I 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.

 

[zonde]

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.

Do you suggest that they where doing some kind of postselection to determine which input state they have prepared? I say you are simply inventing staff.
They clearly state in the text what they do. And they say they manipulated ions individually in the process of preparation (see section II. Experimental setup). So they have no doubt what state they have prepared (at least assuming state preparation independence).

 

[PeterDonis]

Do you suggest that they where doing some kind of postselection to determine which input state they have prepared?

No, they are doing just what the PBR theorem describes: using the measured outcome state to rule out the system having been prepared in the input state that is orthogonal to that outcome state, for that run of the experiment.

I say you are simply inventing staff.

No, I'm just reading the paper, but with more care than you apparently are. See below.

They clearly state in the text what they do.

And one of the things they clearly state is that they are following the protocol described in the original PBR paper. See below.

they say they manipulated ions individually in the process of preparation

Yes; but there are four different possible manipulations (one corresponding to each of the four possible input product states, as shown in Fig. 1 of the paper), and the manipulation for each individual ion is selected at random, without the experimenters knowing or recording which one is selected.

Note that the experimental protocol is not described in full detail in the paper you linked to; the paper references the original PBR paper (Ref. 3), and says it is implementing the protocol described there. The PBR paper is where the random selection of input states (which corresponds in this experiment to random selection of which of four possible manipulations to perform on the ions), without the experimenters knowing or recording which one is selected, is described.

 

[zonde]

No, they are doing just what the PBR theorem describes: using the measured outcome state to rule out the system having been prepared in the input state that is orthogonal to that outcome state, for that run of the experiment.

In order to measure probabilities you have to perform measurement on many identically prepared states and gather statistics. If they would mix up all the preparation states (as single preparation state) there would be no outcome with near zero outcome.

and the manipulation for each individual ion is selected at random, without the experimenters knowing or recording which one is selected.

Can you quote the paper?

 

[DrChinese]

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."

OK, this helps, and I thank you for your patience. I see that in each independently prepared system, there are overlapping probability distributions (as the 0 and + quantum states are not orthogonal).

Is it fair to say: This model, in which 2. is true (psi-epistemic), makes predictions of 1/4 (1/2 * 1/2) for various outcomes. Whereas the quantum predictions are 0 for some, 1/4 or 1/2 for others, and therefore cannot reproduce the predictions of QM.

Below is from the experimental paper: http://de.arxiv.org/abs/1211.0942

 

[PeterDonis]

This model, in which 2. is true (psi-epistemic), makes predictions of 1/4 (1/2 * 1/2) for various outcomes. Whereas the quantum predictions are 0 for some, 1/4 or 1/2 for others, and therefore cannot reproduce the predictions of QM.

No. First, the PBR argument does not make any specific claim about what the psi-epistemic model it describes predicts. It just argues that in that model, there is an ontic state (the one in which each of the two individually prepared qubits has an ontic state in the "overlap" region of the two probability distributions, for $|0\rangle$ and $|+\rangle$) which does not lie within the probability distributions of any of the four outcome quantum states. So if the system is prepared in that ontic state (which they say must happen at least sometimes), it would seem like there would be zero probability of any of the four possible outcomes of the measurement. But of course QM does not predict that.

Second, the graph you show from the experimental paper is for a different preparation procedure than the one PBR describe, as far as I can tell. So the "predictions" it is comparing to for probabilities are not the ones in the PBR paper. I'll address this further in a response to @zonde shortly.

 

[PeterDonis]

In order to measure probabilities you have to perform measurement on many identically prepared states and gather statistics. If they would mix up all the preparation states (as single preparation state) there would be no outcome with near zero outcome.

Yes, I see now that that graph in the paper you linked to (Fig. 2, which @DrChinese showed in his post) is inconsistent with the preparation procedure described in the PBR paper. So the paper you linked to appears to be saying two contradictory things: it says that it implemented the PBR protocol, but it also clearly implies, from Fig. 2, that the actual prepared states for each run were recorded (since, as you say, it would be impossible to produce Fig. 2 if they were not).

 

[PeterDonis]

the quantum predictions are 0 for some, 1/4 or 1/2 for others

The graph you showed, Fig. 2 from the experimental paper, is for probabilities of input states, not outcome states. As I noted in response to @zonde just now, this means that the experimenters could not have been doing the exact protocol that PBR specified in their paper, since they recorded the actual prepared input state for each measurement. The PBR protocol requires that the input state is chosen at random and not recorded. What the experimenters were actually verifying, as far as I can tell, is that, to within experimental accuracy, each of the possible outcome states is in fact orthogonal to one of the four input product states.

The QM prediction given the preparation procedure described by PBR is a probability of 1/4 for each of the four possible outcome states. Mathematically, the quantum state leading to this prediction is a proper mixture of the four product states, each with a coefficient of 1/4.

If you know the specific input product state that was prepared, then of course the QM prediction for outcome states will be different: in that case, I believe it is correct that the probability will be 0 for the outcome state that is orthogonal to the input product state, 1/4 for two of the other outcome states, and 1/2 for the remaining outcome state. But Fig. 2 does not show that, since, as noted above, it shows probabilities of input states, not outcome states.

