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In Chapter 7 of Jackson's book on Classical Electrodynamics, there's the following statement:

Introducing the complex orthogonal unit vectors:

$$\epsilon_{\pm}=\frac{1}{\sqrt{2}}(\epsilon_1\pm\epsilon_2)$$

we can represent an EM traveling in the $\epsilon_3$ direction with arbitrary polarization as:

$$\vec{E}(\vec{r},t)=(E_+\epsilon_++E_-\epsilon_-)e^{i(\vec{k}\cdot\vec{r}-\omega t)}$$

If the amplitudes have the ratio,

$$\frac{E_-}{E_+}=re^{i\alpha}$$

then it can be easily shown that the ratio of semimajor to the semiminor axis is:

$$|\frac{1+r}{1-r}|$$

and that the ellipse is rotated by an angle $(\alpha/2)$.

I tried checking this by myself, by starting with the general field $\vec{E}$ expression, passing to the basis with $(\epsilon_1,\epsilon_2)$ and expressing $E_\pm=A_\pm e^{i\beta_\pm}$, However, so far what I've found is that the $\vec{E}$ in such basis can be expressed as (taking the real components):

$$\vec{E_{1,R}}=\frac{1}{\sqrt{2}}[A_+\cos(\beta_++\vec{k}\cdot\vec{r}-\omega t)+A_-\cos(\beta_-+\vec{k}\cdot\vec{r}-\omega t)]$$

$$\vec{E_{2,R}}=\frac{1}{\sqrt{2}}[-A_+\sin(\beta_++\vec{k}\cdot\vec{r}-\omega t)+A_-\sin(\beta_-+\vec{k}\cdot\vec{r}-\omega t)]$$

which shows that we indeed have an ellipse. Nevertheless, since I can't seem to find the ratio and inclination from Jackson, and I couldn't find online a detailed derivation.

Edit: I had already checked the following post (How can I get the axes of the polarization ellipse from the Jones vector of the light?) which tried to find something similar in terms of the Jones vectors, where all the answers provided are in terms of finding eigenvalues for the circular and linear polarizations only. However, in my case I want to get an explicit derivation of the formulas provided by Jackson, as not only they avoid using Jones vectors, but I think provide a more geometrical aspect of polarization states (hence why I don't consider it duplicate).

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  • $\begingroup$ I have the feeling you are doing it much more complicated than necessary. If the ratio of the two amplitudes is $re^{i\alpha}$, the electric field can be given as: $E(\vec{r},t)=(E_{+}(\epsilon_{+}+re^{i\alpha}\epsilon_{-})e^{i(\vec{k}\vec{r}-wt)})$. Now you can callculate the norm of the term before the wave term and chose $\alpha$ such that it becomes maximal (semi-major axis) and minimal (semi minor axis). $\endgroup$
    – Mechanix
    Commented Mar 9, 2019 at 21:51
  • $\begingroup$ I didn't think about that but it seems a better approach, I'm going to try it. $\endgroup$
    – Charlie
    Commented Mar 9, 2019 at 22:50
  • $\begingroup$ It was marked as duplicated however I had already checked before the link provided and it didn't answer my question, hence why I opened a new one. I include additional information to justify this. $\endgroup$
    – Charlie
    Commented Mar 11, 2019 at 19:46
  • $\begingroup$ It appears my reopen vote has aged away, but there's little to add here to the given duplicate. In short: (1) the ellipse rotation can be seen from the fact that $R(\theta) \epsilon_\pm = e^{\mp i\theta} \epsilon_\pm$, i.e. the circular basis vectors are eigenvectors of rotations about $\epsilon_3$, so that a rotation by $\alpha/2$ can be used to nullify the phase difference between $E_+$ and $E_-$, i.e. rotating the system into the canonical form where both coefficients are positive. $\endgroup$
    – Emilio Pisanty
    Commented Apr 3, 2019 at 20:59
  • $\begingroup$ (2) Once in that canonical form, it is clearer that the major axis of the ellipse is when both components add constructively (so with length $|E_+|+|E_-|$) and the minor axis happens when they interfere destructively (so it has length $|(|E_+|-|E_-|)|$). The identification of the major-to-minor-axis ratio as $(1+r)/(1-r)$ then follows easily. $\endgroup$
    – Emilio Pisanty
    Commented Apr 3, 2019 at 21:05

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