Photon polarization as a two-state system, what is the Hamiltonian?

Question

In chapter III.11-4 of the Feynman lectures, he describes the polarization of a photon (with its momentum in the z direction) as a two-state system with the base states $\{|x\rangle,|y\rangle\}$ or $\{|R\rangle,|L\rangle\}$:

$$ |R\rangle = \frac{1}{\sqrt{2}}(|x\rangle+i|y\rangle)\\ |L\rangle = \frac{1}{\sqrt{2}}(|x\rangle-i|y\rangle) $$

Am I correct that both of these states have the same (definite) energy, the energy $E=\hbar c/\lambda$ of the photon, so the Hamiltonian has a degenerate eigenspace of dimension two?

If that's correct, then all states $|x\rangle,|y\rangle, |R\rangle,|L\rangle$ are stationary (do not change in time), so even in the right-hand circular polarization state $|R\rangle$ there is no rotation of the polarization.

So the polarization must come from interference effects between multiple photons, is that correct?

Or aren't the polarization base states Hamiltonian eigenstates?

Answer 1

so the Hamiltonian has a degenerate eigenspace of dimension two?

Yes.

If that's correct, then all states $|x\rangle,|y\rangle, |R\rangle,|L\rangle$ are stationary (do not change in time),

Yes.

so even in the right-hand circular polarization state $|R\rangle$ there is no rotation of the polarization.

Yes.

---

So the polarization must come from interference effects between multiple photons, is that correct?

That is completely muddled. What makes you think that? The hamiltonian is constant over this space, so every polarization is an eigenstate. Every single photon can have any polarization it wants - linear in any direction, circular, or elliptical - and it does not need other photons to make any linear polarization.

Superposition states like $|R⟩=\tfrac{1}{\sqrt{2}}(|x⟩+i|y⟩)$ are always states of a single particle unless explicitly indicated. This doesn't mean that "a right-circular photon is a mixture of $x$- and $y$-linear photons", which would lead to a pretty paradox since those polarizations are also superpositions of of right- and left-circular states (not photons):

\begin{align} \newcommand{\ket}[1]{|#1⟩} \ket{x}&=\frac{1}{\sqrt{2}}\,(\ket{R}+\ket{L}),\\ \ket{y}&=-\frac{i}{\sqrt{2}}\,(\ket{R}-\ket{L}). \end{align}