Is there an additional assumption in the PBR theorem?

[PeterDonis]

I have been reading up on the PBR theorem, and it seems to me that there might be an additional assumption that is not discussed in the treatments I have read. The most comprehensive treatment of the general subject of theorems of which PBR is an example (theorems which attempt to show that the quantum state ##\psi## must be ontic) is this paper by Leifer:

https://arxiv.org/pdf/1409.1570.pdf

I'll briefly sketch my understanding of the theorem in order to point out the additional assumption that seems to me to be made, but which I don't see discussed. In Example 7.9 in Leifer's paper, we have two copies of a qubit, each of which we can prepare in one of two quantum states ##|0\rangle## and ##|+\rangle##. The preparations are assumed to be independent, so the quantum state of the two-qubit system after both preparations is one of the four product states ##|0\rangle |0\rangle##, ##|0\rangle |+\rangle##, ##|+\rangle |0\rangle##, or ##|+\rangle |+\rangle##. We then make an entangled measurement of the two-qubit system that projects onto one of four possible orthonormal quantum states ##|\phi_{00} \rangle##, ##|\phi_{01} \rangle##, ##|\phi_{10} \rangle##, or ##|\phi_{11} \rangle## (which are written down in the two-qubit basis in the paper).

Everything so far is just a description of a realizable scenario in terms of standard quantum mechanics. However, we now consider a generic model in which there is an ontic state denoted by ##\lambda##, and the set of all possible ontic states is denoted by ##\Lambda##. A quantum state ##\psi## in this model corresponds to a probability distribution ##\mu_\psi (\lambda)## over the possible ontic states. If we have multiple copies of a system, each with some ontic state, each of which is prepared independently, then the probability distribution ##\mu## corresponding to the quantum state of the full system is just the product of the distributions corresponding to the quantum states of each of the individual systems (this is called the Preparation Independence Principle, or PIP). We then say that the model is ##\psi## epistemic if there are some pairs of quantum states ##\psi##, ##\phi## such that the probability distributions ##\mu_\psi## and ##\mu_\phi## overlap. (We assume in what follows that any such pair of states is not orthogonal, so ##\langle \psi | \phi \rangle > 0##.)

The key point in the proof of the theorem is that each of the four possible outcome quantum states of the measurement described above is orthogonal to one of the four possible two-qubit product states, so if that outcome is obtained, it rules out the two-qubit product state that it is orthogonal to. However, since the individual qubit states ##|0\rangle## and ##|+\rangle## are not orthogonal, if our model is ##\psi## epistemic, then the probability distributions corresponding to them must overlap, so there must be some ontic state ##\lambda## that is contained in both, i.e., that is compatible with both quantum states. That means that the ontic state ##(\lambda, \lambda)## for the two-qubit system must be compatible with all four of the possible outcome quantum states. But that means that if the two-qubit system is in that ontic state, which is a possible ontic state for all four of the possible two-qubit product states after the preparation procedure, it gives a nonzero probability for all four of the possible outcome states. This contradicts the QM prediction that each of the possible outcome states rules out one of the four possible two-qubit product states.

What appears to me to be an additional assumption in all this is the assumption that the entangled measurement procedure does not change the ontic state. The theorem relies on properties of the ontic state that follow from it being contained in the probability distribution of one of the four two-qubit product quantum states. But the quantum state of the system after measurement is not any of those states; it is one of the four entangled outcome states. How do we know that this change in the quantum state is not accompanied by a change in the ontic state? And if the ontic state changes during the measurement, clearly we cannot prove the theorem, because we cannot assume that the ontic state after measurement is still one that is contained in the probability distribution of one of the four product states. All we can assume is that the ontic state after measurement is contained in the probability distribution of the entangled state corresponding to the outcome we measured. And that is not enough to establish the theorem. So it seems like the assumption that the ontic state does not change during the measurement is necessary to prove the theorem.

Am I missing something?

 

[PeterDonis]

Having thought about this some more, I think I can sharpen the question I'm asking. Consider the following chain of reasoning, based on my description of the idealized experiment in my previous post:

(1) The ontic state ##(\lambda, \lambda)## lies within the probability distribution ##\mu## for all four of the possible two-qubit product quantum states that can be prepared. Therefore, the ontic state ##(\lambda, \lambda)## can be prepared.

(2) Each of the four possible outcome quantum states from the measurement is orthogonal to one of the four two-qubit product quantum states.

