Quarter-wave plate ($\delta =\pm \pi /2(\mathrm{mod}2\pi )$)

A quarter-wave plate gives a quarter-cycle phase retardation. Equal transmitted wave amplitudes initially in linear polarization at $45^{\circ }$ become circular polarization; unequal wave amplitudes generally give elliptical polarization. Reversing the sign reverses the traversal of the same circle.

Take a half-wave plate with fast axis at $15^{\circ }$ to a linearly polarized input, followed by a quarter-wave plate at a further $60^{\circ }$, so the second axis is at $75^{\circ }$ to the input. With shared quarter-wave retardance error $ϵ$ and half-wave error $2ϵ$, use $Q=\mathrm{diag}(1,i{e}^{iϵ})$ and $H=\mathrm{diag}(1,-e^{2iϵ})$. Rotating each matrix into the input axes and multiplying gives $\mathbf{J}(0)=e^{i75^{\circ }}(1,-i{)}^{\mathsf{T}}/\sqrt{2}$ and the title expansion. Its first-order change is common phase only, so the output remains circular to first order. This input-specific local compensation does not prove a frequency-independent quarter-wave operator for arbitrary input across a finite spectral band.

Suppose the unknown ideal elements are a linear polarizer and a quarter-wave plate, and the two unknown sources are unpolarized and circularly polarized. Test both element orders while rotating their relative axes and monitoring output intensity. Only circular input followed by the quarter-wave plate and then the polarizer gives complete modulation: the plate makes linear light and the analyzer obeys Malus's law. Polarizer-first output is constant half-intensity because the following lossless plate changes polarization but not intensity. Unpolarized input is constant in either order. This identifies both elements and both sources without comparing source brightnesses.

Two rotatable ideal quarter-wave plates can transform a fixed circular polarization into any pure polarization represented by a Jones vector. Use time dependence $e^{-i\omega t}$, $Q(\theta )=R(-\theta )\mathrm{diag}(1,i)R(\theta )$ and input $J_L=(1,i{)}^T/\sqrt{2}$. Every target polarization ellipse can be written, up to a common phase, as $R(-\beta )(\cos \eta ,i\sin \eta {)}^T$, with signed ellipticity angle $-\pi /4\le \eta \le \pi /4$. The first quarter-wave plate at ${\theta }_1=\beta +\eta +\pi /4$ produces linear polarization at $\beta +\eta$. The second, at ${\theta }_2=\beta$, therefore produces the required target. The two angles supply the two independent parameters of pure polarization.

For a reciprocal ideal quarter-wave plate followed by a mirror at normal incidence, use the same fixed transverse laboratory axes for the forward and returning Jones vectors. If the axes coincide with the plate's fast and slow axes, each pass has Jones matrix $J=\mathrm{diag}(1,i)$ up to a common phase factor. A polarization-independent mirror has scalar reflection coefficient $r_m$, so the returned Jones vector is $r_m{J}^2\mathbf{e}=r_m\mathrm{diag}(1,-1)\mathbf{e}$. The plate acts twice as a half-wave plate, rather than undoing its phase retardation.

For input linear polarization at angle $\alpha$ to the fast axis, the returned line is at $-\alpha$ modulo $\pi$. At $\alpha =45^{\circ }$ it is perpendicular to the original line. An axis-aligned input instead returns along the same line. General elliptical polarization requires its full input Jones vector; no unique returned state follows from the word “polarized” alone. The propagation direction reverses, so comparing descriptions of circular polarization must also specify the viewing convention.

Take a linear polarizer $P=\mathrm{diag}(1,0)$ and a reciprocal quarter-wave plate $Q=2^{-1/2}\left(\begin{array}{cc}1& i\\ i& 1\end{array}\right)$ in fixed physical transverse axes. A near-normal isotropic mirror has $M=-I_2$ up to global phase. If the source/detector side encounters P before Q, the round-trip matrix is $PQMQP=-P{Q}^{2}P=0$. Reversing the filter gives $QPMPQ=-QPQ$, which transmits half an initially unpolarized intensity. The result is double-pass rejection, not optical nonreciprocity; changing direction without changing the input polarization does not preserve the same physical input state.