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A reciprocal optical circular polarizer can be assembled from a linear polarizer followed by a quarter-wave plate whose axes make $45^{\circ }$ to the transmitted linear state. Forward passage converts unpolarized light into circular light of half the ideal incident intensity. Reverse passage encounters the retarder first; it leaves unpolarized light unpolarized until the linear polarizer transmits half as linear light. Thus the one-pass intensity can be equal in both directions even though the outgoing polarization differs. Use a coherency matrix, not a single Jones vector, for unpolarized input.

The optical coherency matrix records the second moments of transverse complex amplitudes. Normalize its trace to light intensity. Under a deterministic Jones matrix $M$, $C\mapsto MC{M}^{†}$. Unpolarized light has $C=(I_0/2)I_2$, so its output intensity is $(I_0/2)\mathrm{tr}(M{M}^{†})$; it cannot be represented by one fixed Jones vector.

The degree of polarization quantifies how strongly a transverse optical field departs from an unpolarized mixture. Here ${\mu }_1\ge {\mu }_2\ge 0$ are the eigenvalues of its coherency matrix. For two mutually uncorrelated orthogonal linear components of intensities $I_1,I_2$, $\mathcal{P}=|I_1-I_2|/(I_1+I_2)$. It is undefined when the total light intensity is zero. A signed difference is a separate convention that also identifies a preferred polarization direction.

For propagation along z, write $\mathbf{E}=\mathrm{Re}\{(E_x\hat{\mathbf{x}}+E_y\hat{\mathbf{y}})e^{i(kz-\omega t)}\}$. A Jones vector contains the complex transverse components. Equal phase modulo $\pi$ gives linear polarization; equal magnitudes with phase difference $\pm \pi /2$ give circular polarization. A common overall phase has no effect on the polarization ellipse. A deterministic optical element acts by a Jones matrix, with successive matrices ordered from right to left along the light path.

A circular polarizer uses a linear polarizer transmitting x and a quarter-wave plate whose principal axes are at $\pm 45^{\circ }$. Up to an overall phase, convenient matrices in fixed physical x/y axes areThe plate is lossless, $Q{Q}^{†}=I_2$, and $Q(1,0{)}^T=(1,i{)}^T/\sqrt{2}$ is circular in the supplied convention. Put P on side A and Q on side B: forward passage is $J_{A\to B}=QP$.

Unpolarized light has no single Jones vector. Represent its intensity by the coherency matrix $C_{\mathrm{in}}=(I_0/2)I_2$ and propagate it as $C_{\mathrm{out}}=J{C}_{\mathrm{in}}{J}^{†}$. Forward transmission givesThus the transmitted light is circular with intensity $I_0/2$. The initial polarizer removes one incoherent linear component; the plate changes phase without restoring the lost intensity.

Reverse passage encounters Q before P and has $J_{B\to A}=PQ$ in the same fixed transverse axes. It givesThe output is linearly polarized along x and again has intensity $I_0/2$. A retarder alone leaves an unpolarized coherency matrix unchanged. Consequently an unpolarized scene appears darker from either side, although the transmitted polarization differs. No polarization-dependent response of the observer is needed to explain this reduction.

For the mirror calculations below, use an ideal near-normal isotropic reflector, $M=-I_2$ up to a common phase. In fixed physical transverse axes reflection leaves the polarization ellipse unchanged; it reverses circular helicity because the propagation direction reverses. A reciprocal plate has the same relative phase delay on the return traversal in this fixed-axis convention. In particular $Q^2=i\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$: two passes exchange x and y linear states.

Let ${\mathbf{u}}_{\theta }=(\cos \theta ,\sin \theta {)}^{\mathsf{T}}$ be the transmission direction. An ideal linear polarizer projects the incident electric field onto this direction, so its Jones matrix isSince ${\mathbf{u}}_{\beta }^{\mathsf{T}}{\mathbf{u}}_{\alpha }=\cos (\alpha -\beta )$, the amplitude transmitted by a sequence of polarizers is the product of the projections between successive axes. Here the four directions are $0,\omega t,-2\omega t,0$. After the first polarizer the unpolarized light input has intensity $I/2$ and is in linear polarization along $x$. The transmission through a sequence of linear polarizers therefore follows from successive applications of Malus's lawThe relative angle between the rotating polarizers is $3\omega t$, not $\omega t$. The first factor $1/2$ is an ensemble average over incident polarization: unpolarized light cannot be represented by a single deterministic Jones vector. Equivalently, its coherency matrix is $(I/2)I_2$; the total matrix $P(0)P(-2\omega t)P(\omega t)P(0)$ gives the same output intensity by the trace rule. The quasistatic optical-element approximation assumes the mechanical rotation is slow compared with the optical oscillation.

