Is there an additional assumption in the PBR theorem?

[Demystifier]

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.

I must admit, at the moment it seems to me that you are right. I don't see how to avoid this loophole in the PBR theorem. Would you mind if I present your argument to Matt Leifer (he has written a review on the PBR theorem and my presentation of PBR was based on his presentation of it), to see what he would say about this?

 

[zonde]

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.

Yes, I did the check and it's rather trivial for someone who knows these calculations.

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

In the argument there are four orthogonal measurement states. We might say that there is no ontic state that can contribute more than once to outcomes of orthogonal quantum measurement states. Sum of probabilities for four outcomes of given measurement states is 1. From these two things it follows that each ontic state contributes to measurement outcome at least once and no more than once. This holds for all four prepared states and because there is no overlap in outcomes for all four prepared states it would mean that there is no overlap in ontic states of all four prepared states.

Now what I say is that given particular measurement states and prepared states there are ontic states that contribute more than once to these measurements and there are ontic states do not contribute to any of these four measurements. These later ontic states are the overlap region of four prepared states.
Basically the statement that "there is no ontic state that can contribute more than once to outcomes of orthogonal quantum measurement states" can be false, i.e. orthogonality for ontic state space isn't necessarily the same as for quantum state space.

 

[PeterDonis]

Would you mind if I present your argument to Matt Leifer (he has written a review on the PBR theorem and my presentation of PBR was based on his presentation of it), to see what he would say about this?

Not at all, go ahead!

 

[Demystifier]

Not at all, go ahead!

I've sent him an e-mail, I'll let you know when I get a reply.

 

[PeterDonis]

because there is no overlap in outcomes for all four prepared states it would mean that there is no overlap in ontic states of all four prepared states.

Yes, this follows directly from the assumption that there is no overlap in the probability distributions for orthogonal quantum states. Are you denying that assumption? If so, then your proposed model (which I don't fully understand) is different from mine, since my model accepts that assumption, so if your model could evade the conclusion of the PBR theorem, it would be for a different reason from mine.

 

[zonde]

Yes, this follows directly from the assumption that there is no overlap in the probability distributions for orthogonal quantum states. Are you denying that assumption? If so, then your proposed model (which I don't fully understand) is different from mine, since my model accepts that assumption, so if your model could evade the conclusion of the PBR theorem, it would be for a different reason from mine.

Just to make it clear. In review article that you gave in post #1 there is given Spekkens toy model (2.1. Spekkens toy bit). This is the model I am using to analyze PBR theorem.

About your objection to PBR theorem. PBR considers four measurement states. Let's assume that these four measurements can be performed in one go (input state is split into four outcomes). So we have quantum state at the input and at the output we have four probabilities that sum up to 1. For any given input state state one measurement gives 0 with certainty. So all ontic states from each prepared state are certain to give 0 for one of the measurements. So the ontic state from overlap region should be sure to give 0 for all four measurement outcomes and if follows that there are 0 ontic states from overlap region.
I don't see how you could escape that conclusion if you assume that these four measurements can be performed in one go.

 

[PeterDonis]

PBR considers four measurement states. Let's assume that these four measurements can be performed in one go (input state is split into four outcomes).

PBR considers four possible outcomes from one measurement. Not four measurements. There is only one measurement, so of course it is made in "one go".

For any given input state state one measurement gives 0 with certainty. So all ontic states from each prepared state are certain to give 0 for one of the measurements.

For any given input quantum state, the predicted probability of one of the four measurement outcomes is zero. But you can't deduce from this that all ontic states are certain to give 0 for one of the measurements, unless you add the additional assumption I've been talking about.

 

[zonde]

PBR considers four possible outcomes from one measurement. Not four measurements. There is only one measurement, so of course it is made in "one go".

