Can we violate Bell inequalities by giving up CFD?

[bhobba]

Bertlman socks type model is counterfactualy definite local model that can't create correlations that violate Bell inequality. Your idea was that by giving up CFD but keeping locality it is possible to come up with a model that can violate Bell inequalities. So I don't see that you provided any valid argument for your point.

I gave a model that specifically rejects counter-factual definiteness and predicts the violation of Bell's inequality. Obviously your assertion is wrong.

Oh - I nearly forgot to mention - I make no claim about locality because I don't believe locality applies to correlated systems. But if you do, by a suitable definition of locality, you can reject CFD and keep locality.

Thanks
Bill

 

[bhobba]

A quite bad paper, because it ignores that CFD is derived in Bell's proof. Bell uses locality and the EPR criterion of reality to derive CFD. Thus, all the hopes to "give up CFD" are meaningless, once one does not have to assume it, but can derive it.

So? It uses a definition of CFD and locality and shows its violated by QM.

Thanks
Bill

 

[rubi]

No, relativistic variants of quantum theories are far from "perfectly relativistically covariant". It is far from perfect simply to use the Heisenberg picture because it allows to hide the non-relativistic elements which the Schrödinger picture makes obvious in the wave function. One could name them perfect Lorentz-ether-compatible theories, but not more.

There are no non-relativistic elements in relativistic quantum theories. Relativistic quantum theories carry a unitary representation of the Poincare group and this is all that is needed to call them perfectly relativistically covariant and it is also independent of the Heisenberg or Schrödinger picture. Observers that are moving at relative speed agree on all observable facts. This is really universally agreed upon by all serious physicists who understand relativistic quantum theory, which is the broad majority of the physics community.

Ok, one can say that one has an alternative. Instead of giving up Einstein causality and to return to classical causality in a preferred frame, we can give up causality completely - one does no longer have to try to search for causal explanations of observed correlations, correlations are correlations, such is life, so what. The tobacco industry would be happy.

If you don't want to give up causality at all (not the cheap "signal locality" which I would prefer to name correlarity, but a meaningful notion of causality, which includes Reichenbach's principle of common cause), one is forced to accept a preferred foliation.

There is no need to give up causality. I don't see how you get the idea that one would have to do that. It is certainly wrong.

I disagree. Giving up causality kills an important part of science - the search for realistic causal explanations of observed correlations.

The existence of a perfectly relativistically covariant theory that violates Bell's inequality proves that special relativity (which is a synonym for relativistic covariance) doesn't imply Bell's inequality.

 

[Derek Potter]

One has to deny spacelike separation, because to push the measurement to the end means Alice and Bob measure simultaneously, which is possible in one frame since they are spacelike separated. However, since they are spacelike separated, the measurement will not be simultaneous in another frame, so that means we have not pushed the measurement to the end in all frames. If we choose only the frame in which they measure simultaneously, then we will have a preferred frame, which would negate the point of pushing the measurement to the end. So we have to deny spacelike separation, ie. Bob has to deny that Alice performed the measurement at spacelike separation.

The end is not when Alice and Bob have completed their measurements but when they have shared their measurements with Charles (or each other). This final collation of results is made at time-like separation. In which case it does not matter if Alice measures ahead of Bob. There is no preferred frame, the only criterion is that measurement (collapse) must be postponed until the 4 states have been able to interfere. The fact that Alice and Bob enter Schrödinger Cat states is unfortunate but the conceptual problem for realists was anticipated with Wigner's Friend, here played by Charles.

 

[Derek Potter]

This is a Bertlemann's Socks correlation model in accord with non - local correlations:
The two drawers ( A and B aligned detectors) space like or time like separated contain socks that are in maximally uncertain states of colors red and blue.
While still maintaining 100% correlation between the outcome of A's and B's measurement .It is there from the very beginning. This type of entanglement enables correlations between far distant events for the socks/photons without predetermination of the measured observable.
|φ} = 1/√2 ( | HV } - | VH } ) Blue for V Red for H .
Reference Local QFT. vanhess71
Related question: Does this minimal interpretation allow for Bell inequality violations when A and B detectors are not aligned ?

No. It allows a bilinear correlation curve. QM predicts a cosine correlation curve. This is a common misunderstanding - perfect anticorrelation at 0, perfect correlation at 90 and no correlation at 45 degrees is easily achieved with a variant of the red/blue sock model. Bell inequality violation is maximized at 22.5 degrees etc and cannot be achieved with local properties no matter how complicated you make them.

 

[bhobba]

Giving up causality kills an important part of science - the search for realistic causal explanations of observed correlations.

You consider that an important part of science - others disagree.

For me its what Feynman says:


Thanks
Bill

 

[Ilja]

So? It uses a definition of CFD and locality and shows its violated by QM.

And therefore proves much less than proven by Bell, thus, not worth to be read.

There are no non-relativistic elements in relativistic quantum theories. Relativistic quantum theories carry a unitary representation of the Poincare group and this is all that is needed to call them perfectly relativistically covariant and it is also independent of the Heisenberg or Schrödinger picture. Observers that are moving at relative speed agree on all observable facts. This is really universally agreed upon by all serious physicists who understand relativistic quantum theory, which is the broad majority of the physics community.