 

[DrChinese]

The graph you showed, Fig. 2 from the experimental paper, is for probabilities of input states, not outcome states. ...

The QM prediction given the preparation procedure described by PBR is a probability of 1/4 for each of the four possible outcome states. Mathematically, the quantum state leading to this prediction is a proper mixture of the four product states, each with a coefficient of 1/4.

If you know the specific input product state that was prepared, then of course the QM prediction for outcome states will be different: in that case, I believe it is correct that the probability will be 0 for the outcome state that is orthogonal to the input product state, 1/4 for two of the other outcome states, and 1/2 for the remaining outcome state. But Fig. 2 does not show that, since, as noted above, it shows probabilities of input states, not outcome states.

I realize that by posting the graph from the experimental paper, I have switched somewhat from apples to oranges. Sorry about that (sorta). As I read the graph, the 4 input states are the 4 charts. Each chart then has the 4 possible outcome permutations after the projective measurement. I think that is how you are reading the graphs as well, not sure.

Anyway, maybe I should skip that paper until I get a better handle on the theoretical versions.

 

[PeterDonis]

As I read the graph, the 4 input states are the 4 charts.

Hm. You're right. Their notation is really confusing, at least to someone who is used to the notation of the original PBR paper. So each chart shows how often each outcome state occurs when the given input state is prepared.

I don't think this affects this statement of mine, though:

What the experimenters were actually verifying, as far as I can tell, is that, to within experimental accuracy, each of the possible outcome states is in fact orthogonal to one of the four input product states.

They still aren't running the exact experimental protocol described by PBR. They're just verifying one of the key properties of the QM predictions that underlie the PBR argument.

 

[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.

Going back to earlier comments around the theoretical PBR versions:

8. What are a pair of quantum states per above; and what is the related ontic state space?

I would say that if the (independently prepared) pair of input quantum states were either 0+ or +0, that there would be overlap in the possibility of an outcome entangled state of |$\phi$ 1> = $\frac{1}{\sqrt{2}}(|0\rangle|+\rangle + |+\rangle|0\rangle)$. Or am I representing it backwards? I admit I am confused.

 

[DrChinese]

Hm. You're right. Their notation is really confusing, at least to someone who is used to the notation of the original PBR paper. So each chart shows how often each outcome state occurs when the given input state is prepared.

Mea culpa for changing things...

 

[PeterDonis]

Mea culpa for changing things...

You didn't change the notation, you just used what was in the experimental paper. The authors of that paper are the ones I'd like to have a word with...

What are a pair of quantum states per above; and what is the related ontic state space?

The pair of quantum states that are assumed to have overlapping probability distributions are $|0\rangle$ and $|+\rangle$ in the single qubit Hilbert space. If we use $\lambda$ to denote some ontic state of the one-qubit system that is in the overlap region, then the ontic state $(\lambda, \lambda)$ for the two-qubit system will be contained in the probability distributions for all four of the product states in the two-qubit Hilbert space that are built from those two one-qubit states.

The ontic state space is not specified other than what is stated above.

I would say that if the (independently prepared) pair of input quantum states were either 0+ or +0, that there would be overlap in the possibility of an outcome entangled state of $\frac{1}{\sqrt{2}}(|0\rangle|+\rangle + |+\rangle|0\rangle)$.

No. Start with the ontic state $(\lambda, \lambda)$ described above. The question is, can this ontic state be contained in the probability distributions for any of the four possible outcome quantum states? The answer must be no, because:

(1) The ontic state $(\lambda, \lambda)$ is contained in the probability distributions for all four of the possible input product states (as noted above).

(2) Each of the four possible input product states is orthogonal to one of the four possible outcome quantum states.

(3) If two quantum states are orthogonal, no ontic state can be in the probability distributions for both.

Note that, so far, we have not said anything that requires additional assumptions beyond the ones stated by PBR. We are just working out required implications of the stated PBR assumptions for the model PBR describe.

PBR then argue that:

(A) Since the ontic state $(\lambda, \lambda)$ lies in the probability distributions for all four possible input product states, there is a nonzero probability for it to be prepared.

(B) If the ontic state $(\lambda, \lambda)$ is prepared, there should be a zero probability for any of the four possible measurement outcomes to occur (because that ontic state is not contained in the probability distributions for any of those four states). So the psi-epistemic model being considered should sometimes predict a zero probability for all four of the possible measurement outcomes.

(C) However, according to the predictions of QM, one of the four outcomes must always occur. Therefore, the psi-epistemic model being considered cannot reproduce the predictions of QM.

The additional assumption I've been referring to is in (B) above: the (unstated) assumption that, since the ontic state $(\lambda, \lambda)$ before measurement is not contained in any of the four probability distributions for the possible outcome quantum states, the psi-epistemic model, if the two-qubit system is prepared in that ontic state, must therefore predict a zero probability for all four of those outcome quantum states after measurement. The stated assumptions of the PBR theorem are not sufficient to prove this, so it has to be taken as an additional assumption.