Measurement states are expressed in different bases. Second subsystem is first measurement is expressed in $|0\rangle$, $|1\rangle$ basis but in second measurement it is expressed in $|+\rangle$, $|-\rangle$ basis. It does not make sense from experimental perspective. All four measurements would have to be rewritten in one single basis for them to make sense as four outcomes of single measurement.
But changing the basis in the middle of derivation can be a bit tricky. If we look at entangled state, expressing it in different basis for each of the four measurements needed for Bell inequality measurement we can arrive at conclusion that there shouldn't be Bell inequality violations. And yet there are.

For any given input quantum state, the predicted probability of one of the four measurement outcomes is zero. But you can't deduce from this that all ontic states are certain to give 0 for one of the measurements, unless you add the additional assumption I've been talking about.

Why not? If there are ontic states that do not give 0 for corresponding measurement then ontic model would disagree about predictions with QM. I suppose we consider only those ontic models that give the same predictions as QM.

 

[PeterDonis]

Measurement states are expressed in different bases.

No, they're not. The states $|0>$, $|1>$, $|+>$, $|->$ are not states in the Hilbert space of the two-particle system. They're states in the one-particle Hilbert space. You can write states in the two-particle Hilbert space in terms of any one-particle states you like.

All four measurements would have to be rewritten in one single basis for them to make sense as four outcomes of single measurement.

No, they don't. You can compute all experimental predictions without having to do this.

What does have to be the case is that the four outcome quantum states are orthogonal--and it's easy to confirm that they are by direct computation. Also, for the measurement to be complete, the outcome states must span the Hilbert space--and it's easy to confirm that they do. In other words, the outcome states themselves must form a basis of the Hilbert space, for a complete measurement. And they do.

If there are ontic states that do not give 0 for corresponding measurement then ontic model would disagree about predictions with QM.

Only with the additional assumption I've been talking about.

 

[zonde]

Only with the additional assumption I've been talking about.

Yes, with your additional assumption PBR conclusion can be reached and without it there is a gap in reasoning.
But your assumption can be replaced with weaker assumption. Say we can assume that in ontic model there is a (two particle) system that is the holder of the ontic state. So when measurement changes the ontic state we can pair up all input ontic states with all output ontic states (any input ontic state can be traced to some output ontic state and any output ontic state can be backtraced to some input ontic state without any overlaps). And with that weaker assumption PBR conclusion still goes through.

 

[zonde]

No, they're not. The states $|0>$, $|1>$, $|+>$, $|->$ are not states in the Hilbert space of the two-particle system. They're states in the one-particle Hilbert space. You can write states in the two-particle Hilbert space in terms of any one-particle states you like.

Yes, I can perform measurement that gives output for state $\frac{1}{\sqrt{2}}(|0\rangle|-\rangle + |1\rangle|+\rangle)$. That's not the question.
The question is about performing measurement that gives outputs for states $\frac{1}{\sqrt{2}}(|0\rangle|1\rangle + |1\rangle|0\rangle)$ and $\frac{1}{\sqrt{2}}(|0\rangle|-\rangle + |1\rangle|+\rangle)$
Contrast this with such simpler question:
Can you perform single particle measurement that gives you outputs for $|0\rangle$ and $|+\rangle$ states?

No, they don't. You can compute all experimental predictions without having to do this.

What does have to be the case is that the four outcome quantum states are orthogonal--and it's easy to confirm that they are by direct computation. Also, for the measurement to be complete, the outcome states must span the Hilbert space--and it's easy to confirm that they do. In other words, the outcome states themselves must form a basis of the Hilbert space, for a complete measurement. And they do.

To consider four measurements as four outputs from single measurement they have to be four eigenstates of the same operator.
Yes, you can rewrite four measurement states given in PBR in such a way that they are four eigenstates of the same operator. But in the process you have to relay on a symmetry that is part of QM. But ontic model does not have to respect that symmetry at fundamental level, it only has to simulate that symmetry at statistical level.

 

[DarMM]

Just looking at this now.

Let's say the state is $|00\rangle$ which of course gives no chance to be found in the state $|\phi_{00}\rangle$. However the problem is that $\left(\lambda,\lambda\right)$ (an ontic state shared among the four preparation states) is of course in the support of the probability distribution over ontic states given by $|\phi_{00}\rangle$ and so does seem to predict that state could be detected, in contradiction with QM.

