Is there an additional assumption in the PBR theorem?

[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.

 

[PeterDonis]

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

Then I don't understand the meaning of $d \mu$ in the integral. The integral is over $\Lambda$, which is the ontic state space. That means it should be an integral over $d \lambda$. An integral over $d \mu$ would be an integral over a space of possible probability distributions. But by hypothesis, if we have a particular ontic model, we already know the probability distribution $\mu_\rho$ corresponding to each quantum state $\rho$. We're not integrating over different possible probability distributions.

Equation 9.

Yes, I see it now. But I still have the question I stated above.

 

[DarMM]

Then I don't understand the meaning of $d \mu$ in the integral. The integral is over $\Lambda$, which is the ontic state space. That means it should be an integral over $d \lambda$. An integral over $d \mu$ would be an integral over a space of possible probability distributions. But by hypothesis, if we have a particular ontic model, we already know the probability distribution $\mu_\rho$ corresponding to each quantum state $\rho$. We're not integrating over different possible probability distributions.

That's not what $d\mu$ means, it means integration with respect to a measure that's not necessarily absolutely continuous with respect to the Lebesgue measure. It is still an integration over $\Lambda$ and not probability measures. It's a standard measure theoretic notation. Writing something like:
$$\mu(\lambda)d\lambda$$
assumes the measure is absolutely continuous with respect to the Lebesgue measure, i.e. in informal language is the Lebesgue measure multiplied by a function

 

[PeterDonis]

It's a standard measure theoretic notation.

Hm, ok, sounds like I need to bone up on measure theory.

 

[PeterDonis]

assumes the measure is regular with respect to the Lebesgue measure

So if we assume a particular ontic model whose measure is regular with respect to the Lebesgue measure, I can read $\mu(\lambda) d\lambda$ instead? That might be sufficient for this discussion, since if my argument is correct at all it should be correct for this special case.

 

[DarMM]

So if we assume a particular ontic model whose measure is regular with respect to the Lebesgue measure, I can read $\mu(\lambda) d\lambda$ instead? That might be sufficient for this discussion, since if my argument is correct at all it should be correct for this special case.

Correct.

Very simply, consider a measure like this on $\mathbb{R}^2$:
$$\int_{\mathbb{R}^2}{f(r,\theta)d\mu} = \frac{1}{2\pi R}\int_{S^1}{f(R,\theta)d\theta}$$
i.e. the integral of a function under the measure is its average value on a circle of radius $R$.

This cannot be written as $\mu(r,\theta)rdrd\theta$.