A Is there an additional assumption in the PBR theorem?

[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.

 

[PeterDonis]

the response function doesn't deal with that state after, only the fact of the observed value

Yes, I know, but the requirement you are imposing makes an implicit assumption about the relationship between the ontic state before and the ontic state after. See below.

anything in the support of $\mu_{|00\rangle}(\lambda)$ cannot lead to an observation of $E_0$

This is the implicit assumption. The known requirement is that if $E_0$ is observed, the ontic state after the measurement, which must lie in the support of quantum state $E_0$, cannot lie in the support of $\mu_{00}$, since those two quantum states are orthogonal. But your response function is trying to say that if $E_0$ is observed, the ontic state before measurement cannot lie in the support of $\mu_{00}$. (Actually your response function is saying the contrapositive of this, but the two are logically equivalent.) That is not required by the experimental facts; it's an additional constraint you are imposing.

 

[DarMM]

This is the implicit assumption.

Sorry I really don't understand.

If something in the support of $\mu_{|00\rangle}(\lambda)$ could cause an observation of $E_0$, then you could prepare $|00\rangle$ and observe $E_0$, thus violating the predictions of QM.

 

[PeterDonis]

If something in the support of $\mu_{|00\rangle}(\lambda)$ could cause an observation of $E_0$, then you could prepare $|00\rangle$ and observe $E_0$

This is just another way of stating the additional assumption. It might seem obvious to you, but it's not logically required by the other assumptions of the PBR theorem, so it's an additional assumption. The only thing that is logically required by the other assumptions is that the output ontic state cannot lie in the support of $\mu_{00}$ if $E_0$ is observed. It is not, as far as I can see, logically required by the other assumptions that the input ontic state cannot lie in the support of $\mu_{00}$ if $E_0$ is observed. So requiring that is an additional assumption. It might seem highly implausible that this could be violated, but it does not seem to me to be logically ruled out by the other assumptions of PBR.

 

[DarMM]

It might seem obvious to you, but it's not logically required by the other assumptions of the PBR theorem, so it's an additional assumption

Isn't it just the standard assumption that you want a model that replicates the predictions of QM.

Also why does the input/output distinction matter for a PM-fragment?

$\Gamma(E_0 | \lambda, M) = 0$ refers to if you start the measurement and $\lambda$ is the initial ontic state, this is the chance the $E_0$ outcome is observed for $M$.

No mention of output states, as typical for PM-fragments.

 

[PeterDonis]

Isn't it just the standard assumption that you want a model that replicates the predictions of QM.

No. The predictions of QM say nothing about ontic states, only about quantum states. So by themselves they can't put any constraints on ontic states.

Also why does the input/output distinction matter for a PM-fragment?

Because the claimed contradiction in the PBR theorem implicitly relies on conflating it.

$\Gamma(E_0 | \lambda, M) = 0$ refers to if you start the measurement and $\lambda$ is the initial ontic state, this is the chance the $E_0$ outcome is observed for $M$.

You mean, $\Gamma(E_0 | \lambda, M) = 0$ means that if $\lambda$ is the initial ontic state, there is zero chance that the $E_0$ outcome is observed. But I don't see how that is required to replicate the QM predictions (as above), or how it is logically deduced from the other PBR assumptions.

 

[DarMM]

But I don't see how that is required to replicate the QM predictions (as above), or how it is logically deduced from the other PBR assumptions.

Okay I'll return to the first part next, but it's not logically deduced, you just assume you are attempting to match QM.

No. The predictions of QM say nothing about ontic states, only about quantum states. So by themselves they can't put any constraints on ontic states.

Sorry, I promise I'm trying to understand, but I just don't. If $\lambda$ can be produced by a preparation of $|00\rangle$ it cannot lead to an observation of $E_0$, how could it without violating QM?

Because the claimed contradiction in the PBR theorem implicitly relies on conflating it.

But the concept of output state, either quantum or ontic doesn't exist in a PM-fragment. Look at the PM-fragment of the Beltrametti–Bugajski model, it doesn't have collapse or Lüders rule.