I don't currently see how changing the ontic state would avoid the problem. How would the shift in ontic state know to "avoid" the support of $|\phi_{00}\rangle$? To do that it would have to know the epistemic state $|00\rangle$, but that's purely epistemic/your knowledge.

 

[zonde]

$\left(\lambda,\lambda\right)$ (an ontic state shared among the four preparation states) is of course in the support of the probability distribution over ontic states given by $|\phi_{00}\rangle$

You have to consider all four measurement states, even with PeterDonis assumption granted. Otherwise I can say that maybe $\left(\lambda,\lambda\right)$ is for example in the support of the probability distribution over ontic states given by $|\phi_{0+}\rangle$.

 

[DarMM]

You have to consider all four measurement states, even with PeterDonis assumption granted. Otherwise I can say that maybe $\left(\lambda,\lambda\right)$ is for example in the support of the probability distribution over ontic states given by $|\phi_{0+}\rangle$.

There will be a $\left(\lambda,\lambda\right)$ in the support of $|\phi_{00}\rangle$ though, by the overlap argument, can't we just consider this $\left(\lambda,\lambda\right)$? Sure there are some that might be in the relative complement of $|\phi_{0+}\rangle$'s support with respect to $|\phi_{00}\rangle$, but why does that matter? Unless I'm missing something.

 

[zonde]

There will be a $\left(\lambda,\lambda\right)$ in the support of $|\phi_{00}\rangle$ though, by the overlap argument,

Overlap assumption just says that there is overlap, it does not say where it is. If the support of $|0\rangle\otimes|0\rangle$ is not exactly the same as support of measurement state (i.e. outcome probability is less than 1) then overlap could be in the part that is excluded by the measurement state.

 

[DarMM]

Overlap assumption just says that there is overlap, it does not say where it is. If the support of $|0\rangle\otimes|0\rangle$ is not exactly the same as support of measurement state (i.e. outcome probability is less than 1) then overlap could be in the part that is excluded by the measurement state.

Of course, but there is a set $A$ that is the intersection of the supports of all preparation states. This intersection has to overlap with at least one of the measurement states, or else an ontic state drawn from that sample could not lead to any of the outcomes. So if it overlaps with at least one, just consider that outcome state (regardless of which one it is).

Then there will be one of the preparation states incompatible with that measurement state. The only way to avoid this would be if $\left(\lambda,\lambda\right)$ did "jump" upon measurement out of this measurement state's support. However this would only be possible if it knew about the epistemic state it was incompatible with.

 

[zonde]

Of course, but there is a set $A$ that is the intersection of the supports of all preparation states. This intersection has to overlap with at least one of the measurement states, or else an ontic state drawn from that sample could not lead to any of the outcomes. So if it overlaps with at least one, just consider that outcome state (regardless of which one it is).

Then there will be one of the preparation states incompatible with that measurement state. The only way to avoid this would be if $\left(\lambda,\lambda\right)$ did "jump" upon measurement out of this measurement state's support. However this would only be possible if it knew about the epistemic state it was incompatible with.

Yes, this is PBR argument as I understand it.

Well, have to say I have my own reservations about this argument (different from PeterDonis) which I tried to explain in posts #37 and #41.

 

[DarMM]

Your contention was that there might be ontic states that contribute to, let's say, two outcome quantum states even though they are orthogonal?

 

[zonde]

Your contention was that there might be ontic states that contribute to, let's say, two outcome quantum states even though they are orthogonal?

Yes. I used Spekkens toy model from Leifer's review article (see link in OP) and this was what I got.
Well I can not claim that outcome quantum states are orthogonal. If there would be such four input entangled states (the same as four PBR measurement states) sure they could be transformed into four orthogonal quantum states. But input states are product states and they could remain product states even after these measurements. Maybe that's the difference.

 

[DarMM]

Yes. I used Spekkens toy model from Leifer's review article (see link in OP) and this was what I got.

My understanding would be that this is not possible.

