# Why does a mirror reverse polarization of circularly polarised light?

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

• What about from reflector to glass? Jul 3, 2014 at 0:21
• Yes, both a polished metal mirror and a glass mirror with metal backing ('reflector through glass') both reverse the polarisation of circularly polarised light. The simple test is to look through the circular polariser at either type of mirror; the reflection of the circular polariser appears completely opaque (ie. you cannot see your eye in reflection) Jul 3, 2014 at 2:50

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.

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.

• Thanks Rob, I'll go to the library and get a copy of that paper later today. I think this explanation should be added to the Wikipedia entry. Jul 3, 2014 at 2:53
• I went to add it and someone had beaten me to it. :-)
– rob
Jul 3, 2014 at 15:43

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.

• Or, indeed, just do the right-hand rule in a mirror. Apr 10, 2019 at 16:19

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.

• this means that the reflected electric field in the laboratory frame (the one used for the incident beam) is (1,-i,0) exp(-i(kz+wt)) ? I am getting confused with you introducing two bases with the same noations. Apr 20, 2023 at 4:45
• @yfs Yes, for the case of right circularly polarized light (as defined in the post above) incident upon a mirror, the reflected beam will have an electric field of $\mathbb{R}(1,-\mathrm{i},0)\exp(-\mathrm{i}(kz+\omega t))$, written in the same basis as the incident electric field. However, the jones vector is, by convention, written in a basis with propagation along $+z$, not $-z$. The notation is indeed initially confusing, but it's the convention. Apr 24, 2023 at 16:32

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.