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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.

user36125
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1 Answers1

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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.

Emilio Pisanty
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