- #36
mikeyork
- 323
- 55
stevendaryl: forgive me if I have misunderstood. But don't we already know what ##\lambda## is? Isn't it the eigenvalue of the composite state? So if ##A,B## are individual spins, then ##\lambda## is the composite spin. And your ##\psi(A,B;\alpha,\beta,\lambda)## are essentially Clebsch-Gordon coefficients -- apart from the rotation which takes the orientation of one detector into the other.
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