# 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 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.