When talking about a photon emitted by a laser device, what is the difference between phase and polarization?
Is it redundant to specify both the polarization and the wavefunction's phase?
Does a phase modulator affect both?
Context: This question came up when I was studying QKD and communication implementations which mention rotations of a single emitted photon's phase or polarization by an optical device.
We can write a the electric field of an electromagnetic plane wave polarized in the $x$ direction and propagating in the $z$ direction as \begin{equation} \vec{E}=E_0 \cos\left(\omega t - \frac{\omega z}{c} + \varphi_0\right) \hat{e}_x \end{equation} In this example,
Clearly these are different quantities.
We can also write superpositions of waves of two different linear polarizations. A circularly polarized wave corresponds to two plane waves with orthogonal polarizations, offset by $\pi/2$ in phase, and where each component has the same amplitude. More concretely, we can write the electric field of a circularly polarized electric field propagating in the $z$ direction as \begin{equation} \vec{E}=E_0 \cos\left(\omega t - \frac{\omega z}{c} + \phi_0 \right)\hat{e}_x + E_0 \sin\left(\omega t - \frac{\omega z}{c} + \phi_0\right) \hat{e}_y \end{equation} The polarization vector $\hat{e}_E$ (unit vector in the direction of the electric field) at $z=0$ is \begin{eqnarray} \hat{e}_E &=& \cos\left(\omega t + \phi_0 \right)\hat{e}_x + \sin\left(\omega t + \phi_0\right) \hat{e}_y \\ &=& R[\omega t + \varphi_0] \hat{e}_x \end{eqnarray} where $R[\theta]$ is a rotation matrix that rotates by an angle $\theta$ in the $xy$ plane.
You can see that the direction of the polarization vector rotates around with a frequency $\omega$, the same as the frequency of the light. The phase of the polarization angle is the same as the phase of the light (at $z=0$), and the initial phase of the polarization angle is the same as the initial phase of the sinusoid (at $z=0$) and is equal to $\phi_0$.
If you think of a photon as an electromagnetic wave packet, then the polarization generally refers to the direction of the E field in the wave. In a linearly polarized wave, all of the E fields lie in the same plane. It is also possible to have E field components in two orthogonal directions. If the two are “in phase”, you still have linear polarization. If not, you get “elliptical” polarization in which the field appears to rotate as it arrives at a point.
No, specifying both is most certainly not redundant. I'll borrow the notation of $\S4.4.1$ here to explain why.
Polarization is easiest to understand in the Lorentz gauge $\partial_\nu A^\nu=0$, for which the sourceless Maxwell's equations are $\square A_\mu=0$. This looks very similar to the massless Klein-Gordon equation, but with a spacetime index, so the most general classical solution may be written as$$A_\mu=\int\frac{d^3\mathbf{k}}{(2\pi)^32\omega_\mathbf{k}}\left(a_\mu(k)e^{-ik^\nu x_\nu}+a^\ast_\mu(k)e^{ik^\nu x_\nu}\right).$$That $A_\mu$ is real imposes $a_\mu^\ast=a_\mu$ (quantization promotes $a_\mu$ to an operator, replacing $^\ast$ with $^\dagger$), i.e.$$A_\mu=\int d^3\tilde{k}\:2a_\mu(k)\cos(kx).$$This is a linear combination over one polarization $a_\mu$ per four-frequency, of phase $k^\nu x_\nu$. However, the polarization vector isn't as arbitrary it seems. Our gauge choice is the constraint $k^\mu a_\mu=0$, and its invariance under $a_\mu\mapsto a_\mu-ik_\mu\lambda$ removes another degree of freedom (since $k^\mu k_\mu=0$ is the photon's squared rest mass), so there are two left for polarization. (If the photon had mass, the last argument would fail, so there would be three polarization DOFs.) See Eqs. (4.107)-(4.115) for further details on one way to make these two DOFs more explicit in our formalism.
So far, I've only discussed fields. Classically, this has implications for waves; but the quantum insight is that this applies also to individual photons, each having a four-frequency that appears in $a_\mu(k)\cos(kx)$. Note in particular only phase varies over spacetime when the photon propagates, but its polarization depends only on the four-frequency, which doesn't change so easily. (It can, but then we get into the debate over whether we're dealing with "the same" photons or new ones.)
