A Assumptions of the Bell theorem

[martinbn]

Is there an assumption that on doesn't use a statistical interpretation? Or if this is not off topic how does Bell's theorem go in a statistical intepretation? Bell's inequalities are fine. It's the theorem I am asking about? In fact how does EPR gor for a statistical interpretation?!

 

[Demystifier]

Is there an assumption that on doesn't use a statistical interpretation? Or if this is not off topic how does Bell's theorem go in a statistical intepretation? Bell's inequalities are fine. It's the theorem I am asking about? In fact how does EPR gor for a statistical interpretation?!

Ballentine in his book says explicitly that Bell theorem implies nonlocality. But the Ballentine's version of statistical interpretation is not the only version of statistical interpretation. The problem with the statistical interpretation is that it's very close to "shut up and calculate", in the sense that on many conceptual questions it's agnostic or refuses to give an explicit answer. For that reason, it's hard to tell precisely what the statistical interpretation says related to the Bell theorem.

 

[stevendaryl]

That's a good question. The answer is - yes. But Bell theorem assumes that there is only one outcome, so it should be added to the list. That being said, I have to stress that I don't think that many worlds is a way to save locality, see my https://arxiv.org/abs/1703.08341.

If I understand @eloheim's suggestion, it sounds local, in a weird way. Imagine that associated with each tiny region in spacetime, there are many (maybe infinitely many) versions of what conditions are like in that region. When two neighboring regions interact, the different versions become correlated.

So for example, in the EPR experiment, Alice measures the spin of her particle along the z-axis. Her region of spacetime (not the entire universe) splits into two versions, one where she gets spin-up, one where she gets spin-down. The same thing when Bob measures the spin of his particle along the z-axis. He splits his region into two versions. Nothing nonlocal happening. Then when Alice communicates with Bob to find out what result he got, the two versions line up. The version with Alice getting spin-up lines up with the version with Bob getting spin-down.

I don't know how this could work out, mathematically, but it seems as if it could be local.

 

[martinbn]

Ballentine in his book says explicitly that Bell theorem implies nonlocality. But the Ballentine's version of statistical interpretation is not the only version of statistical interpretation. The problem with the statistical interpretation is that it's very close to "shut up and calculate", in the sense that on many conceptual questions it's agnostic or refuses to give an explicit answer. For that reason, it's hard to tell precisely what the statistical interpretation says related to the Bell theorem.

My question is how does the logical argument run in the statistical interpretation? In the vague language, you insist on using, the argument goes as follows: from the scholarpedia article

The proof of Bell's theorem is obtained by combining the EPR argument (from locality and certain quantum predictions to pre-existing values) and Bell's inequality theorem (from pre-existing values to an inequality incompatible with other quantum predictions).

How do the details of this go in the statistical interpretation language? Or is it that the theorem doesn't apply to the statistical interpretation?

 

[Demystifier]

Imagine that associated with each tiny region in spacetime, there are many (maybe infinitely many) versions of what conditions are like in that region.

The catch is that, at the fundamental level, the many world interpretation does not associate things with regions of spacetime. That's because the wave function, which is the only fundamental thing according to MWI, does not live in spacetime (except for a 1-particle wave function).

 

[Demystifier]

How do the details of this go in the statistical interpretation language?

I cannot tell, because the statistical interpretation is too vague (even for me) to answer such questions.

 

[martinbn]

I cannot tell, because the statistical interpretation is too vague (even for me) to answer such questions.

Ok, how do the details go in any other interpretation? For example the part that EPR implies preexisting values. How does that follow?

 

[Demystifier]

Ok, how do the details go in any other interpretation? For example the part that EPR implies preexisting values. How does that follow?

The idea of the Bell theorem is that it does not depend on details of any specific interpretation. Therefore, the first question does not make sense. Concerning the second question, see e.g. http://www.scholarpedia.org/article...nFWw#The_EPR_argument_for_pre-existing_values

 

[martinbn]

The idea of the Bell theorem is that it does not depend on details of any specific interpretation. Therefore, the first question does not make sense. Concerning the second question, see e.g. http://www.scholarpedia.org/article/Bell's_theorem?fbclid=IwAR2EAAT--l463yaZabl-MDCBnyParUKlQwvpUCAJq-HreA6CP43LflHnFWw#The_EPR_argument_for_pre-existing_values

But that is phrased in therms of particles and their spin.

But without any such interaction, the only way to ensure the perfect anti-correlation between the results on the two sides is to have each particle carry a pre-existing determinate value (appropriately anti-correlated with the value carried by the other particle) for spin along the z-axis.

But in the statistical interpretation the state and the observables are not of a single particle but of the ensemble. My question is how do you phrase it in the statistical interpretation language? I am not asking about any particular statements of the interpretation that you find vague.

