Justo wrote:Joy Christian wrote:The 2 x N spreadsheet of data for 4 settings corresponds to the bound of 4 on the CHSH correlator, not the bound of 2. Nothing can exceed the bound of 4, and no experiment ever has.
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Of course, it is a trivial mathematical truism. But please don't call it Bell inequality when the bound is 4 because is very confusing when you say nothing can violate the Bell inequality and perhaps you are implicitly meaning the bound is 4.
It does not matter. Both inequalities are tautologies due to the rules of arithmetic and nothing can violate either one.
Joy Christian wrote:The 4 x N spreadsheet of data is what CHSH inequality is all about, derived using either CFD or local realism. You just confirmed that no experiment can produce that data, and thus no experiment has ever violated the Bell-CHSH inequality. As I said before, all claims of "violation" of Bell-CHSH are based on the bait-and-switch tactic employed by the experimentalists.
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Of course, when you use CFD the spreadsheet becomes a silly tautology. However, in that case, you all four columns need to have values in it. You just can't do that with real data. Experimental data can only fill two columns leaving blank the other two, it is another way of storing the same data instead of using 4 different 4 x N spreadsheets.
The CHSH inequalities deals with a spreadheet
containing 4 columns of actual data labelled
. There are N rows of data in the spreadsheet. The columns are rearranged in pairs and used to calculate the correlations
, E(a_1,b_2), E(a_2,b_1), E(a_2,b_2))
. Where
 = \lim_{N \rightarrow \infty} -\frac{1}{N}{}\sum_{i=1}^{N} A_i^{a_1} A_i^{b_1})
 = \lim_{N \rightarrow \infty} -\frac{1}{N}{}\sum_{i=1}^{N} A_i^{a_1} A_i^{b_2})
 = \lim_{N \rightarrow \infty} -\frac{1}{N}{}\sum_{i=1}^{N} A_i^{a_2} A_i^{b_1})
 = \lim_{N \rightarrow \infty} -\frac{1}{N}{}\sum_{i=1}^{N} A_i^{a_2} A_i^{b_2})
It is for this scenario that
 - E(a_1,b_2) + E(a_2,b_1) + E(a_2,b_2) \leq 2)
The terms are not independent. This relationship is a mathematical tautology. It is impossible to find a 4xN spreadsheet that will violate this relationship.
You calculate correlations from values, not empty spaces. If you start from four independent 2xN spreadsheets you now have Spreadsheets
you can't use the CHSH. Instead, you have
 = \lim_{N \rightarrow \infty} -\frac{1}{N}{}\sum_{i=1}^{N} A1_i^{a_1} A1_i^{b_1})
 = \lim_{N \rightarrow \infty} -\frac{1}{N}{}\sum_{i=1}^{N} A2_i^{a_1} A2_i^{b_2})
 = \lim_{N \rightarrow \infty} -\frac{1}{N}{}\sum_{i=1}^{N} A3_i^{a_2} A3_i^{b_1})
 = \lim_{N \rightarrow \infty} -\frac{1}{N}{}\sum_{i=1}^{N} A4_i^{a_2} A4_i^{b_2})
And you end up with a correlator:
 - E_2(a_1,b_2) + E_3(a_2,b_1) + E_4(a_2,b_2) \leq 4)
The terms in this expression are independent. This is a mathematical tautology. It is impossible to find four 2xN spreadsheets of data that will violate this relationship.
What Bell believers do is to take experimental data in the form of four 2xN spreadsheets, calculate
 - E_2(a_1,b_2) + E_3(a_2,b_1) + E_4(a_2,b_2))
, and then claim that it violates
.
They are comparing oranges with rocks. By doing this underhanded substitution, they are making an additional assumption. The assumption is simply that it is possible to take the 4 distinct independent spreadsheets
and reduce them to the single spreadsheet
, through permutation of rows. But as I've shown previously. This assumption is false and the key reason why is because of the peculiar use of cyclic recombination of columns used in the original inequality. You will note that the way the settings are chosen for each of the terms follows a cyclic pattern. This is a very important feature of the inequality. Why did Bell use that feature? How come nobody has ever demonstrated so-called "non-locality" by using measurements at pairs of settings that do not have this peculiar cyclic recombination feature? Is it possible that perhaps that's where the magician's trick is hidden? It turns out the cyclic recombinaton is also the undoing of Bell. You can't complete the required re-permutation of rows in a cyclic manner because you will have to undo a previously performed re-permutation.
BTW, I didn't say anything about CFD.
