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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.

Suppose an earlier path selection prepares a photon in path $L$ with vertical polarization $|V⟩$. Conditional on that selection, the normalized Jones vector is $(0,1{)}^{\mathsf{T}}$. For time convention $e^{-i\omega t}$, a quarter-wave plate with fast-axis angle $\eta$ acts asConsequently,A subsequent horizontal/vertical measurement of linear photon polarization in that same path givesIf the earlier path event has probability $p_L$, the joint probabilities are $P(L,H)=p_{L}P(H\mid L)$ and $P(L,V)=p_{L}P(V\mid L)$. Normalizing after a selection changes a joint event into a conditional one; it does not remove the selection's success cost. If paths are coherently recombined instead of selected, their relative amplitudes and phases must be retained in the joint quantum state.

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.

Jones calculus represents a fully polarized monochromatic light field by a two-component Jones vector of transverse complex amplitudes. An ideal linear optical element acts by a complex $2\times 2$ matrix; successive elements act by the ordered product of their matrices. The order matters because the transformations need not commute.

For transverse axes $x,y$, an ideal polarizer transmitting $x$ has matrix $P=\mathrm{diag}(1,0)$, so $P(E_x,E_y{)}^{\mathsf{T}}=(E_x,0{)}^{\mathsf{T}}$. A quarter-wave plate aligned with those axes can be represented, up to a common phase factor and the chosen sign convention, by $Q=\mathrm{diag}(1,i)$. It changes the relative wave phase, enabling conversion between linear polarization and elliptical polarization. Rotating the elements tests these predictions by changes in transmitted light intensity. A single Jones vector does not describe arbitrary partially polarized light; that requires statistics of the field rather than one fixed pair of amplitudes.

For a differentiable Jones vector with $J_x(0)\ne 0$, the ratio $r=J_y/J_x$ fixes polarization independently of common amplitude and phase. The title quotient derivative tests first-order polarization sensitivity. If ${\mathbf{J}}^{\prime }(0)=a\mathbf{J}(0)$ for a complex scalar $a$, then $r^{\prime }(0)=0$. For a lossless normalized output, differentiation of ${\mathbf{J}}^{†}\mathbf{J}=1$ makes $\mathrm{Re}a=0$, so this change is a common phase. A vanishing component requires the complementary ratio or a basis-independent polarization coordinate instead.

A Jones matrix is a complex two-by-two linear transformation of a Jones vector through a deterministic nondepolarizing optical element. Successive elements multiply in reverse reading order. An ideal linear polarizer along a real transverse unit vector $\mathbf{a}$ has matrix $\mathbf{a}{\mathbf{a}}^{\mathsf{T}}$. A common nonzero scalar phase does not change the output polarization of waves.

A Jones vector records the two transverse complex amplitudes of a completely polarized monochromatic wave. Multiplication by a common nonzero complex number preserves its polarization of waves, while changing overall amplitude and phase. With the convention $\mathbf{\Psi }(t)=\mathrm{Re}[\mathbf{J}{e}^{i\omega t}]$, $\mathbf{J}=(1,2i{)}^{\mathsf{T}}$ gives $(\cos \omega t,-2\sin \omega t)$, an ellipse with the longer axis along $y$. Handedness labels require both a viewing direction and a naming convention.

For a vacuum plane electromagnetic wave, a Lorentz boost multiplies both transverse Jones vector amplitudes by the same positive Doppler factor when the basis is carried to the aberrated ray. Their ratio and phase difference therefore remain unchanged. All Stokes parameters gain the same squared factor, leaving normalized parameters, ellipse shape, handedness and degree of polarization unchanged. This does not mean the propagation direction or the components in an arbitrarily fixed pair of spatial axes remain unchanged.

