Past exam of the optics and electrodynamics course of the University of Cambridge / 2021 / Section B / B4 / g / Solution

The circular eigenmode orthogonal projection matrices obey $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$. Define the uniform-medium generatorExpanding its matrix exponential by its power series and using these projection identities givesEquivalently, with $Q=\left(\begin{array}{cc}0& i\\ -i& 0\end{array}\right)$, one has $Q^2=I$, $K=\gamma I+\delta Q$ and $e^{-Kz}=e^{-\gamma z}[I\cosh (\delta z)-Q\sinh (\delta z)]$. This is exactly the corrected matrix of part (d). The entries of $K$ have units of inverse length, and the map satisfies $d\mathcal{J}/dz=-K\mathcal{J}$, $\mathcal{J}(0)=I$ and the homogeneous propagation law $\mathcal{J}(z_1+z_2)=\mathcal{J}(z_1)\mathcal{J}(z_2)$.

The extra ${\mathrm{sech}}^2$ factor in the printed version violates this last law for nonzero $\delta$, so it cannot be generated by one constant $K$ with the stipulated eigenvalues. The constant-$K$ description assumes uniform coefficients and fixed circular eigenmodes.