Imagine you had an ontic state $\lambda$ that corresponded to two orthogonal quantum states, e.g. $\psi_1$ and $\psi_2$. Then you have a device that prepares either $\psi_1$ or $\psi_2$ depending on which button you push. In this case either button could produce $\lambda$.

Now imagine a far distant device that measures the propositions of whether a state is in $\psi_1$ or $\psi_2$.

$\lambda$ allows either of these to be observed, since it lies in both and yet only one will occur given the initial button push according to quantum theory. So $\lambda$ does not fully determine the physics, the measuring device "knows" the quantum state as well.

One minor thing to point out is that the PBR theorem is quite strong in the sense that it doesn't even assume the fundamental dynamics are deterministic, i.e. the $\lambda$s can be states of a stochastic model.

So the quantum state cannot be understood even as partial knowledge of something that is itself random. It needs to be Exotic or Non-Realist.

 

[zonde]

My understanding would be that this is not possible.

Imagine you had an ontic state $\lambda$ that corresponded to two orthogonal quantum states, e.g. $\psi_1$ and $\psi_2$. Then you have a device that prepares either $\psi_1$ or $\psi_2$ depending on which button you push. In this case either button could produce $\lambda$.

Now imagine a far distant device that measures the propositions of whether a state is in $\psi_1$ or $\psi_2$.

$\lambda$ allows either of these to be observed, since it lies in both and yet only one will occur given the initial button push according to quantum theory. So $\lambda$ does not fully determine the physics, the measuring device "knows" the quantum state as well.

In your example you prepare orthogonal states. In PBR prepared states are not orthogonal.

Consider this. You prepare a state $|0\rangle\otimes|0\rangle$. You perform a measurement that you describe with a state $\frac{1}{\sqrt{2}}(|0\rangle\otimes|0\rangle+|1\rangle\otimes|1\rangle)$. Half of the prepared state will pass that measurement. Now, have you got entangled state as an output? Do you have in the output a state $\frac{1}{\sqrt{2}}(|0\rangle\otimes|0\rangle+|1\rangle\otimes|1\rangle)$?
I don't see how this could happen, because the mode $|1\rangle\otimes|1\rangle$ isn't there and it couldn't appear as a result of measurement.

 

[DarMM]

In your example you prepare orthogonal states. In PBR prepared states are not orthogonal.

Of course, because the PBR theorem is necessary to prove the case for non-orthogonal preparation states. Orthogonal preparation states have long been known to have disjoint supports due to the simple argument I gave.

Now, have you got entangled state as an output?
I don't see how this could happen

Why wouldn't it happen? QM predicts it and even in Spekken's model it occurs. If you want a $\psi$-epistemic model to replicate QM you'd need this to happen.

 

[PeterDonis]

I can perform measurement that gives output for state $\frac{1}{\sqrt{2}}(|0\rangle|-\rangle + |1\rangle|+\rangle)$. That's not the question.
The question is about performing measurement that gives outputs for states $\frac{1}{\sqrt{2}}(|0\rangle|1\rangle + |1\rangle|0\rangle)$ and $\frac{1}{\sqrt{2}}(|0\rangle|-\rangle + |1\rangle|+\rangle)$

Why wouldn't you be able to do this? The four output states described in the PBR paper are all orthogonal to each other and span the Hilbert space; therefore they must be eigenstates of some Hermitian operator, so there must be some measurement that has these states as its possible outcome states.

Can you perform single particle measurement that gives you outputs for $|0\rangle$ and $|+\rangle$ states?

No, because those two states are not orthogonal.

To consider four measurements as four outputs from single measurement they have to be four eigenstates of the same operator.

Yes, and they are. See above.

