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

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?

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}

• What puzzles me is the fact that the field caused by the circularly polarized photon is not constant (along the photon's trajecory). It revolves around the direction of motion of the photon. How can it do that if the polarization state of the photon in stationary?
– Bass
Commented Oct 31, 2015 at 14:00
• Same thing does for a linearly polarised photon - the field changes direction over space and over time. They're in the same standing in that respect. Commented Oct 31, 2015 at 18:10
• (That said, it's not quite right to think in those terms. Photon number and the phase of the field are conjugate variables, which is essentially the energy-time uncertainty principle. A state with a well-defined photon number has a completely undefined phase. Similarly, there are states with (fairly) well defined phase, but they're not energy eigenstates and they evolve in time.) Commented Oct 31, 2015 at 18:14
• Wikipedia on light polarization: The oscillation of these fields may be in a single direction (linear polarization), or the field may rotate at the optical frequency (circular or elliptical polarization). How does the field change its direction if it's linearly polarized?
– Bass
Commented Nov 1, 2015 at 14:27
• It changes sign - along +x, say, and then along -x half a period later. This is also a tone dependence and it is as paradoxical as that of linear polarizations. Commented Nov 1, 2015 at 18:28