Thanks for telling me that I'm not a serious physicist and don't understand relativistic quantum theory. If "observers that are moving at relative speed agree on all observable facts" is all you require for a "perfectly relativistically covariant" theory, you simply have different criteria for what it means. IMHO it means manifest Lorentz invariance of all elements of the theory, not only the final observable results. I have never seen such a thing in the Schrödinger picture, with consideration of a measurement process, but it will not be a problem for you to give me a reference for this, not? Or do I first have to become "serious"?

There is no need to give up causality. I don't see how you get the idea that one would have to do that. It is certainly wrong.

There is. Reichenbach's principle of common cause is, together with Einstein causality, all one needs to prove Bell's inequality. You may, of course, continue to name what remains if one rejects Reichenbach's common cause, but I don't think the sad remains deserve this name. The pure "signal locality" certainly does not deserve it.

The existence of a perfectly relativistically covariant theory that violates Bell's inequality proves that special relativity (which is a synonym for relativistic covariance) doesn't imply Bell's inequality.

No, it only means that one can give up causality essentially, by rejecting Reichenbach's principle of common cause, and continue to name the remains "causality" in such a theory without raising much protest. The tobacco industry will be happy if science will no longer search for causal explanations of observed correlations.

 

[Ilja]

You consider that an important part of science - others disagree.
For me its what Feynman says:

Hm, I was unable to localize the disagreement. Instead, I feel nicely supported by his example of astrological influences near 3.20. Slightly reformulated, if it would be true, that the stars could effect something on Earth - which we would observe as a correlation - then all the physics would be wrong, because there is no mechanism - no causal explanation - which would allow to explain this influence.

So, Feynman seems to agree with me that the requirement that for correlations where should be a causal, realistic (even mechanical!) explanation is an important part of science. So important that a theory which does not provide such explanation would be wrong - at least in this example this was quite explicit.

 

[bhobba]

Hm, I was unable to localize the disagreement.

If it disagrees with experiment then its wrong. In that one statement is the essence of science - not the search for realism.

Thanks
Bill

 

[rubi]

Thanks for telling me that I'm not a serious physicist and don't understand relativistic quantum theory. If "observers that are moving at relative speed agree on all observable facts" is all you require for a "perfectly relativistically covariant" theory, you simply have different criteria for what it means. IMHO it means manifest Lorentz invariance of all elements of the theory, not only the final observable results. I have never seen such a thing in the Schrödinger picture, with consideration of a measurement process, but it will not be a problem for you to give me a reference for this, not? Or do I first have to become "serious"?

All elements of convential relativistic quantum theory are manifestly Lorentz invariant. Switching between the Schrödinger and Heisenberg picture is nothing more than the application of the time-translation operator, which exists, because we provably have unitary representations of the Poincare group, which includes time-translation operators (This can be verified by anyone who knows how to calculate commutators, so basically every undergraduate student of quantum mechanics). Almost every textbook on relativistic QM or QFT explains it (for example Weinberg vol. 1). (Being serious definitely helps.)

There is. Reichenbach's principle of common cause is, together with Einstein causality, all one needs to prove Bell's inequality. You may, of course, continue to name what remains if one rejects Reichenbach's common cause, but I don't think the sad remains deserve this name. The pure "signal locality" certainly does not deserve it.

It is well known that QM doesn't satisfy Bell's locality criterion ("Einstein causality"). That doesn't mean it is not causal. The cause for the correlations is of course that the quantum system has been prepared in a state that results in those correlations. If you prepare the system in a different state, you will not see the same correlations. This is a perfect cause and effect relationship. Bell's locality criterion is just too narrow and doesn't capture the meaning of the word causality adequately. So please don't force your personal definition of causality on everybody else.

Instead of continuously pointing to Reichenbach, you should maybe also consider Popper at some point.

No, it only means that one can give up causality essentially, by rejecting Reichenbach's principle of common cause, and continue to name the remains "causality" in such a theory without raising much protest. The tobacco industry will be happy if science will no longer search for causal explanations of observed correlations.

This is really just plain logic: A statement of the form "All X satisfy Y" can be disproven by giving an example of an X that doesn't satisfy Y.

 

[Ilja]

If it disagrees with experiment then its wrong. In that one statement is the essence of science - not the search for realism.

If I would have to summarize the most important point of scientific methodology, I would use a similar formulation. So, no contradiction, only extremal simplification.

And, given that Feynman has not stopped after this first sentence, he thought some more things are worth to be said about science, not?

By the way, at 2.00 he gives another example about what would be scientific to say which contains an explanation of some phenomenon. He does not say "these are simply observations, not related to our theories, thus, science couldn't care less", but gives and explanation. At 2.48 he talks about extrasensory perception - which "cannot be explained by this". He notes that if it could be established that it exists, it would mean physics is incomplete, and it would be extremely interesting to physics.