 

[Demystifier]

But in the statistical interpretation the state and the observables are not of a single particle but of the ensemble. My question is how do you phrase it in the statistical interpretation language?

Ah, that should be easy. In the first step, instead of saying "each particle carry a pre-existing determinate value", you make it a bit more precise by saying "each particle in the pair carry a pre-existing determinate value". In the second step, you generalize the pair to an ensemble of pairs and say "in every member of the ensemble of pairs, each particle in the pair carry a pre-existing determinate value".

 

[stevendaryl]

But in the statistical interpretation the state and the observables are not of a single particle but of the ensemble. My question is how do you phrase it in the statistical interpretation language? I am not asking about any particular statements of the interpretation that you find vague.

I don't understand this business of the "ensemble" having properties, independent of the systems making up the ensemble.

In the particular case of EPR, Alice and Bob repeatedly do the same thing:

1. A third party prepares an anti-correlated electron/positron pair.
2. Alice chooses an axis $\vec{\alpha}$.
3. Bob chooses an axis $\vec{\beta}$.
4. They each measure the spin of one of the particles relative to their chosen axis.

Then after doing this many many rounds, for many different combinations of axes, they compute statistics for their measurements.

So what does it mean to say that quantum mechanics applies to the ensemble, and not to the individual rounds of this experiment?

 

[Demystifier]

So what does it mean to say that quantum mechanics applies to the ensemble, and not to the individual rounds of this experiment?

I think it is just the frequentist probabilistic interpretation of the wave function. It says that the wave function is just a tool to compute the probability, where probability, in the frequentist interpretation, cannot be associated with an individual round of an experiment.

 

[stevendaryl]

I think it is just the frequentist probabilistic interpretation of the wave function. It says that the wave function is just a tool to compute the probability, where probability, in the frequentist interpretation, cannot be associated with an individual round of an experiment.

But that's not an interpretation. That seems to me to be just leaving it without an interpretation.

 

[Demystifier]

But that's not an interpretation. That seems to me to be just leaving it without an interpretation.

Well, many interpretations claim that the wave function is something more than just a tool to compute the probability. So claiming the opposite, that the wave function is not something more, is an interpretation too. More precisely, it's a denial of a large class of interpretations, but denial of an interpretation is an interpretation, isn't it? In fact, I argued elsewhere that the statistical interpretation is the maximal denial interpretation https://www.physicsforums.com/threa...al-nor-statistical.998661/page-4#post-6450421

 

[martinbn]

I'll try to explain what I mean. Of course, it is possible that it is just terminology, but it bothers me.

I receive one particle at a time and measure the spin along the z-axis. Half of the time I get "up", half of the time I get "down". In a statistical interpretation the spin-z observable doesn't have a specific value, because there isn't a single value that I get on this equivalence class of measurements. In a non statistical interpretation it is still possible to have the case, at least apriori, that each particle was in a state |up> or |down> and that the spin-z observable has a value for each of them. But in the statistical interpretation it makes no sense to even make the statement.

Going back to EPRB, the quantum mechanic system is the equivalence class of equally prepared pairs of particles and the Bohm state is the sate of that ensemble. The possible outcomes of a measurement are (up, down) or (down, up). And half of the time one happens, the other half the other one. So I don't understand what it means (let alone how it follows from EPR) that the spin had a preexisting value! What is the value of the spin?

 

[Demystifier]

I'll try to explain what I mean. Of course, it is possible that it is just terminology, but it bothers me.

I receive one particle at a time and measure the spin along the z-axis. Half of the time I get "up", half of the time I get "down". In a statistical interpretation the spin-z observable doesn't have a specific value, because there isn't a single value that I get on this equivalence class of measurements. In a non statistical interpretation it is still possible to have the case, at least apriori, that each particle was in a state |up> or |down> and that the spin-z observable has a value for each of them. But in the statistical interpretation it makes no sense to even make the statement.

Going back to EPRB, the quantum mechanic system is the equivalence class of equally prepared pairs of particles and the Bohm state is the sate of that ensemble. The possible outcomes of a measurement are (up, down) or (down, up). And half of the time one happens, the other half the other one. So I don't understand what it means (let alone how it follows from EPR) that the spin had a preexisting value! What is the value of the spin?

The spin, of course, does not have the value in the whole ensemble. But it does have a value (at least a postexisting value, if not a preexisting one) in each individual member of the ensemble. The statistical interpretation does not deny the existence of individual members. It only denies that the wave function describes an individual member. But then what describes an individual member? The statistical interpretation remains agnostic (at best), inconsistent (at worst) or vague (at somewhere in between best and worst) on that.

