I Understanding PBR's Additional Assumption

[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.