Understanding PBR's Additional Assumption

[DrChinese]

This thread is inspired by PeterDonis' thread about "hidden" assumptions in the PBR Theorem. He referenced Matt Leifer's in depth article on PBR (original referenced). And Demystifier has provided an excellent summary of PBR as a PDF file, which I have attached.

Leifer's "Is The Quantum State Real?" (comprehensive discussion of the subject)
https://arxiv.org/pdf/1409.1570.pdf

PBR Original (a bit more technical)
https://arxiv.org/abs/1111.3328

I am trying to follow PeterDonis' thread, but I realize I have several things I am not sure I fully understand. So I want to ask a few questions around my understanding of PBR.

Please reference Demystifier's PDF, page 17 (diagram); which is the same as Fig.2 from the PBR Original paper. Also Demystifier's PDF, page 18 (4 states ##|\phi_{1} \rangle##, ##|\phi_{2} \rangle##, ##|\phi_{3} \rangle##, or ##|\phi_{4} \rangle##); which are the same 4 states Leifer has on page 35 (his ##|\phi_{00} \rangle##, ##|\phi_{01} \rangle##, ##|\phi_{10} \rangle##, ##|\phi_{11} \rangle##).

1. The 4 states are all entangled states, correct? No specific mechanism is provided to get to this, we just assume that this can be accomplished. (I might compare it to the 4 entangled Bell states that can result in the BSA projection during entanglement swapping.)
2. For that to work, the 2 prepared systems must have their quantum particles arrive at the measurement apparatus at times which do not allow them to be distinguished, correct?
3. The measurement apparatus projects the input particles (each in either 0 or + pure state) into one of the 4 states, but does not allow the experimenter to select which specific state is to result. Correct?

 

[PeterDonis]

The 4 states are all entangled states, correct?

The four outcome states are all entangled, yes.

No specific mechanism is provided to get to this, we just assume that this can be accomplished. (I might compare it to the 4 entangled Bell states that can result in the BSA projection during entanglement swapping.)

As far as I can tell, the four outcome states are Bell states.

For that to work, the 2 prepared systems must have their quantum particles arrive at the measurement apparatus at times which do not allow them to be distinguished, correct?

I would agree with this.

The measurement apparatus projects the input particles (each in either 0 or + pure state) into one of the 4 states

It entangles the two particles (which were previously not entangled) and projects the two-particle system into one of the four outcome states. The final state is entangled, so neither particle has a definite state by itself, only the two-particle system does.

but does not allow the experimenter to select which specific state is to result.

I would agree with this.

Note that all of the above is talking about the quantum states only.

 

[DrChinese]

Peter,

Thanks. So with your help, I have a few follow-on questions. Here is one:

4. The 2 independently prepared systems, each in either the 0 or + state: I assume these must be in a superposition? Let's say one system is prepared by Alice, and the other by Bob. Alice and Bob cannot know whether a particular quantum particle is coming out as 0 (or +), correct? Alice simply prepares her system so that the output stream is either 0 or + (but never 1 or -), and she has no control over which emerges.

 

[PeterDonis]

The 2 independently prepared systems, each in either the 0 or + state: I assume these must be in a superposition?

No. The two-particle system is definitely in one of the four separable states ##|0>|0>##, ##|0>|+>##, ##|+>|0>##, or ##|+>|+>##. We just don't know which. In terms of quantum states, this would be a proper mixture, described mathematically as a density matrix adding together the four possible pure states with probability ##1/4## for each.

 

[DrChinese]

No. The two-particle system is definitely in one of the four separable states ##|0>|0>##, ##|0>|+>##, ##|+>|0>##, or ##|+>|+>##. We just don't know which. In terms of quantum states, this would be a proper mixture, described mathematically as a density matrix adding together the four possible pure states with probability ##1/4## for each.

Right, I see the Alice+Bob pure states must combine to be Product States (separable). But Alice's must individually be in a superposition of 0 and +, and Bob's must individually be in a superposition of 0 and +. That would yield equal probability for the 4 cases.

Or are you saying Alice and Bob could know which they produced? They're the ones doing the preparing. But I don't think that could be the case. They must not have any way to know whether they produced 0 or +.

[For purposes of a physical illustration: I imagine that Alice is producing a stream of photons which are either polarized at 0 degrees or at 45 degrees. So 0 degrees corresponds to the "0" state, and 45 degrees corresponds to the "+" state. A "+" photon would have a 50-50 chance of being 0 ("0") or 90 degrees ("1"), and a "0" photon would have a 50-50 chance of being 45 degrees ("+") or 135 degrees ("-").]