Use $\mathbf{E}(z,t)=\mathrm{Re}[\mathbf{J}{e}^{i(kz-\omega t)}]$. A Jones vector $\mathbf{J}=(J_x,J_y{)}^{\mathsf{T}}$ records the two transverse complex amplitudes. The amplitude ratio and relative phase determine the polarization ellipse; a common nonzero scalar does not change the polarization. A fixed Jones vector describes fully polarized monochromatic light, not an unpolarized statistical mixture.

Linear birefringence gives orthogonal principal polarizations different refractive indices. Put $n_y-n_x=\Delta n>0$, so $x$ is the fast axis. A plate of thickness $d$ gives slow-minus-fast phase retardationThe thinnest quarter-wave plate hasHere ${\lambda }_0=2\pi c/{\omega }_0$ is the vacuum wavelength. For linear input at angle $\theta$ to the fast axis, the output at design frequency is $(\cos \theta ,i\sin \theta {)}^{\mathsf{T}}$ up to common phase. Axis-aligned input remains linear; equal nonzero components at $45^{\circ }$ modulo $90^{\circ }$ give circular polarization; other input angles give elliptical polarization. There is no need to spatially separate the two components.^[ref]

For waveplate retardance detuning, defineNeglecting material dispersion reduces this to $ϵ=(\pi /2)(\omega -{\omega }_0)/{\omega }_0$. Remove the common fast-axis phase. Then $Q=\mathrm{diag}(1,i{e}^{iϵ})$. A similar half-wave plate of twice the thickness has twice the retardation, giving $H=\mathrm{diag}(1,-e^{2iϵ})$. The shared error parameter presumes the same material and matched thickness ratio.

For the half-wave and quarter-wave achromatic circularizer, the first fast-axis angle is $\alpha =15^{\circ }$ and the second is $\gamma =\alpha +\beta =75^{\circ }$. With the stated passive rotation convention,An explicit expression valid for the given shared error is obtained by putting $z=e^{iϵ}$ and definingAt $ϵ=0$, the half-wave plate rotates the linear input to $30^{\circ }$, which is $-45^{\circ }$ relative to the second fast axis. ThusThis is circular polarization. With our time convention its fixed-point electric vector is proportional, after a time-origin shift, to $(\cos \omega t,-\sin \omega t)$; this specifies the rotation without relying on conflicting handedness conventions.

Expanding the exact product to first order givesThese coefficients are rounded to three significant figures. Before rounding, the derivative is exactly ${\mathbf{J}}_0^{\prime }=i\kappa {\mathbf{J}}_0$, with $\kappa =(3-\sqrt{3})/2$. The Jones-component ratio sensitivity therefore vanishes:Only common phase changes at first order. This proves compensation for the prescribed input near the design frequency, not exact achromatic performance throughout the visible spectrum for arbitrary input polarization.

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.

Polarization under a boost. Lorentz invariance of normalized plane-wave polarization preserves the polarization ellipse's shape, handedness and degree of polarization when its transverse basis is carried to the aberrated ray. The propagation direction changes; equality of a fixed laboratory Cartesian pair of field components is not the appropriate comparison.

Rotate the $yz$ axes so that a general incident vacuum plane wave has unit direction $\hat{\mathbf{n}}=(\cos \vartheta ,\sin \vartheta ,0)$. Choose the transverse basis ${\mathbf{e}}_p=(-\sin \vartheta ,\cos \vartheta ,0)$, ${\mathbf{e}}_s=(0,0,1)$, and arbitrary complex amplitudes $E_p,E_s$. ThenPutting $\beta =v/c$, the given transformation givesThe four-wavevector transforms with Doppler factor $\mathcal{D}=\gamma (1-\beta \cos \vartheta )>0$, givingIn the corresponding basis ${\mathbf{e}}_p^{\prime }=(-\sin {\vartheta }^{\prime },\cos {\vartheta }^{\prime },0)$, ${\mathbf{e}}_s^{\prime }={\mathbf{e}}_s$, the transformed field isBoth components of the Jones vector acquire the same positive real multiplier. Their amplitude ratio and relative wave phase are unchanged, proving preservation of arbitrary linear, circular or elliptical polarization. All Stokes parameters scale by ${\mathcal{D}}^2$, so their normalized values and the degree of polarization are unchanged. An unpolarized light ensemble remains unpolarized under this single-ray boost. A mixture of different ray directions requires separate transformations and is outside this statement.

