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 [itex]\lambda[/itex] determines the outcomes of all measurements, but that is a mathematical conclusion.
You might start assuming that
- [itex]P_A(a,\lambda) =[/itex] the probability of Alice getting spin up, given that the hidden variable has value [itex]\lambda[/itex], and that the orientation of Alice's detector is [itex]a[/itex]
- [itex]P_B(b,\lambda) =[/itex] the probability of Bob getting spin up, given that the hidden variable has value [itex]\lambda[/itex], and that the orientation of Bob's detector is [itex]b[/itex].
If we assume that every particle is detected, and that the spin is either spin-up or spin-down, then
- [itex]1 - P_A(a,\lambda) =[/itex] the probability of Alice getting spin down, given that the hidden variable has value [itex]\lambda[/itex], and that the orientation of Alice's detector is [itex]a[/itex]
- [itex]1 - P_B(b,\lambda) =[/itex] the probability of Bob getting spin down, given that the hidden variable has value [itex]\lambda[/itex], and that the orientation of Bob's detector is [itex]b[/itex].
But in the case of the anti-correlated spin-1/2 twin pairs, you know that: If Alice measures spin-up at angle [itex]a[/itex], then Bob will definitely not measure spin-down at that angle. So there is zero probability that they both measure spin-up at angle [itex]a[/itex]. That implies:
(1) [itex]P_A(a,\lambda) \cdot P_B(a, \lambda) = 0[/itex]
But also, if Alice measures spin-down, then Bob definitely will NOT measure spin-down. That implies:
(2) [itex](1-P_A(a,\lambda)) \cdot (1 - P_B(a, \lambda)) = 0[/itex]
Together, (1) and (2) imply that
[itex]P_A(a,\lambda) = 0\ \&\ P_B(a,\lambda) = 1[/itex]
or [itex]P_A(a,\lambda) = 1\ \&\ P_B(a,\lambda) = 0[/itex]
That implies that the spin Alice measures for angle [itex]a[/itex] is completely determined by [itex]\lambda[/itex], and similarly for Bob.