## Simple violation of Bell inequalities

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

### Simple violation of Bell inequalities

Here is a simple derivation of (some) Bell inequality (top) - for any probability distribution among these 8 possibilities for 3 binary variables ABC, the inequality is fulfilled.
Bottom: example of its violation assuming just Born rule: probabilities being normalized squares of amplitudes:

Hence we just need to explain Born rule to understand violation of Bell inequalities - and its natural understanding brings MERW diffusion: https://en.wikipedia.org/wiki/Maximal_E ... andom_Walk
So MERW shows why standard diffusion has failed (e.g. wrongly predicting that semiconductor is a conductor) - because it has used only an approximation of the (Jaynes) principle of maximum entropy (required by statistical physics), and if using the real entropy maximum (MERW), there is no longer discrepancy - e.g. its stationary probability distribution is exactly as in the quantum ground state.

In fact MERW turns out just assuming uniform or Boltzmann distribution among possible paths - exactly like in Feynman's Eulclidean path integrals (there are some differences), hence the agreement with quantum predictions is not a surprise (while still MERW being just a (repaired) diffusion).
Including the Born rule: probabilities being (normalized) squares of amplitudes - amplitude describes probability distribution at the end of half-paths toward past or future in Boltzmann ensemble among paths, to randomly get some value in a given moment we need to "draw it" from both time directions - hence probability is square of amplitude:

Here is my just rewritten paper about connection between QM and MERW, also e.g. the Shor's algorithm I would gladly discuss:
https://arxiv.org/pdf/0910.2724v2.pdf
Jarek

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### Re: Simple violation of Bell inequalities

I don't think what you have posted is a valid Bell inequality. At any rate, it is mathematically impossible to violate a Bell inequality.
FrediFizzx
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### Re: Simple violation of Bell inequalities

I have taken this Bell inequality from http://www.johnboccio.com/research/quan ... /paper.pdf
which says its source is: J. Preskill, lecture notes at http://www.theory.caltech.edu/people/preskill/ph229

Anyway, it is mathematically impossible to violate also this one if assuming a probability distribution among such 8 possibilities ( {0,1}^3 for binary variables ABC) - as in the diagram, sum over all P(A=B) + P(A=B) + P(A=C) cases covers all 8 possibilities, hence sums to at least 1.

However, as in QM or MERW ( https://en.wikipedia.org/wiki/Maximal_E ... andom_Walk ), for violation I have assumed Born rules: probability is normalized square of amplitude.
This square is the problem - it leads to violation of Bell, but MERW shows natural intuition where it comes from: assuming Boltzmann distribution among paths, probability toward one direction (past or future) is the amplitude, to randomly get a given value we need to "draw it" from both directions - getting the squares, violating Bell.
Jarek

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### Re: Simple violation of Bell inequalities

Jarek wrote:I have taken this Bell inequality from http://www.johnboccio.com/research/quan ... /paper.pdf
which says its source is: J. Preskill, lecture notes at http://www.theory.caltech.edu/people/preskill/ph229

Anyway, it is mathematically impossible to violate also this one if assuming a probability distribution among such 8 possibilities ( {0,1}^3 for binary variables ABC) - as in the diagram, sum over all P(A=B) + P(A=B) + P(A=C) cases covers all 8 possibilities, hence sums to at least 1.

However, as in QM or MERW ( https://en.wikipedia.org/wiki/Maximal_E ... andom_Walk ), for violation I have assumed Born rules: probability is normalized square of amplitude.
This square is the problem - it leads to violation of Bell, but MERW shows natural intuition where it comes from: assuming Boltzmann distribution among paths, probability toward one direction (past or future) is the amplitude, to randomly get a given value we need to "draw it" from both directions - getting the squares, violating Bell.