 

[DarMM]

You mean, Γ(E0|λ,M)=0\Gamma(E_0 | \lambda, M) = 0 means that if λ\lambda is the initial ontic state, there is zero chance that the E0E_0 outcome is observed. But I don't see how that is required to replicate the QM predictions (as above), or how it is logically deduced from the other PBR assumptions.

That's equivalent to what I said, no?

 

[PeterDonis]

If $\lambda$ can be produced by a preparation of $|00\rangle$ it cannot lead to an observation of $E_0$, how could it without violating QM?

By the ontic state being $\lambda$ before measurement, but some other state, which does lie in the support of the measured outcome quantum state $E_0$, after measurement.

I'm not saying I have an explicit model in my back pocket that has this property. I'm just saying that, as far as I can tell, such a model is not logically ruled out by the other assumptions of PBR. So ruling it out requires an additional assumption, which I've tried to state in various ways in this and related threads.

the concept of output state, either quantum or ontic doesn't exist in a PM-fragment.

The system is always in an ontic state, after measurement as well as before. The PM-fragment does not describe the ontic state space, but the complete model that's required to prove the PBR theorem assumes more than just a PM fragment. It assumes that there is an ontological model, as described in section 4.2 of the Leifer paper. It just doesn't assume much about the ontological model, other than the basic properties described by Leifer that all ontological models must have.

 

[PeterDonis]

That's equivalent to what I said, no?

I don't think so, but I'm not sure to which statement of yours you are referring.

 

[DarMM]

I don't think so, but I'm not sure to which statement of yours you are referring.

It doesn't matter too much, just that $\Gamma(E_0 | \lambda, M) = 0$ being the probability is equivalent to saying the probability is zero.

 

[PeterDonis]

$\Gamma(E_0 | \lambda, M) = 0$ being the probability is equivalent to saying the probability is zero.

Ah, ok. I agree that's what $\Gamma(E_0 | \lambda, M) = 0$ means. I don't agree that $\Gamma(E_0 | \lambda, M) = 0$ is required in order to match the predictions of QM, for the reason given in the first part of my post #78.

 

[DarMM]

The system is always in an ontic state, after measurement as well as before. The PM-fragment does not describe the ontic state space, but the complete model that's required to prove the PBR theorem assumes more than just a PM fragment. It assumes that there is an ontological model, as described in section 4.2 of the Leifer paper. It just doesn't assume much about the ontological model, other than the basic properties described by Leifer that all ontological models must have.

Of course, what I mean is that ontological models of PM-fragments don't discuss output states, that's why I mentioned the Beltrametti–Bugajski model as an example. It doesn't matter how the ontic state evolves after or during measurement in the modelling of a PM-fragment. All that matters is the input state $\lambda$ and the chance it will cause the observation of an outcome $E_i$ as encoded in $\Gamma(E_i | \lambda, M)$.

By the ontic state being λ\lambda before measurement, but some other state, which does lie in the support of the measured outcome quantum state, after measurement.

But then you would prepare $|00\rangle$ and see $E_0$.

 

[PeterDonis]

then you would prepare $|00\rangle$ and see $E_0$.

No. You would prepare some quantum state other than $|00\rangle$ which also had $\lambda$ contained in its support, and see $E_0$. The response function does not take the input quantum state as input; it only takes the input ontic state as input. So we can have $\Gamma(E_0|\lambda, M) \neq 0$ while still having zero probability of measuring $E_0$ if we prepare the quantum state $|00\rangle$.

 

[PeterDonis]

Ah, I see that I misspoke before when I responded to this:

If $\lambda$ can be produced by a preparation of $|00\rangle$ it cannot lead to an observation of $E_0$, how could it without violating QM?

I should have said that yes, it's true that if $|00\rangle$ is prepared, we cannot observe $E_0$. But saying that the quantum state $|00\rangle$ is prepared is not the same as saying the ontic state $\lambda$ is prepared.

Let me take a step back and try to describe in one place what is entailed by the kind of psi-epistemic model I am thinking of, which meets all of the stated assumptions of the PBR theorem but, because it violates what I am saying is an additional unstated assumption, evades the conclusion of the PBR theorem.

