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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.

Suppose the two input beams share frequency, electromagnetic polarization and overlapping transverse profiles at each output, with stable relative wave phase during the photodetector's averaging interval. For the ideal beam splitter scattering matrix above, expanding the squared complex amplitudes givesThe imaginary part terms describe wave interference. Applying the single-input power split to each input separately and adding the results would discard these terms. Their opposite signs ensure $P_1+P_2=|a_1{|}^2+|a_2{|}^2$: conservation of energy survives even when one output is dark.

For an original example, let $a_1=\sqrt{P}$, $a_2=\sqrt{P}{e}^{i\delta }$ and $P=1\mathrm{mW}$. Then $P_1=P(1-\sin \delta )$ and $P_2=P(1+\sin \delta )$:

At $+\pi /2$, $a_2=i\sqrt{P}$ makes $b_1=(\sqrt{P}-\sqrt{P})/\sqrt{2}=0$ and $b_2=i\sqrt{2P}$; at $-\pi /2$ the dark output swaps. Thus the two inputs can feed all their combined power into one output without gain. At zero relative input wave phase, the splitter's own relative wave phase leaves the contributions in quadrature, so neither output is dark.

For fluctuating fields, average the cross term as $\mathrm{Im}⟨a_1^{\ast }{a}_2⟩$. It vanishes for mutually uncorrelated inputs, making the independent power-splitting rule valid for the averaged output. Partial first-order optical coherence reduces it. Orthogonal linear polarizations eliminate the cross term in a polarization-insensitive power readout, and spatial mismatch reduces the overlap. Detector averaging can also erase rapid relative wave phase changes. A fixed relation between two inputs can be prepared by splitting one laser beam and recombining the two arms.

An actual teaching protocol in Michelson Interferometer & Precision Wavelength Measurement — Week 1, University of Colorado Boulder, §§4–6, starts with an ideal-model prediction, then builds a Michelson interferometer using a helium-neon laser, a nonpolarizing beam splitter, adjustable mirrors, a micrometer translation stage and a photodetector. Steering mirrors align beam overlap at near and far planes; far-plane interference fringes refine alignment. Moving the arm mirror changes the returning relative wave phase; detector voltage is watched with a live acquisition display or oscilloscope. The guide compares measured arm powers with 50:50 splitting and asks for a revised field model. It tests electromagnetic polarization dependence, optionally changing one arm with a quarter-wave plate, and estimates noise from repeated readings at fixed mirror position. These controls test whether fringe changes reflect coherent recombination rather than splitter imbalance or misalignment. The guide supplies a protocol, not a numerical dataset. Its detector noise, imperfect overlap and electromagnetic polarization mismatch limit displacement inference.^[ref]

Additional background subtraction controls follow from the field model: block both arms to record the detector offset and ambient light, then unblock each arm separately to establish its detected power at fixed gain. With both open, scan the mirror and compare extrema after subtracting that background. Keep the detector below saturation and repeat the scan to check drift. A changing signal alone cannot distinguish source optical coherence from motion or changing overlap. For a vacuum round trip, a mirror displacement $x$ changes relative wave phase by $4\pi x/\lambda$; the useful signal is its modulation after detection, not an optical-frequency oscillation that the detector must resolve.

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.

The electro-optic Kerr effect produces linear birefringence proportional to the square of a static applied electric field. Defining the Kerr effect coefficient $K$ by $\Delta n={\lambda }_{0}K{E}^2$, a length $L$ produces phase retardation $\delta =2\pi KL{E}^2$. A switchable quarter-wave plate requires $|\delta |=\pi /2$ modulo $2\pi$, with fixed principal axes. This differs from the Faraday effect, whose eigenpolarizations are circular polarizations.

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.

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.

