Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Jarek Duda on 26/08/2018 06:39:09
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While the original Bell inequality (https://en.wikipedia.org/wiki/Bell's_theorem) 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/preskill/ph229/notes/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.
We get such Born rule (https://arxiv.org/pdf/0910.2724) if considering 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 (https://en.wikipedia.org/wiki/Path_integral_formulation#Euclidean_path_integrals) or its thermodynamical analogue: MERW (Maximal Entropy Random Walk: https://en.wikipedia.org/wiki/Maximal_entropy_random_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:
(https://image.ibb.co/dMuRXF/born_res.png)
Is ensemble of trajectories the proper way to understand violation of this obvious inequality?
Comparing with local realism (https://en.wikipedia.org/wiki/Bell%27s_theorem#Local_realism) from Bell theorem, path ensemble has realism and is non-local in standard "evolving 3D" way of thinking ... however, it is local in 4D view: spacetime, Einstein's block universe (https://en.wikipedia.org/wiki/Eternalism_(philosophy_of_time)) - where particles are their trajectories.
What other models with realism allow to violate this inequality?
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Except that the probability of union of disjoint events is the product of their probabilities, not the sum. Boolean arithmetic has + = "OR" , x= "AND". Union requires AND.
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alancalverd, we have 8 basic disjoint events here (flipping 3 coins):
p000 + p001 + p010 + p011 + p100 + p101 + p110 + p111 = 1
Pr(A=B) = p000 + p001 + p110 + p111 corresponds to union/alternative of four of them, leading to above inequality.
We have product for probability of multiple independent events happening simultaneously.
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P(A=B) =0.5, obviously.
Union (AND) is not the same as alternative (OR).
If A≠B then either A = C or B = C
So there will be at least two identical coins. No inequality or mystery.
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As written, the possibility to violate such obvious inequality requires 3 variables/coins.
If you don't like my explanation, one was in earlier link, here is another: https://arxiv.org/pdf/1212.5214.pdf
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a few points:
• What is the meaning of a probability >1 ?
• In the example of the three binary events, Pr(A=B), Pr(A=C), and Pr(B=C) are not independent, so you cannot just add them up like that.
As you point out, 000, 001, 010, 011, 100, 101, 110, and 111 have equal probabilities of occurring. It is true that Pr(A=B) is 0.5 (000, 001, 110, and 111 of the 8 possible outcomes satisfy) and Pr(A=C) is 0.5 (000, 010, 101, and 111) and Pr(B=C) is 0.5 (000, 011, 100, and 111). But 000 and 111 appear in all cases, and you can't double count those outcomes, so Pr(A=B or B=C or A=C) is exactly 1.
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Once again, the inequality is: Pr(A=B) + Pr(A=C) + Pr(B=C) >= 1
Each of these 3 probabilities represent alternative of 4 out of 8 possibilities.
No probability can exceed 1, but their sum can.
The inequality is true for any probability distribution among 8 possibilities, not only uniform you are focusing on.
... but somehow it is violated by QM - the question is how to understand it?
More complete explanation: https://arxiv.org/pdf/1212.5214.pdf
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The outcomes A=B, A=C and B=C are not independent. If A≠B, we know A = C or B = C, but if A = B we know nothing about C.
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Sure, nobody says they are.