Graduate Photon Spin and Polarization filters

[PavanKumar]

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?

 

[Geofleur]

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.

 

[vanhees71]

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

 

[Geofleur]

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!

 

[George Jones]

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).

 

[vanhees71]

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 $h \in \{-1,1\}$.