With circularly polarised waves, the effect can only be observed interferometrically, because the Faraday rotation simply becomes the imposition of different phase delays on the two circularly polarised components. To understand this, witness that the Faraday rotation on a general polarisation state expressed with linear polarisation state basis is:

$$\left(\begin{array}{c}x\\y\end{array}\right)\mapsto \left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)\left(\begin{array}{cc}x\\y\end{array}\right)$$

and then we transform to circular polarisation basis states $f_+,\,f_-$ by:

$$\left(\begin{array}{c}f_+\\f_-\end{array}\right)= \frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)\left(\begin{array}{cc}x\\y\end{array}\right)$$

so that our Faraday rotation through angle $\theta$ above becomes:

$$\left(\begin{array}{c}f_+\\f_-\end{array}\right)\mapsto \left(\begin{array}{cc}1&i\\1&-i\end{array}\right)\left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)\left(\begin{array}{cc}1&i\\1&-i\end{array}\right)^{-1}\left(\begin{array}{cc}f_+\\f_-\end{array}\right) = \left(\begin{array}{cc}e^{i\,\theta}&0\\0&e^{-i\,\theta}\end{array}\right)\left(\begin{array}{cc}f_+\\f_-\end{array}\right)$$

Indeed, this is the essence of the spin-1 particle: its eigenstates pick up a phase of $e^{i\,\theta}$ on rotation through angle $\theta$ (unlike an electron, which would pick up a phase of $e^{i\,\frac{\theta}{2}}$).

**Edit:** Your understanding expressed in your question seems to be right. In the above I give the Jones matrix, *i.e.* the linear transformation wrought on the vector of superposition weights, for the two cases where (1) linear polarisation states and (2) circular polarisation states are used as the state space basis: in the former case, the effect is a rotation of the plane of polarisation through $\theta$ radians, in the latter the right handed circular polarised component's phase is increased by $\theta$ radians, the left hand component's phase is decreased by $\theta$ radians. In all cases, this $\theta$ is given by $\theta = V\, B_\parallel \,L$ where $L$ is the crystal's length, $B_\parallel$ the magnetic induction component along the optical axis and $L$ the crystal's length.

**Edit 2**: Well, the results for circularly polarised loght follow from: (1) experimental measurements for linearly polarised light and (2) experimental confirmation of the system's linearity; once we have (2) then circular / linear description is simply a change of basis. Moreover, you *can* do the experiments for circularly polarised light, you're simply looking for changes in phase. So you can do the experiment interferometrically. Or, you can do an experiment as below with waveplates and the rotator.

The rotator works exactly the same with linearly polarised input light whether or not the waveplates are in place, confirming the linearity of the system. So it is acting on the circularly polarised light states exactly as my basis trasnformation equations above foretell. At a deeper level, the waveplate's action as a transformer to and from circularly polarised light can be checked by the following experiment in 1936 by Richard Beth:

Richard A. Beth, "Mechanical Detection and Measurement of the Angular Momentum of Light" *Phys.* *Rev.*** 50** July 15 1936

This isn't enough information to fully specify the state of the light, since you have yet to specify the relative strength between the two components. It seems, however, that you're thinking of light composed of a superposition of $x$- and $y$-polarized light, and that you want to have the ellipse's axes line up with the coordinate frame, but that constrains the relative phase between the $x$ and $y$ components to be $\pi/2$.

(If you *don't* do this, then you will have an ellipse with axes at some arbitrary angle, which introduces some gratuitous complications to the geometry, for essentially zero gain. If you do have two arbitrary orthogonal polarizations at some arbitrary relative phase, then the first thing to do is to extract a frame where the same light is decomposed as two orthogonal linear polarizations at a relative phase of $\pi/2$ $-$ a frame which *always* exists, and which can be found via the method in this answer of mine $-$ and then carry on with the analysis on that frame.)

Thus, the simplest nontrivial elliptical polarization has the form
$$
\mathbf E(t) = \frac{E_0}{\sqrt{1+\varepsilon^2}}\begin{pmatrix}\cos(\omega t) \\ \varepsilon \sin(\omega t)\end{pmatrix},
$$
where $\varepsilon$ is the (signed) ellipticity of the light. If you take this form, it is then easy to calculate the equivalent for Malus' law, by taking the inner product with the unit vector $\hat{\mathbf u}=(\cos(\theta),\sin(\theta))$, squaring, and time-averaging the result:
\begin{align}
I(\theta)
&=
\left<(\mathbf E(t) \cdot \hat{\mathbf u})^2\right>
\\&=
\frac{E_0^2}{1+\varepsilon^2}\left<\left(\cos(\theta)\cos(\omega t) + \varepsilon \sin(\theta)\sin(\omega t)\right)^2\right>
\\&=
\frac{E_0^2}{1+\varepsilon^2}\left<\left(
\cos^2(\theta)\cos^2(\omega t)
+ 2\varepsilon \sin(\theta)\cos(\theta)\sin(\omega t)\cos(\omega t)
+ \varepsilon^2 \sin^2(\theta)\sin^2(\omega t)
\right)\right>
\\&=
\frac12
\frac{E_0^2}{1+\varepsilon^2}\left(\cos^2(\theta) + \varepsilon^2 \sin^2(\theta)\right).
\end{align}
That's it, pretty much. You can rephrase it in a number of ways, and you can choose several other representations of the initial elliptical light, but they're all equivalent.

## Best Answer

There is no conventional choice, the Jones vector formalism must be attached to

right-handedcoordinate system.You do have a choice between the increasing and decreasing phase conventions, which will change signs all over the place, and can make calculations incorrect if you mix them up. Be very careful of the wikipedia article on Jones vectors and matrices, they do not use a consistent phase convention (even though they state that they do at the top).

Since the Jones vector is attached to a (possible changing) local coordinate frame, a good way to do a polarization ray trace is to use a global three dimensional coordinate system. This has been accomplished by Yun, Crabtree, McClain and Chipman in their papers "Three-dimensional polarization ray-tracing calculus": "definition and diattenuation" (

Appl. Optics50no. 18, pp. 2855-2865 (2011), doi:10.1364/AO.50.002855), and "retardance" (Appl. Optics50no. 18, pp. 2866-2874 (2011), doi:10.1364/AO.50.002866).Take note that because of the inherent rotation of the coordinate system through/from interfaces, the Jones matrices will appear to have retardances that are

not physically created,but are just a geometrical artifact of the change of local coordinate system.Poynting vector vs $\bf{k}$-vector definitionsAs user17581 pointed out, some people have defined the Poynting vector as the direction of the $+z$-axis. This makes sense because ${\bf{S}}={\bf{E}}\times{\bf{H}}$ so the electric field is always perpendicular to ${\bf{S}}$ and ${\bf{H}}$. The question is then, why isn't it defined that way?

I believe it is because in a material, like a crystal, there are two or three indices of refraction, in two or three particular directions. If a Jones matrix is described for a

particularinput angle via the Poynting vector in a crystal then a particular Jones matrix is obtained (i.e., it really doesn't describe the entire space of possible inputs to the crystal). This matrix, however, doesn't really make sense because the underlying material properties are anisotropic. If you, however, define the Jones matrices for each index of refraction (and the associated $\bf{k}$-vectors), and model the propagation as two (or more) rays through the crystal, then recombine them at the end, then you basically have a "basis" of Jones matrices for the crystal, which you do not have to recompute each time.