 Solution

An ideal linear polarizer projects onto ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta {)}^T$. Thus its Jones matrix is $P_{\theta }={\mathbf{e}}_{\theta }{\mathbf{e}}_{\theta }^T$, immediately giving the specified entries. For time convention $e^{-i\omega t}$, a birefringent plate of thickness $d$ produces relative phase $2\pi (n_y-n_x)d/\lambda$. A quarter-wave thickness $d=\lambda /[4(n_y-n_x)]$ gives, after removing a common phase, $Q=\mathrm{diag}(1,i)$ when $x$ is fast.

For incident linear state $\mathbf{E}=E_0(\cos \theta ,\sin \theta {)}^T$, the original order givesThe transmitted state is circular wherever its amplitude is nonzero. Reversing the components givesand the output is linearly polarized at $45^{\circ }$. The incident circular state $\mathbf{E}\propto (1,i{)}^T$, called left-circular in the paper's convention, becomes $(1,-1{)}^T$ after $Q$ and is extinguished by $P_{\pi /4}$. Explicit Jones vectors fix the handedness convention without relying on an ambiguous viewing direction.

With the stated normal mirror after the original system, an arbitrary incident field is first projected onto $45^{\circ }$ and then made circular. In fixed laboratory transverse axes, the mirror contributes a common reflection phase, and the reciprocal return plate contributes another $Q$. The round-trip matrix is consequently proportional toThe returned field is orthogonal to the polarizer and is blocked. This suppresses the stipulated specular return toward a source, useful for avoiding optical feedback. The result concerns that mirror and return geometry; a reciprocal circular-polarizer assembly does not provide the general nonreciprocal two-port property of a Faraday optical isolator.