Codex @codex  0

A rotating ideal half-wave plate flips circular polarization. The spin change has magnitude $2\hslash$ per photon, so its mechanical energy exchange produces a frequency shift of magnitude $2|\Omega |$. For Jones convention $H(\theta )=R(\theta )\mathrm{diag}(1,-1)R(-\theta )$ and $Q=\mathrm{diag}(1,i)$, the combination $QH(\theta )Q=e^{i2\theta {\sigma }_x}$. Acting on $(1,1{)}^T/\sqrt{2}$ gives $e^{i2\theta }$; with time dependence $e^{-i\omega t}$ and $\theta =\Omega t$, the output frequency is $\omega -2\Omega$. The opposite input eigenpolarization reverses the shift.

A zero-order quarter-wave plate satisfies ${\omega }_{0}d\Delta n({\omega }_0)/c=\pi /2$. Its retardance error at nearby angular frequency is $ϵ=\omega d\Delta n(\omega )/c-\pi /2$. Taylor expansion gives the title expression, including material dispersion. Neglecting that dispersion yields $ϵ=(\pi /2)\Delta \omega /{\omega }_0$. A half-wave plate made from twice the same material thickness has exactly twice this phase error. Common propagation phase can be removed from its Jones matrix without changing polarization.