 

[andrewkirk]

Instead of continuously pointing to Reichenbach, you should maybe also consider Popper at some point.

I was intrigued that at least half of Feynman's thoughts in that video were pure Popper. I would think of this as a case of 'great minds think alike' if not for the strange fact that Feynman was well-known to regard philosophy of science as useless ("Philosophy of science is about as useful to scientists as ornithology is to birds").

Popper wrote of falsifiability in 1934, when Feynman was only 16. Most probably Feynman was aware of Popper's ideas - although he may not have been aware that they belonged to Popper, or that Popper was a philosopher of science. Or maybe Feynman came upon them independently and later. I regard Feynman as a good philosopher as well as a brilliant scientist, notwithstanding his ostensible disdain for philosophy.

 

[bhobba]

I regard Feynman as a good philosopher as well as a brilliant scientist, notwithstanding his ostensible disdain for philosophy.

He was.

Which makes his view on philosophy interesting - anti philosophy is also a philosophy.

Trouble with me is I agree with him - for me philosophy is mostly semantic waffle. Sorry - but I simply can't warm to it even though I gave it a fair go by starting a postgraduate certificate in it - although it turned out more a historical analysis of it rather than the ideas itself. It's simply not my bag.

Thanks
Bill

 

[Ilja]

All elements of convential relativistic quantum theory are manifestly Lorentz invariant. Switching between the Schrödinger and Heisenberg picture is nothing more than the application of the time-translation operator, which exists, because we provably have unitary representations of the Poincare group, which includes time-translation operators (This can be verified by anyone who knows how to calculate commutators, so basically every undergraduate student of quantum mechanics). Almost every textbook on relativistic QM or QFT explains it (for example Weinberg vol. 1).

Reading for example Fulling, Aspects of Quantum Field Theory in Curved Spacetime p.19, I have a slightly different impression.

"The Schrödinger formalism gives time a privileged role. The Heisenberg point of view permits t and the spatial coordinates to be treated on the same footing, hence permits a geometrically covariant formulation in keeping with the spirit of relativity theory." IOW, the Schrödinger formalism does not permit a manifestly covariant formulation, not? The note "As previously remarked, it is far from clear that a Schrödinger formulation should even make sense for a field system — especially with explicitly time-dependent field equations — because of the difficulty of constructing a Hamiltonian operator" seems also interesting.

It is well known that QM doesn't satisfy Bell's locality criterion ("Einstein causality"). That doesn't mean it is not causal. The cause for the correlations is of course that the quantum system has been prepared in a state that results in those correlations.

I know that if somebody insists on naming these poor remains "causality" it is hopeless to convince him, so be it.

Instead of continuously pointing to Reichenbach, you should maybe also consider Popper at some point.

I have no problem considering Popper. Popper tell's me that a theory which accepts Reichenbachs principle of common cause has a much greater predictive power, because it predicts zero correlation for everything which is not causally connected in the theory.

 

[rubi]

Reading for example Fulling, Aspects of Quantum Field Theory in Curved Spacetime p.19, I have a slightly different impression.

"The Schrödinger formalism gives time a privileged role. The Heisenberg point of view permits t and the spatial coordinates to be treated on the same footing, hence permits a geometrically covariant formulation in keeping with the spirit of relativity theory." IOW, the Schrödinger formalism does not permit a manifestly covariant formulation, not?

Not. The Schrödinger formalism is perfectly equivalent to the Heisenberg formalism. You can rewrite any Lorentz covariant theory into a form that doesn't look Lorentz covariant on first sight. Even Maxwell's equations are usually presented in a form the hides the Lorentz covariance and "gives time a preferred role". That doesn't change the fact that they are Lorentz invariant, as can be seen by rewriting them in tensor notation and these formulations are equivalent. It is exactly the same thing for the Schrödinger and Heisenberg picture and it's absolutely trivial to prove the equivalence and it is done in every textbook.

The note "As previously remarked, it is far from clear that a Schrödinger formulation should even make sense for a field system — especially with explicitly time-dependent field equations — because of the difficulty of constructing a Hamiltonian operator" seems also interesting.

In QFT on CST, the situation is more difficult, since one doesn't have a representation of the Poincare group anymore (obviously, since general spacetimes are usually not Poincare invariant). Anyway, we don't need QFT on CST to prove the existence of a manifestly Lorentz covariant quantum theory. There are lots of trivial examples. Just consult Reed & Simon if you're looking for rigorous proofs.

I know that if somebody insists on naming these poor remains "causality" it is hopeless to convince him, so be it.

It is the standard notion of causality that every scientist acknowledges. And guess what? None of the horror scenarios that you portrayed actually occurred. Science is doing fine and progress is made every day.

I have no problem considering Popper. Popper tell's me that a theory which accepts Reichenbachs principle of common cause has a much greater predictive power, because it predicts zero correlation for everything which is not causally connected in the theory.

The common cause for the correlations is the preparation of the state, which is causally connected to the event of observation. QM is actually in full agreement with your beloved principle of common cause.