Reflection from the moving slab. Take the upper interface as $z=0$, the material below it, and the incident ray along $-\hat{\mathbf{z}}$. This fixes the sign left implicit by “along the $z$-axis”; choosing the opposite surface convention changes no result below. Treat $n$ as the real refractive index in the slab's rest frame, with unit relative magnetic permeability. In the laboratory,Transform to the slab rest frame:The incidence is now oblique, in the $xz$ plane, at angle $i$ to the surface normal:The incoming ensemble remains unpolarized light. In this frame the ordinary stationary-interface Fresnel equations give the amplitude coefficients:For generic oblique incidence, $R_s=|r_s{|}^2$ and $R_p=|r_p{|}^2$ differ. Equal incident $s$ and $p$ intensities therefore produce unequal reflected intensities: the reflection is partially polarized. Its degree of linear polarization isThe inverse boost preserves this degree, so the same conclusion holds in the laboratory even though the laboratory incident ray is normal to the surface. The moving material cannot be treated as an isotropic stationary dielectric in that frame.

The reflected rest-frame wavevector is $(\omega /c)(\gamma ,-\gamma \beta ,0,+1)$. Boosting back gives $(\omega /c)(1,0,0,+1)$: the reflected laboratory ray is normal to the surface and has the incident frequency. There is no contradiction with the oblique rest-frame geometry. Translation tangential to a uniform interface does not itself introduce a moving normal mirror.

Complete linear polarization. For Brewster reflection from a tangentially moving dielectric, at Brewster's angle, $r_p=0$ and $\tan i_B=n$. Combining this with the aberration relation givesOnly the $s$ component is reflected. The plane of incidence in the slab frame is $xz$, so the reflected electric field is along $y$, perpendicular to the slab velocity. Boosting this $s$ field back keeps its electric direction along $y$. For example, the reflected rest-frame $s$ field has $B_z^{\prime }=-\beta E_y^{\prime }/c$; hence $E_y=\gamma (E_y^{\prime }+v{B}_z^{\prime })=E_y^{\prime }/\gamma$ and $E_x=E_z=0$.

This exact extinction assumes a transparent nonmagnetic dielectric with $n\ne 1$. At $v=0$ the reflected intensities are equal and the reflected beam is unpolarized. If $n=1$, both reflection coefficients vanish and there is no reflected beam whose polarization can be assigned. With dispersion, evaluate $n$ at the shifted frequency ${\omega }^{\prime }=\gamma \omega$; the Brewster-speed equation is then implicit. absorption generally prevents a real incidence angle from making the complex $r_p$ exactly zero.

Using the needle definitions in the question-level solution, $\mathbf{u}\cdot \mathbf{e}=u_z=\cos \theta$. The scattered field is parallel to $\mathbf{v}=u_x\hat{\mathbf{x}}+u_z\hat{\mathbf{z}}$, andThe light is linearly polarized along the needle's projection into the $xz$ observation plane. Its Jones vector in the $(x,z)$ basis is $(u_x,u_z{)}^{\mathsf{T}}/\sqrt{1-u_y^2}$ whenever the intensity is nonzero. A needle perpendicular to the incident electric field, or pointing directly at the observer, has zero leading-order signal.

