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(06052016, 07:45 PM)Schmelzer Wrote: I don't understand your question. In my mathematics, +1 (1) + 1 + 1 is always 4, and this has nothing to do with CHSH and whatever ranges or expectation terms. So this sounds like a trick question. But so what, I will have a look at the trick.
It's not a trick question. I just want to know if you think that result for CHSH is possible. Yes or no?
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Depends on what you consider. If you consider a localrealistic theory which fulfills the conditions of the theorem, and talk about the sum
\[ S=E(a,b)E(a,b^{\prime })+E(a^{\prime },b)+E(a^{\prime },b^{\prime }),\]
then S=4 is not possible. If you talk about the sum +1 (1) + 1 + 1, then it is not only possible, but holds always. If you talk not about computing the sum for the same \(\lambda\), but for four different particular values, without any averaging, then it is possible. If you average enough so that you have
\[E(a,b) = \int A(a,\lambda)B(b,\lambda) \rho(\lambda) d\lambda\]
with sufficient accuracy and the same functions \(A, B, \rho\) for all four expressions, then not.
So, a typical trick question, you do not give enough specification for a unique answer and hope nonetheless for an answer.
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06052016, 10:54 PM
(This post was last modified: 06062016, 03:55 AM by FrediFizzx.)
(06052016, 09:26 PM)Schmelzer Wrote: Depends on what you consider. If you consider a localrealistic theory which fulfills the conditions of the theorem, and talk about the sum
\[ S=E(a,b)E(a,b^{\prime })+E(a^{\prime },b)+E(a^{\prime },b^{\prime }),\]
then S=4 is not possible. If you talk about the sum +1 (1) + 1 + 1, then it is not only possible, but holds always. If you talk not about computing the sum for the same \(\lambda\), but for four different particular values, without any averaging, then it is possible. If you average enough so that you have
\[E(a,b) = \int A(a,\lambda)B(b,\lambda) \rho(\lambda) d\lambda\]
with sufficient accuracy and the same functions \(A, B, \rho\) for all four expressions, then not.
So, a typical trick question, you do not give enough specification for a unique answer and hope nonetheless for an answer.
Ok, you say it "depends" so no real answer to the question. But I did perfectly specify what I meant. All that I specified was the CHSH string of expectation terms that can range from 1 to +1. Nothing else was specified. Now, it is pretty easy to see from what I did specify that the answer has to be yes that a result of +1  (1) + 1 +1 = 4 is possible. What this boils down to is that if the CHSH string of terms are independent from each other, then we can have an inequality,
\[ E(a,b)E(a,b^{\prime })+E(a^{\prime },b)+E(a^{\prime },b^{\prime }) \leq 4.\]
And that is the inequality that both QM and all experiments to date have used and have never violated. All one has to do is to look and see that all experiments and QM have always used this inequality with independent expectation terms. I have never seen a counter example to this simple demonstration by mathematical inspection. If what Bell says is right, then there ought to be one. And note that local realistic theories don't even need to enter into this demonstration. Just show how QM or the experiments have ever violated one of Bell's inequalities. You can't; it is impossible.
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Sorry, no, nobody has ever been interested in this trivial inequality. And therefore it is completely irrelevant what you tell us about this inequality. Name it FrediFizzxinequality and be happy with it.
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06062016, 05:03 AM
(This post was last modified: 06062016, 06:15 AM by FrediFizzx.)
(06062016, 04:18 AM)Schmelzer Wrote: Sorry, no, nobody has ever been interested in this trivial inequality. And therefore it is completely irrelevant what you tell us about this inequality. Name it FrediFizzxinequality and be happy with it.
If nobody has been interested in that inequality, then why is it that it is the one that QM and the experiments all use?
Anyways, you should just admit that you can't demonstrate exactly how QM or the experiments violate any of the Bell inequalities. It is quite simple math to show that it is impossible for anything including QM, experiments and LHV models to violate the Bell inequalities.
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06062016, 07:15 AM
(This post was last modified: 06062016, 07:15 AM by Schmelzer.)
(06062016, 05:03 AM)FrediFizzx Wrote: If nobody has been interested in that inequality, then why is it that it is the one that QM and the experiments all use?
Nobody uses it. What is used is the CHSH inequality which is \(S\le 2\). Which can be derived for Einsteincausal realistic theories.
(06062016, 05:03 AM)FrediFizzx Wrote: Anyways, you should just admit that you can't demonstrate exactly how QM or the experiments violate any of the Bell inequalities. It is quite simple math to show that it is impossible for anything including QM, experiments and LHV models to violate the Bell inequalities.
What? It is sufficient to compute the QM predictions for the particular experiment. Or to use the experimental results. Of course, they violate the CHSH inequality \(S\le 2\), not the trivial FrediFizzx inequality \(S\le 4\).
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(06062016, 07:15 AM)Schmelzer Wrote: (06062016, 05:03 AM)FrediFizzx Wrote: If nobody has been interested in that inequality, then why is it that it is the one that QM and the experiments all use?
Nobody uses it. What is used is the CHSH inequality which is \(S\le 2\). Which can be derived for Einsteincausal realistic theories.
(06062016, 05:03 AM)FrediFizzx Wrote: Anyways, you should just admit that you can't demonstrate exactly how QM or the experiments violate any of the Bell inequalities. It is quite simple math to show that it is impossible for anything including QM, experiments and LHV models to violate the Bell inequalities.
What? It is sufficient to compute the QM predictions for the particular experiment. Or to use the experimental results. Of course, they violate the CHSH inequality \(S\le 2\), not the trivial FrediFizzx inequality \(S\le 4\).
They do use the inequality with the bound of 4. And since it is impossible for anyone to show that Bell's inequalities have ever been violated, it is easy to prove that the inequality with a bound of 4 is being used. Just look at the Wikipedia entry,
https://en.wikipedia.org/wiki/Bell%27s_t...redictions
When they do the final calculation for the CHSH string, each expectation term is independent. Thus they are using the inequality with the bound of 4. If those expectation terms are dependent on each other like in the BellCHSH inequality with a bound of 2, it is impossible to get anything greater than 2. It really is just simple mathematics.
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Sorry, but this makes no sense at all. At least I'm unable to give this text a meaningful interpretation.
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(06062016, 08:03 AM)Schmelzer Wrote: Sorry, but this makes no sense at all. At least I'm unable to give this text a meaningful interpretation.
It is really quite simple. In the Wikipedia entry, when they calculated the CHSH string of expectation terms there was no dependency between the terms like there is in BellCHSH. Therefore they used the inequality with the bound of 4 which I easily demonstrated via simple mathematical inspection. IOW, they cheated when they claim there is a violation of BellCHSH. It's apples and oranges. Pure and simple. Believe me, it is mathematically impossible for anything to violate any of the Bell inequalities. You can't do it is why you are not even trying to demonstrate that it is possible. If you think you can do it, then let's see a rigorous demonstration mathematically. And then I will show you where you make your mistake.
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No, they used no inequality at all, but computed what quantum theory predicts. The result was something greater than 2, thus, violated the BellCHSH inequality, which would hold in a Einsteincausal realistic theory.
The whole point is that QT is not equivalent to any Einsteincausal realistic theory. So, of course, if they compute the QT prediction, they use the QT rules and not some inequality which holds for Einsteincausal realistic theories.
Of course, nothing can violate mathematical proofs, which is a triviality. But you can have theories which violate the assumptions of the BellCHSH theorems, thus, can violate the resulting inequalities.
