## Simple violation of Bell inequalities

Foundations of physics and/or philosophy of physics, and in particular, posts on unresolved or controversial issues

### Re: Simple violation of Bell inequalities

Jarek

Posts: 158
Joined: Tue Dec 08, 2015 1:57 am

### Re: Simple violation of Bell inequalities

Jarek wrote:Can you violate "tossing 3 coins, at least two are equal": Pr(A=B) + Pr(A=C) + Pr(B=C) >=1 inequality without Born rule/time symmetry?

Jarek, I just don't understand what you mean. If *three coins* are tossed, at least two are equal. The inequality Pr(A=B) + Pr(A=C) + Pr(B=C) >=1 follows from the fact that the union of the three events {A=B}, {A=C},{B=C} is certain.

You can't violate that inequality in the context in which it is valid. You are not talking about three coins being tossed. You are talking about three different sub-experiment in each of which only two coins are tossed. So you should have given "Pr" a subscript. There is Pr_11(A=B), Pr_13(A=C), Pr_23(B=C). The subscripts refer to which coins are tossed, and symbols A, B, C refer to the outcomes of the three coins which are being tossed, in different combinations, in different sub-experiments.

Your language is typical, I'm afraid, of the language of physicists. Your notation is inadequate. It is misleading. It does not make it easy to communicate with mathematicians or statisticians. Of course, most physicists feel no need whatsoever to communicate with mathematicians or statisticians. So it doesn't matter. But that is why there is so much *noise* in fifty years of physicists talking about Bell's theorem. They don't have an adequate language to talk about it, they are too arrogant to realise this, and hence they go round and round in circles.
gill1109
Mathematical Statistician

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Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

### Re: Simple violation of Bell inequalities

Yes, alternative "(A=B) OR (A=C) OR (B=C)" is always true (is tautology) for binary variables - what can be seen as example of Dirchlet's principle.

Or from probabilistic perspective, assume there is some probability distribution among 8 possibilities: sum_ABC pABC = 1
Pr(A=B) = p000 + p001 + p110 + p111
Pr(A=C) = p000 + p010 + p101 + p111
Pr(B=C) = p000 + p100 + p011 + p111
Pr(A=B) + Pr(A=C) + Pr(B=C) = 2p000 + 2p111 + sum_ABC pABC >= 1

Absolutely obvious, but violated e.g. by QM or MERW. No "local realism", but just standard probabilistics ( https://en.wikipedia.org/wiki/Probability_axioms ):
- first axiom: assumption of existence of probability distribution among all possibilities (that unknown = unmeasured), second axiom: Pr(Omega) = 1,
- third axiom - standard rule: "probability of alternative of disjoint events is sum of their probabilities" Pr(A=B) = p000 + p001 + p110 + p111

To violate it, we need to replace these two natural assumption with Born rule:
- instead of probabilities of events, there exist their amplitudes,
- Born rule: "probability of alternative of disjoint event is proportional to sum of squares of their probabilities": Pr_11 (A=B) ~ ((psi000 + psi001)^2 + (psi110 + psi111)^2)
Indeed formally there should be added indexes of which variables are measured - to distinguish just being unknown, from being unmeasured: the Born rule summation is over unmeasured variables.

Then we can violate inequalities derived using standard probability, like above.
for example assuming amplitude:
psi000 = psi111 = 0
psi001 = psi010 = psi011 = psi100 = psi101 = psi110 = 1
we get
Pr_11 (A=B) = ((psi000 + psi001)^2 + (psi110 + psi111)^2) /(sum_AB (psiAB0 + psiAB1)^2 ) = 1/5
Pr_11(A=B) + Pr_13(A=C) + Pr_23(B=C) = 3/5 violating the inequality.

Please write if I can elaborate on something.
Jarek

Posts: 158
Joined: Tue Dec 08, 2015 1:57 am

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