# Bell's inequality when efficiency < 1

1. Aug 21, 2013

### naima

Hi PF

could you help me to understand why (9) has to be inserted in (7) when the efficiency $$\eta < 1?$$

2. Aug 21, 2013

### naima

I think that a part of the answer is here
but i do not understand eq (6):
$$<ab>_{coinc} = <ab>/(\eta_A \eta_B)$$
why do this conditional expectation must grow?

3. Aug 28, 2013

### naima

I found the algebric origin of the fact that in Bell's theorem we replace 2 by $4/\eta - 2$
suppose that a detector has a probability $\eta$ 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:
$$(1 - \eta)^2$$
$$\eta (1 - \eta)$$
$$(1 - \eta)\eta$$
$$\eta^2$$
the observer cannot count the first case events
We have 2) + 3) + 4) = $1 - (1 - \eta)^2) = (2 - \eta) \eta$
the conditional probability of 4) knowing that 1) did not occur is
$$\frac{\eta^2 }{ (2 - \eta) \eta} = \eta/(2 - \eta)$$
the expected value Bell = $\eta/(2 - \eta)<=2$ implies $Bell <= (4/\eta) - 2$