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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.

The Faraday effect rotates the plane of linear polarization when light propagates through a material with a magnetic field component along the ray. In the weak-field transparent regime,with a stated propagation and transverse-basis convention; $V$ is the Verdet constant. For a uniform longitudinal field and sample length $L$, the magnitude is $|VBL|$. A scalar intensity change without a corresponding polarization analysis would not establish this effect.

The microscopic field breaks the equivalence of the two circular polarization modes: the Lorentz force and the Zeeman shifts of optical transitions produce different susceptibilities for their opposite senses of rotation. A linearly polarized wave is an equal-amplitude superposition of these two modes. If their real refractive refractive indices are $n_{+},n_{-}$, their relative wave phase after the sample is $\Delta \varphi =(\omega L/c)(n_{+}-n_{-})$. Recombination rotates the line byEqual transmitted circular amplitudes preserve linear polarization. Unequal absorption produces circular dichroism and generally elliptical polarization as well, so pure rotation is an approximation appropriate away from strong absorption.

A modern motivating experiment sends polarized laser light through a glass rod inside a solenoid, then through a rotatable analyzer onto a photodiode. Scan the analyzer first to calibrate its angular response. Driving the solenoid sinusoidally and recording the detector synchronously separates a small rotation signal from unsynchronized noise; the conversion to angle still requires the analyzer slope and the magnetic-field integral through the rod. A first-party William & Mary laboratory manual specifies this apparatus and measurement procedure.^[ref]

Additional controls are equal-magnitude field reversals, zero-field readings before and after each pair, and a blocked-beam pickup test. The odd-in-field rotation is $[{\theta }_{\mathrm{obs}}(B)-{\theta }_{\mathrm{obs}}(-B)]/2$; a fixed analyzer offset cancels. Heating, stressed-glass birefringence, source drift and field-dependent transmission need separate checks. A single reading near crossed extinction loses the sign because its leading response is quadratic; an analyzer near $45^{\circ }$ gives a linear small-rotation response.

Faraday rotation is nonreciprocal. With the magnetic bias held fixed, retracing the path adds rotation in the same laboratory sense rather than undoing it, unlike reciprocal natural optical activity. This permits an optical isolator with a $45^{\circ }$ Faraday rotator between appropriately oriented polarizers. One must compare the returned polarization in a consistently defined laboratory basis; merely reversing the sign of $\mathbf{B}\cdot \hat{\mathbf{k}}$ while retaining the old ray-basis convention gives an incorrect cancellation.

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.

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.

Birefringence gives two orthogonal linear eigenpolarizations with different refractive indices, $n_f$ and $n_s$. They accumulate different wave phases through the same physical thickness $d$. Their relative phase retardation isHere ${\lambda }_0$ is the vacuum wavelength. For the thinnest positive-retardance plate, $d={\lambda }_0/[4(n_s-n_f)]$. The plate uses a difference in phase velocity, rather than differential wave absorption. Equal wave amplitudes along the two axes, obtained from input linear polarization at $45^{\circ }$ to them, acquire phase quadrature and become circular polarization. Reversing that conversion gives linear polarization from the corresponding circular input. A different input angle generally gives elliptical polarization on one pass.

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.

When ${\gamma }^{\pm }={\alpha }^{\pm }$ are real, $\gamma$ and $\delta$ are real. The two opposite circular polarizations retain their wave phases but are attenuated at different wave amplitude rates. This circular dichroism changes the ellipticity without rotating the principal axis of the ellipse.

For input $E_0\hat{\mathbf{x}}$, the corrected Jones matrix givesThe common factor $e^{-\gamma z}$ describes the mea wave amplitude attenuation. The $\cosh (\delta z)$ factor reflects the sum of the two transmitted modal wave amplitudes; $\tanh (\delta z)$ fixes their imbalance and hence the signed ellipticity. The real field has components proportional to $\cosh (\delta z)\cos \omega t$ and $\sinh (\delta z)\sin \omega t$, soThe ellipse's principal axis stays along the original input direction. For large differential optical depth it approaches the less attenuated circular eigenmode. A passive medium has ${\alpha }^{\pm }\ge 0$, or $\gamma \ge |\delta |$, so the growing hyperbolic function factor does not imply net optical gain. Using the printed $\mathrm{sech}$ prefactor would give a spurious extra attenuation inconsistent with the modal laws.