(3) The probability distributions ##\mu## for orthogonal quantum states cannot overlap (i.e,. no ontic state can be contained in both).

(4) Therefore, the ontic state ##(\lambda, \lambda)## cannot be contained in the probability distribution for any of the four possible outcome quantum states from the measurement.

(5) Therefore, any of the four possible measurement results rules out the ontic state ##(\lambda, \lambda)##.

(6) However, the four possible measurement result quantum states form a complete basis of the two-qubit Hilbert space, so the measurement is a complete measurement and any ontic state should be in at least one of the four probability distributions.

(7) Therefore, the hypothesized model under which the ontic state ##(\lambda, \lambda)## can be prepared leads to a contradiction.

The issue I see is with #6 above; what it should actually say is

(6') However, the four possible measurement result quantum states form a complete basis of the two-qubit Hilbert space, so the measurement is a complete measurement and any ontic state that is a possible outcome of the measurement should be in at least one of the four probability distributions.

The italicized addition makes clear the additional assumption being made: that the ontic state after measurement is the same as the ontic state before measurement, or, to put it another way, that any ontic state that can be produced by the preparation process must also be an ontic state that can be produced as the outcome of the measurement process.

On its face, it at least seems worth considering that the measurement process might change the ontic state, or more generally might change the set of possible ontic states, since the measurement is an entangled measurement--it entangles the two qubits, which were not entangled before the measurement. In other words, a model in which the ontic states that were possible were different for entangled quantum states vs. non-entangled quantum states would seem to evade the conclusion of the PBR theorem.

 

[Demystifier]

But the quantum state of the system after measurement is not any of those states; it is one of the four entangled outcome states. How do we know that this change in the quantum state is not accompanied by a change in the ontic state?

Am I missing something?

I think you are. When you say that the quantum state changes, do you mean "changes" in the ontic sense, or "changes" in the epistemic sense?

If you mean change in the ontic sense, then you already assume that the quantum state is ontic, which indeed is what the theorem is supposed to prove. So in this way you don't disprove the conclusion of the theorem.

If you mean change in the purely epistemic sense, then such changes are not relevant for the study of ontic properties of the system. Hence you must ignore those changes because they are not relevant for the theorem. The theorem concerns only ontic changes. It says that if there are ontic changes at all, then changes of the quantum state are ontic too.

 

[ftr]

Does psi ontic mean that particle is everywhere at the same time. That is and not or

 

[Demystifier]

Does psi ontic mean that particle is everywhere at the same time.

No, because particle may not be the same as its wave function. See e.g. Bohmian mechanics.

 

[ftr]

I meant psi ontic as in PBR. It seems to imply it.

 

[Demystifier]

I meant psi optic as in PBR. I seems to imply it.

In this sense, PBR implies that the wave function is everywhere at the same time. But it's not the same as stating that the particle is everywhere at the same time.

 

[DarMM]

Does psi ontic mean that particle is everywhere at the same time. That is and not or

Very roughly, ##\psi##-ontic means that some fundamental object in the theory obeys the Schrodinger equation.

 

[Demystifier]

Very roughly, ##psi##-ontic means that some fundamental object in the theory obeys the Schrodinger equation.

Very roughly. The proof that psi is ontic does not imply that psi is fundamental. For instance, ##\psi({\bf x},t)## with ##{\bf x}## being the position of a neutron is certainly not fundamental, as the neutron consists of 3 quarks glued by undefined number of gluons.

 

[DarMM]

Very roughly. The proof that psi is ontic does not imply that psi is fundamental. For instance, ##\psi({\bf x},t)## with ##{\bf x}## being the position of a neutron is certainly not fundamental, as the neutron consists of 3 quarks glued by undefined number of gluons.

You're right of course. I should just stick with the accurate statement, that ##\psi##-ontic means:
$$\Lambda = \mathcal{A} \times \mathcal{H}$$
With ##\Lambda## the ontic state space and ##\mathcal{H}## the quantum Hilbert space.

 

[PeterDonis]

When you say that the quantum state changes, do you mean "changes" in the ontic sense, or "changes" in the epistemic sense?

I mean according to the math of QM. According to the math of QM, once a measurement is performed and you know the outcome, the quantum state is the state corresponding to that outcome.

In the idealized model I was describing, this change would be interpreted epistemically; but the interpretation is not needed to tell you what the quantum state is after the measurement. That's interpretation independent; it's part of basic QM.

