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

There are errors in the original PDF at this point, not merely in the TeX transcription. The second exponential must denote the eigenvalue ${\lambda }_{-}(z)=e^{-{\gamma }^{-}z}$, not the propagation coefficient itself. More substantially, the printed prefactor $\mathrm{sech}(\delta z)$ is inconsistent with the stated exponential eigenvalues. It must be $\cosh (\delta z)$. The original's primed Jones-matrix label distinguishes the distributed component from the earlier static matrix; below write $\mathcal{J}(z)$ to avoid confusing that prime with a derivative.

Complex eigenvalues have no intrinsic largest/smallest ordering, so $+$ and $-$ are modal labels. Keep the circular eigenvectors fixed, put $\gamma =({\gamma }^{+}+{\gamma }^{-})/2$ and $\delta =({\gamma }^{+}-{\gamma }^{-})/2$, and use the spectral decomposition from part (c):Its diagonal entries are $(e^{-{\gamma }^{+}z}+e^{-{\gamma }^{-}z})/2=e^{-\gamma z}\cosh (\delta z)$, and its upper off-diagonal entry is $i(e^{-{\gamma }^{+}z}-e^{-{\gamma }^{-}z})/2=-i{e}^{-\gamma z}\sinh (\delta z)$. ThereforeWhere $\cosh (\delta z)\ne 0$, this can be factored asThe unfactored expression is valid even at zeros of the cosine-like factor when $\delta$ is imaginary. The printed expression is ${\mathrm{sech}}^2(\delta z)$ times this correct matrix; its eigenvalues are consequently $e^{-{\gamma }^{\pm }z}{\mathrm{sech}}^2(\delta z)$ rather than those specified. A second check is $\det \mathcal{J}=e^{-2\gamma z}$, whereas the printed version has determinant $e^{-2\gamma z}{\mathrm{sech}}^4(\delta z)$. Thus the requested printed identity cannot be proved except in the degenerate case $\delta =0$; the following parts use the consistent propagator.