Ok. What you have done is shifted to an inequality P(A=B) + P(A=B) + P(A=C) >= 0 for your supposed violation. Same type of mistake everyone makes.
.
FrediFizzx
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### Re: Simple violation of Bell inequalities

No, the original inequality, which is easy to prove (just the diagram) is P(A=B) + P(A=B) + P(A=C) >= 1
Using Born rules I was able to get down to 0.6, which seems the best we can get - here is pdf of Mathematica file where I have got it: https://www.dropbox.com/s/62ce2fhu7r852m5/bell.pdf

I haven't used the original Bell inequalities ( https://en.wikipedia.org/wiki/Bell%27s_ ... inequality ) because I couldn't find proof I could present in so simple form.
C(a,c) - C(a,b) - C(b,c) <= 1
But to clear any doubts, here is analogous calculation for them, violated if assuming Born rules: https://www.dropbox.com/s/suikigpmnsjzpmp/bell_org.pdf

Here are plots for the sum in both cases using many random vectors of amplitudes - we can see that their violation is rather difficult, however might happen for Born rules:
Jarek

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### Re: Simple violation of Bell inequalities

Sorry but that is just not mathematically possible. You must be using a different inequality for the "violation".
FrediFizzx
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### Re: Simple violation of Bell inequalities

Could you elaborate?
We know well (also from experiments) that Bell inequalities are violated in QM - because of the Born rule (squares).
The violations I have written also use exactly the same squares for this purpose.
If you don't trust these extremely sophisticated calculations (denominator in example is 1^2 + 2^2 + 2^2 + 1^2 = 10), just run calculations which I have attached ... or point something suspicious there?
Jarek

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### Re: Simple violation of Bell inequalities

Jarek wrote:Could you elaborate?
We know well (also from experiments) that Bell inequalities are violated in QM - because of the Born rule (squares).
The violations I have written also use exactly the same squares for this purpose.
If you don't trust these extremely sophisticated calculations (denominator in example is 1^2 + 2^2 + 2^2 + 1^2 = 10), just run calculations which I have attached ... or point something suspicious there?

Hi Jarek,

I will let Fred answer your question, since I agree with him, but let me point out a linguistic issue that can be confusing between what Fred is saying and what most of the physics community mean by "Bell inequalities are violated in quantum mechanics."

Nothing at all can violate any mathematical inequality, so there is no question of any "violation" of Bell inequalities in either quantum mechanics or experiments. To be sure, Bell inequalities are exceeded in both quantum mechanics and experiments, but not violated in either of them. What you are showing has nothing to do with any violation either. You are switching to a completely different inequality, with higher bounds, just as what is done in quantum mechanics and in actual experiments.

Thanks, by the way, for your email to me.

***
Joy Christian
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### Re: Simple violation of Bell inequalities

Jarek wrote:Could you elaborate?
We know well (also from experiments) that Bell inequalities are violated in QM - because of the Born rule (squares).
The violations I have written also use exactly the same squares for this purpose.
If you don't trust these extremely sophisticated calculations (denominator in example is 1^2 + 2^2 + 2^2 + 1^2 = 10), just run calculations which I have attached ... or point something suspicious there?

If you believe (which you should) that mathematics is logically correct, then only one of two things could be going on with your supposed violation. Either the original inequality is not mathematically true or you are using a different inequality when showing a "violation" of the original inequality. Sorry, but I am not going to slog through your non-standard treatment to explain this further. But I do suspect the latter case of the two options so it is up to you to figure out what inequality you have switched to. The bound in your "violation" case is either 0 or perhaps it may be 1/2. So in fact there is no violation at all.

QM and the experiments do the same thing; they actually switch to a different inequality and then falsely claim a violation of the original inequality. Bad physics if you ask me. Just trying to prevent you from falling into the same trap.
FrediFizzx
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### Re: Simple violation of Bell inequalities

Hi Fred and Joy,
Whatever we will call it: violation or exceeding, I have also shown it for the original Bell inequalities a few posts up.
Sure inequalities as a mathematical theorem cannot be violated, unless we change the assumptions - and this is exactly what is done here or in QM: inequalities are derived for standard probability theory, while example of their violation is for a nonstandard one: with these very controversial squares - exactly as in QM Born rules.

My example is not very long or complex (bottom of the diagram at the top) and I would gladly add explanations if it's not clear - beside the general impossibility remarks, could you maybe refer to it, what do you think is the problem there?
Jarek

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### Re: Simple violation of Bell inequalities

Jarek wrote:Hi Fred and Joy,
Whatever we will call it: violation or exceeding, I have also shown it for the original Bell inequalities a few posts up.
Sure inequalities as a mathematical theorem cannot be violated, unless we change the assumptions - and this is exactly what is done here or in QM: inequalities are derived for standard probability theory, while example of their violation is for a nonstandard one: with these very controversial squares - exactly as in QM Born rules.