Alfred Kastler's 1966 lecture recalls his 1950 proposal and distinguishes the early ${}^{23}\mathrm{Na}$ atomic beam work from later sealed-vapour experiments. Its Fig. 5 shows subsequent sodium radio wave resonances, including multiple-photon transitions.^[ref]

The 1952 demonstration by Jean Brossel, Alfred Kastler and Jacques Winter used a sodium atomic beam, lamp light with circular polarization and fluorescence readout. Jean Brossel was a central experimental contributor; Alfred Kastler received the 1966 Nobel Prize in Physics alone.^[ref][ref]

The lecture also describes a ${}^{199}\mathrm{Hg}$ experiment: a pumping beam with circular polarization prepares the vapour, while a perpendicular probe converts transverse magnetic dipole moment Larmor precession into modulated absorption at a photodetector. Interrupting pump and RF together exposes decaying free Larmor precession. Its envelope measures transverse relaxation under that model; it is distinct from the sodium atomic beam fluorescence signal.^[ref]

MIT's later rubidium experiment pumps natural ${}^{85}\mathrm{Rb}$/${}^{87}\mathrm{Rb}$ with a discharge lamp filtered to D1 ($5S_{1/2}\to 5P_{1/2}$, near $795\mathrm{nm}$). A polarizer and quarter-wave plate precede a heated rubidium/neon cell. Helmholtz coils set a weak static magnetic field. Separate transverse RF mixing increases absorption, measured by photodiode transmission.^[ref]

Compare resonant and detuned RF with the optical beam fixed, then change the calibrated static magnetic field: a true magnetic sublevel resonance should move with its Zeeman splitting. Lamp drift, stray magnetic fields, imperfect circular polarization, optical attenuation and RF power broadening can change the observed signal. Consequently, a transmission dip alone does not determine every sublevel population. These are measurement-design considerations, not additional reported historical observations.

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.

Use $Z|H⟩=|H⟩$, $Z|V⟩=-|V⟩$. Choose the eigenstate phasesThe inverse relations are $|H⟩=(|H^{\prime }⟩+|V^{\prime }⟩)/\sqrt{2}=(|R⟩+|L⟩)/\sqrt{2}$ and $|V⟩=(|H^{\prime }⟩-|V^{\prime }⟩)/\sqrt{2}=-i(|R⟩-|L⟩)/\sqrt{2}$. The X states are linearly polarized at $+45^{\circ }$ and $-45^{\circ }$ to horizontal, while the Y states have equal orthogonal amplitudes in quadrature and hence opposite circular polarizations. We define R as the $+Y$ state; descriptions as clockwise or anticlockwise must specify the observation direction and time convention.

A polarizing beam splitter and two single-photon detectors directly measure Z, routing H and V to separate detectors. For X, place a half-wave plate with its axis at $22.5^{\circ }$ before that splitter; it implements a Hadamard transformation up to a global phase. For Y, first apply a quarter-wave retardance $Q=\mathrm{diag}(1,-i)$ in the H/V axes and then the same half-wave plate. The total analyzer transformation is $HQ$, with $HQ|R⟩=|H⟩$ and $HQ|L⟩=|V⟩$. The physical fast/slow-axis assignment is selected to realize this stated retardance.

A transverse static electric field can induce linear birefringence through the electro-optic Kerr effect. Define its coefficient by $\Delta n={\lambda }_{0}K{E}_{\mathrm{dc}}^2$, with the principal axes parallel and perpendicular to the applied field. A length $d_K$ then gives ${\delta }_K=2\pi K{d}_K{E}_{\mathrm{dc}}^2$.

For input polarized at $45^{\circ }$ to those axes, switching from zero retardation to $|{\delta }_K|=\pi /2$ switches linear to circular polarization. Other input angles generally become elliptical, and axis-aligned input stays linear. The required field magnitude for the lowest-order setting isA Kerr cell is suitable as a switchable quarter-wave plate. The field can be changed electronically without mechanically rotating the axes; actual switching time depends on the material response and driving circuit. This is induced linear birefringence, not polarization-selective absorption.^[ref]

The longitudinal Faraday effect produces circular birefringence: right- and left-circular components have different phase velocities. Linear input contains equal amplitudes of both circular eigenpolarizations. A relative phase ${\delta }_F$ rotates its linear polarization through ${\delta }_F/2$, preserving linearity in an ideal transparent sample. Conventionally the signed rotation is ${\vartheta }_F=V{B}_{\parallel }{d}_F$, with Verdet constant $V$.

