Why does adding an $i$ to a Jones vector produce a circular polarization? It's common to define Jones vectors to describe polarization. So, for instance, we would have $$\begin{bmatrix} 1 \\0 \end{bmatrix} \quad \textrm{represents a horizontal electric field}.$$
And just like $\begin{bmatrix} 0 \\ 1 \end{bmatrix} $ would represent a vertical electric field for polarization. In this representation we also use matrices to represent polarizers.
My question basically is why does the vector
$$\begin{bmatrix} 1 \\ i \end{bmatrix}$$
describe a circular polarization?
Note: $i^2 = -1$.
 A: A Jones vector $\tilde{\mathbf{E}}$ relates to a physical electric field $\mathbf E(t)$ via
$$
E(t) = \mathrm{Re}\bigg[\tilde{\mathbf{E}} e^{-i\omega t}\bigg].
$$
If you put in $\tilde{\mathbf{E}} = (1,i)$ as your Jones vector, this gives you 
$$
\mathbf E(t) 
= \mathrm{Re}\bigg[\tilde{\mathbf{E}} e^{-i\omega t}\bigg]
= \mathrm{Re}\bigg[\begin{pmatrix}1\\i\end{pmatrix} e^{-i\omega t}\bigg]
= \mathrm{Re}\bigg[\begin{pmatrix}e^{-i\omega t}\\ie^{-i\omega t}\end{pmatrix} \bigg]
= \begin{pmatrix}\cos(\omega t) \\ \sin(\omega t)\end{pmatrix},
$$
which traces out a circle in the $(E_x,E_y)$ plane.
A: This is my opinion, I am not sure whether it's correct or not.
That's just because the matrix for circular polarization can be written as a linear combination of a horizontal and vertical field vectors. The coefficients being 1 and i to them respectively.This is just a way to point out that you cannot add the two shm's.
Another reason might be that the two shm's are out of phase by 90 degrees and so are 1 and i.
Another thing I noticed was their resemblance to spin eigenvectors in x, y and z directions.
