In the PBR-Theorem they try to prove that $\psi$-epistemic models of quantum theory are impossible. The argument goes something like this: Suppose we have two nonorthogonal states \begin{equation}|\psi\rangle=|\uparrow\rangle \text{ and } |\phi\rangle=|+\rangle=\frac{1}{\sqrt{2}}(|\uparrow\rangle+|\downarrow\rangle) \end{equation} in two identical physical systems. The combined system then is in one of these physical states \begin{aligned} & \left|\omega_1\right\rangle_{12}=|\uparrow\rangle_1 \otimes|\uparrow\rangle_2 \\ & \left|\omega_2\right\rangle_{12}=|\uparrow\rangle_1 \otimes|+\rangle_2 \\ & \left|\omega_3\right\rangle_{12}=|+\rangle_1 \otimes|\uparrow\rangle_2 \\ & \left|\omega_4\right\rangle_{12}=|+\rangle_1 \otimes|+\rangle_2. \end{aligned} Then, there exists a measurement that projects onto the basis vectors \begin{aligned} & \left|\chi_1\right\rangle_{12}=\frac{1}{\sqrt{2}}\left(|\uparrow\rangle_1 \otimes|\downarrow\rangle_2+|\downarrow\rangle_1 \otimes|\uparrow\rangle_2\right) \\ & \left|\chi_2\right\rangle_{12}=\frac{1}{\sqrt{2}}\left(|\uparrow\rangle_1 \otimes|-\rangle_2+|\downarrow\rangle_1 \otimes|+\rangle_2\right) \\ & \left|\chi_3\right\rangle_{12}=\frac{1}{\sqrt{2}}\left(|+\rangle_1 \otimes|\downarrow\rangle_2+|-\rangle_1 \otimes|\uparrow\rangle_2\right) \\ & \left|\chi_4\right\rangle_{12}=\frac{1}{\sqrt{2}}\left(|+\rangle_1 \otimes|-\rangle_2+|-\rangle_1 \otimes|+\rangle_2\right) \end{aligned} for which we can easily calculate that \begin{aligned} & \left\langle\omega_1 \mid \chi_1\right\rangle=0 \\ & \left\langle\omega_2 \mid \chi_2\right\rangle=0 \\ & \left\langle\omega_3 \mid \chi_3\right\rangle=0 \\ & \left\langle\omega_4 \mid \chi_4\right\rangle=0 . \end{aligned} PBR conclude from this that $\psi$-epistemicism is wrong, but i dont understand why. Where exactly is the premise of $\psi$-epistemicism mathematically implemented in the theorem. And why does an ontic model not have the same conclusion? I would like to understand how the theorem would play out in an ontic model.\ I know this is related to overlapping probability distributions in the ontic state space. But i guess i dont understand where exactly the overlap is contradicted.


1 Answer 1


The extract you have shown is the $\psi$-ontic model. The $\psi$-epistemic model is described in the preceding paragraphs of Pusey, Barrett, & Rudolph.

The interpretation of the quantum wavefunction $|\psi\rangle$ can be categorized as either $\psi$-ontic if "every complete physical state or ontic state in the theory is consistent with only one pure quantum state" and $\psi$-epistemic "if there exist ontic states that are consistent with more than one pure quantum state." (From Wikipedia.)

PBR consider an experiment where we can prepare a system in one of two ways, leading (according to quantum mechanics) to pure quantum states $|\uparrow\rangle$ and $|+\rangle$ respectively. The real ontic state is labelled $\lambda$ and is allowed to include components not described by quantum mechanics.

The $\psi$-epistemic model has situations where preparations of both pure states can lead to the same real, ontic state. This is assumed to happen with at least non-zero probability $q$.

The figure below illustrates this. We have two systems that we can prepare either as $|\uparrow\rangle$ (top row) or $|+\rangle$ (bottom row) independently. Shading indicates where these correspond to real ontic states corresponding to the columns in the two tables. (Each block shown can represent many possible columns/states.) There will be some real states that can only arise when prepared as $|\uparrow\rangle$, and some real states that can only arise from $|+\rangle$, and if the $\psi$-epistemic model is valid, some that can arise from either. These occur with probability $q$.

enter image description here

Then the measurement apparatus makes a joint measurement of both states. It is assumed that the behaviour of the apparatus depends only on the actual ontic states $\lambda_1$, $\lambda_2$ of the two systems, post-preparation. It does not know how they were prepared. Thus, there is a probability $q^2$ that both systems will be in a state such that the measurement apparatus cannot know which pure state was prepared. Whatever answer it picks, there will be a non-zero probability that, given the preparation methods, this outcome is impossible, with zero probability. It will therefore violate the predictions of quantum mechanics some of the time.

If the predictions of quantum mechanics are correct, then $q$ must be zero, and the pure quantum state is a real property of $\lambda$. For any given ontic state $\lambda$, it has a well-defined pure quantum state $\psi$.

The section you quote sets out the predictions of quantum mechanics, asserting that the probability of certain combinations of set-up and outcome are impossible. In the $\psi$-ontic case, they are. In the $\psi$-epistemic case, they can't be - thus the $\psi$-epistemic model contradicts these zero probability predictions.

  • $\begingroup$ Thanks a lot for the detailed explanation, i think i understand it now a lot better! For clarification: The measurement device measures what preparation could impossibly have been prepared prior to the measurement, e.g. if i measure |X_1> (see above), the pointer will show "not |up>|up>" because this could not possibly have been the prepared state. If we assume overlap, at least sometimes the device will show a impossible result, because it will randomly point on one of the measurement outcomes, which might be orthogonal to the prepared state? Sorry if I am a bit slow here $\endgroup$
    – Luca
    Commented Dec 29, 2023 at 22:22
  • 1
    $\begingroup$ Yes, that's the intention. Because the measurement device only sees the real state, it can't reliably determine the quantum state prepared in the epistemic case (so the pointer labels are a bit misleading). It has to do something (the measurement gives a complete set of basis states), but every answer will be wrong (according to QM) some of the time. If the real state implies the quantum state so the measurement device can give the right answer, that implies the quantum state is ontic. $\endgroup$ Commented Dec 29, 2023 at 22:34
  • $\begingroup$ Awesome! Thanks a lot for your very helpful explanations :)! $\endgroup$
    – Luca
    Commented Dec 29, 2023 at 22:52

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