Alien8 said:
Solid curve is the prediction based on Malus' law. It's 1/2 from QM prediction. To get that I think you work out independent probabilities where Pa(+) = Pb(+) and Pab(+ and +) = Pa(+)Pb(+).
I think to check Malus' law all we need is to look at Alice and Bob's readings individually. Total number of "+" should be about the same as "-" no matter what angle settings is at either Alice or Bob's analyzer. I don't see why would that not be easy to verify, but I also don't see that would change anything because it says nothing about how the two readings are supposed to match against each other. In local theory there is no any connection between the two events so the rest is in the hands of probability theory, and because the two events are supposed to be independent we use the equation for independent probabilities: P(A and B) = P(A)P(B), which leads to that solid cure. Or something like that, I'm not quite sure how to work out the integral.
No, a local theory doesn't imply independence of the results, and it does not imply P(A and B) = P(A)P(B). The reason why not is that even though A can't influence B, and B can't influence A, there might be a third cause that influences both. That's what the "local hidden variables" idea is all about: whether the correlations can be explained by assuming that there is a cause (the hidden variable) that influences both measurements.
A locally realistic model based on Malus' law is this: assume that in the twin-photon version of EPR, two photons are created with the same random polarization angle \phi. If Alice's filter is at angle \alpha then she detects a photon with probability cos^2(\alpha - \phi). Similarly, if Bob's filter is at angle \beta, then he detects a photon with probability cos^2(\alpha - \phi). The correlation E(\alpha, \beta) would then be:
E(\alpha, \beta) = P_{++} + P_{--} - P_{+-} - P_{-+}
where P_{++} is the probability both Alice and Bob detect a photon, P_{+-} is the probability Alice detects one and Bob doesn't, etc.
For this model,
P_{++} = \frac{1}{\pi} \int d\phi cos^2(\alpha - \phi) cos^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}cos^2(\alpha - \beta)
P_{+-} = \frac{1}{\pi} \int d\phi cos^2(\alpha - \phi) sin^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}sin^2(\alpha - \beta)
P_{-+} = \frac{1}{\pi} \int d\phi sin^2(\alpha - \phi) cos^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}sin^2(\alpha - \beta)
P_{--} = \frac{1}{\pi} \int d\phi sin^2(\alpha - \phi) sin^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}cos^2(\alpha - \beta)
So
E(\alpha, \beta) = \frac{1}{2}(cos^2(\alpha - \beta) - sin^2(\alpha - \beta)) = \frac{1}{2} cos(2 (\alpha - \beta))
That's exactly 1/2 of the QM prediction.