 

[Ilja]

It is the standard notion of causality that every scientist acknowledges. And guess what? None of the horror scenarios that you portrayed actually occurred. Science is doing fine and progress is made every day.

A horror scenario would appear only if one would take the rejection of Reichenbach's principle seriously and apply it to science in general. This is nothing we should be afraid of. So it simply prevents progress in the explanation of the violations of Bell's inequality - thus, progress toward a more fundamental theory beyond quantum theory.

If such a theory will not be found, because of such rejections, this is not very problematic. It will probably be found some hundred years later anyway. Until this happens, there is enough room yet where a lot of progress can and will be made, in particular by applying Reichenbach's principle. So, no, I do not portray any horror scenario, because I'm sure that scientists will be inconsistent in the rejection of Reichenbach's principle.

And, no, what every scientist acknowledges is only that causality contains also those poor remains. But I doubt that even a large minority of scientists would accept that Reichenbach's principle of common cause could be simply rejected, and that this would not be important for them, because their notion of causality anyway does not contain Reichenbach's principle.

Anyway, we don't need QFT on CST to prove the existence of a manifestly Lorentz covariant quantum theory. There are lots of trivial examples. Just consult Reed & Simon if you're looking for rigorous proofs.

Trivial examples, yes - free particle theories without interactions, as far as I know, and some very special low-dimensional examples. AFAIK, Haag's theorem is yet relevant, not?

The common cause for the correlations is the preparation of the state, which is causally connected to the event of observation. QM is actually in full agreement with your beloved principle of common cause.

Decide what you want to claim:
1.) The principle of common cause holds in QFT.
2.) The relativistic causal structure holds in QFT.
3.) The Bell inequalities are violated in QFT.

Given that the principle of common cause, together with the relativistic causal structure, gives the Bell inequalities, believing all three seems problematic. See for example
E.G. Cavalcanti, R. Lal -- On modifications of Reichenbach’s principle of common cause in light of Bell's theorem, J. Phys. A: Math. Theor. 47, 424018 (2014), arxiv:1311.6852v1 for this.

 

[Derek Potter]

The common cause for the correlations is the preparation of the state, which is causally connected to the event of observation.

?

 

[Ilja]

Not. The Schrödinger formalism is perfectly equivalent to the Heisenberg formalism. You can rewrite any Lorentz covariant theory into a form that doesn't look Lorentz covariant on first sight. Even Maxwell's equations are usually presented in a form the hides the Lorentz covariance and "gives time a preferred role". That doesn't change the fact that they are Lorentz invariant, as can be seen by rewriting them in tensor notation and these formulations are equivalent. It is exactly the same thing for the Schrödinger and Heisenberg picture and it's absolutely trivial to prove the equivalence and it is done in every textbook.

Sorry, but I think you mingle the equivalence of the two formalisms regarding observable predictions with manifest covariance. Manifest covariance means that all parts, even the unobservable parts of the mathematical apparatus, have covariance. A non-covariant formalism may be equivalent to a manifestly covariant one - that means, the predictions about observables will be the same. But this does not make above formalisms manifestly covariant.

The equivalence is indeed quite trivial (if we ignore all the subtleties of field theory) if we fix a time coordinate. But in the Schrödinger formalism this time coordinate plays a very different role than the space coordinates, and nothing in this formalism is manifestly covariant.

I can understand that if you have field operators Phi(x,t) defined for all points of spacetime, then you can define a meaningful way the Poincare group acts on these operators. I can also understand that if you consider complete solutions phi(x,t), say, for a free particle, that an action of the Poincare group on these solutions may be defined.

But if I define a state Psi(Q,t), where Q denotes the configuration of a field, thus, a whole function phi(x), I do not see a simple manifest way to define a nontrivial Lorentz transformation for it.

 

[RUTA]

I quoted these post from other thread. I don't want to distract discussion in other thread so I'm starting a new one about statements in these posts.

Basically the question is if we can violate Bell inequalities by two separated but correlated systems that can be as non-classical as we like (as long as we can speak about paired "clicks in detectors") i.e. if we give up counter factual definiteness (CFD) but keep locality.
Bhobba and Haelfix are making bold claim that this can be done. But this is just handwaving. So I would like to ask to demonstrate this based on model. Say how using correlated qubits at two spacelike separated places can lead to violation of Bell inequalities in paired detector "clicks"?

Here is an example http://www.ijqf.org/archives/2402. Also, note that it is a realist theory without CFD.

 

[rubi]

Trivial examples, yes - free particle theories without interactions, as far as I know, and some very special low-dimensional examples. AFAIK, Haag's theorem is yet relevant, not?

So you finally acknowledge the fact that there exist perfectly relativistically covariant quantum theories, contrary to your inital claim? (Free QED is already enough to correctly predict the Bell tests by the way.)
Haags theorem is not relevant to the existence of interacting quantum field theories. It just states that they can't be unitarily equivalent to free theories, which is neither necessary nor expected. It is strongly believed that interacting 4d QFT's exist (otherwise the Clay institute wouldn't have put a million dollar bounty on it). It's just that it is mathematically non-trivial and if you look at the details of the interacting phi^4_3 theory, you will see why.

