# Bell's ansatz and knowledge

considering the eprb experiment, the correlation is written in terms of local hidden variable :

The density of probability at the source is uniform it cannot depend on the measurement angles. Rho(v)

The measurement are made at the two places a(ta,v) b(tb,v)

The datas are the recollected at the same point.

In this last operation it is clear that the integration 'knows' about the measurement angles !

Hence the conclusion is that the correlation at the end should be written as $$\int_{\Omega[ta,tb]}a(ta,v)b(tb,v)\rho_{end}(ta,tb,v)$$ ?

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considering the eprb experiment, the correlation is written in terms of local hidden variable :

The density of probability at the source is uniform it cannot depend on the measurement angles. Rho(v)

The measurement are made at the two places a(ta,v) b(tb,v)

The datas are the recollected at the same point.

In this last operation it is clear that the integration 'knows' about the measurement angles !

Hence the conclusion is that the correlation at the end should be written as $$\int_{\Omega[ta,tb]}a(ta,v)b(tb,v)\rho_{end}(ta,tb,v)$$ ?
Strictly speaking, the measurement angles are not variables as far as the integration is concerned. They are embeded into the functions which generate the outcomes and are fixed.

To continue i had the following analysis of bell-chsh : we consider the measurement results of the operator chsh : AB-AB'+A'B'+A'B

Consdering first the a side we get result a for the forst term. Now the wavefunction is an eigenstate of A hence the measurement for the second term gives again a. The same reasoning applies for a'. Then i consider the b side and i got the following problem : after measuring the third term which gives b' i measure again with B but after measuring B' the wavefunction is not an eigen state of B hence the measurement result for chsh could be : ab-ab'+a'b'+\-a'b ?

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Add: the later result if it is a minus sign would mean that we have for the measurement results (a+a')(b-b')=4,0 or -4 separately.