# Jones formalism for calculating quarter waveplate angle for circular polarized light

I am planning to change the polarization of a vertically linear polarized laser to circular polarized light with the help of a quarter waveplate.

I know the final result: I have to rotate the fast axis of the waveplate 45° to the incident polarization axis in order to achieve left handed circular polarized light. A rotation of -45° results in right handed polarization.

My task is know to use the Jones formalism to calculate this result. Vertically linear polarized light can be written in Jones matrices as

\begin{pmatrix}1 \\ 0\end{pmatrix}

right handed circular polarized light can be written in Jones matrices as

$$\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -i\end{pmatrix}$$

left handed circular polarized light can be written in Jones matrices as

$$\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ i\end{pmatrix}$$

Now I don't know how to proceed from here. Has anyone a good advise for me?

## migrated from electronics.stackexchange.comJul 18 '16 at 15:15

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• The Jones matrix for e.g. right circular polarized light is: $$\frac{1}{2} \begin{pmatrix} 1 & i \\ -i & -1 \end{pmatrix}$$ To see this try applying this matrix to your linearly polarized Jones vector. It will give a right circularly polarized Jones vector.
• The Jones matrix for a quarter wave plate is:$$\begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/2} \end{pmatrix}$$ You can see this e.g. by applying it to the vectors $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ separately and observing the phase difference that is introduced between the two.