Decide what you want to claim:
1.) The principle of common cause holds in QFT.
2.) The relativistic causal structure holds in QFT.
3.) The Bell inequalities are violated in QFT.

Given that the principle of common cause, together with the relativistic causal structure, gives the Bell inequalities, believing all three seems problematic. See for example
E.G. Cavalcanti, R. Lal -- On modifications of Reichenbach’s principle of common cause in light of Bell's theorem, J. Phys. A: Math. Theor. 47, 424018 (2014), arxiv:1311.6852v1 for this.

The causal relationship I'm talking about is that whenever we prepare the system in a specific entangled state, we will see the correlations and whenever we prepare it in a different state, we don't see the correlations (or see different correlations). Therefore, we can say that the cause for the appearance of the correlations is our preparation of the state. So quantum theory explains the correlations, even if you don't like it, and this is all a scientist needs. If this doesn't satisfy Reichenbachs principle, then Reichenbachs principle is just not a relevant principle, because it is way too strict. And the fact that the only way to save it seems to be to introduce an ether and come up with essentially a conspiracy theory is really more than enough evidence for its rejection.

?

See above.

Sorry, but I think you mingle the equivalence of the two formalisms regarding observable predictions with manifest covariance. Manifest covariance means that all parts, even the unobservable parts of the mathematical apparatus, have covariance. A non-covariant formalism may be equivalent to a manifestly covariant one - that means, the predictions about observables will be the same. But this does not make above formalisms manifestly covariant.

The equivalence is indeed quite trivial (if we ignore all the subtleties of field theory) if we fix a time coordinate. But in the Schrödinger formalism this time coordinate plays a very different role than the space coordinates, and nothing in this formalism is manifestly covariant.

I can understand that if you have field operators Phi(x,t) defined for all points of spacetime, then you can define a meaningful way the Poincare group acts on these operators. I can also understand that if you consider complete solutions phi(x,t), say, for a free particle, that an action of the Poincare group on these solutions may be defined.

But if I define a state Psi(Q,t), where Q denotes the configuration of a field, thus, a whole function phi(x), I do not see a simple manifest way to define a nontrivial Lorentz transformation for it.

The Lorentz covariance is exactly as manifest as it is in Maxwell's equations. In the Heisenberg picture, you have a non time-dependent state $\left|\Psi\right>$ and operators $\phi(x,t)$ and you get the Schrödinger picture by defining $\left|\Psi(t)\right> = U(t) \left|\Psi\right>$ and $\phi(x) = U(t)^\dagger \phi(x,t) U(t)$. The state will satisfy the Schrödinger equation defined by the generator of $U(t)$, as can be easily checked by applying a time-derivative. The time coordinate plays exactly the same role in both pictures. All the expectation values are the same. This is not even specific to quantum theory. You can also formulate classical relativistic theories in an initial-value formulation with a preferred time-coordinate. Even GR has such a formulation (the ADM formalism). There is nothing wrong about rewriting equations in an equivalent way. And even if it were (which it isn't), then free QED in the Heisenberg picture would still be a perfectly manifestly Lorentz covariant quantum theory, which you claim doesn't exist.

 

[atyy]

The measurement problem is only a problem to a physicist, who is convinced that there must be some theory of everything that also includes himself. I would argue that this physicist has no basis for his conviction.

Yes, only to a physicist who believes that there is some theory that also includes him or at least his measurement apparatus. One can certainly take your view, as Bohr and Heisenberg did. But as I mentioned, many others including Landau and LIfshitz, Dirac, Weinberg, Bell and Tsirelson did not.

There is certainly always new physics to be discovered, but I see no reason to believe that this new physics must be a classical description, so I don't agree that the violation of Bell's inequality implies that nature must be non-local. I also think that the measurement problem and the violation of Bell's inequality are not necessarily related.

Yes, the next theory beyond quantum theory may also have a measurement problem. However, it the usual sense of the word, "nature" refers to a theory without a measurement problem, and classical theories like general relativity do not have a measurement problem. So Bell's theorem says that if we use such a theory, then it is nonlocal. The measurement problem and Bell's inequality violation are related, because one way of solving the measurement problem is to introduce hidden variables, eg. Bohmian Mechanics. Bell's theorem says that such a theory that reproduces quantum mechanics will be nonlocal.

I don't see how pushing the measurement to the end implies Alice and Bob measure simultaneously. What I'm saying is that textbook derivation of the violation of Bell's inequality with conventional quantum mechanics never references the collapse. All probabilities are calculated with the pre-collapsed state.

By definition, pushing the measurement to the end means Alice and Bob measure simultaneously - is there another option?

 

[atyy]

So you finally acknowledge the fact that there exist perfectly relativistically covariant quantum theories, contrary to your inital claim? (Free QED is already enough to correctly predict the Bell tests by the way.)

It is conventional to say it either way that free QED is or is not manifestly relativistically covariant. Wave function collapse means that the wave function evolution is neither covariant nor invariant. However it doesn't matter, since the predictions are relativistically invariant.

