- #36
zonde
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YesDrChinese said:Fine, an example for us to work with. A few questions about the terminology:
1. Do unentangled photons have POL, PH hidden variables too? If so, this provides additional testable constraints. You imply the answer is yes, but I want to be sure.
The same as in Malus law I'=I*cos^2(theta). It's angle between polarization axis of photons and polarization axis of polarizer.DrChinese said:2. I get POL. But you say that PH (phase) is "distributed" according to the function abs(sin(2theta)). What is theta here?
abs(sin(2theta)) is just derivative of cos^2(theta) with dropped sign.
Hmm, not sure how to address this question. Let's try it this way.DrChinese said:3. Then you say that the probability a photon has a given value of the hidden variables is abs(sin(2*ph)). Can you give an example? The indicated hidden variable is PH and I do not believe this function sums to 100% across a suitable range (as it would need to).
If we take out of the sample (of one side) all photons with the same polarization they have to follow Malus law so that when we use "ph" to calculate "passing" and "absorbing" intervals of photons they together should produce characteristic cos^2(alpha) picture (with alpha being angle between polarization axis of photons and polarization axis of polarizer)
About the sum. Integral of function abs(sin(2*ph)) in interval 0 - Pi/2 is 1. So if you take interval 0 - 2Pi then it's 4. But isn't this question about sum being 100% just the matter of normalization? If I give description of distribution I define that integral of this function is whole sample.
But maybe you have on mind the thing about relating two distributions in non-symmetric fashion (when we consider Alice's and Bob's samples together)? In that case yes they do not mach for 100%. So there are always unpaired photons with given setup of photon source.