Spherical-particle frequency scaling. For a dielectric sphere of radius $a\ll \lambda$, the quasistatic electric polarizability isThe electric dipole radiation formula and $I_0={ϵ}_{0}c|E_0{|}^2/2$ giveThus Rayleigh scattering scales as ${\omega }^4$, or ${\lambda }^{-4}$, when the polarizability varies slowly with frequency. Material resonances and absorption can change this simple scaling. For the incident $z$ polarization, the dipole is along $z$: intensity is proportional to ${\sin }^2\chi$, where $\chi$ is the angle from $z$. There is no radiation on the dipole axis and a maximum in the $xy$ plane. In a direction $\hat{\mathbf{n}}$ the scattered electric field is along $\hat{\mathbf{z}}-(\hat{\mathbf{z}}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}$, hence linearly polarized wherever nonzero.^[ref]

Needle model used in parts (a)–(d). Let its axis be $\mathbf{u}=(u_x,u_y,u_z)=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$. A very slender, high-permittivity needle has dominant longitudinal response, so use the rank-one needle polarizability $\mathbf{\alpha }\simeq {\alpha }_{\parallel }\mathbf{u}{\mathbf{u}}^{\mathsf{T}}$. Its transverse response is neglected at leading order. For observation along $y$, define $\mathbf{v}=u_x\hat{\mathbf{x}}+u_z\hat{\mathbf{z}}$, $s=|\mathbf{v}{|}^2=1-u_y^2$ andFor a unit incident Jones vector $\mathbf{e}$, the common far-field relation is ${\mathbf{E}}_s\propto \mathbf{v}(\mathbf{u}\cdot \mathbf{e})$, and $I_s=K{I}_{0}s|\mathbf{u}\cdot \mathbf{e}{|}^2$. Here $I_s$ is irradiance at distance $R$; multiplying by $R^2$ gives power per solid angle. No absolute value of ${\alpha }_{\parallel }$ is given or needed for the requested relative intensities.

Diffuse random cloud, with the incident polarization of (a). The single-particle amplitudes through $x$ and $z$ analysers are proportional to $u_x{u}_z$ and $u_z^2$. The isotropic orientation fourth moment gives the needle-scattering analyzer momentsFor example, $⟨u_z^4⟩=\frac{1}{2}{\int }_{-1}^1{\mu }^{4}d\mu =1/5$; symmetry and $u_x^2+u_y^2+u_z^2=1$ give $2⟨u_x^2{u}_z^2⟩=1/3-1/5$. Since the particles scatter mutually incoherently, intensities add. For $N$ illuminated particles,The cross correlation vanishes by reflection symmetry. The degree of linear polarization is $(I_z-I_x)/(I_z+I_x)=\boxed{1/2}$: random anisotropic scatterers produce partially depolarized light, even though every individual needle produces a single linear polarization.

Continuous homogeneous dielectric. One must sum amplitudes coherently and include their spatial phases, rather than retain the preceding sum of intensities. Random orientations give $⟨\mathbf{u}{\mathbf{u}}^{\mathsf{T}}⟩=\mathbf{I}/3$, so the averaged response is isotropic and its induced polarization is along $z$. Its coherent radiation toward $y$, when allowed by the sample geometry, is therefore purely $z$ polarized, with $I_x=0$. The independent-needle prefactor is no longer a quantitative dense-medium model; local-field and collective effects determine the effective permittivity.

There is also coherent continuum phase matching: a uniform bulk driven by a plane wave has no arbitrary outgoing $90^{\circ }$ plane-wave component. In an ideal infinite homogeneous medium there is no diffuse side scattering at all; propagation, refraction and reflection are determined by the medium and its interfaces. A finite sample's coherent side signal depends on its shape and boundaries. Thus the old $1:3$ diffuse ratio is replaced by a coherent isotropic-medium response, not by a universal nonzero $N^2$ side-scattering intensity.

General pure state. Use time dependence $e^{i(kz-\omega t)}$. The most general nonzero Jones vector can be writtenThere are four real parameters in the full complex field. Removing overall intensity $A^2$ and the unobservable common phase $\chi$ leaves two real degrees of freedom for pure polarization, $\alpha$ and $\delta$. Their endpoint degeneracies simply reflect the fact that the phase of a vanishing component is irrelevant. A single Jones vector cannot represent partially polarized light.