 

[PeterDonis]

are you saying Alice and Bob could know which they produced?

No. The experiment is explicitly set up so that Alice and Bob cannot control which particular pure state they produce. All they know is that each of them has an equal probability of producing ##|0>## or ##|+>##.

 

[DrChinese]

No. The experiment is explicitly set up so that Alice and Bob cannot control which particular pure state they produce. All they know is that each of them has an equal probability of producing ##|0>## or ##|+>##.

OK, that's how I understood it. But I wasn't so sure! Thanks. And another...

5. We have on the input side the 4 product states (from 2 independently prepared sources), and on the output side 4 entangled states. This is the quantum mechanical prediction, that entangled states will result from a suitable projective measurement.

Let's say the output state is "Not 00", which is one of the 4 possible output entangled states. You can't get that entangled state if the actual input states were only one of 0+, +0, or ++ (eliminating the 00 input case). The output would instead be some kind of Product state instead, and it would not pass the entanglement test (presumably violating some Bell Inequality or no-go).

And similarly for the other 3 entangled states. So that is the contradiction with the idea that there was a specific value prepared by Alice (or Bob), but that we just did not know it. Do I have that correct?

 

[PeterDonis]

We have on the input side the 4 product states (from 2 independently prepared sources), and on the output side 4 entangled states. This is the quantum mechanical prediction, that entangled states will result from a suitable projective measurement.

Yes.

Let's say the output state is "Not 00", which is one of the 4 possible output entangled states.

For clarity, this is the entangled state that happens to be orthogonal to ##|00>##, so it rules out ##|00>## as the input state.

You can't get that entangled state if the actual input states were only one of 0+, +0, or ++ (eliminating the 00 input case).

Why not? That's perfectly consistent with standard QM, as far as I can see. Standard QM does not preclude having an entangled output state with a separable input state.

 

[zonde]

No. The two-particle system is definitely in one of the four separable states ##|0>|0>##, ##|0>|+>##, ##|+>|0>##, or ##|+>|+>##. We just don't know which. In terms of quantum states, this would be a proper mixture, described mathematically as a density matrix adding together the four possible pure states with probability ##1/4## for each.

(I'm commenting on sentence in bold.) If we perform the experiment we have to know in which input state the system is prepared because we have to check that this particular input state gives 0 for particular output state. That's the prediction on which the reasoning is built. So we definitely have to verify this prediction. And there is no reasoning in PBR that depends on mixing up input states.

 

[DrChinese]

Why not? That's perfectly consistent with standard QM, as far as I can see. Standard QM does not preclude having an entangled output state with a separable input state.

I'm not questioning that. I am saying that you cannot get the entangled "Not 00" state from the trio of input states: 0+, +0, or ++ (i.e. eliminating the 00 input case).

If you could, we would not have the contradiction.The contradiction being that the output entangled state of "Not 00" requires that the input state of "00" was possible.

 

[DrChinese]

(I'm commenting on sentence in bold.) If we perform the experiment we have to know in which input state the system is prepared because we have to check that this particular input state gives 0 for particular output state. That's the prediction on which the reasoning is built. So we definitely have to verify this prediction. And there is no reasoning in PBR that depends on mixing up input states.

I don't think so. The setup must preclude knowledge of which of the 4 input states were presented.

 

[zonde]

The setup must preclude knowledge of which of the 4 input states were presented.

Why?

 

[PeterDonis]

I am saying that you cannot get the entangled "Not 00" state from the trio of input states: 0+, +0, or ++ (i.e. eliminating the 00 input case).

Why not? I'm still not understanding your objection here. You seem to be saying that we can't get an entangled output state from a separable input state. But if that were actually true, the entire experimental setup described in the PBR paper would be impossible, and since the proof of the theorem assumes that such a setup is possible, the theorem would be easily refuted. Is that what you're saying?

 

[PeterDonis]

If we perform the experiment we have to know in which input state the system is prepared

No, we don't. Go read the PBR paper; you evidently don't understand the experimental setup it's describing.

 

[PeterDonis]

You seem to be saying that we can't get an entangled output state from a separable input state.

I'm not questioning that.

I missed this statement of yours before; I now see that you're not questioning that we can get an entangled output state from a separable input state. But I'm still not sure I understand what objection you are making. Let me restate what the PBR paper says about the quantum states:

After preparation but before measurement, the two-particle system is in one of the four product states ##|0>|0>##, ##|0>|+>##, ##|+>|0>##, ##|+>|+>##. We just don't know which (because the preparation procedure does not make this information available).

