A paper that I was reading wanted to transform the jones matrix of linearly horizontal polarization(LHP) to right circular polarization(RCP).

The paper states:

Consider... $J_{LHP}\to J_{RCP}$... In order to get to RCP from an initial LHP we must first rotate down to the equator to L+45. This transformation takes its path all the way around the sphere in a helical, descending manner and is physically accomplished by rotating a Waveplates in the plane of polarization.

The author doesn't provide information on how we finally reach RCP. From my understanding we reached a linear state($L+45$) through a path that traces elliptical states. How do we arrive to RCP or am I missing something?


1 Answer 1


I don't quite understand the geometry proposed by the author but consider the following:

The left-handed and right handed circular light are described by the Jones vector \begin{align} \vert LHP\rangle&=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\ i\end{array}\right)\, ,\\ \vert RHP\rangle&=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\ -i\end{array}\right) \end{align} so a rotation about the $\hat z$ axis $R_z(\theta)=e^{-i\theta \sigma_z/2}$ by and angle $\theta=\pi$ \begin{align} R_z(\pi)=\left(\begin{array}{cc} e^{i\pi/2} & 0 \\ 0&e^{-i\pi/2} \end{array}\right) =\left(\begin{array}{cc} i& 0 \\ 0&-i \end{array}\right)=i \left(\begin{array}{cc} 1 & 0 \\ 0&-1 \end{array}\right) \end{align} acting on $\vert LHP\rangle$ will yield $\vert RHP\rangle$ up to global inessential phase.

Basically the Jones vectors for the left- and right-handed polarization are eigenstates of $\sigma_y$ so lie opposite on the equator of the Poincaré sphere; a rotation about $R_z$ tracing a path along the equator should transform one into the other.

  • $\begingroup$ Thanks! That helps a lot. I knew some things but reading the paper confused me a lot and as a result it led me to question my knowledge on the matter. $\endgroup$
    – Perfectoid
    Jun 16 at 19:23

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