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Let's imagine we have Right Circularly Polarized Light propagating in the $+\hat{z}$ direction toward a perfectly reflecting mirror. Before reflection, the light has the electric field:

$$\vec{E}(z)=E_0\cos(kz-\omega t)\hat{x}+E_0\sin(kz-\omega t)\hat{y}.$$

After reflection, it is well known that the reflected light will now be Left Circularly Polarized, but what will the expression be for the electric field? The light now travels in the $-\hat{z}$ direction.

The electric field for Left Circularly Polarized light is $$\vec{E}(z)=E_0\cos(kz-\omega t)\hat{x}-E_0\sin(kz-\omega t)\hat{y},$$ but this does not take into account that the direction of propagation has changed.

What is the proper electric field expression after reflection, and why (i.e. how did you come up with the expression)?

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For a perfectly reflecting mirror, the reflected field of an incident plane wave is obtained by changing the sign of $k$ and of the amplitude. We can analyze each component separately: $$E_x^i = E_0\cos(kz-\omega t) \implies E_x^r = -E_0\cos(-kz-\omega t) = -E_0\cos(kz+\omega t) $$ $$E_y^i = E_0\sin(kz-\omega t) \implies E_y^r = -E_0\sin(-kz-\omega t) = E_0\sin(kz+\omega t) $$

Hence, the reflected electric field is $\vec{E}{}^r = -E_0\cos(kz+\omega t) \hat{x} + E_0\sin(kz+\omega t)\hat{y} $. The argument of the $\cos$ and $\sin$ functions implies that the wave now propagates in the opposite direction compared with the incident one. The reflected wave has left circular polarization because at any given value of $z$, if we let the field evolve in time, the arrow describes a counter clock-wise circular motion, looked as though the new (reflected) propagation direction pointed at you. This last point is the key to realize that the polarization has been reversed.

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  • $\begingroup$ But your solution is not a wave, no? $\endgroup$
    – garyp
    Commented Oct 20, 2023 at 0:35
  • $\begingroup$ @garyp I'm not sure what you mean by "not a wave". It is definitely a solution of the wave equation. $\endgroup$
    – Diego
    Commented Oct 20, 2023 at 13:25
  • $\begingroup$ "...is obtained by changing the sign of $k$ and of the amplitude". Why did you change the sign of both the amplitude and $k$? Can this phenomena be described with a more rigorous formalism (like a Jones matrix) or an explicit phase shift applied to each component? Is there a way of explaining this besides "we flip the signs because we flip the signs". Also, your final expression does not exactly match the traditionally known electric field expression for LCP light. Why the difference? $\endgroup$
    – Rydberg
    Commented Oct 20, 2023 at 14:41
  • $\begingroup$ The "traditional" expression you mention is for LCP propagating towards positive $z$, whereas mine propagates towards negative $z$. You have to change the sign of k, because the sign of k determines the direction of propagation. Also, the flipping of the amplitudes is a result of applying the boundary condition that the total $\vec{E}=0$ at the boundary between the propagation medium and the perfect mirror. See page 3 of waves.utoronto.ca/prof/svhum/ece422/notes/17-reflrefr.pdf for more details (culminating in eq. (18)) $\endgroup$
    – Diego
    Commented Oct 20, 2023 at 16:06
  • $\begingroup$ @Rydberg Any function $f(\vec k\cdot\vec x-\omega t)$ is a solution to the wave equation which propagates in the direction of $\vec k$ with speed $\omega/|k|$. You can model the sign change as a phase difference; see my answer. $\endgroup$
    – rob ♦
    Commented Oct 20, 2023 at 20:42
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The electric field is continuous everywhere in space, except at the locations of point charges, line charges, or surface charge densities. A conductor may maintain a charge density on its surface in order to enforce zero electric field inside the conductor, but that affects only the normal component of the field. For propagating light, you care about the transverse field, which varies continuously from the sum of the incident plus reflected fields and zero.

In a real conductor the field goes to zero exponentially, instead of instantly, with a "skin depth" which depends on the frequency and the conductivity.

The free charges in the conductor therefore conspire to produce an electric field at the surface which is equal and opposite to the incident light. This new conductor-created field then propagates away as the reflected light.

Remember that you can change the sign of the coefficients by introducing a phase offset to your angle functions, like $\cos(kz-\omega t+\delta)$. To demonstrate that the reflected wave has linear momentum going the other way, but unchanged angular momentum, also apply the continuity requirements to the magnetic field.

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  • $\begingroup$ What would the phase shift be in this case? I think it would be $\pi$, right? If you impose a $\pi$ phase shift, how does that change the sign of BOTH the $cos$ and $k$ for example? $\endgroup$
    – Rydberg
    Commented Oct 20, 2023 at 20:51
  • $\begingroup$ If the conducting surface is at $z=z_0$, you must have $(kz_0)-(-kz_0+\delta)=(2n+1)\pi$, for integer $n$. The phase shift $\delta$ is (the way I've written it here) the phase shift at the origin of the coordinate system. If the incident wave went instead like $\cos\left(\vec k\cdot(\vec z-\vec z_0)-\omega t\right)$, the reflected wave would go like $\cos\left(-\vec k\cdot(\vec z-\vec z_0)-\omega t+\pi\right)$. $\endgroup$
    – rob ♦
    Commented Oct 21, 2023 at 3:52

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