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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?

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migrated from electronics.stackexchange.com Jul 18 '16 at 15:15

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Hints:

  • 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.
  • Furthermore you can always build in rotation matrices, which corresponds to physically rotating e.g. your quarter waveplate relative to the incoming polarization.

Now you can try to build the required matrix (see first point above) from a quarter waveplate and rotations, which will solve the problem.

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