/sci/ - Science & Math

>>10779247
You know, this Anon >>10779344 is right, the longer the thread goes on, somebody sometimes has to refute your bullshit, might as well be me. Why do you think this matter is settled? It can be shown trivially, that quantum mechanics, indeed, does not satisfy Bell's inequalities. I won't bother to explain all the details, since I'm sure you've seen the inner works if you claim to refute them, but try to at least see the gist of it. Assume a wave function of the form

[math] |\psi \rangle = \frac{1}{\sqrt{2}}(a^{+}_{1x}a^{+}_{2y} - a^{+}_{1y} a^{+}_{2x})|\emptyset \rangle[/math]

as I'm sure you know, this represents a simple bi-photon wave function of two photons travelling into opposite directions, with polarisations (or spin, if you wish) in the perpendicular directions, as given in the treatment of the quantization of the free electromagnetic field. A trival state arising from the annihilation of a electron and positron, for example.

Both spins in both directions have the same probability weights, as can be seen if you evaluate the average number of photons with a given polarisation on each side

[math] \overline{n_{1x}} = \langle \psi | a^{+}_{1x}a_{1x} | \psi \rangle = \frac{1}{2} = \overline{n_{1y}} [/math]

If we place a pair of linear polarizers, each being rotated to polarize in arbitrary angles [math] \vartheta _{1} , \vartheta _{2} [/math] then we can through the calculation of the intensity correlation functions derive the fhe following auxillary function (nowhere do we assume any orthogonality whatsoever)

[math] C(\vartheta _{1} , \vartheta _{2}) = \langle A_{1} A_{2} \rangle = P_{++} + P_{--} - P_{+-} - P_{-+} = -\cos(2\vartheta _{1} - 2\vartheta _{2}) [/math]

Where P functions represent the probabilty that detectors on both side will detect 1) 1 photon on each side, 2) 1 on left, zero on right, 3) zero on left, 1 on right and lastly 4) None on each side.

cont.