Three linear polarizers. Define the transmission axis ${\mathbf{a}}_{\theta }=(\cos \theta ,\sin \theta {)}^T$. An ideal linear polarizer has Jones matrix $P_{\theta }={\mathbf{a}}_{\theta }{\mathbf{a}}_{\theta }^T$. For definiteness label $(1,-i{)}^T/\sqrt{2}$ as right-circular in the chosen convention; reversing the handedness convention changes no intensity below. The first polarizer produces $(1,0{)}^T/\sqrt{2}$. The intermediate polarizer and final $y$ analyzer then giveso Malus's law givesThe maximum is $1/8$ at $\theta =\pi /4+m\pi /2$, and the transmission vanishes at $\theta =m\pi /2$. The first polarizer accounts for an initial factor $1/2$; retaining that factor is essential when quoting transmission relative to the incident circularly polarized beam.

Waveplate and quarter-wave matrix. A lossless waveplate has linear birefringence: its orthogonal fast and slow axes have indices $n_f<n_s$. A thickness $L$ produces slow-minus-fast phase retardationResolving the incident Jones vector into the plate axes, advancing the two phases and transforming back gives, up to a common phase,A quarter-wave plate has $\delta =\pi /2$ modulo $2\pi$. With $c_{\theta }=\cos \theta$ and $s_{\theta }=\sin \theta$, its Jones matrix is thereforeIn particular $Q(0)=\mathrm{diag}(1,i)$. Equal incident components along the two axes acquire a relative quarter-cycle phase, converting $45^{\circ }$ linear polarization to circular polarization; an incident component along either principal axis stays linearly polarized.

Half-wave replacement. A half-wave plate has $\delta =\pi$ and henceIt converts the $x$-polarized output of the first polarizer to a linear state at angle $2\theta$. The final $y$ analyzer givesThe maximum is $1/2$, because the ideal half-wave plate loses no light after the first polarizer.

Quarter-wave replacement. The transmitted $y$ amplitude is $(1-i)c_{\theta }{s}_{\theta }/\sqrt{2}$, givingMore generally, the retarder gives $I_{\mathrm{out}}/I_{\mathrm{in}}=\frac{1}{2}{\sin }^2(\delta /2){\sin }^2(2\theta )$. All three requested curves have the same zeros and angular period, but different heights:

Preparing an arbitrary ellipse. Yes: two rotatable quarter-wave plates already suffice. Let the input left-circular state be ${\mathbf{J}}_L=(1,i{)}^T/\sqrt{2}$. Up to an overall phase, any target pure polarization can be parameterized by its polarization ellipse:Here $\beta$ is the major-axis orientation, and $\eta$ sets the signed minor-to-major semiaxis ratio $\tan \eta$. First note directly from $Q(\theta )$ thatChoose the two fast-axis angles asThe first quarter-wave plate produces linear light at $\beta +\eta$. In the axes of the second, its components are $(\cos \eta ,\sin \eta )$; the second gives precisely ${\mathbf{J}}_{\mathrm{target}}$. This proves pure-polarization preparation with two quarter-wave plates, including the linear and circular limiting cases. A half-wave plate can also rotate the axes, but is not required. The phrase about left-handed input is understood as left-circular light in this construction; the explicit vector fixes the convention.

Switchable retardation. The electro-optic Kerr effect induces linear birefringence quadratic in a static transverse electric field. Defining the coefficient $K$ by $\Delta n={\lambda }_{0}K{E}_{\mathrm{dc}}^2$, its phase retardation isThe two eigenpolarizations are linear, parallel and perpendicular to the applied field. Thus an appropriate field produces a switchable quarter-wave plate:The sign of $K$ determines which axis is fast. Switching the field off removes the induced retardation, assuming no residual birefringence.