 

[atyy]

The end is not when Alice and Bob have completed their measurements but when they have shared their measurements with Charles (or each other). This final collation of results is made at time-like separation. In which case it does not matter if Alice measures ahead of Bob. There is no preferred frame, the only criterion is that measurement (collapse) must be postponed until the 4 states have been able to interfere. The fact that Alice and Bob enter Schrödinger Cat states is unfortunate but the conceptual problem for realists was anticipated with Wigner's Friend, here played by Charles.

Yes, that's a slightly more general version of what I said. In either case, there is no violation of the Bell inequalities at spacelike separation, so no implication of nonlocality via the Bell inequalities.

 

[atyy]

It is well known that QM doesn't satisfy Bell's locality criterion ("Einstein causality"). That doesn't mean it is not causal. The cause for the correlations is of course that the quantum system has been prepared in a state that results in those correlations. If you prepare the system in a different state, you will not see the same correlations. This is a perfect cause and effect relationship. Bell's locality criterion is just too narrow and doesn't capture the meaning of the word causality adequately. So please don't force your personal definition of causality on everybody else.

Ilja's definition is the conventional definition throughout science. I think many would agree that maybe there is some other definition that makes what you say correct, but so far there are none widely agreed upon. For example, there is brief commentary by Cavalcanti and Lal on proposals like yours (considering the entangled state to be the cause), but these are not yet widely accepted. (At any rate, if the entangled state is the cause, the formalism is manifestly not relativistically invariant).

http://arxiv.org/abs/1311.6852 (p11): "Another way of dropping FP while keeping PCC would be to point out that correlations do not need to be explained in terms of a factorisability condition, but that the quantum state of the joint system in its causal past can itself be considered as the common cause of the correlations. An objection to this point of view, however, is that the precise correlations cannot be determined without knowledge of the measurements to be performed (the inputs x and y in Fig. 1), and these may be determined by factors not in the common past of the correlated events. A similar criticism may be made of the L-S approach. However, an advantage of the latter is that it does give an analogue of the factorisation condition (rather than simply dropping it), and thus could allow for a generalisation of Reichenbach’s Principle of Common Cause in understanding the implication of causal structure for probabilistic correlations, and be of potential application in areas such as causal discovery algorithms."

 

[Ilja]

So you finally acknowledge the fact that there exist perfectly relativistically covariant quantum theories, contrary to your inital claim? (Free QED is already enough to correctly predict the Bell tests by the way.)

What I question is your "perfectly". And I continue to question it, because all the conceptual issues with the measurement problem and so on are simply ignored.

Haags theorem is not relevant to the existence of interacting quantum field theories. It just states that they can't be unitarily equivalent to free theories, which is neither necessary nor expected. It is strongly believed that interacting 4d QFT's exist (otherwise the Clay institute wouldn't have put a million dollar bounty on it).

As if this would matter. I think this only shows that they don't exist - there are enough clever guys who would have found one if it exists given such a price.

Anyway, it would be useless - gravity is at best an effective field theory, and field theory in combination with gravity will not fare better. But effective field theories may have a Lorentz symmetry in their large distance limit, but are conceptually not Lorentz-covariant.

The causal relationship I'm talking about is that whenever we prepare the system in a specific entangled state, we will see the correlations and whenever we prepare it in a different state, we don't see the correlations (or see different correlations).

Yes, fine, but this is not all what Reichenbach's principle is about. The point is that the correlation should be explained by the common cause. In a quite precise sense of probability theory, P(A and B|cc) = P(A|cc) P(B|cc).

Therefore, we can say that the cause for the appearance of the correlations is our preparation of the state. So quantum theory explains the correlations, even if you don't like it, and this is all a scientist needs.

For those who don't like mathematics and formulas, this may be sufficient as an "explanation". Scientists have usually higher requirements for this. The tobacco industry would be, again, very happy if such a verbal description would be all what scientists need to explain correlations.

And the fact that the only way to save it seems to be to introduce an ether and come up with essentially a conspiracy theory is really more than enough evidence for its rejection.

LOL. A big problem - the Lorentz ether interpretation coming back. And, no, the Lorentz ether does not need any conspiracy, this is an old fairy tale for schoolboys. The Poincare group is the symmetry group of a wave equation, and if everything follows the same wave equation, you obtain it automatically without any conspiracy.

The Lorentz covariance is exactly as manifest as it is in Maxwell's equations. In the Heisenberg picture, you have a non time-dependent state $\left|\Psi\right>$ and operators $\phi(x,t)$ and you get the Schrödinger picture by defining $\left|\Psi(t)\right> = U(t) \left|\Psi\right>$ and $\phi(x) = U(t)^\dagger \phi(x,t) U(t)$. The state will satisfy the Schrödinger equation defined by the generator of $U(t)$, as can be easily checked by applying a time-derivative. The time coordinate plays exactly the same role in both pictures. All the expectation values are the same. This is not even specific to quantum theory. You can also formulate classical relativistic theories in an initial-value formulation with a preferred time-coordinate. Even GR has such a formulation (the ADM formalism). There is nothing wrong about rewriting equations in an equivalent way. And even if it were (which it isn't), then free QED in the Heisenberg picture would still be a perfectly manifestly Lorentz covariant quantum theory, which you claim doesn't exist.