The theorem concerns only ontic changes.

I find this statement confusing. The theorem explicitly describes a model in which the quantum state is viewed epistemically, and then purports to show that this model cannot in fact reproduce the predictions of quantum mechanics. A theorem that concerned only ontic changes could not do that.

 

[PeterDonis]

Does psi ontic mean that particle is everywhere at the same time. That is and not or

The paper I linked to describes what "psi-ontic" means mathematically for purposes of the PBR theorem. Please read it.

 

[ftr]

The paper I linked to describes what "psi-ontic" means mathematically for purposes of the PBR theorem. Please read it.

I skimmed it and from what they say I understood that psi is "real" because we have to take it as a whole in interaction( I myself had said that before). For me that implies the the electron for example "IS" the whole wavefunction. The detection at certain point is just a consequence of interacting with a "classical" system. I was asking for clarification of my understanding.

 

[PeterDonis]

I skimmed it

This is an "A" level thread. Skimming won't do. You need to read through the paper in detail and understand what it's saying. If you find that difficult, bear in mind that you need to have a graduate level understanding of the subject matter of an "A" level thread in order to make useful posts in it. I marked it "A" level for that reason.

I was asking for clarification of my understanding.

That won't really be possible because, based on your post, you appear to lack too much background knowledge to be able to receive a clarification at the "A" level that would fit in a PF discussion thread. If you want to ask about terms like "psi-ontic" and "psi-epistemic" more generally, you can start a new thread at the "I" level (or the "B" level, but at that level those concepts really can't be discussed very much).

 

[ftr]

Skimming won't do.

I did read quite a bit of it, skimming can have different depths. But I find your suggestion of reading each paper in depth to participate in the discussion sound impractical. So I think responses like Demystifier(and replies) can be useful to me and whether how much I can understand of them does not harm anybody in any way, people can simply ignore my post, which happens a lot.

 

[PeterDonis]

I find your suggestion of reading each paper in depth to participate in the discussion sound impractical.

As I said, I marked the thread "A" level for a reason. Someone who already has a graduate level background in the subject matter won't find it "impractical" to read and understand the paper. So if you do, then, as I said, you probably don't have the requisite background for this thread.

 

[ftr]

Ok, no problem. Your post helped me to find better sources to understand the issue like this one.

https://www.perimeterinstitute.ca/videos/why-i-am-not-psi-ontologist

 

[Demystifier]

I mean according to the math of QM. According to the math of QM, once a measurement is performed and you know the outcome, the quantum state is the state corresponding to that outcome.

Look at my own presentation of the PBR theorem (attached here). Can you explain your objection on the PBR theorem by referring to my presentation of it? Or does perhaps my presentation resolves the issue that bothered you?

 

[PeterDonis]

Can you explain your objection on the PBR theorem by referring to my presentation of it?

On slide 18, you say that the consequence of the assumed overlap is that there is a probability greater than zero that the outcome will be consistent with all four possibilities for the initial preparation. That is where you are putting in (implicitly) the additional assumption.

What you know, without additional assumptions, from the overlap is that proposition (1) in my post #2 in this thread is true. But that's not what your slide #18 is saying; it's saying something different, which does not follow from the assumed overlap unless you implicitly add a chain of reasoning something like what I gave in post #2 (and I pointed out where I see the additional assumption coming in there).

Another way of putting it would be: the statement on slide #18 assumes, not just that the ontic state ##\lambda## determines the outcome (which I would agree with), but that it does so in a particular way, such that the ontic state before measurement takes place must be contained in the probability distribution for the quantum state representing the outcome after the measurement. This assumes that the measurement does not change the ontic state.

 

[Imager]

Look at my own presentation of the PBR theorem

Very helpful!

 

[zonde]

I agree with @PeterDonis about additional assumption. But I would like to question assumption (3) as well (as I see it is along the same lines as PeterDonis reasoning).

(3) The probability distributions ##\mu## for orthogonal quantum states cannot overlap (i.e,. no ontic state can be contained in both).

Let's say we consider setup where two inputs interfere on BS or PBS. Let's consider two possible outcomes where all detections happen behind one of the two outputs. Two outcomes are orthogonal to each other and there is no overlap of ontic states in both outcomes.
But ontic states in inputs certainly overlap for these two orthgonal outputs and from perspective of QM description is the same whether we describe outputs or inputs, right? If so then one should be very careful even with statement (3) and make distinction whether we talk about inputs or outputs.

