Should/can one describe the angular momentum carried by a Photon as a 2 dimensional Hilbert space analogous to an electron spin?

Question

I am currently trying to wrap my head around electron photon interaction processes in Atoms, as for example described here. In particular I am interested in the dependency of the angle between photon movement-direction and angular momentum component-quantization axis. I am a bit confused by all the different stuff being mentioned at different places in that context in particular about the mixture of classical electrodynamic approaches with non relativistic Qm-approaches, which both feel like they should be insufficient for the task in that case. So thinking a bit about it I came up with the following pretty approach and would like to know if it has anything to do with reality.

Naive application of my Intuitions spotting an analogy So lets say we have an electron bound in a coulomb potential with energy quantum number $n$ absolute angular momentum number $l$ and z-direction angular momentum number $m$. Looking at photon interactions lets start with the most simple case of a photon with circular polarization $\sigma_+$ or $\sigma_-$ hitting our system from the direction of the z-axis. I then would expect $m$ to change by $+1$ for $\sigma_+$ or $-1$ for $\sigma_-$ and the $l$ and $n$ quantum-numbers accordingly depended on the sign of $m$ before the absorption. That is also how its described here.

Now as I understand it the spin-state of a photon can be described as a Vector in a 2-dimensional Hilbert space spanned by the states $\sigma_+$ or $\sigma_-$ (, because its a spin 1-particle and the 0 spin-component-state is forbidden for some QED reasons, that I never understood). What we would like to do now is to map the photon spins onto angular momenta. Since we know the photon-spin-space is isomorphic to the electron-spin-space and we already identified 2 states in both spaces that correspond to carrying sharply defined angular momentum in the direction of the $z-$axis, it would seem rather intuitive to just map those spaces onto each other by mapping $\sigma_+$ and $\sigma_-$ to the spin states in $z$-direction, which would be the eigenstates of Pauli-matrix $ \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}. $ After doing that I would assume that the eigenstates of angular momentum in $x-$direction of the photon would be the ones generated by throwing that Isomoprhism on the spin eigenstates in $x-$direction of the electron, so the eingenstates of $ \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} $, which would lead to $\frac{1}{\sqrt{2}} (\sigma_+ + \sigma_-)$ and $\frac{1}{\sqrt{2}} (\sigma_+ - \sigma_-)$, which look kind of like linear polarized light-states.

So my concrete question would be: "Are these states $\frac{1}{\sqrt{2}} (\sigma_+ + \sigma_-)$ and $\frac{1}{\sqrt{2}} (\sigma_+ - \sigma_-)$ actually photon states, that carry sharply defined angular momentum in a direction perpendicular to the movement direction of the photon or is this naive mapping just nonsense?" and "Do these states actually correspond to linearly polarized light or is this classical analogy too far fetched?"

Answer 1

No, the photon polarization space is not the same as the spin-1/2 representation with the Pauli matrices as the generators of rotation, though it is also two-dimensional.

By Wigner's classification (see also this answer of mine), the relativistic theory of photons results in the identity component of the Poincaré group acting on photons in terms of a one-dimensional irreducible representation labeled by the helicity $h$. Parity flips this to $-h$, so the massless particles transform in a two-dimensional representation $h\oplus -h$ ($h=1$ for a photon).

More concretely: We have two independent polarization states that span the polarization space, the left-circular polarization $\lvert L\rangle$ and the right-circular polarization $\lvert R\rangle$ (or clockwise-circular and anti-clockwise-circular, whatever you want to call them). These states transform just by a phase under whatever rotation you apply to them, $\lvert L\rangle \mapsto \mathrm{e}^{\mathrm{i}\phi h}\lvert L\rangle, \lvert R\rangle \mapsto \mathrm{e}^{-\mathrm{i}\phi h}\lvert R\rangle$, where $\phi$ is the angle of a rotation in the plane perpendicular to the photon's direction of motion. You can form from this the linear polarizations $\lvert L\rangle \pm \lvert R\rangle$, which transform by sines and cosines of $\phi$, i.e. transform under 2d rotations in that plane as you would expect vectors to.

Now, you probably ask yourself what happens for rotations that are not in that perpendicular plane. Well, those change the direction of motion of the photon, and you can't really say how they work in this 2-dimensional space because this is only a non-relativistic slice of the "true" photon state space, which forms an infinite-dimensional irreducible representation of the Poincaré group. The full rotation group (as a subgroup of the Poincaré group) acts on that space, not on the 2-dimensional polarization space described above.

Photons just are not amenable to a fully non-relativistic description because they lack a rest frame - for massive particles, the full 3d rotation group leaves the momentum in the rest frame invariant, and there is no issue with recovering the usual description in terms of non-relativistic spin-1/2 in that case, but for massless particles, the little group is only the 2d Euclidean group including the rotations in the plane perpendicular to the direction of movement we saw above. If you want to describe the action of the full rotation group, you have to go to the relativistic description in terms of the infinite-dimensional representation of the Poincaré group.