I repeat, I have a little bit sharper criteria than you for perfection of a theory. I have not seen yet any consistent Lorentz-covariant description of the measurement process. At least none which would be comparable in clarity with the description of the measurement process in de Broglie-Bohm theory, which is clearly non-covariant.

Given that it has also a free QED variant, from its start in Bohm's article, one can compare them with your Lorentz-covariant form. I think that the latter is far from perfect, except in the quite trivial variant of perfectness which simply removes all imperfect things from the consideration.

 

[atyy]

As if this would matter. I think this only shows that they don't exist - there are enough clever guys who would have found one if it exists given such a price.

Haag's theorem doesn't prevent relativistic QFTs from existing, since these have already been constructed in 1+1D and 2+1D.

I repeat, I have a little bit sharper criteria than you for perfection of a theory. I have not seen yet any consistent Lorentz-covariant description of the measurement process. At least none which would be comparable in clarity with the description of the measurement process in de Broglie-Bohm theory, which is clearly non-covariant.

But Bell's theorem doesn't rule them out - ie. is it possible for nonlocal Lorentz covariant hidden variables to exist? Maybe http://arxiv.org/abs/1111.1425?

 

[Ilja]

But Bell's theorem doesn't rule them out - ie. is it possible for nonlocal Lorentz covariant hidden variables to exist? Maybe http://arxiv.org/abs/1111.1425?

I think if one relies on causality (requiring Reichenbach's principle and no causal loops) covariant theories are ruled out.

BI excludes common cause, so what remains is or A->B or B->A in arbitrary small environments of A, B. Not above, because no causal loops. Then A->B defines a classical causality connected with a preferred foliation.

 

[rubi]

Yes, only to a physicist who believes that there is some theory that also includes him or at least his measurement apparatus. One can certainly take your view, as Bohr and Heisenberg did. But as I mentioned, many others including Landau and LIfshitz, Dirac, Weinberg, Bell and Tsirelson did not.

I agree that one can also take the other point of view and I accept that people do so. I just wanted to explain that one doesn't need to and quantum theory can be a very satisfactory theory if one doesn't.

Yes, the next theory beyond quantum theory may also have a measurement problem. However, it the usual sense of the word, "nature" refers to a theory without a measurement problem, and classical theories like general relativity do not have a measurement problem. So Bell's theorem says that if we use such a theory, then it is nonlocal. The measurement problem and Bell's inequality violation are related, because one way of solving the measurement problem is to introduce hidden variables, eg. Bohmian Mechanics. Bell's theorem says that such a theory that reproduces quantum mechanics will be nonlocal.

Well, I would say that "nature" refers to nature and a theory is just a representation of some ideas about nature in the language of mathematics. We can't read off what nature is by looking at a mathematical theory.

By definition, pushing the measurement to the end means Alice and Bob measure simultaneously - is there another option?

Pushing the measurement to the end means that you are describing the situation from the outside, i.e. you have a third observer. But as a matter of fact, Alice and Bob perform measurements at spacelike separated intervals and the results are consistent with the statistics that is predicted by the pre-collapsed state.

Ilja's definition is the conventional definition throughout science. I think many would agree that maybe there is some other definition that makes what you say correct, but so far there are none widely agreed upon. For example, there is brief commentary by Cavalcanti and Lal on proposals like yours (considering the entangled state to be the cause), but these are not yet widely accepted. (At any rate, if the entangled state is the cause, the formalism is manifestly not relativistically invariant).

I believe that the majority of quantum physicists would agree that Reichenbachs criterion is too strong for application in quantum theory.

What I question is your "perfectly". And I continue to question it, because all the conceptual issues with the measurement problem and so on are simply ignored.

Perfectly means that the theory is invariant under the Poincare group. This is the definition of a Lorentz covariant theory.

As if this would matter. I think this only shows that they don't exist - there are enough clever guys who would have found one if it exists given such a price.

Anyway, it would be useless - gravity is at best an effective field theory, and field theory in combination with gravity will not fare better. But effective field theories may have a Lorentz symmetry in their large distance limit, but are conceptually not Lorentz-covariant

So the remaining 5 unsolved millenium problems also unsolvable, since nobody has solved them yet? This is a hilarious claim. Anyway, I can only tell you that you will not find a single person working in the area of rigorous QFT who seriously believes that 4d Yang-Mills doesn't exist. It's seen about as unlikely as assuming that P=NP will turn out right. But of course you are invited to submit your refutation of the remaining millenium problems and collect the 5 million dollars. We can talk about it again, when I read about it in the news.

Loop quantum gravity provides a rigorous potential theory of quantum gravity coupled to all known standard model matter. It would really be helpful if you didn't randomly mention all these subjects that you clearly don't really understand as if it would be in favour of your argument.

