Bell's inequality when efficiency < 1
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naima
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I found the algebric origin of the fact that in Bell's theorem we replace 2 by [itex]4/\eta - 2[/itex]
suppose that a detector has a probability [itex]\eta[/itex] to detect a particle
we have four possibilities when a pair is created:
1) left and right not detected
2) left detected but not right
3) right detected but not left
4) left and right detected
the probabilities are:
[tex](1 - \eta)^2[/tex]
[tex]\eta (1 - \eta)[/tex]
[tex](1 - \eta)\eta[/tex]
[tex]\eta^2[/tex]
the observer cannot count the first case events
We have 2) + 3) + 4) = [itex]1 - (1 - \eta)^2) = (2 - \eta) \eta[/itex]
the conditional probability of 4) knowing that 1) did not occur is
[tex]\frac{\eta^2 }{ (2 - \eta) \eta} = \eta/(2 - \eta)[/tex]
the expected value Bell = [itex]\eta/(2 - \eta)<=2[/itex] implies [itex]Bell <= (4/\eta) - 2[/itex]
suppose that a detector has a probability [itex]\eta[/itex] to detect a particle
we have four possibilities when a pair is created:
1) left and right not detected
2) left detected but not right
3) right detected but not left
4) left and right detected
the probabilities are:
[tex](1 - \eta)^2[/tex]
[tex]\eta (1 - \eta)[/tex]
[tex](1 - \eta)\eta[/tex]
[tex]\eta^2[/tex]
the observer cannot count the first case events
We have 2) + 3) + 4) = [itex]1 - (1 - \eta)^2) = (2 - \eta) \eta[/itex]
the conditional probability of 4) knowing that 1) did not occur is
[tex]\frac{\eta^2 }{ (2 - \eta) \eta} = \eta/(2 - \eta)[/tex]
the expected value Bell = [itex]\eta/(2 - \eta)<=2[/itex] implies [itex]Bell <= (4/\eta) - 2[/itex]
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