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In 1976, John Bell proved that any locally causal theory can't account for certain observed correlations, he formulated the local causality hypotesis in terms of "local beables".

In particular, he argued that this was necessary to distinguish between theories that are relativistic and theories that aren't. For instance, in the Coulomb gauge of Electromagnetism, the scalar potential instantaneously changes depending on the location of all charges in the universe, but the theory is considered locally causal because the physically relevant quantities are the EM fields.

In quantum mechanics, the wave function (like the scalar potential) also depends on the configuration space of all particles, and therefore can't be considered physically relevant. On the other hand, if you don't introduce any relevant quantity in the theory, it seems to be impossible to impose/discuss a relativistic extension of it, because Bell's notion would be meaningless.

  • Can the notion of local causality be relaxed so that it doesn't depend on local beables?
  • Would this new notion of local causality allow us to decide if a theory is relativistic or not? If it doesn't contain local beables, in which sense would EM be relativistic according to this notion?
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  • $\begingroup$ The reference for the question is: cds.cern.ch/record/980036/files/197508125.pdf Bell says in the abstract: "...local causality: correlations between physical events in different spacetime regions should be explicable in terms of physical events in the overlap of their backwards light cones." He also says "quantum mechanics is not embeddable in a locally causal theory." (end of section 5). $\endgroup$
    – DrChinese
    Commented Jan 10 at 15:54

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Bell did not prove that locally causal theories are ruled out. He proved a theorem. If and only if the assumptions of that theorem are true its conclusions are inevitable. There are two such assumptions:

  1. Locality

  2. The detector settings and the hidden variables are independent

If assumption 2 is false you can have a local theory that accounts for Bell experiments. We also have good reasons to believe that 2 is indeed false since the detectors and the particle source interact both electromagnetically and gravitationally (they consist of atoms - electrons and nuclei which are massive, charged particles). Due to those interactions one cannot mathematically describe the state of the detector independently of the state of the source since the combined system (detectors + source) must satisfy the equations of electromagnetism and gravity.

I don't see how a theory can be locally causal without local beables. But one can in principle get a non-local theory to be consistent with relativity by making it consistent with Lorentz ether theory (the theories are mathematically equivalent). I think the team behind Bohmian mechanics are working in that direction.

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  • $\begingroup$ If 2) is false, a local theory would be able to explain GHZ correlations only if the detector orientations were themselves pre-determined with the emission event. If this is true in general, then you are forced to conclude that all correlations are just coincidental, and you can't draw empirical conclusion from any experiment, because your settings are not statistically independent. $\endgroup$
    – Davyz2
    Commented Jan 10 at 13:45
  • $\begingroup$ @Davyz2, I did not just assume that 2 is false. I have provided an argument (splitting the combined system into independent systems is mathematically impossible). Can you point out what is wrong with it? There are situations where 2) is true, when the quantities you measure are macroscopic/statistical. If you care about the positions/momenta of the center of mass of two billiard balls then yes, you can assume them to be independent since the EM interactions cancel out. But in Bell tests you measure microscopic quantities (spin of a single particle) so 2) is false in this case. $\endgroup$
    – Andrei
    Commented Jan 10 at 13:56
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    $\begingroup$ @DrChinese, it does not matter if you use random number generators, etc. As long as the experiment uses charged particles (electrons and nuclei) they interact and, the combined system must include all charges. An electron at A produces an electric field at B, and that electric field must be as described by Maxwell's equations, so the states at A and B cannot be independent. There is nothing you can do about it. The so-called independent sources are based on pure speculation (how inflation occurred, etc.). It's far from being an established fact. We do not have a theory for the Big-Bang event. $\endgroup$
    – Andrei
    Commented Jan 11 at 9:57
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    $\begingroup$ @DrChinese, all cahrged particles interact, both in classical and quantum electrodynamics. At macroscopic level, as pointed out in my previous comments, electromagnetism is very well approximated by Newtonian mechanics with contact forces. This is why it may appear to you that those interactions do not matter. But microscopically they are there. About the Big-Bang, I assumed you were speaking about Bell tests performed with photons from distant quasars. In order to claim that those photons are correlated or not you need to know the initial state of the universe and the evolution law. $\endgroup$
    – Andrei
    Commented Jan 12 at 16:27
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    $\begingroup$ @DrChinese Unfortunately we do not know either. We have no theory of the Big-Bang itself (we only have some idea of what happened after the event) so we do not know the initial state. We do not have a theory of gravity+QM so we do not know the evolution law. Claiming that the states of those photons are independent is pure speculation. $\endgroup$
    – Andrei
    Commented Jan 12 at 16:48
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One option is to consider the possibility that we might actually use the unmodified equations of motion of quantum mechanics to work out if quantum mechanics is local.

The Heisenberg picture observables of a system only depend on the observables of systems with which it has interacted and in that sense the theory is local. There is an account of how Bell correlations arise in quantum mechanics as a result of locally inaccessible quantum information being carried in decoherent systems:

https://arxiv.org/abs/quant-ph/9906007

https://arxiv.org/abs/1109.6223

https://arxiv.org/abs/2008.02328

If the relevant equation of motion is Lorentz invariant then the theory is also relativistic:

https://arxiv.org/abs/0909.2673

https://arxiv.org/abs/2207.09020

https://arxiv.org/abs/quant-ph/0511188

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  • $\begingroup$ Aren't the Schrodinger's and Heisenberg's picture unitarily equivalent? Clearly they cannot have different notions of locality, two observers describing the dynamics in these representations must agree on a single notion of relativity, my question was what exactly is this notion, and how you can impose it if the theory only contains functionals of configuration space like states and observables. $\endgroup$
    – Davyz2
    Commented Jan 10 at 13:37
  • $\begingroup$ Reaching the conclusion that quantum physics is local in the SP is possible but less straightforward arxiv.org/abs/2312.04701 $\endgroup$
    – alanf
    Commented Jan 10 at 14:20
  • $\begingroup$ I don't understand what you mean by "local" in the Heisenberg's picture, that condition only shows that no information can be transmitted via local operations on entangled states, this doesn't depend in any way on the fact that the speed of light must be constant in every frame, it is a completely different and non-relativistic notion of locality, it is in fact true without requiring any particular geometry for space and time. $\endgroup$
    – Davyz2
    Commented Jan 10 at 14:55
  • $\begingroup$ I have added a couple of references that specifically address the issue of relativistic quantum theory. $\endgroup$
    – alanf
    Commented Jan 11 at 8:30
  • $\begingroup$ Do you have any references by authors that are not named Deutsch or Rubin, or are not obviously proponents of MWI? None of these are common quantum references. On the other hand, if you are pushing a specific interpretation, it is only fair to disclose that to the reader in your answer. PS I did not downvote you. $\endgroup$
    – DrChinese
    Commented Jan 11 at 21:42

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