My example is not very long or complex (bottom of the diagram at the top) and I would gladly add explanations if it's not clear - beside the general impossibility remarks, could you maybe refer to it, what do you think is the problem there?

Hi Jarek,

It seems to me that what you are showing in your paper is inconsequential to the Bell inequality. Are you not just comparing MERW to QM? If you come to this forum and claim "Simple violation of Bell inequalities", you are making a false claim since we have successfully shown that Bell's theory is wrong in more ways than one. There is in fact a classical local-realistic explanation of the strong correlations of QM by Joy. So when you say "violation" in your paper, you are helping to propagate more mathematical madness. I suspect that you don't even need to mention the Bell inequalities to make your case. Everyone already knows about the strong correlations of QM.
.
FrediFizzx
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### Re: Simple violation of Bell inequalities

Hi Fred,
Indeed, the question is to understand the "strong correlations of QM": the difference between:
1) standard probabilistic models where probability of a set of possibilities is sum of their individual probabilities,
2) and theories with Born rule like QM and MERW: where probability of a set of possibilities is proportional to square of their sum of amplitudes.

Bell inequalities are proven for 1), but like in the example from my first post, can be violated in essentially different 2), which has a bit different correlations.

We know 2) from QM formalism.
MERW is just uniform or Boltzmann distribution among entire paths (like in Euclidean path integrals), what leads to essentially different stationary probability distribution than standard diffusion.
For example for a lattice with defects below: all but the marked positions have additional self-loop (degree 5 or 4), making it a a simple model of semiconductor.
Standard diffusion (GRW) leads to nearly uniform probability distribution of electrons, hence a potential would cause electron flow - what is wrong as we know it is not a conductor.
In contrast, both QM and MERW predict exactly the same stationary probability: of the quantum ground state, with strong localization (e.g. Anderson) preventing conductance:

As in the second diagram in my first post above, this Boltzmann distribution among paths also brings natural intuitions for the Born rules 2): that amplitudes correspond to probability distribution at the end of past or future half-paths (identical), to get probability we have to multiply them - getting the squares ... in contrast to 1) which leads to Bell inequalities.
Jarek

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### Re: Simple violation of Bell inequalities

Preskill lecture: http://www.theory.caltech.edu/people/pr ... /chap4.pdf
Page 9 has the inequality I have originally used with "text derivation" (slightly modified notation):

"Now suppose that there are actually local hidden variables that provide a complete description of this system, and the quantum correlations are to arise from a probability distribution governing the hidden variables. Then, in this context, the Bell inequality is the statement
P(A=B) + P(B=C) + P(A=C) >= 1
This is satisfied by any probability distribution for the three coins because no matter what the values of the coins, there will always be two that are the same. But in quantum mechanics,
P(A=B) + P(B=C) + P(A=C) = 3/4"

The example of violation (to 3/4) uses states which are hard to translate to MERW picture, but here is the one I have used in the diagram which can be directly used for MERW picture:

psi = (|001> + |010> + |100> + |011> + |010> + |110>) / sqrt(6)

Thanks to Born rules, measuring 2 out of 3 such variables, we get violation to 3/5:
P(A=B) + P(B=C) + P(A=C) = 3/5
Jarek

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### Re: Simple violation of Bell inequalities

Jarek wrote:"Now suppose that there are actually local hidden variables that provide a complete description of this system, and the quantum correlations are to arise from a probability distribution governing the hidden variables. Then, in this context, the Bell inequality is the statement
P(A=B) + P(B=C) + P(A=C) >= 1

Here is the key you are missing: The outcomes of the A coin used in the P(A=B) calculation is the very same outcome of the A coin used in the P(A=C) calculation. Similarly, the B outcomes used in the P(A=B) are the very same ones used in the P(B=C) calculation, and the same applies for the C coin outcomes in P(A=C). This is the ONLY scenario in which the above inequality is valid.

b]This is satisfied by any probability distribution for the three coins because no matter what the values of the coins, there will always be two that are the same[/b]. But in quantum mechanics,
P(A=B) + P(B=C) + P(A=C) = 3/4"

But this expression is very misleading. If you investigate more carefully what the QM calculation actually means, and what experimental measurements actually do, you realize that it is not what I described above. They toss the A and B coins a large number of times, then gather the results and calculate P(A=B). To calculate P(B=C) Instead of reusing the B and just tossing C only, they repeat the whole thing and gather new results by tossing both B and C together, and then same again for A and C. As you see, the cyclicity that was present in the data used to calculate the original inequality is now lost.