 

[Demystifier]

On slide 18, you say that the consequence of the assumed overlap is that there is a probability greater than zero that the outcome will be consistent with all four possibilities for the initial preparation. That is where you are putting in (implicitly) the additional assumption.

What you know, without additional assumptions, from the overlap is that proposition (1) in my post #2 in this thread is true. But that's not what your slide #18 is saying; it's saying something different, which does not follow from the assumed overlap unless you implicitly add a chain of reasoning something like what I gave in post #2 (and I pointed out where I see the additional assumption coming in there).

Another way of putting it would be: the statement on slide #18 assumes, not just that the ontic state ##\lambda## determines the outcome (which I would agree with), but that it does so in a particular way, such that the ontic state before measurement takes place must be contained in the probability distribution for the quantum state representing the outcome after the measurement. This assumes that the measurement does not change the ontic state.

For definiteness, suppose that the measurement outcome is ##|\phi_1\rangle##. The PBR theorem (page 18 in my presentation) takes for granted that this outcome eliminates the possibility that the system was prepared in ##|00\rangle##. In this case, I think that your objection can be interpreted in two different ways. Let me discuss both.

Interpretation 1:

One possible interpretation of your objection is this: No, the outcome ##|\phi_1\rangle## does not eliminate possibility that the system was prepared in ##|00\rangle##. Perhaps the system was really prepared in ##|00\rangle##, but later the state ##|00\rangle## changed to something else. This change could have been caused by the act of measurement.

My response to this objection is the following. Yes, such a change is a logical possibility, but such a change would not be consistent with the rules of QM. Loosely speaking, the PBR theorem states that if the rules of QM are true, then ##\psi## must be ontic.

Of course, I do not say that the act of measurement in QM does not change ##\psi##. It does, in a well known way described either by a unitary evolution (governed by the measurement-interaction Hamiltonian) or the projection postulate. But it seems to me that Interpretation 1 requires some additional change of ##\psi## that is not described by the standard rules of QM.

Interpretation 2:

Another possible interpretation of your objection is this: The act of measurement changes ##\lambda## in a way that does not introduce any additional change of ##\psi## (except the change described by QM).

My response to this objection is the following. Yes, such a change is a logical possibility, and it is perfectly consistent with the rules of QM. (An example of this possibility is Bohmian mechanics.) But such a change of ##\lambda## that does not introduce any additional change of ##\psi## is not relevant to the PBR theorem, as the PBR theorem is only a theorem about properties of ##\psi##.

EDIT: Did I reach DH7 here? http://commonsenseatheism.com/?p=13435

 

[PeterDonis]

In this case, I think that your objection can be interpreted in two different ways.

No, it can't; there is only one way. You are failing to see this by using language that obfuscates what is going on.

Here is a description in better language of what might happen during the given experiment, if the additional assumption I am referring to is dropped:

Observing the outcome corresponding to the quantum state ##|\phi_1\rangle## does not eliminate the possibility that the system was prepared in an ontic state ##\lambda_0## that lies in the probability distribution ##\mu## corresponding to the quantum state ##|00\rangle##, because the ontic state ##\lambda_0## after preparation but before measurement might not be the same as the ontic state ##\lambda_1## after measurement.

The PBR theorem does not consider this possibility; if it is possible, then a case like this evades the conclusion of the PBR theorem. The only way to rule it out, so the argument of the theorem goes through, is to assume that the ontic state ##\lambda## does not change during the measurement: that is what allows you to use the observation of the outcome ##|\phi_1\rangle## to rule out the possibility that the preparation process produced the quantum state ##|00\rangle##, since there cannot be any single ontic state that lies in both probability distributions.

Your two interpretations only appear to be different because you obfuscate the above by talking as though there is some way the quantum state might or might not "change", separately from the ontic state ##\lambda## changing (or not). That's wrong. The quantum state in this model is epistemic (remember this is a model that PBR are constructing in order to derive a contradiction from it). So the only "change" in the quantum state, by construction, is the one explicitly described: it is (in the particular case we are discussing now) ##|00\rangle## after preparation but before measurement, and ##|\phi_1 \rangle## after measurement. That describes the change in our knowledge about the system as a result of the measurement, and that's the only change there can be in the quantum state, because it's epistemic.

But such a change of ##\lambda## that does not introduce any additional change of ##\psi## is not relevant to the PBR theorem, as the PBR theorem is only a theorem about properties of ##\psi##.