A pure Faraday rotator cannot serve as the required switchable quarter-wave plate. It rotates a line into another line rather than producing the required relative phase between fixed orthogonal linear components. Applying a field that gives a nominal quarter-cycle circular-eigenmode retardation simply rotates linear polarization by $45^{\circ }$. Circular dichroism or a separate linear retarder would change the device, and is not the stated pure Faraday mechanism.

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.

For two-element polarization identification, assume ideal elements and use an intensity detector. For each source, test both element orders while rotating their relative axes. The circular source followed by the quarter-wave plate and then the polarizer is the combination that gives full intensity modulation. The plate converts circular light to linear light, and the downstream polarizer then obeys Malus's law.

If the polarizer comes first, it transmits half the circular source's intensity and the downstream lossless plate cannot change that intensity. Unpolarized light also gives constant half-intensity in either order: a wave plate leaves it unpolarized, while a polarizer followed by a wave plate has no final analyzer. Thus the modulating source is circular, its first element is the quarter-wave plate, and its second is the polarizer; the remaining source is unpolarized. Testing both orders avoids needing to know either source's original brightness.

Birefringence is polarization-dependent propagation in an optically anisotropic medium. Its permittivity tensor relates $\mathbf{D}$ to $\mathbf{E}$. In a lossless uniaxial dielectric, take the optic axis along $z$ and the principal relative permittivities as $(n_o^2,n_o^2,n_e^2)$. For a wave vector making angle $\vartheta$ with that axis, the ordinary mode has index $n_o$, while the extraordinary phase-index angular dependence isThe ordinary electric field is perpendicular to the plane containing the wave vector and optic axis. The extraordinary electric displacement field lies in that plane and is perpendicular to the wave vector, but its electric field need not be. Its energy-flow direction can therefore differ from the wave-normal direction: birefringent walk-off. Along the optic axis the two indices coincide. Refraction generally splits an incident beam into two differently polarized rays, producing double images, as in calcite.

A waveplate instead uses the relative phase of two eigenpolarizations. With the axes chosen for the propagation direction, the retardance is $\Delta =2\pi L(n_2-n_1)/{\lambda }_0$. A quarter-wave plate gives $|\Delta |=\pi /2$ modulo $2\pi$ and converts equal-amplitude orthogonal components between linear and circular polarization. A half-wave plate gives $|\Delta |=\pi$ and rotates a linear polarization. This is a phase effect; unequal absorption is a separate property.^[ref]

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.

For propagation along $z$, a transverse wave has two perpendicular displacement components. With $\varphi =kz-\omega t$, write $u_x=A\cos \varphi$, $u_y=B\cos (\varphi +\delta )$. Their trajectory at a fixed point obeysThe general nondegenerate trajectory is an ellipse: elliptical polarization. For $\delta =0$ or $\pi$, the components remain in phase or antiphase and the trajectory collapses to a line. A single nonzero component is also linearly polarized. Equal amplitudes $A=B$ with $\delta =\pm \pi /2$ give circular polarization. Unequal quadrature amplitudes give an ellipse aligned with the chosen axes, while general phase differences rotate its axes.

For the displayed convention and $\delta =+\pi /2$, at $z=0$ the vector starts at $+x$ and moves towards $+y$ as time increases. Reversing $\delta$ reverses its rotation. Right- and left-handed labels require a stated viewing and phase convention, so rotation is defined explicitly here. A fixed ellipse is a coherent relationship between transverse components; unpolarized light instead has fluctuating orientation and phase relationships on the measurement timescale.

A linear polarizer transmits a linearly polarized beam with intensity proportional to ${\cos }^2\theta$, allowing its direction to be measured by an analyser. circular polarization transmits equal intensity through any linear analyser, but that test alone does not distinguish it from unpolarized light. A quarter-wave plate converts a suitable coherent circular state to linear polarization, allowing the combination to discriminate them. For light the electric field defines the usual polarization; the magnetic field remains perpendicular to it and to propagation. Longitudinal sound in a homogeneous gas lacks the two transverse degrees of freedom needed for these polarization states.

