In Quantum Mechanics, two different wavefunctions can have a non-zero probability of finding a particle at a position $x$.

According to hidden variable theories, if a particle is found at $x$, it was there before the measurement. $x$ was the hidden variable.

According to PBR theorem, a single hidden variable $x$ can not be consistent with multiple wavefunctions. The assignment of hidden variables to wavefunctions cannot be many-to-one. This contradicts the first paragraph of the post

Since the first paragraph is true, doesn't this mean that hidden variables, including Bohmian mechanics, are disproved?

  • $\begingroup$ x is not the position of the quantum. It is always the position of the detector. There are a lot of misunderstandings going on about that. A supporter of a hidden variable theory only has to do one thing: to convince us: he or she has to tell us when the next quantum will be absorbed by the detector at position x. That should be an easy thing to do, right? $\endgroup$ Nov 2, 2022 at 2:26
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    $\begingroup$ Wouldn't the idea of hidden variables be that there are other variables we can't see that affect the state? I don't think $x$ could be a hidden variable since you can measure it. $\endgroup$
    – Andrew
    Nov 2, 2022 at 2:52
  • $\begingroup$ @Andrew I think the idea is the same as in classical mechanics : $x$ and $p$ are always well defined but hidden before measurement. But you're right. According to my post's logic, the hidden variable $x$ completely specifies the state before measurement. But that's not true because there's also $p$. I shall add this as the answer $\endgroup$
    – Ryder Rude
    Nov 2, 2022 at 3:15
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    $\begingroup$ @FlatterMann The particle becomes eigenstate of eigenvalue $x$ after detection. This can be verified with further measurements and has been verified. It is wrong to say that $x$ has nothing to do with the state of the particle. You always try to make measurement devices sound like the fundamental entities of the theory. $\endgroup$
    – Ryder Rude
    Nov 2, 2022 at 3:41
  • $\begingroup$ @RyderRude States belong to the ensemble. A single quantum is simply an irreversible energy transfer between two systems. It doesn't exist as an object. Measurement devices are the things that I used to build for a living. The theory only has operators that pretend to work like measurement devices, but don't. $\endgroup$ Nov 2, 2022 at 7:01

2 Answers 2


It does not "disprove" hidden variables, but rather says that, under a suitable formulation and assumptions about what form the hidden variable theory takes, that the hidden variables must be at least as complex as the quantum Hilbert space. That is to say, there is a(n almost) distinct class of hidden-variable assignments for each and every of the uncountably many, $\beth_1$, possible vectors in the Hilbert space or, if you like, that the hidden variables can be described or labeled in such a way that at least one of those variables must be equivalent to the quantum vector itself.

The idea behind the PBR theorem is this. Consider a coin in classical probability theory - not one we're going to flip, but one we're going to have some other device give us in either the heads or tails conditions, randomly, according to some settings. If that device is set up to choose heads with some probability $p$, then a person ignorant of exactly what the coin has been flipped to, but knows the setting on the device, must say that it is heads only with probability $p$. Moreover, by setting $p$ suitably on the device, we can convey to someone else a "heads" with any real-number probability $p$ even though the coin itself has only two states.

Then, under suitable combining assumptions for how multiple coins' states combine, if we have a preparation system for a quantum "coin", the theorem tells us we cannot treat the whole continuum of quantum probabilities about it as simply being due to that it has a few states and the preparation system generates those probabilities by mixing them together. Instead, it is like the coin really has not only infinitely many states, but so many there is at least one for each analogue of the real number $p$ in the classical coin case (for a qubit, this means a point on the Bloch sphere) or to say it another way, it is as though the coin carried the whole probability setting that was on the preparer. Or else, either the combining assumptions are wrong (though I've also heard that isn't "strong enough" to defeat the theorem), or, in a result reminiscent of Bell's Theorem but for the preparer, multiple preparers must somehow "spookily" correlate their preparations non-locally.

  • $\begingroup$ So, if we proved that the phase space cannot be partitioned, without overlap, among the wavefunctions, then would it disprove the hidden variable theories which take the phase space as their hidden variable space? The set of wavefunctions have a much higher cardinality than the set of points on the phase space. $\endgroup$
    – Ryder Rude
    Nov 2, 2022 at 6:07

In the post, I assumed that $x$ completely specifies the hidden variable state. But that's not true because the two states can have the same $x$ but differ in $p$. Hence the combined states $(x, p) $ are different and there's no contradiction with PBR.

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    $\begingroup$ Where did you prove that they must differ in p? $\endgroup$ Nov 2, 2022 at 12:40
  • $\begingroup$ @FlatterMann I'm just saying that the particular logic of my post doesn't work. Indeed, someone needs to prove that they differ in $p$ for hidden variable theories to survive, at least the ones which take $(x, p) $ as their hidden variables. I find this impossible to prove because the set of wavefunctions has a much larger cardinality than the points on the phase space. So I don't think the phase space hidden variable theories can survive the PBR theorem $\endgroup$
    – Ryder Rude
    Nov 2, 2022 at 13:20
  • $\begingroup$ Wave functions for spin 1/2 systems have large cardinality? Two is a large number now? $\endgroup$ Nov 2, 2022 at 13:21
  • $\begingroup$ @FlatterMann I already agreed that wavefunctions have a higher cardinality. But the problem is that theories like Bohmian mechanics do not take "spin" as a hidden variable. Spin is just part of the Quantum Potential that determines particle trajectories. Someone needs to do the proof for phase space hidden variables. $\endgroup$
    – Ryder Rude
    Nov 2, 2022 at 13:23
  • $\begingroup$ There are no trajectories. $\endgroup$ Nov 2, 2022 at 13:24

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