This seems obviously false to me. The argument of the theorem explicitly makes use of properties of the ontic state ##\lambda##, specifically which probability distributions, corresponding to which quantum states, a single ontic state can be contained in. See above.

 

[PeterDonis]

EDIT: Did I reach DH7 here? http://commonsenseatheism.com/?p=13435

I don't think so, since you still don't seem to understand my central point.

 

[PeterDonis]

I would like to question assumption (3)

I personally don't see an issue with assumption (3); it seems natural that, in a psi-epistemic model, two quantum states that are orthogonal should not have any ontic states in common (i.e., contained in both of their probability distributions). But it's true that a psi-epistemic model in which this was not the case would also evade the conclusion of the PBR theorem. Whether it would have other problems I don't know.

from perspective of QM description is the same whether we describe outputs or inputs, right?

No. The quantum state after measurement, in the model PBR use in their theorem, is explicitly stated to be different from the quantum state before measurement. That means different probability distributions over the ontic states for the outputs than for the inputs.

 

[PeterDonis]

Observing the outcome corresponding to the quantum state ##|\phi_1\rangle## does not eliminate the possibility that the system was prepared in an ontic state ##\lambda_0## that lies in the probability distribution ##\mu## corresponding to the quantum state ##|00⟩##

Your two interpretations only appear to be different because you obfuscate the above by talking as though there is some way the quantum state might or might not "change", separately from the ontic state ##\lambda## changing (or not). That's wrong.

Some further thoughts on the above:

According to basic QM, if a particular quantum state is prepared, and nothing else is done to the system before that system is measured, then the measurement cannot result in an outcome described by a quantum state orthogonal to the quantum state that is prepared. By contraposition, if a measurement yields an outcome described by a particular quantum state, then a preparation process just before that measurement (with nothing else being done to the system) could not have prepared a quantum state that is orthogonal to the state describing the observed measurement outcome.

This, all by itself, i.e., without any talk about ontic states, is sufficient to ground the claim that, for example, if the outcome ##|\phi_1>## is observed, then the quantum state ##|00>## could not have been prepared, because the latter state is orthogonal to the former. So we have to be careful not to describe the experiment in any way that allows the possibility for the quantum state ##|00>## to have been prepared.

What I described in previous posts was an ontic state, which I'll call ##\lambda_0##, being prepared, which is contained in the probability distributions for all four of the quantum states ##|00>##, ##|01>##, ##|10>##, ##|11>##. That means this ontic state cannot be contained in the probability distribution for the outcome quantum state ##|\phi_1>## (or for any of the other possible outcome states). So the ontic state after measurement, which I'll call ##\lambda_1##, must be different from ##\lambda_0##, since it must lie within the probability distribution for the quantum state ##|\phi_1>## (since we observed that outcome). If the PBR theorem is to hold, this must somehow not be possible. How?

I don't see any stated assumption in any statement of the theorem that rules out the above. So it seems to me that there must be an additional assumption, unstated, which rules out what I described just above. The easiest way to state that assumption seems to me to be that the ontic state must not be allowed to change during the measurement.

However, there might be another way of stating it that is more precise. What I described above is a situation in which we find out, after the measurement, that the ontic state in which the system was prepared was contained in the probability distribution of a quantum state which was orthogonal to the quantum state corresponding to the observed outcome of the measurement. So another assumption that would rule this out would be something like: the ontic state of the system before measurement cannot be contained in the probability distribution of any quantum state which is orthogonal to the quantum state that describes the observed measurement outcome.

So basically, what we are looking at here is two possibilities for the relationship between the ontic state before measurement and the quantum state after measurement:

(A) If the PBR theorem holds, the ontic state before measurement cannot be contained in the probability distribution of any quantum state which is orthogonal to the quantum state after measurement.

(B) If the PBR theorem is to be violated, the ontic state before measurement will, in some cases, be contained in the probability distribution of a quantum state which is orthogonal to the quantum state after measurement.

Note, however, that in the case discussed above, the ontic state before measurement is still contained in the probability distribution of at least one quantum state which is not orthogonal to the quantum state after measurement. It seems to me that this weaker assumption, consistent with B (which is not sufficient to make the PBR theorem hold), should still be true.