 

[eloheim]

If I understand @eloheim's suggestion, it sounds local, in a weird way. Imagine that associated with each tiny region in spacetime, there are many (maybe infinitely many) versions of what conditions are like in that region. When two neighboring regions interact, the different versions become correlated.

Yes this is the idea (and I think Demystifier did understand correctly too). The point being that the super-classical correlations only become apparent when the two results have been brought back into contact and compared, so until that happens the possibilities can sort of ride along in their bubbles without fear of violating causality. Like I said I saw it presented as a demonstration of a type of theory that could side-step bell's theorem but it wasn't anything more than that.

 

[martinbn]

The spin, of course, does not have the value in the whole ensemble. But it does have a value (at least a postexisting value, if not a preexisting one) in each individual member of the ensemble. The statistical interpretation does not deny the existence of individual members. It only denies that the wave function describes an individual member. But then what describes an individual member? The statistical interpretation remains agnostic (at best), inconsistent (at worst) or vague (at somewhere in between best and worst) on that.

It also says that the observables are for the whole ensemble not the individual representatives. And the theory is concerned with the values of the observables. Now my question is probably clearer. Bell's theorem, vaguely stated, says that under some assumptions any theory that has the prediction of QM is nonlocal. My question is: is one of the assumption that the theory needs to apply to the individual objects, not the ensembles?

 

[Demystifier]

My question is: is one of the assumption that the theory needs to apply to the individual objects, not the ensembles?

Yes, I would say so. After all we measure individual events and the theory is supposed to explain the measurements. But (to avoid a frequent misunderstanding), it does not mean that the theory needs to be deterministic, a theory of individual events may also be stochastic.

 

[martinbn]

Yes, I would say so. After all we measure individual events and the theory is supposed to explain the measurements. But (to avoid a frequent misunderstanding), it does not mean that the theory needs to be deterministic, a theory of individual events may also be stochastic.

So, the theorem does not prove that QM is nonlocal, given that there is at least one interpretation that the theorem does not apply to.

 

[gentzen]

So, the theorem does not prove that QM is nonlocal, given that there is at least one interpretation that the theorem does not apply to.

Even stochastic interpretations of QM are nonlocal. Or do you know one which is not?

 

[Demystifier]

So, the theorem does not prove that QM is nonlocal, given that there is at least one interpretation that the theorem does not apply to.

But as I said, I don't think that statistical interpretation denies individual events. And Ballentine explicitly says that QM is nonlocal. So it's not clear why do you think so.

 

[Fra]

I am confused again, but as we are talking abot EPR here, I assume we by local means "bell-local" right? or which version of locality are we talking about?

There was at some paper ealier in the thred this definition

"A physical theory is EPR-‐local iff according to the theory procedures carried out in one region do not immediately disturb the physical state of systems in sufficiently distant regions in any significant way"

This looks like a reasonable definition to me.

But I really suspect there is a blurring of the concept of locality and various presumptions about causality. I see its' often said that the "bell locality condition" is the factorizability or partitioning of the sum or the hidden variable paths.

Ie it seems sometimes the partition
$$ P(A |O_ {A}) = \sum_{\lambda} P(A|\lambda|O_ {A}) P(\lambda|O_ {A}) $$
is what some seem to label "bell locality condition", is this what some talk about? If so, I object to that notion as this contains also implicit assumptions of causation.

So what kind of locality definition do you use, when you say physics is non-local? One can hardly claim that we proved that Alice immediately disturbs Bobs lab, can we?

Can we step back and just sort out the basic concept?

/Fredrik

 

[PeterDonis]

My question is: is one of the assumption that the theory needs to apply to the individual objects, not the ensembles?

The theorem is about correlations between individual measurement results, not ensembles, so I would say yes. However:

the theorem does not prove that QM is nonlocal, given that there is at least one interpretation that the theorem does not apply to.

The theorem is not about QM interpretations, or indeed about QM directly at all. It is a mathematical theorem that applies to the predictions of any theory that satisfies the assumptions. The point about QM in connection with Bell's Theorem is that QM's predictions violate the Bell inequalities, so QM, as a theory that makes predictions (and QM's predictions are independent of any interpretation), must violate at least one of the assumptions of Bell's theorem. And when you look at the math of how QM makes predictions, it is obvious that QM violates the factorizability assumption, the one that says, roughly, that the joint probability distribution for results of two measurements as a function of the settings for each measurement, must factor into two distributions, each of which is only a function of the settings for one of the two measurements. Since the factorizability assumption is usually called the "locality" assumption, the fact that QM violates it means that QM is "nonlocal" in the sense that it violates that assumption.

