Is there an additional assumption in the PBR theorem?

[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 $\mu(r,\theta)rdrd\theta$.

 

[PeterDonis]

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

Is this because the average for the measure is taken at just one value $r = R$, so, heuristically, the measure is a delta function in $r$ and so is not normalizable?

 

[DarMM]

Is this because the average for the measure is taken at just one value $r = R$, so, heuristically, the measure is a delta function in $r$ and so is not normalizable?

Basically yes. There are more complex examples, but in most cases measures not absolutely continuous with respect to the Lebesgue measure involve sums of delta-functions of some kind (which is technically not actually a function and thus writing it under an integral in the way physicists do is strictly speaking invalid).

 

[DarMM]

A more complex set of examples would be:
$$\int_{\mathbb{R}}{f(x)d\mu} = 0, supp(f) \not\subset \mathcal{C}$$
That is the integral of a function under this class of measures is zero for functions without support on the Cantor set.

This isn't a unique measure as I haven't defined its action on the Cantor set, but I hope it conveys the point.

EDIT: Previous example I posted doesn't properly define a measure, but a distribution

 

[Demystifier]

Not at all, go ahead!

I have received the answer from Matt Leifer, which I quote without discarding anything:

"What does the ontic state after the measurement have to do with anything? The argument is entirely about the state of affairs between preparation and measurement."

Not very impressive, IMHO.

 

[martinbn]

I believe that the standard terminology is that a measure is absolutely continuous with respect to another measure, not regular with respect to.

 

[DarMM]

"What does the ontic state after the measurement have to do with anything? The argument is entirely about the state of affairs between preparation and measurement."

This is essentially what I've been arguing above, the proof only refers to input states (quantum and ontic) and response functions. The contradiction can be derived entirely in that context.

 

[DarMM]

I believe that the standard terminology is that a measure is absolutely continuous with respect to another measure, not regular with respect to.

You're right of course.

@PeterDonis , "regular" above should be "absolutely continuous", i.e. "absolutely continuous with respect to the Lebesgue measure".

EDIT: Edited previous posts to reflect this for anybody reading it later.

 

[PeterDonis]

Not very impressive, IMHO.

This is essentially what I've been arguing above

I agree that Leifer's response seems to be the same general response that @DarMM has been giving in this thread. Let me try to restate what I think this response is saying.

We have an ontic state $\lambda$ that, by construction, lies in the support of all four of the possible input product quantum states. Since each input product state is orthogonal to one of the four possible outcome quantum states, this ontic state must lead to a prediction of zero probability for all four of the possible outcome quantum states, which of course cannot be consistent with the predictions of QM. The reason for this is not, as I had thought, that the ontic state $\lambda$ itself cannot lie in the support of any of the four possible outcome quantum states. It is that Equation 9 in the Leifer paper, which is said to be required for predictions of an ontic model to be consistent with QM, has as an obvious consequence that, if an ontic state lies in the support of some quantum state $\psi$, the response function for that ontic state must predict zero probability for any quantum state that is orthogonal to $\psi$. This does not make any assumption about the dynamics or what the outcome ontic state is.

I still want to think some more about the implications of this and what it would mean for an ontic model to violate Equation 9, but it does seem clear after looking through the Leifer paper again that Equation 9 is an explicit assumption of the PBR theorem (not just that theorem, it applies much more generally, but that theorem does have it as an explicit assumption), and that Equation 9 does the same logical work as what I thought was an unstated assumption of the theorem. So I'm now convinced that there isn't an additional unstated assumption required to prove the PBR theorem--or at least not the one I thought.

 

[DarMM]

I still want to think some more about the implications of this and what it would mean for an ontic model to violate Equation 9

One possibility is that you could change:
$$\mu_{\rho}(\lambda)$$
to
$$\mu_{\rho,M}(\lambda)$$

That is the likelihood of an ontic state depends on the future measurement, i.e. retrocausal or acausal theories. This escapes the PBR result.

You might also reject the PIP via the presence of wormholes or some other omnipresent nontriviality in spacetime that permits things to be interdependent at preparation. Similarly this escapes the result. However Leifer has proven bounds that make this one unlikely.

 

[zonde]

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.

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.

So I test this approach.
I rewrite these four measurement states:
$|\xi_1\rangle=\frac{1}{\sqrt{2}}(|0\rangle\otimes|1\rangle+|1\rangle\otimes|0\rangle)$
$|\xi_2\rangle=\frac{1}{\sqrt{2}}(|0\rangle\otimes|-\rangle+|1\rangle\otimes|+\rangle)$
$|\xi_3\rangle=\frac{1}{\sqrt{2}}(|+\rangle\otimes|1\rangle+|-\rangle\otimes|0\rangle)$
$|\xi_4\rangle=\frac{1}{\sqrt{2}}(|+\rangle\otimes|-\rangle+|-\rangle\otimes|+\rangle)$

as:
$|\xi_1\rangle=\frac{1}{\sqrt{2}}(|0\rangle\otimes|1\rangle+|1\rangle\otimes|0\rangle)$
$|\xi_2\rangle=\frac{1}{2}(|0\rangle\otimes|0\rangle+|1\rangle\otimes|1\rangle-0\rangle\otimes|1\rangle+|1\rangle\otimes|0\rangle)$
$|\xi_3\rangle=\frac{1}{2}(|0\rangle\otimes|0\rangle+|1\rangle\otimes|1\rangle+0\rangle\otimes|1\rangle-|1\rangle\otimes|0\rangle)$
$|\xi_4\rangle=\frac{1}{\sqrt{2}}(|0\rangle\otimes|0\rangle-|1\rangle\otimes|1\rangle)$

But know I have another problem, measurements $|\xi_2\rangle$ and $|\xi_3\rangle$ are not operationally meaningful at least for photons. There is no measurement that can perform such a four way interference. Two way interference (like in $|\xi_1\rangle$ and $|\xi_4\rangle$) can be performed by swapping measurement contexts of the two photons i.e. $|0_A\rangle$ mode is measured against $|1_B\rangle$ mode. But that approach is meaningless for four way interference.
Of course $|\xi_2\rangle$ and $|\xi_3\rangle$ could be measured if I express them in different basis, but that brings me back to starting point.

 

[DarMM]

The projector $|\xi_2\rangle\langle \xi_2|$ can be expressed in different bases, but that wouldn't affect its measurability, you either have device that can measure it or not.

 

[zonde]

The projector $|\xi_2\rangle\langle \xi_2|$ can be expressed in different bases, but that wouldn't affect its measurability, you either have device that can measure it or not.

The question is not about measurability of any separate measurement state. The question is about combining all four measurement states into single measurement with four outcomes.

 

[DarMM]

The question is not about measurability of any separate measurement state. The question is about combining all four measurement states into single measurement with four outcomes.

Wouldn't the measurability of such a PVM again not depend on the basis? Basically I don't understand how the basis matters.

 

[zonde]

Wouldn't the measurability of such a PVM again not depend on the basis? Basically I don't understand how the basis matters.

I don't know if there is some general rule about possibility to implement some measurement.
In PBR four measurement states are expressed in different bases. The way I see it's sort of obvious that experimental equipment can't make one photon mode interact with two incompatible other photon modes at the same time. The way out would be to express all four measurement states in the same basis. The choice of basis should not matter as long as it's the same for all four measurements.

 

[PeterDonis]

The way out would be to express all four measurement states in the same basis.

The four measurement states are a basis. They are all orthogonal, and there are four of them, which is sufficient to span the Hilbert space.