Born rules and Bell violation from ensemble of trajectories

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Born rules and Bell violation from ensemble of trajectories

Postby Jarek » Fri Aug 24, 2018 2:00 am

While the original Bell inequality might leave some hope for violation, here is one which seems completely impossible to violate - for three binary variables A,B,C:

Pr(A=B) + Pr(A=C) + Pr(B=C) >= 1

It has obvious intuitive proof: drawing three coins, at least two of them need to give the same value.
Alternatively, choosing any probability distribution pABC among these 2^3=8 possibilities, we have:
Pr(A=B) = p000 + p001 + p110 + p111 ...
Pr(A=B) + Pr(A=C) + Pr(B=C) = 1 + 2 p000 + 2 p111
... however, it is violated in QM, see e.g. page 9 here: http://www.theory.caltech.edu/people/pr ... /chap4.pdf

If we want to understand why our physics violates Bell inequalities, the above one seems the best to work on as the simplest and having absolutely obvious proof.
QM uses Born rules for this violation:
1) Intuitively: probability of union of disjoint events is sum of their probabilities: pAB? = pAB0 + pAB1, leading to above inequality.
2) Born rule: probability of union of disjoint events is proportional to square of sum of their amplitudes: pAB? ~ (psiAB0 + psiAB1)^2
Such Born rule allows to violate this inequality to 3/5 < 1 by using psi000=psi111=0, psi001=psi010=psi011=psi100=psi101=psi110 > 0.

I have just refreshed https://arxiv.org/pdf/0910.2724 (previous thread) adding section III about violation of this inequality using ensemble of trajectories: that proper statistical physics shouldn't see particles as just points, but rather as their trajectories to consider e.g. Boltzmann ensemble - it is in Feynman's Euclidean path integrals or its thermodynamical analogue: MERW (Maximal Entropy Random Walk: https://en.wikipedia.org/wiki/Maximal_e ... andom_walk ).

For example looking at [0,1] infinite potential well, standard random walk predicts rho=1 uniform probability density, while QM and uniform ensemble of trajectories predict different rho~sin^2 with localization, and the square like in Born rules has clear interpretation:

Image

Do you know a different way to understand violation of this inequality?
Jarek
 
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Re: Born rules and Bell violation from ensemble of trajector

Postby Mikko » Fri Aug 24, 2018 5:28 am

Jarek wrote:While the original Bell inequality might leave some hope for violation, here is one which seems completely impossible to violate - for three binary variables A,B,C:

Pr(A=B) + Pr(A=C) + Pr(B=C) >= 1

This could be generalized a little by writing it like a triangle inequality:

Pr(A≠B) + Pr(A≠C) ≤ Pr(B≠C)

This way it is not necessary to restrict to binary variables. In addition, the analogy with the triangle inequality of geometry helps to understand why this inequality makes sense. These considerations are of course unimportant to current discussion.
Do you know a different way to understand violation of this inequality?

Of course there is the standard quantum mechanics and e.g. the Copenhagen interpretation.
Mikko
 
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Re: Born rules and Bell violation from ensemble of trajector

Postby Jarek » Fri Aug 24, 2018 6:20 am

Mikko, substituting Pr(A≠B) = 1 - Pr(A=B) we get:
Pr(A≠B) + Pr(A≠B) + Pr(A≠B) <= 2
It is a bit different than what you have written - I am not certain about its status (?)

Anyway, the general problem is that violation of such looking obvious inequalities (by QM and physics) is believed to require giving up local realism.
Using ensemble (uniform, Boltzmann) of paths also allows to violate it in similar way (through Born rules) - this is realistic model, and in fact required if we e.g. think of general relativity: where we need to consider entire spcatime, particles are their paths.
It is not local in "evolving 3D" picture, but it is local in 4D spacetime/Einstein's block universe view - where particles are their trajectories, ensembles of such objects we should consider.

Ok, I should specify my question: what other realistic models (with some objective situation) allow to violate this looking obvious inequality?
Do you agree that considering ensembles of paths allows for that?
Jarek
 
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Re: Born rules and Bell violation from ensemble of trajector

Postby Mikko » Sat Aug 25, 2018 12:07 am

Jarek wrote:Mikko, substituting Pr(A≠B) = 1 - Pr(A=B) we get:
Pr(A≠B) + Pr(A≠B) + Pr(A≠B) <= 2
It is a bit different than what you have written - I am not certain about its status (?)

Apparently you meant to write Pr(A≠B) + Pr(A≠C) + Pr(B≠C) <= 2. There are two ways to compare these inequalities. One may regard them as two different inequalities about the same (or different) variables. They can be inferred from the same general assumptions by largely similar reasoning and used for similar purposes. Another way is to regard them as the same inequality but using different variables:

Start with Pr(A=B) + Pr(A=C) + Pr(B=C) >= 1.
Define X, Y, Z as X = A, Y = B, Z ≠ C where the last definition is unambiguous because C is a binary variable.
Now we have Pr(X=Y) + Pr(X≠Z) + Pr(Y≠Z) >= 1.
Move the first term to other side: Pr(X≠Z) + Pr(Y≠Z) >= 1 - Pr(X=Y).
Rewrite the right side: Pr(X≠Z) + Pr(Y≠Z) >= Pr(X≠Y).
Mikko
 
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Re: Born rules and Bell violation from ensemble of trajector

Postby Jarek » Sat Aug 25, 2018 12:28 am

Indeed, they are equivalent - true if we choose any probability distribution among 2^3=8 possibilities, but can be violated by physics, QM ... and other models with Born rule: pAB? ~ (psiAB0 + psiAB1)^2, like realistic and "4D local": ensemble of trajectories.
Jarek
 
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