After measurement, the two-particle system is in one of the four entangled states, each of which is orthogonal to one of the four product states above. Therefore, according to PBR, knowing the measurement result allows us to deduce that, after preparation but before measurement, the two-particle system could not have been in the product state that is orthogonal to the outcome state. In the particular example I've been using, the measurement outcome state is orthogonal to the product state ##|0>|0>##, and therefore, according to PBR, we can deduce that, after preparation but before measurement, the two-particle system was not in quantum state ##|0>|0>##.

My argument about the additional assumption in the PBR theorem does not question any of the above, since the above only talks about quantum states and does not make any claims about whether or not the quantum state ##\psi## is ontic or not. It's all just standard QM, interpretation-independent. So if you are saying that something in the above is not correct, you would need to explain why it's not correct using just standard QM, independent of any interpretation. (And if you could do so, that would, as I said, invalidate the PBR theorem by showing that the experimental setup it assumes is not possible.)

 

[DrChinese]

Why not? I'm still not understanding your objection here. You seem to be saying that we can't get an entangled output state from a separable input state.

I don't have an objection, I agree that you can get entangled states from separable states. I am just saying that there must be "something" that is unknown or in a superposition in order to get to something entangled. For example: if we knew that the input state was 00, we can't get the 4 state PBR entanglement from that. If the inputs were 00 or ++, we can't get the 4 state PBR entanglement from that. If we had 3 of the 4 possible input states, say 00 or ++ or 0+: we might get some PBR entanglement, but it would not be very high quality.

You need all 4 input states to occur about equally, and with no possibility of knowing which of those 4 occurred. That is, if you want to get the PBR entanglement. Which I think is what you said... I hope I understood that.

EDIT: I see now we both crossed each other's replies...

 

[PeterDonis]

You need all 4 input states to occur about equally, and with no possibility of knowing which of those 4 occurred. That is, if you want to get the PBR entanglement. Which I think is what you said...

I don't know if this is what I said, because I don't know what you mean by "all 4 input states to occur about equally". Does PBR's description of the experimental setup, as I described it in post #15, meet this definition?

 

[DrChinese]

I don't know if this is what I said, because I don't know what you mean by "all 4 input states to occur about equally". Does PBR's description of the experimental setup, as I described it in post #15, meet this definition?

Sure. They (and you) don't specifically mention that the 4 input cases should occur about equally, but I think it is safe to assume. Otherwise you'd have some knowledge about the input states being presented. To the extent you have a degree of that knowledge, you would get less PBR entanglement.

I'm not suggesting anything that remotely invalidates PBR, nor anything to directly address your comments in the other thread ("additional assumption"). Mostly I am trying to understand the PBR theorem and formulate it in my head.

 

[PeterDonis]

Sure.

Ok.

They don't specifically mention that the 4 input cases should occur about equally

Yes, they do, because they specify that Alice and Bob each have a 50-50 chance of preparing ##|0>## or ##|+>##. That means the proper mixed state that describes the outcome of both preparations must be an equal mixture of the four product states, which means the four possible input cases do occur equally.

 

[PeterDonis]

if we knew that the input state was 00, we can't get the 4 state PBR entanglement from that. If the inputs were 00 or ++, we can't get the 4 state PBR entanglement from that. If we had 3 of the 4 possible input states, say 00 or ++ or 0+: we might get some PBR entanglement, but it would not be very high quality.

I'm not sure what you're trying to say here. The outcome of the PBR measurement is not "4 state PBR entanglement". It's one of the four states (each of which is entangled, as an individual state of the two-particle system). If we knew the input state was ##|0>|0>##, that would rule out one of the four outcome states--we would know it couldn't be the one that's orthogonal to ##|0>|0>##. But it could still be one of the other three, and that would still count as a PBR measurement.

 

[DrChinese]

Yes, they do, because they specify that Alice and Bob each have a 50-50 chance of preparing ##|0>## or ##|+>##. That means the proper mixed state that describes the outcome of both preparations must be an equal mixture of the four product states, which means the four possible input cases do occur equally.

OK great, that makes perfect sense. So if one makes the main PBR psi-epistemic assumption (that one of the 4 input cases occurred, but we don't know which), then a contradiction occurs. That contradiction does not occur on the psi-ontic side, because the assumption is not present.

 

[DrChinese]

The outcome of the PBR measurement is not "4 state PBR entanglement".