 

[PeterDonis]

the problem is that $\left(\lambda,\lambda\right)$ (an ontic state shared among the four preparation states) is of course in the support of the probability distribution over ontic states given by $|\phi_{00}\rangle$

No, it isn't. That's the problem, according to PBR: the ontic state $\left(\lambda,\lambda\right)$ is not in the support of the probability distribution of any of the outcome quantum states; it can't be, because it lies in the support of the probability distributions for all four preparation states, and each of those preparation states is orthogonal to one of the outcome states, and the probability distributions of orthogonal quantum states cannot overlap (that's one of the assumptions of the PBR theorem). So we have an ontic state that can be prepared--it's consistent with the preparation quantum state--but apparently predicts zero probability for all four outcome quantum states (because it doesn't lie within the support of the probability distributions for any of them), which is not consistent with the predictions of QM.

I don't currently see how changing the ontic state would avoid the problem.

Because the above argument, which purports to derive a contradiction, depends on the assumption that the ontic state before measurement, $\left(\lambda,\lambda\right)$, has to be within the support of the probability distribution of an outcome quantum state, in order for a nonzero probability for that outcome to be predicted. Which at least appears to assume that the ontic state after measurement is the same as the ontic state before measurement.

 

[zonde]

Why wouldn't you be able to do this? The four output states described in the PBR paper are all orthogonal to each other and span the Hilbert space; therefore they must be eigenstates of some Hermitian operator, so there must be some measurement that has these states as its possible outcome states.

I think I got it. If I wish I can write all four measurements in the same basis and then do the PBR reasoning. So my objections do not matter.

 

[DarMM]

and the probability distributions of orthogonal quantum states cannot overlap (that's one of the assumptions of the PBR theorem)

Does PBR assume this, it's easy enough to derive in the ontological models framework.

according to PBR: the ontic state is not in the support of the probability distribution of any of the outcome quantum states

has to be within the support of the probability distribution of an outcome quantum state

These seem to contradict one another, can you explain what you mean. How can $\left(\lambda,\lambda\right)$ according to PBR not be in the support of any of the outcome states and yet assume it has to be within the support of an outcome state?

 

[PeterDonis]

Does PBR assume this, it's easy enough to derive in the ontological models framework.

It might be a previously proved theorem, yes. I just wanted to make clear that the PBR theorem proof makes use of this proposition.

These seem to contradict one another

The first is talking about what has to be true given that orthogonal quantum states cannot have overlapping probability distributions.

The second is talking about what would have to be true for the psi-epistemic model considered by PBR to match the predictions of QM--which, according to the PBR theorem, it can't, precisely because the second contradicts the first. In other words, the PBR theorem proof basically consists of showing that the first is true, then showing that for a psi-epistemic model to match the predictions of QM, the second would also have to be true; but of course the second contradicts the first so it can't be true.

 

[PeterDonis]

It seems a simple way out of this would be to drop the state preparation independence assumption. If measurements can be entangled then so can be state preparations.

The state preparation independence assumption, as used by PBR, applies only to the preparation of product states, i.e., non-entangled states. For product states that assumption seems to be highly reasonable. Leifer's paper contains a fairly detailed discussion of this assumption and arguments against it.

 

[DarMM]

No, it isn't. That's the problem, according to PBR: the ontic state $\left(\lambda,\lambda\right)$ is not in the support of the probability distribution of any of the outcome quantum states; it can't be, because it lies in the support of the probability distributions for all four preparation states, and each of those preparation states is orthogonal to one of the outcome states, and the probability distributions of orthogonal quantum states cannot overlap (that's one of the assumptions of the PBR theorem)

I don't think this matters too much, it's just the particular way of deriving the contradiction. Leifer uses one way and the original paper uses another, it's just saying which subset of overlaps contradicts others.

I'm using that since all four preparation distributions must overlap and to give a result at all to the measurement must overlap with one of the outcome distributions you get a contradiction, i.e. I'm saying what leads to your statement. Trying to assume they overlap with one at all is contradictory.

PBR go another way and show that obeying QM forces the single particle distributions $|0\rangle$ and $|+\rangle$ to not overlap.

Because the above argument, which purports to derive a contradiction, depends on the assumption that the ontic state before measurement, $\left(\lambda,\lambda\right)$, has to be within the support of the probability distribution of an outcome quantum state, in order for a nonzero probability for that outcome to be predicted. Which at least appears to assume that the ontic state after measurement is the same as the ontic state before measurement.