Birefringence means that two independent polarization eigenmodes have different refractive indices at the same frequency and propagation direction. The rotatory-power formula relevant here requires circular birefringence: the eigenmodes are opposite circular polarizations. General linear birefringence instead introduces phase retardation between linearly polarized components and need not rotate every incident linear polarization through one common angle.

Use the complex amplitude convention $\mathrm{Re}[\mathbf{E}{e}^{-i\omega t}]$ and transverse unit vectors ${\mathbf{e}}_1,{\mathbf{e}}_2$. DefinePropagation through thickness $d$ adds phases ${\varphi }_{\pm }=(2\pi /{\lambda }_0)n_{\pm }d$, where ${\lambda }_0$ is the vacuum wavelength. Recombining the two components givesThus the rotation and the specific rotatory power per unit length areThe units of $\rho$ are $\mathrm{rad}{\mathrm{m}}^{-1}$. The sign follows the specified circular basis and viewing convention. Equal attenuation of the two eigenmodes has been assumed; unequal attenuation also produces elliptical polarization.

Flint glass is approximately optically isotropic without an applied magnetic field. A longitudinal magnetic field changes the motion of its bound electrons: the magnetic part of the Lorentz force acts with opposite sense on the two circularly rotating components of a driven electronic oscillation. Their optical response, and hence their refractive indices, split. Equivalently, the transverse permittivity tensor acquires a field-odd antisymmetric contribution whose eigenvectors are opposite circular polarizations.

The resulting circular birefringence gives Faraday rotation. For a weak longitudinal bias it is linear in the field:Here ${\rm Y}$ is the Verdet constant and $L$ the path length. The magnetic field breaks time-reversal symmetry, which is why this rotation is nonreciprocal. Reversing the bias reverses the rotation; reversing propagation while keeping the bias fixed does not undo it when physical polarization directions are compared in fixed laboratory axes. A transverse magnetic bias can have different, higher-order effects; the setup here is the longitudinal Faraday effect.

A carrier in a locally parabolic isotropic band obeys $m^{\ast }\dot{\mathbf{v}}=q\mathbf{v}\times \mathbf{B}$, soIlluminate a high-mobility crystal with a transverse alternating electric field, sweep a calibrated static magnetic field at known microwave or far-infrared frequency, and detect resonant absorbed power. Cooling lengthens the relaxation time; resolving the resonance generally needs ${\omega }_c\tau \gg 1$. Varying field direction measures band anisotropy, while circular polarization distinguishes the two senses of rotation for opposite carrier charges. This is orbital motion rather than electron spin resonance.

The classic silicon/germanium measurements used single crystals near $4\mathrm{K}$, light modulation and circularly polarized microwaves, with directional scans to infer anisotropic carrier-energy surfaces.^[ref] Carrier generation, calibration, scattering linewidth and inhomogeneity must be separated from a genuine mass shift.

In a general band the cyclotron mass is an orbit property, $m_c={\hslash }^2(\partial A_k/\partial E)/(2\pi )$ for the appropriate field-normal orbit. It need not equal every component of the band-curvature mass tensor. For an ellipsoidal parabolic band with field along a principal $z$ direction it is $\sqrt{m_x{m}_y}$. Thus a single resonance does not determine a whole effective-mass tensor without angular information.

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.

The permittivity tensor is the linear material-response matrix relating electric displacement field to electric field at a specified frequency. An isotropic local medium has ${ϵ}_{ij}=ϵ{\delta }_{ij}$. A magnetically biased transparent medium can have an imaginary antisymmetric transverse contribution; its eigenmodes can have opposite circular polarizations and different refractive indices.