It's one of the four states (each of which is entangled, as an individual state of the two-particle system).

I am saying the exact same thing, perhaps my wording is not as clear as yours.

The result of the projective measurement on the prepared system (having 1 of 4 possible input Product states) is to be 1 of 4 possible entangled states (which is not controllable by the experimenter). That is the PBR entanglement I am referring to. (I am simply contrasting that to traditional Bell test entanglement, which is always 1 entangled state.)

 

[PeterDonis]

So if one makes the main PBR psi-epistemic assumption (that one of the 4 input cases occurred, but we don't know which), then a contradiction occurs.

No, this is not the psi-epistemic assumption. The fact that one of the 4 input cases occurred, but we don't know which, is part of the specification of the scenario in the theorem; it is not supposed to require either psi-epistemic or psi-ontic to be true, but to be consistent with standard QM, independent of any interpretation.

The definition of psi-epistemic in the PBR theorem is that there must be at least one ontic state that is contained in the probability distributions of more than one quantum state, i.e,, that the probability distributions of at least one pair of quantum states over the ontic state space must overlap. But the proof of the theorem does not require that this must be true for the experimental setup described by PBR to be possible, in which one of the 4 input cases occurred, but we don't know which. The statement that one of the 4 input cases occurred, but we don't know which, is a statement purely about quantum states; such statements are interpretation independent.

 

[lodbrok]

If I may ask, so why do you need quantum states at all. It appears you could apply the PBR logic to an experiment involving two coins tossed once, with 4 possible outcomes HH, HT, TH, TT, having probability 1/4 when you don't know which one occurred. But the actual occurrence of any of them rules out the other possibilities. What am I missing?

 

[PeterDonis]

why do you need quantum states at all

Because the point of the PBR theorem is to show that the quantum state must be ontic, not epistemic, at least given a particular set of assumptions. Or, to put it another way, the point is to show that, given a particular set of assumptions, only a model in which the quantum state is ontic (under the PBR definition of "ontic") can reproduce the experimental predictions of quantum mechanics. So you have to look at what QM predicts, and QM uses quantum states to make its predictions.

 

[romsofia]

Let's apply this to a simpler system: Say we prepare an atom in some state, and throw it through an analyzer on the z-axis and it gives me spin up. I don't know what my input state was, but I know it wasn't spin-down on the z-axis. It could've been spin up on any axis, spin down on an axis but z, a mixture, etc.

I don't understand where the contradiction occurs, is the contradiction that statistically there is the possibility that my input state was spin down?

Essentially, is the conclusion that because we know that our input state couldn't be spin down, that state vectors have the physical things encoded into them, so they must be physical themselves?

 

[PeterDonis]

is the conclusion that because we know that our input state couldn't be spin down, that state vectors have the physical things encoded into them, so they must be physical themselves?

No. The particular statement that, once we know the measurement outcome, one of the four possible input states (in the scenario described in the PBR theorem proof) is ruled out, is not the conclusion of the PBR theorem. It's just a particular step in the reasoning. Please read the papers linked to by @DrChinese in the OP of this thread; the one by Matt Leifer, in particular, discusses in detail the PBR argument and all of the assumptions and concepts that go into it.

 

[zonde]

If we perform the experiment we have to know in which input state the system is prepared

No, we don't. Go read the PBR paper; you evidently don't understand the experimental setup it's describing.

I believe I understand the setup well enough. Here is reference: https://arxiv.org/abs/1211.0942
In fig.2 there are measured probabilities for each input state separately. And detailed description of experimental state preparation procedure includes performing well controlled manipulations of separate ion states.

 

[DrChinese]

The definition of psi-epistemic in the PBR theorem is that there must be at least one ontic state that is contained in the probability distributions of more than one quantum state, i.e,, that the probability distributions of at least one pair of quantum states over the ontic state space must overlap. But the proof of the theorem does not require that this must be true for the experimental setup, in which one of the 4 input cases occurred, but we don't know which, to be possible. The statement that one of the 4 input cases occurred, but we don't know which, is a statement purely about quantum states; such statements are interpretation independent.

I must still be confused about the definition of psi-epistemic, ontic states and a few other things. You say:
The definition of psi-epistemic in the PBR theorem is that there must be at least one ontic state that is contained in the probability distributions of more than one quantum state, i.e,, that the probability distributions of at least one pair of quantum states over the ontic state space must overlap.

Demystifier says:
A probability distribution μ(##\lambda##) is ontic (i.e., corresponds to something real) iff it can be determined uniquely from the fundamental ##\lambda##. Otherwise μ(##\lambda##) is called epistemic.

