Editing Circular polarization (section)

==Mathematical description==
The [[Classical physics|classical]] [[Sine wave|sinusoidal]] plane wave solution of the [[electromagnetic wave equation]] for the [[Electric field|electric]] and [[Magnetic field|magnetic]] fields is:
:<math>\begin{align}
 \mathbf{E} ( \mathbf{r}, t ) &= \left| \,\mathbf{E}\, \right| \mathrm{Re} \left\{ \mathbf{Q} \left|\psi\right\rangle \exp \left[ i \left( kz - \omega t \right) \right] \right\} \\
 \mathbf{B} ( \mathbf{r}, t ) &= \dfrac{1}{c} \hat{ \mathbf{z} } \times \mathbf{E} ( \mathbf{r} , t )
\end{align}</math>

where k is the [[wavenumber]];
:<math> \omega = c k</math>

is the [[angular frequency]] of the wave; <math> \mathbf{Q} = \left [ \hat{ \mathbf{x}}, \hat{\mathbf{y}} \right ] </math> is an orthogonal <math> 2 \times 2</math> matrix whose columns span the transverse x-y plane; and <math> c </math> is the [[speed of light]].

Here,
:<math> \left| \,\mathbf{E}\, \right| </math>

is the [[amplitude]] of the field, and
:<math>  |\psi\rangle \ \stackrel{\mathrm{def}}{=}\ \begin{pmatrix} \psi_x \\ \psi_y  \end{pmatrix} =  \begin{pmatrix} \cos\theta \exp \left ( i \alpha_x \right )  \\ \sin\theta \exp \left ( i \alpha_y \right )  \end{pmatrix}  </math>

is the normalized [[Jones calculus|Jones vector]] in the x-y plane.

If <math>  \alpha_y  </math> is rotated by <math>  \pi / 2  </math> radians with respect to <math>  \alpha_x  </math> and the x amplitude equals the y amplitude, the wave is circularly polarized. The Jones vector is:
:<math>  |\psi\rangle = {1\over \sqrt{2}} \begin{pmatrix} 1  \\ \pm i \end{pmatrix} \exp \left ( i \alpha_x \right )  </math>

where the plus sign indicates left circular polarization, and the minus sign indicates right circular polarization. In the case of circular polarization, the electric field vector of constant magnitude rotates in the ''x''-''y'' plane.

If basis vectors are defined such that:
:<math>  |\mathrm{R}\rangle \ \stackrel{\mathrm{def}}{=}\  {1\over \sqrt{2}}\begin{pmatrix} 1  \\ -i \end{pmatrix}  </math>

and:
:<math>  |\mathrm{L}\rangle \ \stackrel{\mathrm{def}}{=}\  {1\over \sqrt{2}}\begin{pmatrix} 1  \\ i \end{pmatrix}  </math>

then the polarization state can be written in the "R-L basis" as:
:<math>  |\psi\rangle = \psi_\mathrm{R} |\mathrm{R}\rangle + \psi_\mathrm{L} |\mathrm{L}\rangle </math>

where:
:<math>\begin{align}
 \psi_\mathrm{R} ~&\stackrel{\mathrm{def}}{=}~ \frac{1}{\sqrt{2}} \left( \cos\theta + i\sin\theta \exp \left( i \delta \right) \right) \exp \left( i \alpha_x \right) \\
 \psi_\mathrm{L} ~&\stackrel{\mathrm{def}}{=}~ \frac{1}{\sqrt{2}} \left( \cos\theta - i\sin\theta \exp \left( i \delta \right) \right) \exp \left( i \alpha_x \right)
\end{align}</math>

and:

:<math> \delta= \alpha_y - \alpha_x. </math>