What is the wave form when the left hand circular polarized wave passes through 1/8 wave plate?

Question

Consider an optical beam with left-handed circular polarization (LHCP), passing through an 1/8 wave plate.

I know that LHCP is written as $$ \tag{1}\left( \begin{array}{c} 1 \\ i \end{array} \right) $$ in term of a Jones vector. So, I can calculate this problem by using the matrix form of the 1/8 wave plate. $$\tag{2} \left( \begin{array}{cc} 1 & 0 \\ 0 & \exp(i\pi/4) \end{array} \right) \left( \begin{array}{c} 1 \\ i \end{array} \right) = \left( \begin{array}{c} 1 \\ i\exp(i\pi/4) \end{array} \right) = \left( \begin{array}{c} \exp(i 0) \\ \exp(i3\pi/4) \end{array} \right) . $$

But, I don't know what is this result of wave form. Is this circular polarization or elliptical polarization?

Please let me know. Thank you.

Answer 1

One can always consider the effect of a wave plate as a rotation on the Poincare sphere. A quarter wave plate would rotate a circular state of polarization one quarter of a revolution of the sphere, from the pole to the equator. Therefore it would turn circular polarization into a linear polarization. A 1/8 wave plate would rotate it only half as much, ending halfway between the pole and the equator. So the result would be elliptical polarization.

Another way to see this is to use the left- and right-handed circular states of polarizations as an orthogonal basis. Then one can use dot products of your final state of polarization with the complex conjugates of these basis vectors to determine what the coefficients are to expand your final state of polarization in terms of this basis. I'll leave this for you as an exercise.