In this model, it is possible to prepare an ontic state $\lambda$ which, by construction, does not lie in the support of any of the four possible outcome quantum states of the measurement described by PBR. Therefore, the ontic state must change during the measurement, at least if this ontic state is prepared as the input.

There are four different ways to prepare this ontic state, corresponding to the four possible input quantum states. Each of these ways of preparing this ontic state precludes obtaining one of the four outcome quantum states as a result of the measurement (the one the prepared input state is orthogonal to).

Since there is no ontic state that is in the support of more than one of the four outcome quantum states (since they are all orthogonal), the above entails that the model cannot be deterministic: the same input ontic state cannot always lead to the same outcome ontic state, since the same input ontic state can lead to different outcome quantum states.

I'll leave it at that for now.

 

[DarMM]

No. You would prepare some quantum state other than $|00\rangle$ which also had $\lambda$ contained in its support, and see $E_0$. The response function does not take the input quantum state as input; it only takes the input ontic state as input. So we can have $\Gamma(E_0|\lambda, M) \neq 0$ while still having zero probability of measuring $E_0$ if we prepare the quantum state $|00\rangle$.

Of course the response functions depend only on the ontic state, my point is more that all $\lambda$ in the support of $\mu_{|00\rangle}(\lambda)$ must obey $\Gamma(E_0|\lambda, M) = 0$. Do you disagree with this?

 

[DarMM]

In this model, it is possible to prepare an ontic state λ\lambda which, by construction, does not lie in the support of any of the four possible outcome quantum states of the measurement described by PBR. Therefore, the ontic state must change during the measurement, at least if this ontic state is prepared as the input.

Is it possible to state your contention purely in terms of input ontic states and response functions? I find the talk of output states in general confusing as the theorem is proven in the context of PM-fragments. The whole point of ontological models of PM-fragments is that you don't consider output states or the dynamics. The various Bell states are simple ways of labeling the outcomes, they're not considered as post observation states.

 

[zonde]

I found another objection to PBR derivation. This time I guess it might be the same thing @PeterDonis is talking about.
PBR assumes Preparation Independence Postulate (PIP). However when working out QM prediction for measurement outcomes, two particle system is treated as closed system (Hilbert space contains only these two particles and nothing else) making them entangled. This as I see violates PIP.
Say we can treat two prepared particles as part of four particle system in order to satisfy PIP. In that case switching from $|0\rangle$, $|1\rangle$ basis to $|+\rangle$, $|-\rangle$ basis makes predictions uncertain as we have to know the state of other two particles but we don't. In other words in two particle entanglement we can know the state of one particle given the state of the other particle, but in four particle GHZ entanglement (or entanglement swapping) we have to know the state of three particles to be certain about the state of fourth particle and if we look at only two particle subsystem we can not make definite predictions about measurements (in the basis that is diagonal to preparation basis).
This seems similar to findings of Bell theorem - if we assume ontic states corresponding to QM states in one basis, then measurements in diagonal basis need to invoke non-local constraints (on changes in ontic states) in order to match QM predictions.

 

[DarMM]

I found another objection to PBR derivation. This time I guess it might be the same thing @PeterDonis is talking about.
PBR assumes Preparation Independence Postulate (PIP). However when working out QM prediction for measurement outcomes, two particle system is treated as closed system (Hilbert space contains only these two particles and nothing else) making them entangled. This as I see violates PIP

PIP only refers to product states not requiring nonlocal effects at the ontic level. In the PBR case the input states are product states, so the particles aren't entangled.

 

[zonde]

PIP only refers to product states not requiring nonlocal effects at the ontic level. In the PBR case the input states are product states, so the particles aren't entangled.

So let's take product state $|0\rangle\otimes|0\rangle$ and let's look at it in diagonal basis $|+\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ and $|-\rangle=\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$:

$|0\rangle\otimes|0\rangle=\frac{1}{2}(|++\rangle+|--\rangle)+\frac{1}{2}(|+-\rangle+|-+\rangle)$

It looks entangled now. If I will measure it with measurement state $\frac{1}{\sqrt{2}}(|++\rangle-|--\rangle)$ it will give zero outcome, right?