Unpolarized incident light has equal, mutually incoherent $y$- and $z$-polarized intensities. Both components scatter into the same direction $\mathbf{v}$, so their coherency matrices add toThis matrix has rank one when the signal is nonzero. The light is therefore fully linearly polarized along $\mathbf{v}$, withA fixed highly anisotropic needle can act as a polarizing scatterer even for unpolarized input. The equality of total intensities with (c) does not mean the incident states were identical. These conclusions use the leading longitudinal-polarizability approximation; finite transverse response can change the polarization purity.

Work in the dilute, single-scattering limit and add the diffuse scattered light intensities of randomly positioned particles. The incident unpolarized light has equal, mutually uncorrelated $x$ and $y$ electric components. Each particle therefore has an induced electric dipole moment along its permitted axis, with the same mean squared magnitude proportional to $|\alpha {|}^2$.

For an observer in direction $\hat{\mathbf{n}}=(\sin \psi ,0,\cos \psi )$, choose transverse unit directions ${\mathbf{e}}_{\parallel }=(\cos \psi ,0,-\sin \psi )$ in the scattering plane and ${\mathbf{e}}_{\perp }=\hat{\mathbf{y}}$ perpendicular to it. The electric dipole radiation field is proportional to the transverse projection of the electric dipole. Hence an $x$ electric dipole contributes a wave amplitude factor $\cos \psi$ along ${\mathbf{e}}_{\parallel }$, while a $y$ electric dipole contributes its full wave amplitude along ${\mathbf{e}}_{\perp }$.

Let $C>0$ collect the common incident light intensity, electric polarizability, distance and frequency factors. ThenThe two components are uncorrelated, so the coherency matrix is diagonal in this basis. The degree of polarization isA signed degree of linear polarization, defined positive perpendicular to the scattering plane, has the same expression without the absolute value. The dominant direction is $\hat{\mathbf{y}}$ when $N_2>N_1{\cos }^2\psi$, and ${\mathbf{e}}_{\parallel }$ when the inequality reverses. If the denominator vanishes there is no scattered signal and its degree of polarization is undefined. A coherent exactly forward component, including the transmitted incident beam, is not part of this diffuse-intensity calculation; random particle positions do not by themselves destroy forward optical coherence.

Let $I_0$ be the incident light intensity. The first ideal polarizer transmits $I_0/2$ and produces linear polarization along ${\mathbf{e}}_1$. Faraday rotation turns that direction through $\theta ={\rm Y}L{B}_{\parallel }$, and Malus's law at the second polarizer givesEquivalently, use the input coherency matrix $C_{\mathrm{in}}=(I_0/2)I_2$ and $I_f=\mathrm{tr}(J{C}_{\mathrm{in}}{J}^{†})$. This produces the same one-half factor required by unpolarized light.

Maxima satisfy $\theta =\pi /4+m\pi$, with $m\in \mathbb{Z}$. For $L=1\mathrm{m}$ and ${\rm Y}=9\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$, the smallest field magnitude isIts direction is the axial direction that rotates the polarization passed by $P_1$ toward the pass axis of $P_2$ through the shorter $45^{\circ }$. In the right-handed ray-basis convention where $\rho =+{\rm Y}{B}_{\parallel }$ and the depicted second axis is $+45^{\circ }$, this is $B_{\parallel }>0$, along $S\to O$. Reversing that field at the same magnitude gives $\theta =-\pi /4$ and zero transmission. A signed Verdet constant is meaningful only together with its rotation convention, so the polarization-alignment description fixes the physical direction without relying on ambiguous clockwise labels in a perspective diagram.

Pure polarization means that the transverse components of a light field have a fixed complex amplitude ratio. A single Jones vector describes the state; its coherency matrix has rank one. Removing overall intensity and a common phase leaves two real parameters, which determine its polarization ellipse.