Let $R(\theta )$ be the real rotation matrix with columns along the plate's fast and slow axes. Expressing the input in those axes, applying the principal retardance, then rotating back gives the rotated half-wave plate matrixEach fixed quarter-wave plate has $Q=\mathrm{diag}(1,i)$ in the supplied convention. Propagation through the three plates therefore givesThe order matters: this is a product of three propagating plates, not a similarity transformation $QH{Q}^{-1}$. For input Jones vector ${\mathbf{e}}_{+}=(1,1{)}^T/\sqrt{2}$, the Pauli matrix has eigenvalue $+1$, so $J{\mathbf{e}}_{+}=e^{i2\theta }{\mathbf{e}}_{+}$. The final polarization remains at 45 degrees, but its phase depends on the rotating angle.

Take the physical field as the real part of $\mathbf{E}{e}^{-i\omega t}$ and let $\theta =\Omega t$. ThenThus the rotating half-wave plate frequency shift hasThe magnitude is the stated $2\Omega$ shift. Its sign follows the explicit time and rotation conventions above; reversing the physical rotation or using the orthogonal linear input reverses the shift. A time-phasor convention must be changed consistently with the plate matrices, rather than used alone to change a physical redshift into a blueshift.

The physical origin is a rotational Doppler effect and angular-momentum exchange. In the stated convention, the first fixed quarter-wave plate changes the 45-degree input into circular polarization of spin $+\hslash$ per photon. The rotating half-wave plate flips it to spin $-\hslash$, transferring $2\hslash$ to the rotating plate. Maintaining its rotation takes mechanical energy $2\hslash \Omega$ from each photon, agreeing with the redshift. The final fixed plate restores linear polarization; fixed plates can exchange angular momentum with their supports without the rotating element's corresponding mechanical work. The instantaneous Jones treatment assumes slow rotation compared with optical and transit-time scales and ideal retardances over the small shifted bandwidth.

For the overlapping monochromatic wave components, let $d$ be the thickness and ${\lambda }_0$ the vacuum wavelength. Define $\Delta n=n_s-n_f>0$ using the slow and fast refractive indices. With complex amplitude convention $e^{i(kz-\omega t)}$, the accumulated slow-minus-fast phase difference isThis is the optical path difference $\Delta nd$ expressed as a wave phase. It is a signed accumulation convention; with the opposite time convention the sign reverses. The outgoing polarization of an electromagnetic wave depends on $\delta$ modulo $2\pi$. Here the positive slow-minus-fast $\Delta n$ differs from the signed principal-index $n_e-n_o$ used in the parent birefringence entry when $n_e<n_o$.

For an original worked example, stipulate $\Delta n=0.0100$ at ${\lambda }_0=600\mathrm{nm}$ and a uniform thickness $d=15.0\mu \mathrm{m}$. These are design assumptions, not measured material data. The optical path difference is $150\mathrm{nm}$, soIt is a quarter-wave plate. At the same wavelength and refractive indices, doubling $d$ to $30.0\mu \mathrm{m}$ gives $\delta =\pi$, a half-wave plate. A $1.0\mu \mathrm{m}$ thickness increase changes $\delta$ by $\pi /30$, or $6^{\circ }$; retaining the same birefringence at $500\mathrm{nm}$ would instead give $\delta =3\pi /5$ for the $15.0\mu \mathrm{m}$ plate. Real refractive indices also vary with wavelength, so this last calculation explicitly neglects optical dispersion.

For these equal transmitted wave amplitudes, orthogonal projection onto a polarizer crossed with the initial $45^{\circ }$ linear polarization gives a simple independent readout. After removing a common wave phase, the component complex amplitudes are proportional to $(1,e^{i\delta })/\sqrt{2}$; orthogonal projection onto $(1,-1)/\sqrt{2}$ gives $(1-e^{i\delta })/2$. Squaring its modulus yields $I_{\perp }/I_{\mathrm{in}}={\sin }^2(\delta /2)$: zero at $d=0$, one half at $15\mu \mathrm{m}$, and one at $30\mu \mathrm{m}$. This assumes ideal polarizers, negligible specular reflection and wave absorption, overlapping components and stable relative wave phase. It measures a periodic function of phase retardation, so one brightness measurement cannot determine the thickness uniquely.