The Faraday effect instead gives circular birefringence for a longitudinal magnetic field. The two circular eigenmodes acquire different phases, rotating a linear polarization throughwith sign set by the handedness convention and the Verdet constant $V$. Its Jones matrix in the linear basis is a real rotation, up to a common phase. It preserves the ellipticity of a pure state, so a linear input remains linear. A $\pi /2$ phase difference between circular eigenmodes rotates a linear state by $\pi /4$; it does not make it circular. The Kerr device supplies the switchable quarter-wave plate; a Faraday rotator alone does not.

Let the two arms deliver intensities $I_1,I_2$ with unit Jones vectors ${\mathbf{e}}_1,{\mathbf{e}}_2$. For a perfectly coherent input, the recombined intensity under relative phase $\varphi$ isConsequently the polarization overlap in interferometric visibility isFor equal returned intensities, two linear polarizations at angle $\theta$ have overlap $\cos \theta$ and thereforeChoose one linear direction as ${\mathbf{e}}_L=(1,0{)}^T$ and a circular polarization as ${\mathbf{e}}_C=(1,\pm i{)}^T/\sqrt{2}$. Their overlap has magnitude $1/\sqrt{2}$, givingA non-polarizing beam splitter does not by itself guarantee equal arm losses; if $I_1\ne I_2$, multiply either boxed polarization factor by $2\sqrt{I_1{I}_2}/(I_1+I_2)$. Perfect temporal coherence does not compensate for polarization or intensity imbalance.

Use the same fixed laboratory $x,y$ axes for the incident and returning Jones vectors, with $x$ along the fast axis. Remove a common phase factor so that the one-pass Jones matrix is $J=\mathrm{diag}(1,i)$. The reverse pass through this reciprocal plate accumulates the same relative phase retardation, not its inverse. The mirror has scalar reflection coefficient $r_m$ in these fixed axes. ThereforeThe effective retardance is $\pi$: the complete return acts as a half-wave plate. For initial linear polarization at angle $\alpha$ to the fast axis,In particular, the $45^{\circ }$ input used for linear-to-circular conversion returns in linear polarization perpendicular to its original line: $(1,1)/\sqrt{2}\mapsto (1,-1)/\sqrt{2}$. An input along either plate axis returns along that same axis.

The wording does not specify the initial polarization of waves. Thus “orthogonal linear return” requires the usual $45^{\circ }$ linearly polarized input; it is not a universal answer for arbitrary polarized input. The Jones matrix above supplies the general answer. A non-normal metallic reflection may introduce different complex reflection coefficients for the two components and would require additional geometry and optical data. Descriptions of the intermediate circular polarization must also account for the reversed propagation direction when assigning handedness.

The characteristic polynomial of the given Jones matrix isso a convenient choice of eigenvalue labels isDirect multiplication gives $J{\mathbf{u}}_{\pm }={\lambda }_{\pm }{\mathbf{u}}_{\pm }$, and ${\mathbf{u}}_{+}^{†}{\mathbf{u}}_{-}=0$. They represent opposite circular polarizations. Since $a,b$ are real, $J$ is a Hermitian matrix and both eigenvalues are real, but they need not be positive. A negative eigenvalue adds a wave phase reversal to that mode. The labels are algebraic: ${\lambda }_{+}$ is the larger numerical eigenvalue only when $b\ge 0$. If $b=0$, the eigenvalue is degenerate and every nonzero Jones vector is an eigenvector; the circular pair remains a valid basis.

Take an incident field along $\hat{\mathbf{x}}$, without loss of generality because this component has the same circular eigenmodes for every direction of input linear polarization. Resolve its Jones vector into equal circular components:Using the complex amplitude convention $\mathrm{Re}[\mathbf{E}{e}^{-i\omega t}]$, the real circular contributions areThey counter-rotate and add toIn general the tip traces an ellipse with semiaxis lengths $|a|E_0$ and $|b|E_0$. Equal transmitted circular wave amplitudes produce a line; removal of one circular mode, ${\lambda }_{+}=0$ or ${\lambda }_{-}=0$, produces a circle. The special zero-output case has no defined polarization of waves.