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Bell's inequality when efficiency < 1

  1. Aug 21, 2013 #1

    naima

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    Hi PF

    Somebody gave me this link
    could you help me to understand why (9) has to be inserted in (7) when the efficiency [tex]\eta < 1?[/tex]
     
  2. jcsd
  3. Aug 21, 2013 #2

    naima

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    I think that a part of the answer is here
    but i do not understand eq (6):
    [tex] <ab>_{coinc} = <ab>/(\eta_A \eta_B)[/tex]
    why do this conditional expectation must grow?
     
  4. Aug 28, 2013 #3

    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]
     
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