Photon Spin and Polarization filters

  • #1
I understand how polarization can be explained using EM waves. However, I am unable to understand how to explain how polarization filters work when we use the concept of photon spins. Can someone help me with that?
 

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  • #2
Geofleur
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There's a really good video over at 3Blue1Brown that explains the quantum mechanics of photons passing through polarization filters:


They don't mention it, but when light is circularly polarized, each photon has a spin angular momentum of ##+\hbar## or ##-\hbar##, depending on whether the light is left or right circularly polarized.
 
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  • #3
vanhees71
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Be careful! A photon has no spin in the usual sense. Massless quanta have to be treated separately from massive ones. That's why in standard QFT massless particles have only 0 (for scalar and pseudo scalar fields) or 2 (for fields with spin ##\geq 1/2##) spin-like degrees of freedom (which most intuitively can be chosen in terms of the single-free-particle momentum-helicity basis ##|\vec{p},h \rangle## with ##h=\pm s## and ##\vec{p}## with the dispersion relation ##p \cdot p=0##, i.e., ##E=|\vec{p}|##).

For a first qualitative explanation of polaroids (absorptive polarization filters), see

https://en.wikipedia.org/wiki/Polarizer
 
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Geofleur
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Does that mean, instead of saying that the photon has spin 1, we should instead say that it has helicity 1? I didn't realize that the representations of the Poincare group for massless particles cannot be labeled by spin!
 
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George Jones
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Does that mean, instead of saying that the photon has spin 1, we should instead say that it has helicity 1? I didn't realize that the representations of the Poincare group for massless particles cannot be labeled by spin!
This is because the little group of a massive particle, SU(2), is different than the little group of a photon, E(2).
 
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  • #6
vanhees71
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Does that mean, instead of saying that the photon has spin 1, we should instead say that it has helicity 1? I didn't realize that the representations of the Poincare group for massless particles cannot be labeled by spin!
The usual terminology is to say that the photon has spin 1 (in the sense of ##\vec{J}^2## has the lowest eigenvalue ##1 \cdot (1+1)=2##). Since the photon is massless this implies that there are two spin-degrees in freedom. A natural choice for a single-photon basis is to take momentum eigenvectors and eigenvectors of the angular momentum component in direction of the photon's mopmentum, i.e., the helicity, and this helicity ##h \in \{-1,1\}##.
 

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