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This question already has an answer here:

While obtaining the density of states of photons one multiple it with 2 for two polarization states, My question is that the whole calculation is based on quantum phenomena of light so why one uses wave phenomena (Polarization) in here.

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marked as duplicate by John Rennie, Kyle Kanos, AccidentalFourierTransform, Jon Custer, sammy gerbil Jan 5 '17 at 20:55

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  • $\begingroup$ polarization is not only a classical phenomenon. en.wikipedia.org/wiki/Photon_polarization after all we call it a "wave function". It is all about probabilities at the quantum mechanical frame $\endgroup$ – anna v Jan 5 '17 at 7:09
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Polarization is not more "classical" than frequency or wavelength is. These characteristics are not fundamentally impacted when quantizing. Every EM wave can be decomposed in two waves, one that has left-hand circular polarization and one that has a righ-hand circular polarization. When you quantize a classical wave, you quantize the amplitude of the wave, but you don't touch the polarization (see for example https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field ), just like you don't touch frequency or wavelength.

It turns out that polarization can be associated with the spin of the photon, however there is a one to one relationship between polarization states and spin states, so basically they are both describing the same physical phenomenon.

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