Yes, fine, but this is not all what Reichenbach's principle is about. The point is that the correlation should be explained by the common cause. In a quite precise sense of probability theory, P(A and B|cc) = P(A|cc) P(B|cc).

Well, Reichenbachs principle needs to be rejected then if it forces us to give up a perfectly satisfactory theory. It's not like Reichenbachs principle is something that nature must necessarily obey. Nature can behave as she may and we have to accept that. religious believes like yours have no place in science.

For those who don't like mathematics and formulas, this may be sufficient as an "explanation". Scientists have usually higher requirements for this. The tobacco industry would be, again, very happy if such a verbal description would be all what scientists need to explain correlations.

Real scientists will give up a theory if it can't be rescued in a reasonable way. And Reichenbachs principle is such a theory.

LOL. A big problem - the Lorentz ether interpretation coming back. And, no, the Lorentz ether does not need any conspiracy, this is an old fairy tale for schoolboys. The Poincare group is the symmetry group of a wave equation, and if everything follows the same wave equation, you obtain it automatically without any conspiracy

Bohmian mechanics definitely needs a conspiracy to explain why the Lorentz violation cannot be observed (arXiv:1208.4119). Even the paper you quoted earlier comes to this conclusion. And introducing an ether with all its consequences when there is really no need for it is just not reasonable.

I repeat, I have a little bit sharper criteria than you for perfection of a theory. I have not seen yet any consistent Lorentz-covariant description of the measurement process. At least none which would be comparable in clarity with the description of the measurement process in de Broglie-Bohm theory, which is clearly non-covariant.

Given that it has also a free QED variant, from its start in Bohm's article, one can compare them with your Lorentz-covariant form. I think that the latter is far from perfect, except in the quite trivial variant of perfectness which simply removes all imperfect things from the consideration.

Every phenomenon has an equivalent discription in every inertial frame and they are connected by Lorentz transformations. This is what Lorentz covariance means. If you don't agree that this is the definition of Lorentz covariance, then I'm wasting my time here.

 

[stevendaryl]

To me, the details of QFT or QM are not so relevant to the issue of locality/causality/etc. as the bare facts of EPR correlations. To me, that's the core question, is whether Bell's-inequality-violating correlations are somehow nonlocal (in the sense of SR).

The perfect correlations in EPR seem to imply a strong connection between distant experiments. As a correlation, it's nonlocal. But it doesn't violate SR's prohibition against FTL propagation of information. But those two facts together are strange. Why doesn't it?

The distinction is between something I would call "factorizability" and "signal locality". Factorizability is just the claim that the evolution of distant subsystems proceeds independently. Roughly speaking, what happens at Alice's location should depend only on conditions local to Alice, and what happens at Bob's location should depend only on conditions local to Bob. That is, facts about Bob's situation shouldn't tell us anything about Alice's future results, unless those results are somehow determined by conditions local to Alice. EPR violates the principle of factorizability. But this violation is not visible in the equations of QFT or QM. Those equations are perfectly factorable, it is only when you interpret the amplitudes as probability distributions for measurements that non-factorizability comes in.

Signal locality is weaker than factorizability, but in a strange way (or it seems strange to me). There is a failure of factorizability if knowing facts about Bob's situation reveals something about Alice's future results (in a way that local knowledge of Alice's situation doesn't). But Bob's situation has two sort-of independent components:

1. Choices made by Bob.
2. Choices made by "nature"--that is, random events.

Signal locality basically amounts to factorizability for Bob's choices. If all you know is what Bob's choices are, that'll tell you nothing about what's going to happen to Alice that couldn't be predicted using Alice's local conditions. So Bob's choices can't influence Alice's results.

This distinction between factorizability and signal locality causes philosophical problems for me, no matter what interpretation of "locality" you're using. For some people, signal locality is all that's important, so they're perfectly happy with saying QM is local (or is not nonlocal, to make a fine distinction). But I have problems with that. What is special about "choices made by agents"? Why should physics particular care about those sources of unpredictability?

On the other hand, saying that the violation of Bell's inequality implies that nature is nonlocal is unsatisfying for other reasons. If interactions are nonlocal at the fundamental level (as they are in the Bohmian interpretation of QM), then why can't it be used for FTL signalling? I certainly understand the proof that it can't be, but it seems very ugly and suspicious to have a fundamental fact about the universe (such as the rest frame relative to which these nonlocal interactions are instantaneous) be inherently undetectable.

 

[RUTA]

But there does seem to be a sense in which Bob's setting and outcome in a particular trial *do* influence Alice's result in EPR-Bell. For example, consider the Mermin device (Mermin, N.D.: Bringing home the atomic world: Quantum mysteries for anybody. American Journal of Physics 49, 940-943 (1981)) where Bob chooses setting 1 and finds an outcome of R (say). As Mermin shows, it can't be the case that conditions local to Alice already existed for outcomes in each of the three possible settings (no CFD), so knowing Bob's setting and outcome tells you something that couldn't have been know otherwise before Alice actually chooses her setting and obtains her result.