I don't think so. The ontic state might change after measurement, it doesn't matter I think. What matters is the response functions. There is a quantity $M$ with values $E_i$ that correspond to each of the outcome states, in the sense that they are the conditional probabilities you use upon seeing $E_i$, but it's silent on if $\lambda$ is altered as it does so.

All that matters is that each $\lambda$ is associated with a response function $\Gamma(E_i | \lambda, M)$, the chance of observing $E_i$ given the ontic state was $\lambda$ at the moment of measurement and $E_i$ is an outcome of $M$.

The problem here is that $\Gamma(E_i | \lambda, M)$ must have $\Gamma(E_j | \lambda, M) = 0$ for a given $j$ for ontic states drawn from one of the preparation states. However the existence of an overlap region contradicts this, as the ontic state would have to satisfy all four constraints on the response function, which is impossible.

 

[PeterDonis]

I don't think this matters too much, it's just the particular way of deriving the contradiction. Leifer uses one way and the original paper uses another, it's just saying which subset of overlaps contradicts others.

Yes, I agree, there are a number of ways of looking at the contradiction and how to arrive at it. They're all logically equivalent.

The problem here is that $\Gamma(E_i | \lambda, M)$ must have $\Gamma(E_j | \lambda, M) = 0$ for a given $j$ for ontic states drawn from one of the preparation states.

I agree that this is true if the additional assumption I've been talking about is true. You are simply giving a different way of phrasing the additional assumption: basically, that $\Gamma(E_j | \lambda, M) = 0$ must be the case if $\lambda$ is contained in the probability distribution of a quantum state that is orthogonal to the outcome state $E_j$. In other words, the constraints on the response function that you refer to are the additional assumption in this formulation.

 

[DarMM]

In other words, the constraints on the response function that you refer to are the additional assumption in this formulation.

Aren't those constraints necessary to agree with QM?

 

[PeterDonis]

Aren't those constraints necessary to agree with QM?

No. The QM predictions only deal with quantum states. The constraints are constraints on response functions associated with ontic states, which are not the same as quantum states (at least, not in the kind of model PBR is considering here).

 

[DarMM]

No. The QM predictions only deal with quantum states. The constraints are constraints on response functions associated with ontic states, which are not the same as quantum states (at least, not in the kind of model PBR is considering here).

Of course, but the response function gives the chance that you'll observe a quantity in that ontic state. If $|00\rangle$ is prepared, then the chance of observing $E_0$ is zero, thus to agree with that experimental fact (predicted by QM), you have to have $\Gamma(E_0 | \lambda, M) = 0$ for any ontic states $\lambda$ drawn from the support of $\mu_{|00\rangle}(\lambda)$

 

[PeterDonis]

If $|00\rangle$ is prepared, then the chance of observing $E_0$ is zero

Yes, this is required by standard QM.

to agree with that experimental fact (predicted by QM), you have to have $\Gamma(E_0 | \lambda, M) = 0$ for any ontic states $\lambda$ drawn from the support of $\mu_{|00\rangle}(\lambda)$

Not without the additional assumption I have been talking about. The experimental fact only relates the input and outcome quantum states. The constraint you are imposing on the response function is a constraint on the relationship between the input ontic state and the outcome quantum state. But the property of probability distributions that you are basing the constraint on can, in itself, only constrain the relationship between the input ontic state and the input quantum state, or between the outcome ontic state and the outcome quantum state. The constraint you are imposing is an additional constraint above and beyond that.

 

[DarMM]

I don't understand, the response function doesn't deal with that state after, only the fact of the observed value. It's a relation between the ontic state and the value observed for the quantity $M$ and anything in the support of $\mu_{|00\rangle}(\lambda)$ cannot lead to an observation of $E_0$, thus $\Gamma(E_0 | \lambda, M) = 0$. I don't see how the outcome ontic state or quantum state really matter.

I mean the PBR theorem takes place in a PM-fragment of QM, like most theorems in the ontological models framework, so it doesn't really consider outcome states in general.