I am having trouble to understand the reasoning in the following paper,

On the reality of the quantum state. MF Pusey, J Barret and T Rudolph. Nature Phys. 8, 475–478 (2012); arXiv:1111.3328.

From a few general assumptions, it claims to have proved the quantum wave functions must represent the physical states, rather than the knowledge about the physical systems. I didn't get the reasoning from page 2 to page 3, for the simplest case. The argument goes as following.

With two copies of the same device independently, each of which can prepare its quantum states in either $|0\rangle$ or $|+\rangle$:

$$ |+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}} $$

for which the distributions of physical states are $\mu_0(\lambda)$ and $\mu_+(\lambda)$. Suppose that $|0\rangle$ and $|+\rangle$ represent the knowledge states, which means $\mu_0(\lambda)$ and $\mu_+(\lambda)$ can overlap for $\lambda \in \Delta$. "This means that the physical state of the two systems is compatible with any of the four possible quantum states $|0\rangle \otimes |0\rangle$, $|0\rangle \otimes |+\rangle$, $|+\rangle \otimes |0\rangle$ and $|+\rangle \otimes |+\rangle$"

Then two systems are brought together and measured by projecting onto four orthogonal states $|\xi_1\rangle$, $|\xi_2\rangle$, $|\xi_3\rangle$ and $|\xi_4\rangle$, such that

\begin{align} \langle0,0 | \xi_1\rangle & = 0 \\ \langle0,+ | \xi_2\rangle & = 0 \\ \langle+,0 | \xi_3\rangle & = 0 \\ \langle+,+ | \xi_4\rangle & = 0 \end{align}

From here, the authors stated that, for the overlap region $\lambda \in \Delta$ "it runs the risk of giving an outcome that quantum theory predicts should occur with probability 0". It is this last step reasoning that got me lost. How could one reach the conclusion of all four possible states, namely $|0\rangle \otimes |0\rangle$, $|0\rangle \otimes |+\rangle$, $|+\rangle \otimes |0\rangle$ and $|+\rangle \otimes |+\rangle$, having probability of 0, based on the above four orthogonal relations and $\lambda \in \Delta$? Please help me understand this step of the proof.

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    $\begingroup$ After a cursory look at it, it seems to be yet another paper of the sort that makes a scientifically unnecessary and unwarranted assumption about reality and then "proves" that this assumption is not covered by quantum mechanics. Absolutely nothing relevant to science can be learned this way. Is this good philosophy? Maybe... if you think that a philosophy paper that asks "Is Man a Turnip?" and then comes to the conclusion that man is not a turnip is a relevant philosophy paper. What do you think? $\endgroup$
    – CuriousOne
    Commented Feb 8, 2016 at 22:21
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    $\begingroup$ @CuriousOne I am not sure what you really means "this assumption is not covered by quantum mechanics". Do you think people all clearly understand and agree on what the quantum wave-function represents? The problem is, it is not clear at all. It is relevant to what the Von Neumann's assumption of "wave-function reduction" really means. Is it simply a state of knowledge change, or is it really indicating some "new" physical phenomenon we haven't really understand, which I would doubt by just common sense. $\endgroup$
    – user36125
    Commented Feb 8, 2016 at 23:41
  • $\begingroup$ @CuriousOne Not really - the PBR theorem does contain some new material (it made a big splash some three years ago), though the charge is probably that it only really formalized stuff that people intuitively knew, and it does not rule out any $\psi$-epistemic theory that was being seriously considered by anyone. What it shows is that if there is at all a "state of the world" $\lambda$, then the wavefunction $\psi$ must be a physical descriptor and not just "pure information"; the only real alternative left are forms of 't Hooftian superdeterminism. Not a paper to read lightly. $\endgroup$ Commented Feb 8, 2016 at 23:45
  • $\begingroup$ This problem is certainly relevant to the quantum measurement and all related physical processes. What differentiates the science and general philosophy is the realization of experimental verification, or falsification. I am seeking an answer that can be experimentally tested. I am wondering whether the proof in this paper can, or cannot, be experimentally implemented. But first I need to understand what authors are talking about, before I could make that judgement. This is the motivation of my question about their proof. $\endgroup$
    – user36125
    Commented Feb 8, 2016 at 23:45
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1 Answer 1


The chain of reasoning runs something like this:

  • Assume that the distributions $\mu_0(\lambda)$ and $\mu_+(\lambda)$ overlap in $\Delta\subset\Lambda$, so with probability $q$ preparing $|0⟩$ will produce a state $\lambda$ consistent with $|+⟩$ and vice versa.
  • Assume that the two systems can be prepared independently.
  • Therefore, with probability $q$ the preparation will produce states $\lambda_1$ and $\lambda_2$ consistent with both $|0⟩$ and $|+⟩$ for the two systems.
  • $|\xi_1⟩$ is orthogonal to $|0,0⟩$, so it is incompatible with any state that could be produced when preparing $|0,0⟩$, such as $(\lambda_1,\lambda_2)$. The only way that $(\lambda_1,\lambda_2)$ could be compatible with $|\xi_1⟩$ is that the system knew, when I prepared $|0,0⟩$, that I was going to measure $|\xi_1⟩$, and then subselected from the support in $\Lambda^2$ of $|0,0⟩$, and that is a superdeterministic assumption.
  • Similarly none of $|\xi_2⟩$, $|\xi_3⟩$ or $|\xi_4⟩$ can happen.
  • ... so the probability of measurement is zero? Contradiction.

Hopefully that is clear enough.

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    $\begingroup$ Is this the proof that QM I is correct, using nothing but statements from QM I? Sorry for prying, but what did we learn here? $\endgroup$
    – CuriousOne
    Commented Feb 8, 2016 at 23:58
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    $\begingroup$ As I said: If there is an objective state of the world, and if we're allowed to prepare systems independently, then given the objective state of the world $\lambda$ we can always infer the wavefunction $\psi$, i.e. it "exists" as a property of the objective state of the world. This is as opposed to $\psi$-epistemic models where $\psi$ is exclusively a state of information in the observer's head which gets updated after measurement results are discovered. If you don't see the need for such models, then you didn't need convincing in the first place. $\endgroup$ Commented Feb 9, 2016 at 0:24
  • $\begingroup$ But the outcome of the future (isn't that what dynamics is about?) isn't just the state (objective or not), it's also what we decide to measure about it. Nature can't know what information we will extract from it, so "the state" necessarily has to leave that open because it is ignorant of what is between "it" and the final outcome of what we call "a measurement". I don't see a particular need to dice that ever finer, as long as we don't start seeing contradictions between what the theory predicts and what measurements return. $\endgroup$
    – CuriousOne
    Commented Feb 9, 2016 at 0:30
  • $\begingroup$ @EmilioPisanty Thanks! It only proved that the quantum theory contradicts the way how the authors formalized the "state of the world" λ and its relation to quantum states. Why is this way of formalizing the "state of the world" λ so general that it should cover the quantum physical world? $\endgroup$
    – user36125
    Commented Feb 9, 2016 at 0:34
  • $\begingroup$ Apologies about the key slip... my mind is failing, but not that badly. :-) $\endgroup$
    – CuriousOne
    Commented Feb 9, 2016 at 0:36

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