Also note that the weaker assumption, consistent with B above, still requires the ontic state to change during measurement, because it is still true that the ontic state before measurement cannot be contained in the probability distribution of the quantum state after measurement. Assumption A, by contrast, allows models in which the ontic state does not change during measurement; but it also allows models in which the ontic state does change during measurement.

 

[PeterDonis]

Another comment on all this: I think there is language in some of the PBR papers talking about how the ontic state should "determine the outcome of the measurement", which is another way of getting at the intuition behind an assumption like assumption A in my previous post. If this is taken to mean that the ontic state after the measurement should be determined by the ontic state before the measurement, I see no problem with this. But in a psi-epistemic model, the ontic state before the measurement should not, in general, determine the quantum state after the measurement, because the whole point is that the quantum state describes what is known about the system, and the ontic state of the system will not be the only thing that determines that.

 

[zonde]

I don't understand why PBR conclusion is valid at all. They imply that ontic state from overlap region should give non zero probability for all of the four measurements (which is inconsistent with predictions of QM). But why this ontic state from overlap region can't give zero probability for all four measurements. Such possibility obviously would be consistent with predictions of QM.

 

[PeterDonis]

They imply that ontic state from overlap region should give non zero probability for all of the four measurements (which is inconsistent with predictions of QM).

Why?

why this ontic state from overlap region can't give zero probability for all four measurements.

If it gave zero probability for all four measurement results, and those four measurement results span the entire Hilbert space, what measurement result would be possible from this ontic state, if your statement here were correct?

 

[zonde]

Why?

Well, they claim there is contradiction because for any prepared state QM predicts zero outcome for at least one measurement out of four.

If it gave zero probability for all four measurement results, and those four measurement results span the entire Hilbert space, what measurement result would be possible from this ontic state, if your statement here were correct?

Ok, makes sense. But how one can check that these four measurements span the entire Hilbert space? This seems important but they give no justification for that statement so it is left as a homework for the reader.

 

[Demystifier]

The quantum state in this model is epistemic (remember this is a model that PBR are constructing in order to derive a contradiction from it). So the only "change" in the quantum state, by construction, is the one explicitly described: it is (in the particular case we are discussing now) ##|00\rangle## after preparation but before measurement, and ##|\phi_1 \rangle## after measurement. That describes the change in our knowledge about the system as a result of the measurement, and that's the only change there can be in the quantum state, because it's epistemic.

If that was the only possible change of ##\psi##, it would contradict QM. Surely, QM allows also a change due to unitary Hamiltonian evolution not related to measurement. It seems to me that you confuse your intuitive understanding of the concept of "epistemic" with the technical definition of "epistemic" used in the PBR theorem. The technical definition of "epistemic" in the PBR theorem differs from what most people intuitively understand by that concept. The PBR theorem rules out epistemic ##\psi## in this technical sense, but not necessarily in the usual intuitive sense.

 

[zonde]

If it gave zero probability for all four measurement results, and those four measurement results span the entire Hilbert space, what measurement result would be possible from this ontic state, if your statement here were correct?

I tried to investigate your question using some realistic experimental setup and some toy model. So I came up with such answer: ontic state from overlap region would produce non-zero outcomes for measurement state that has non-zero projections onto all preparation states.

 

[PeterDonis]

how one can check that these four measurements span the entire Hilbert space?

It's obvious. The Hilbert space has four degrees of freedom, so any four orthogonal vectors span it. And those four particular vectors are the Bell states, which appear in many, many papers on the subject.

This seems important but they give no justification for that statement so it is left as a homework for the reader.

That's because it's obvious and well known to anyone with knowledge of the field. Peer-reviewed papers don't bother deriving or justifying results of that sort.

 

[PeterDonis]

QM allows also a change due to unitary Hamiltonian evolution not related to measurement.

In general, yes, but that is specifically excluded between the preparation and the measurement for the particular case we're discussing, the one described in the PBR paper.

It seems to me that you confuse your intuitive understanding of the concept of "epistemic" with the technical definition of "epistemic" used in the PBR theorem.

I'm using the same definition the PBR paper uses: ##\psi## is epistemic if quantum states correspond to probability distributions over ontic states ##\lambda##, and at least some ontic states appear in the probability distributions for more than one quantum state. (Or, as PBR state it, the probability distributions for at least one pair of quantum states overlap.)

 

[PeterDonis]

ontic state from overlap region would produce non-zero outcomes for measurement state that has non-zero projections onto all preparation states.

I'm not sure how this addresses the issue under discussion. Can you elaborate?