Why does a mirror reverse polarization of circularly polarised light?

Question

A glass mirror (with metal backing layer) will reverse the polarisation of circularly polarised light upon reflection.

A polished piece of metal will also reverse the polarisation of circularly polarised light upon reflection. (I have tested and confirmed this for myself).

wikipedia states the reason a mirror will reverse the polarisation of circularly polarised light is:

...[A]s a result of the interaction of the electromagnetic field with the conducting surface of the mirror, both orthogonal components are effectively shifted by one half of a wavelength.

However, my understanding of mirrors is that only a polished piece of metal will phase shift a wavelength by half a wavelength, whereas a glass mirror (with metal backing layer) will not produce a phase shift. For example wikipedia which states:

According to Fresnel equations there is only a phase shift if n2 > n1 (n = refractive index). This is the case in the transition of air to reflector, but not from glass to reflector

Answer 1

While rob is correct about the quantum mechanical picture I think that this case is at least as easy to understand as in the classical description.

Classically circular polarization can be described in terms of a time-varying linear polarization, so let's just look at two points on a wave.

I'm going to chose a beam in the $+z$ direction to examine two points on the wave: one where the polarization currently points along $+\hat{x} - \epsilon \hat{y}$, and a very short time later where the polarization is in the $+\hat{x} + \epsilon \hat{y}$ direction. The wave has right-handed circular polarization.

Now we let the beam bounce off a mirror in the $x\text{--}y$ plane. This reverses the direction of propagation but leaves the time-order in which are two points of interest pass any given point unaffected. A little thought suffices to show that the reflected wave has left-handed circular polarization.

Answer 2

From a quantum-mechanical perspective, circularly-polarized light is made of photons with their spins parallel to their momentum. The mirror reverses the photons' momentum but does not affect their spins, so the dot product $\sigma\cdot p$ changes sign.

Both the quantum and classical approaches are examined in Beth's 1936 measurement of the angular momentum of light, one of my favorite underrated classic papers.

Answer 3

Another way to understand how reflection of circularly polarized light causes reversal of the polarization direction is to hold a screw perpendicular to a mirror and look at its reflection, and then think about what you see. The field vectors in circularly polarized light trace out a corkscrew.

Answer 4

To supplement the other answers, I'd like to add a note about how this emerges mathematically using notation and conventions from Jones calculus.

For a polarized (linear or circular) ray of light propagating along the $+z$ axis, the electric field can be written as the real part of oscillating complex phases: $$\begin{bmatrix}E_x(z,t)\\E_y(z,t)\\E_z(z,t)\end{bmatrix}=\Re\left(\,\begin{bmatrix}E_{0x}\mathrm{e}^{\mathrm{i}\phi_x}\\E_{0y}\mathrm{e}^{\mathrm{i}\phi_y}\\0\end{bmatrix}\mathrm{e}^{\mathrm{i}(kz-\omega t)}\right).$$ The Jones vector is $\begin{bmatrix}E_{0x}\mathrm{e}^{\mathrm{i}\phi_x}\\E_{0y}\mathrm{e}^{\mathrm{i}\phi_y}\end{bmatrix},$ which is essentially the electric field in the two nontrivial spatial dimensions, with the time-dependence and spatial oscillation stripped out. This vector is completely general for all polarized light states (it cannot describe unpolarized light). For instance, $x$- and $y$- linearly polarized light are expressed with Jones vectors $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$ respectively. Right- and left-circularly polarized light are given (upto normalization and amplitude coefficients) by $\begin{bmatrix}1\\-\mathrm{i}\end{bmatrix}$ and $\begin{bmatrix}1\\+\mathrm{i}\end{bmatrix}$ respectively.

For the case of reflection, we will consider right-circularly polarized light as an example. The electric field of the beam incident upon the mirror is $$\Re\left(\,\begin{bmatrix}1\\-\mathrm{i}\\0\end{bmatrix}\mathrm{e}^{\mathrm{i}(kz-\omega t)}\right).$$ Upon reflection, the polarization state stays the same but the wavenumber is changed $k\to -k$. However, the Jones vector is defined in a right-handed coordinate system in which the ray propagates along $+z$. So, we'll need to transform the old Jones vector such that it's in a basis where $+z$ is reversed. The options are a 180° rotation around the $y$ or $x$ axes (remember mirror operations destroy handedness), which would transform the Jones vector from $\begin{bmatrix}1\\-\mathrm{i}\end{bmatrix}$ to either $\begin{bmatrix}-1\\-\mathrm{i}\end{bmatrix}$ or $\begin{bmatrix}1\\+\mathrm{i}\end{bmatrix}$ respectively. These are the same upto a minus sign; they are both the Jones vector for left-circularly polarized light. The same procedure can be used for the case of left-circularly polarized light incident on a mirror.

Answer 5

The reversal of circular polarization on reflection can be understood, using the law of conservation of angular momentum. Circularly polarized light carries angular momentum, which is parallel to the wave vector for right-hand circular polarization, and antiparallel for left-hand one. For normal incidence, reflection reverses the light wave vector, but not its angular momentum, because no angular momentum is transferred to the ideal mirror. Thus upon reflection the wave vector changes sign, whereas the angular momentum vector does not, hence circular polarization is reversed. However, circular polarization reversal occurs only for normal incidence. For example, for grazing incidence, by the same reasoning of angular momentum conservation, upon reflection the circular polarization maintained, i.e., no reversal of circular polarization occurs.