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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: $i^2 = -1$.

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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.

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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.

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