for the coin. But a complete list of physical properties λ is analogous to a list of coin properties, such as position, momentum, etc. Just as “heads up” is compatible with either flipping method, a particular value of λ might be compatible with either preparation method.

In this analogy the probability of coin landing heads (which is taken as a canonical example of something that is ‘statistical’) is analogous with the quantum state. The exact physical properties of the coin (position, momentum, etc) are analogous with exact physical properties (λ) of the quantum state. But don’t the exact physical properties of the coin uniquely determine the probability of it landing heads? To me the original definition of 'statistical' above has the implication going in the wrong direction. If the quantum state (probability of heads) uniquely determines the systems physical properties (coins physical properties) then it should be considered not statistical. Or is landing “heads up” supposed to be identified with λ in this analogy? Shouldn’t landing “heads up” be analogous to the outcome of a specific quantum measurement? What am I missing?

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  • $\begingroup$ @EmilioPisanty that is a good answer but I guess I am mostly still confused about the specific classical analogy. $\endgroup$ Sep 22 '20 at 17:14
  • $\begingroup$ Consider to spell out acronym in title. $\endgroup$
    – Qmechanic
    Sep 22 '20 at 17:18
  • $\begingroup$ Note that the text you're quoting comes from the original preprint, which was substantially edited for clarity before publication. You should read carefully the final version, as there's a substantial chance that it can resolve your query directly. $\endgroup$ Sep 23 '20 at 9:56
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    $\begingroup$ @Qmechanic "PBR Theorem" is much more recognizable as the name of the theorem than the spelt-out names; it makes little sense to spell it out explicitly in the title here. $\endgroup$ Sep 23 '20 at 9:57
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Your classical analogy is extremely tricky to use in this context (and is likely to fail completely), because classical mechanics is a completely deterministic theory.

To recap from my answer to What are the differences between a $\psi$-epistemic ontological model and a $\psi$-ontic model of quantum mechanics, exactly?, a $\psi$-epistemic model (the correct technical term for what the paper refers to as 'statistical interpretation' in more hand-wavy language) is any ontic model where the ontic state does not determine the quantum state, i.e. where there are multiple quantum states $\psi$ that are compatible with each ontic state $\lambda$:

To have something like that in your classical coin-flipping theory, you would have to have a dynamical state of the coin (position, momentum, angular momentum, orientation) which was compatible with two separate outcomes. This does not happen in classical mechanics, because classical mechanics is different to quantum mechanics.

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  • $\begingroup$ Thanks. The most recent version of the paper is much clearer and throws out the whole 'classical analogy' entirely. $\endgroup$ Sep 23 '20 at 16:51

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