It turns out that the cyclicity in the data is responsible for the original inequality. By destroying the cyclicity, you are effectively using a different inequality which has a different bound. Therefore the claimed violation by QM and experiments is simply a mathematical error due to misapplication of inequalities and shoddy notation. Fixing the notation by adding subscripts to correspond to the data being used we get:

P(A₁=B₁) + P(B₁=C₁) + P(A₁=C₁) >= 1

For the first case, and

P(A₁=B₁) + P(B₂=C₂) + P(A₃=C₃) = 3/5

For the second case. There is absolutely no violation!!! And it is sad that this mathematical error is still being propagated by those who should know better.
At least, now you know better.
minkwe

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### Re: Simple violation of Bell inequalities

Thanks for the good explanation. Do you think this mathematical madness will ever end? Doesn't seem like it if people keep propagating in paper after paper.
FrediFizzx
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### Re: Simple violation of Bell inequalities

FrediFizzx wrote:Thanks for the good explanation. Do you think this mathematical madness will ever end? Doesn't seem like it if people keep propagating in paper after paper.

As always, minkwe has provided a very clear explanation. It is not difficult to understand even by "non-experts." But the "experts" (or rather Bell-supremacists) have a misleading way-out of minkwe's argument to justify the continuing madness. This way-out too is a complete baloney. But it is more difficult to dispose off. It relies on the statistical notion of large-N limit, or equivalently on the integration in the derivation of the bound of 2 rather than a summation. Some ardent Bell-believers claim that if we perform many experiments, or trials, then in the large-N or infinite limit the bound of 2 should hold even for the non-cyclic case described by minkwe, but in actual experiments it is "violated." Bell-believers use well known mathematical theorems of statistics, or a simple property of integration, to justify this baloney. To counter it, a somewhat more subtle argument must be presented. This is what I have done in the Corollary presented in this paper: https://arxiv.org/abs/1704.02876.

***
Joy Christian
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### Re: Simple violation of Bell inequalities

minkwe wrote:Here is the key you are missing: The outcomes of the A coin used in the P(A=B) calculation is the very same outcome of the A coin used in the P(A=C) calculation. Similarly, the B outcomes used in the P(A=B) are the very same ones used in the P(B=C) calculation, and the same applies for the C coin outcomes in P(A=C). This is the ONLY scenario in which the above inequality is valid.

Assume any (hidden) probability distribution on 8 scenarios of values for 3 binary variables.
Now measure two of them (example of taking https://en.wikipedia.org/wiki/Marginal_distribution ), using natural:

(*) probability of alternative of (non-overlapping) events is sum of individual probabilities.

Here Pr(AB) = sum_C Pr(ABC).
This way, as in the diagram in my first post, you get:

P(A=B) + P(B=C) + P(A=C) = 1 + 2P(A=B=C=0) + 2P(A=B=C=1)

so it is at least 1
However, QM can get lower values, if we want to reproduce such violation/exceeding, we need a different rule, e.g. Born's:

(**) probability of alternative of (non-overlapping) events is proportional to sum of square of amplitudes (being square roots of probabilities of individual events)

|psi| = sqrt(probabitiy)
Pr(union) ~ (sum of sqrt(probability) )^2

I don't know why do you think it is related to some "cyclicity"?
It is just about marginalization: assuming a probability distribution on a larger set, then marginalizing it to a smaller set, e.g. using (*): Pr(AB) = sum_C Pr(ABC).

