Jones calculus

Remarks

A lossless element has a unitary Jones matrix, since it preserves the intensity $J^* J$. Jones vectors that differ by a nonzero complex factor describe the same polarization state. The matrices depend on the choice of axes and of the sign convention for phases.

Example: the linear polarizer with transmission axis at angle $\theta$ to the $x$ axis has $$P_\theta = \begin{pmatrix} \cos^2\theta & \cos\theta \sin\theta \\ \cos\theta\sin\theta & \sin^2\theta \end{pmatrix}.$$ On horizontal light $J = (1, 0)^T$ it gives $P_\theta J = \cos\theta \, (\cos\theta, \sin\theta)^T$, of intensity $\cos^2\theta$. A quarter-wave plate with fast axis along $x$ has, up to a global phase, the matrix $\operatorname{diag}(1, i)$. It turns light polarized at $45°$, with $J = (1, 1)^T/\sqrt2$, into $(1, i)^T/\sqrt2$, which is circularly polarized.

Non-example: an ideal depolarizer turns every input into unpolarized light. Unpolarized light has no Jones vector, so a depolarizer has no Jones matrix; the Mueller calculus is needed for it.