He later says something like your statement:
To prove that psi is NOT ontic, it is sufficient to prove that there is at least one pair of [quantum states] for which [their probability distributions] do overlap.

6. So what are a pair of quantum states, in the PBR scenario, which do NOT have overlapping probability distributions (but can have a common ontic state)? It seems as if any pair of the 4 entangled states can share at least 2 of the 4 possible input states. Therefore there is some overlap on those. What doesn't overlap?

 

[PeterDonis]

In fig.2 there are measured probabilities for each input state separately.

These are measured probabilities for each input state separately after the measurement is made and the outcome is known--i.e., using the result of the measurement to rule out the input state orthogonal to the measured outcome state.

They are not measured probabilities for each input state separately after preparation but before measurement, which is what you would need in order to know exactly which input state was prepared on each run, as you were claiming.

 

[PeterDonis]

So what are a pair of quantum states, in the PBR scenario, which do NOT have overlapping probability distributions (but can have a common ontic state)?

There are none. If the probability distributions of two quantum states do NOT overlap (which will be true for any pair of quantum states which are orthogonal), then they cannot have any ontic state in common. That is what "do not overlap" means.

The definition of psi-epistemic is that there is at least one pair of quantum states whose probability distributions DO overlap, i.e., have an ontic state in common. Such quantum states cannot be orthogonal to each other, but the single-qubit quantum states ##|0\rangle## and ##|+\rangle## meet this requirement: they are not orthogonal, and so in a psi-epistemic model they can have an ontic state in common.

The first quote you gave from @Demystifier is saying the same thing, just looked at from the opposite direction, so to speak. If no two quantum states have probability distributions that overlap, which means we have a psi-ontic model, then any ontic state can only be contained in the probability distribution of one quantum state, which means that the ontic state uniquely determines the quantum state. In a psi-epistemic model, the ontic state does not uniquely determine the quantum state, since there are at least some ontic states which are contained in the probability distributions of more than one quantum state.

 

[DrChinese]

1. There are none. If the probability distributions of two quantum states do NOT overlap (which will be true for any pair of quantum states which are orthogonal), then they cannot have any ontic state in common. That is what "do not overlap" means.

2. The definition of psi-epistemic is that there is at least one pair of quantum states whose probability distributions DO overlap, i.e., have an ontic state in common. Such quantum states cannot be orthogonal to each other, but the single-qubit quantum states ##|0\rangle## and ##|+\rangle## meet this requirement: they are not orthogonal, and so in a psi-epistemic model they can have an ontic state in common.

You say above in 1. that there are none. Yet, the PBR theorem is that 2. is not true for their case, so there must be "at least one pair of quantum states whose probability distributions do [NOT] overlap [for some single ontic state]". I am trying to label each element of the argument properly. I can see that the 4 entangled states each preclude 1 input possibility (anti-distinguishable). But I am failing to see what overlap is NOT occurring that would occur in your 2.

 

[PeterDonis]

You say above in 1. that there are none.

There are none for orthogonal quantum states.

the PBR theorem is that 2. is not true for their case

The PBR theorem is that, given their assumptions (which I am saying must include one additional assumption that they do not address), a psi-epistemic model cannot reproduce the predictions of QM. It is not that 2. is not true for the model they consider; the model they consider is psi-epistemic, and 2. is true for it. The theorem purports to prove that this model cannot reproduce the predictions of QM; but that does not mean 2. is not true for that model.

 

[PeterDonis]

The PBR theorem is that, given their assumptions (which I am saying must include one additional assumption that they do not address), a psi-epistemic model cannot reproduce the predictions of QM. It is not that 2. is not true for the model they consider; the model they consider is psi-epistemic, and 2. is true for it. The theorem purports to prove that this model cannot reproduce the predictions of QM; but that does not mean 2. is not true for that model.

Maybe it will help to restate this more schematically thus: the form of the PBR theorem argument is not

"This model cannot reproduce the predictions of QM if 2. is true, therefore 2. is not true for this model."

The form of the PBR argument is

"This model, in which 2. is true, cannot reproduce the predictions of QM."

 

[PeterDonis]

there must be "at least one pair of quantum states whose probability distributions do [NOT] overlap [for some single ontic state]"

This doesn't make sense. If two probability distributions do not overlap, they cannot have any ontic state in common; that is what "do not overlap" means. So "do not overlap for some single ontic state" is a contradiction in terms.