naima said:
Thank you for the link Truecrimson.
I read in page 3 that the locality property
##P(a,b,\lambda) = P(a,\lambda) P(b,\lambda) ## leads the Bell's inequalities without requiring that the same lambda give outputs for noncommuting observations.
I understand now why Shyan could not give me a reference with this property of hidden variables. (nobody here reacted)
It's not necessary to ASSUME that \lambda determines the outcomes of all measurements, but that is a mathematical conclusion.
You might start assuming that
- P_A(a,\lambda) = the probability of Alice getting spin up, given that the hidden variable has value \lambda, and that the orientation of Alice's detector is a
- P_B(b,\lambda) = the probability of Bob getting spin up, given that the hidden variable has value \lambda, and that the orientation of Bob's detector is b.
If we assume that every particle is detected, and that the spin is either spin-up or spin-down, then
- 1 - P_A(a,\lambda) = the probability of Alice getting spin down, given that the hidden variable has value \lambda, and that the orientation of Alice's detector is a
- 1 - P_B(b,\lambda) = the probability of Bob getting spin down, given that the hidden variable has value \lambda, and that the orientation of Bob's detector is b.
But in the case of the anti-correlated spin-1/2 twin pairs, you know that: If Alice measures spin-up at angle a, then Bob will definitely not measure spin-down at that angle. So there is zero probability that they both measure spin-up at angle a. That implies:
(1) P_A(a,\lambda) \cdot P_B(a, \lambda) = 0
But also, if Alice measures spin-down, then Bob definitely will NOT measure spin-down. That implies:
(2) (1-P_A(a,\lambda)) \cdot (1 - P_B(a, \lambda)) = 0
Together, (1) and (2) imply that
P_A(a,\lambda) = 0\ \&\ P_B(a,\lambda) = 1
or P_A(a,\lambda) = 1\ \&\ P_B(a,\lambda) = 0
That implies that the spin Alice measures for angle a is completely determined by \lambda, and similarly for Bob.