 

[DarMM]

So let's take product state $|0\rangle\otimes|0\rangle$ and let's look at it in diagonal basis $|+\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ and $|-\rangle=\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$:

$|0\rangle\otimes|0\rangle=\frac{1}{2}(|++\rangle+|--\rangle)+\frac{1}{2}(|+-\rangle+|-+\rangle)$

It looks entangled now.

Maybe I'm missing the point, but it's still a product state and not entangled.

 

[PeterDonis]

all $\lambda$ in the support of $\mu_{|00\rangle}(\lambda)$ must obey $\Gamma(E_0|\lambda, M) = 0$. Do you disagree with this?

It's not a matter of agreement or disagreement. I am saying that this claim of yours does not logically follow from the stated assumptions of PBR. So it, or something logically equivalent to it, needs to be adopted as an additional assumption for the theorem to be proved.

If I'm incorrect, the way to show me is to show me how the claim above logically follows from the stated assumptions of PBR.

Is it possible to state your contention purely in terms of input ontic states and response functions?

See just above.

 

[PeterDonis]

when working out QM prediction for measurement outcomes, two particle system is treated as closed system (Hilbert space contains only these two particles and nothing else) making them entangled.

This is not correct; the fact that the two-particle Hilbert space is the entire Hilbert space considered does not mean the two particles must be entangled. There are plenty of states in the two-particle Hilbert space which are not entangled states; the four possible input product states are examples of such non-entangled states.

 

[PeterDonis]

It looks entangled now.

No, it's not. "Entangled" is not the same as "superposition". You changed basis so that the state is a superposition, but it's still not an entangled state; you can still factor it into a product of one-particle states. You've just obfuscated the latter fact by the notation you chose.

 

[DarMM]

If I'm incorrect, the way to show me is to show me how the claim above logically follows from the stated assumptions of PBR.

I don't know exactly which assumptions you are including in the assumptions of the PBR theorem, but it follows from the requirement to match quantum mechanics. To match QM it must be the case that all states in the support of $\mu_{|00\rangle}(\lambda)$ have $\Gamma(E_0 | \lambda, M) = 0$, i.e. it ultimately follows from the assumption of:
$$\int_{\Lambda}{\Gamma(E_i | \lambda, M)d\mu_{\rho}(\lambda)} = Tr(\rho E_i)$$

 

[PeterDonis]

I don't know exactly which assumptions you are including in the assumptions of the PBR theorem

The Leifer paper I linked to in the OP gives a good discussion of them.

it ultimately follows from the assumption of:

I'll think about this.

 

[DarMM]

The Leifer paper I linked to in the OP gives a good discussion of them.

Leifer's account doesn't use the concept of outcome states, so I was wondering if you were looking at an alternate formulation.

 

[PeterDonis]

it ultimately follows from the assumption of:

$$
\int_{\Lambda}{\Gamma(E_i | \lambda, M)d\mu_{\rho}(\lambda)} = Tr(\rho E_i)
$$

Shouldn't this be slightly different? I think it should be:

$$
\int_{\Lambda}{\Gamma(E_i | \lambda, M) \mu_{\rho}(\lambda)} d \lambda = Tr(\rho E_i)
$$

 

[PeterDonis]

it ultimately follows from the assumption of

Does this assumption (with the correction I proposed in post #97, if that correction is valid) appear somewhere in the Leifer paper?

 

[DarMM]

Shouldn't this be slightly different? I think it should be:

No, because it could be any measure (e.g. a point mass measure) not necessarily one that is absolutely continuous with respect to the Lebesgue measure, i.e. Lebesgue measure with weight

Does this assumption (with the correction I proposed in post #97, if that correction is valid) appear somewhere in the Leifer paper?

Equation 9.

 

[DarMM]

The only difference with Leifer's notation is I put the $\rho$ subscript on the measure to indicate it is one of the measures associated with $\rho$, Leifer puts it on the set of measures itself, but this is an unimportant difference.