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Polarization rotation: Jones Matrix that maps Horizontal to right circular

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I am looking at the Poincaré sphere and I am trying to compute a Jones matrix for a particular rotation. Specifically, I would like it to perform the following maps:

$O :|H \rangle \rightarrow |R \rangle$

$O :|V \rangle \rightarrow |L \rangle$

$O :|L \rangle \rightarrow |H \rangle$

$O :|R \rangle \rightarrow |V \rangle$

where $|R \rangle$, $|L \rangle$ are right circular and left circular light. Is this possible? I should mention that I would also accept the same equations with L,R replaced by linearly polarized light $|D \rangle$, $|A \rangle$. If it is possible, what would be common plates that could do it?

This post imported from StackExchange Physics at 2017-10-08 21:12 (UTC), posted by SE-user Ben Sprott
asked Feb 22, 2012 in Theoretical Physics by Ben Sprott (30 points) [ no revision ]

1 Answer

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@wsc answer is interesting but misses a key point. Jones vector are defined upto a global phase, which gives us enough degree of freedom to solve your problem. Since your operation corresponds to a $\frac\pi2$-rotation around the $Y$ axis in the Poincaré sphere, it is physically doable.

Algebraically, after the first to equations, the matrix is determined to be $$\frac1{\sqrt2}\begin{bmatrix}1&e^{i\phi} \\ -i&i e^{i\phi}\end{bmatrix}.$$ The third condition imposes $\phi=-\frac\pi2$, which gives the final matrix $$M=\frac1{\sqrt2}\begin{bmatrix}1&-i \\ -i&1\end{bmatrix}.$$ $M$ is fully determined and consistent with the fourth condition.

Edited to add: A little linear agebra will show you that this matrix corresponds to a quarter wave plate rotated with a $\frac\pi4$ angle relatively to the vertical direction. Of course, it is easy to give physical intuition after I deduced it from the algebra:

  • a quarter wave plate is needed to transform a circular polarization into a linear polarization and vice-versa;
  • Applying twice the transformation swaps $|H\rangle$ and $|V\rangle$. This is what a half-wave plate at a $\frac\pi4$-angle does. And a half-wave plate is nothing more that 2 stacked quarter-wave plates (at least in theory). This gives the $\frac\pi4$-angle needed for our quarter-wave plate. QED without algebra.
This post imported from StackExchange Physics at 2017-10-08 21:12 (UTC), posted by SE-user Frédéric Grosshans
answered Feb 22, 2012 by Frédéric Grosshans (250 points) [ no revision ]
Haha, oops. Forgot about that.

This post imported from StackExchange Physics at 2017-10-08 21:12 (UTC), posted by SE-user wsc

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