I have recently worked on the details of getting (**) instead of (*) in MERW for the violation and it has turns out a bit more complicated: https://www.dropbox.com/s/ax35hvxrorx72ff/bell_MERW.pdf
Jarek

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### Re: Simple violation of Bell inequalities

Jarek wrote:
minkwe wrote:Here is the key you are missing: The outcomes of the A coin used in the P(A=B) calculation is the very same outcome of the A coin used in the P(A=C) calculation. Similarly, the B outcomes used in the P(A=B) are the very same ones used in the P(B=C) calculation, and the same applies for the C coin outcomes in P(A=C). This is the ONLY scenario in which the above inequality is valid.

Assume any (hidden) probability distribution on 8 scenarios of values for 3 binary variables.
Now measure two of them (example of taking https://en.wikipedia.org/wiki/Marginal_distribution ), using natural:

(*) probability of alternative of (non-overlapping) events is sum of individual probabilities.

Here Pr(AB) = sum_C Pr(ABC).
This way, as in the diagram in my first post, you get:

P(A=B) + P(B=C) + P(A=C) = 1 + 2P(A=B=C=0) + 2P(A=B=C=1)

so it is at least 1
However, QM can get lower values, if we want to reproduce such violation/exceeding, we need a different rule, e.g. Born's:

(**) probability of alternative of (non-overlapping) events is proportional to sum of square of amplitudes (being square roots of probabilities of individual events)

|psi| = sqrt(probabitiy)
Pr(union) ~ (sum of sqrt(probability) )^2

I don't know why do you think it is related to some "cyclicity"?
It is just about marginalization: assuming a probability distribution on a larger set, then marginalizing it to a smaller set, e.g. using (*): Pr(AB) = sum_C Pr(ABC).

I have recently worked on the details of getting (**) instead of (*) in MERW for the violation and it has turns out a bit more complicated: https://www.dropbox.com/s/ax35hvxrorx72ff/bell_MERW.pdf

Are you a broken record or something? This is exactly what I responded to, it is just as wrong as before.

How can Pr(ABC) be meaningful if A,B,C are pair-wise mutually exclusive events? You either measure AB, or AC or BC. Pr(ABC ) is undefined. You are comparing apples and oranges. Don't you see that the right-hand-side of the expression P(A=B) + P(B=C) + P(A=C) = 1 + 2P(A=B=C=0) + 2P(A=B=C=1) is meaningless, thus the LHS is also meaningless, the event ABC is impossible, not just in QM. You've taken the coin analogy too far, nobody seriously believes that QM predicts 3/4 for COINS. Pr(ABC ) is undefined in the context of Bell's theorem.
minkwe

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### Re: Simple violation of Bell inequalities

Why do you think it is different from having 3 coins: 8 possibilities (000,001,010,011,100,101,110,111) with some probability distribution among them (P(ABC)) ?
Measuring two of them, there is still some probability distribution for the third one: P(AB) = P(AB0) + P(AB1).

The above assumption implies e.g. P(A=B) + P(B=C) + P(A=C) >=1, violated e.g. in QM.
Indeed the problem is to pin down the issue with the above "3 coin" picture with some hidden probability distribution P(ABC).

The crucial hint is the quantum measurement process, which is quite invasive/destructive, e.g. like its idealization: Stern-Gerlach experiment, taking continuous space of initial spin directions, and enforcing final discrete: parallel or anti-parallel alignment to the strong magnetic field.
This way "measuring 2 coins", we somehow influence the third one - leading to Bell violation.

But understanding the details is far from trivial - we need a simple model with some resemblance to QM, like Born rules - also leading to Bell violation.
And MERW fits perfectly here - it is just uniform probability distribution among paths on a graph, allowing to derive Born rules in just 2 lines: https://en.wikipedia.org/wiki/Maximal_E ... derivation
Specifically, we ask for the number of length 2L paths with vertex i in the center (sum_{jk} (A^L)_{ji} (A^L)_{ik}), which in the L -> infinity limit gives Born rule: Pr(i) ~ psi_i^2, where psi turns out exactly the quantum ground state amplitude.
... and such Born rules lead to violation of Bell inequalities: https://www.dropbox.com/s/ax35hvxrorx72ff/bell_MERW.pdf
Jarek

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### Re: Simple violation of Bell inequalities

***
Pr(ABC) is identically zero for any EPR-Bohm type experiment, regardless of what any theory (classical or quantum) predicts.

***
Joy Christian
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