It is, of course, possible to have QM interpretations that imply violations of other assumptions of the theorem, such as the assumption that measurements have single outcomes, which is violated by the MWI. But none of those interpretations can change the fact that QM, as a theory (independent of any interpretation), violates the factorizability assumption. So none of those interpretations can change the fact that QM is "nonlocal" in the sense of violating that assumption. Refusing to use the term "nonlocal" to describe violating the factorizability assumption does not change the fact that QM does violate it.

 

[Demystifier]

But none of those interpretations can change the fact that QM, as a theory (independent of any interpretation), violates the factorizability assumption.

What about factorizability assumption in classical physics? For instance, do Bertlmann socks violate the factorizability assumption? If they do, then Bell nonlocality is more than violation of factorizability assumption.

 

[martinbn]

Even stochastic interpretations of QM are nonlocal. Or do you know one which is not?

I am not saying that QM isn't nonlocal. I am saying that I don't follow the reasoning in the theorem. So I don't see how the theorem implies it.

But as I said, I don't think that statistical interpretation denies individual events. And Ballentine explicitly says that QM is nonlocal. So it's not clear why do you think so.

Nobody denies individual events. But EPR implies that preexisting values of observable. So, let me ask you this. Alice receives particle, one at a time, and measures the spin along the z-axis. Half of the results are "up", half "down". What is the value of spin-z? You say it is "up" or "down" for each individual particle. But that seems like amusing non-statistical description.

It is, of course, possible to have QM interpretations that imply violations of other assumptions of the theorem, such as the assumption that measurements have single outcomes, which is violated by the MWI. But none of those interpretations can change the fact that QM, as a theory (independent of any interpretation), violates the factorizability assumption. So none of those interpretations can change the fact that QM is "nonlocal" in the sense of violating that assumption. Refusing to use the term "nonlocal" to describe violating the factorizability assumption does not change the fact that QM does violate it.

The theorem in question is a bit more than the inequities and their violations. It is the combinations of the inequalities and EPR + perfect correlations implies preexisting values. All that implies that the locality assumption is wrong, hence nonlocality. Just to point out that this nonlocality is not just the factorazability in Bell's inequalities. It is the Einstein's locality.

 

[Demystifier]

Nobody denies individual events. But EPR implies that preexisting values of observable. So, let me ask you this. Alice receives particle, one at a time, and measures the spin along the z-axis. Half of the results are "up", half "down". What is the value of spin-z? You say it is "up" or "down" for each individual particle. But that seems like amusing non-statistical description.

Before answering your question, let me tell that EPR implies preexisting values if one assumes locality.

Now your question. Yes, the value of spin is either up or down for each individual particle. But it's not any less statistical than the fact that each individual classical coin is either heads or tails.

 

[martinbn]

Before answering your question, let me tell that EPR implies preexisting values if one assumes locality.

Yes, I understand what this says. And here locality is locality and not Bell's locality, right?

Now your question. Yes, the value of spin is either up or down for each individual particle. But it's not any less statistical than the fact that each individual classical coin is either heads or tails.

So, the theorem with statement "every theory that reproduces QM predictions, is nonlocal" has the additional assumption, that it is a theory which describes the dynamics of individual objects and not ensembles, right? So you can add this to your list of assumption.

 

[Demystifier]

Yes, I understand what this says. And here locality is locality and not Bell's locality, right?

Well, I'm not sure what do you mean by "locality" here. The locality used in the EPR argument is not signal locality, nor operators-commuting-at-spatial-separations locality, nor local-Hamiltonian locality. Technically it's also not Bell locality (because EPR did it much before Bell entered the scene), but it's vary akin to Bell locality.

So, the theorem with statement "every theory that reproduces QM predictions, is nonlocal" has the additional assumption, that it is a theory which describes the dynamics of individual objects and not ensembles, right? So you can add this to your list of assumption.

I would add this to the assumptions if there was at least one interpretation of QM which claims to avoid nonlocality by not dealing with individual objects. But I am not aware of any such interpretation.

 

[vanhees71]

Before answering your question, let me tell that EPR implies preexisting values if one assumes locality.

Now your question. Yes, the value of spin is either up or down for each individual particle. But it's not any less statistical than the fact that each individual classical coin is either heads or tails.

In standard QM the spin component in a given direction of a spin-1/2 particle is either determined or undetermined. If it is determined the particle is prepared in an eigenstate of this spin component's representing self-adjoint operator and the spin component then takes the corresponding eigenvalue. If the particle is not prepared in such an eigenstate the spin-component's value is indetermined, and Born's rule gives you the probabilities to get each of the possible eigenvalues when measuring this spin component.

I don't know, what EPR are thinking. The more often I try to understand this paper the less I succeed ;-).