The polarization ellipse is the curve traced by the transverse electric field of a monochromatic wave at a fixed point. Its orientation and signed ratio of minor to major semiaxis specify a pure polarization. A zero minor semiaxis gives linear polarization, and equal semiaxes give circular polarization.

Viewed perpendicular to the magnetic field, the normal Zeeman effect has a $\pi$ component with linear polarization parallel to the magnetic field and two $\sigma$ components polarized perpendicular to it. Along the magnetic field, $\pi$ vanishes and the $\sigma$ pair has opposite circular polarizations. Pieter Zeeman described these geometry-dependent patterns in his lecture.^[ref]

The geometry can be derived from electric dipole radiation. Let the observation unit vector be $\mathbf{n}=(\sin \theta ,0,\cos \theta )$, with $\theta$ measured from the magnetic field. Radiation retains only the transverse part of the oscillating electric dipole moment $\mathbf{d}$:For $\pi$, take $\mathbf{d}\parallel \hat{\mathbf{z}}$; its angular light intensity factor is ${\sin }^2\theta$, so it vanishes on axis. For ${\sigma }^{\pm }$, use normalized complex dipoles $(\hat{\mathbf{x}}\pm i\hat{\mathbf{y}})/\sqrt{2}$. Their angular factors are $(1+{\cos }^2\theta )/2$. Projection gives circles on axis, lines at $\theta =\pi /2$, and generally elliptical polarization between. These factors assume equal dipole magnitudes; actual line light intensities also depend on initial quantum state populations, transition strengths and collection geometry. No absolute handedness name is needed: it depends on the viewing and phase conventions.

A transverse polarizer can separate $\pi$ from $\sigma$ even before the spectral line components are fully resolved. Along the magnetic field, a quarter-wave plate followed by a polarizer converts the two opposite circular polarizations into distinguishable linear settings. This explains why polarization analysis carried information beyond a small increase in line width.

For perfectly temporally coherent beams with unit Jones vectors ${\mathbf{e}}_1,{\mathbf{e}}_2$, their interference term is proportional to the complex inner product ${\mathbf{e}}_1^{†}{\mathbf{e}}_2$. Their interferometric visibility is the displayed expression. Equal beam light intensities reduce it to the magnitude of the polarization overlap. Two linear polarizations at relative angle $\theta$ give $|\cos \theta |$; a linear and a circular polarization give $1/\sqrt{2}$. Perfect coherence does not remove loss of contrast from unequal intensities or orthogonal polarizations.

Polarization of waves describes the direction and time-dependent geometry of a transverse wave's vector during its oscillation at a fixed position. In three spatial dimensions, a transverse wave can have two independent directions of oscillation, both perpendicular to its direction of propagation. For polarization of an electromagnetic wave, the relevant vector is the transverse electric field; for a mechanical transverse wave, it can be the transverse displacement. A purely longitudinal wave has no choice between these transverse directions, and the orientation of a static electric field is not polarization of waves.

Choose transverse axes $x,y$ in a Cartesian coordinate system. A monochromatic wave with a fixed phase difference has the real electric fieldHere $A_x,A_y$ are wave amplitudes, $\omega$ is the angular frequency, and $\delta$ is the phase difference of the $y$ oscillation relative to the $x$ oscillation. The vector's tip traces its path during the wave period $T=2\pi /\omega$. Both plotted coordinates are real electric field components; neither axis represents the imaginary part of a complex amplitude.

For nonzero wave amplitudes, $\delta =0$ or $\pi$ modulo $2\pi$ gives linear polarization: the vector moves back and forth along a fixed line. Equal nonzero wave amplitudes and $\delta =\pm \pi /2$ modulo $2\pi$ give circular polarization: the tip traces a circle of constant radius. Unequal nonzero wave amplitudes with $\delta =\pm \pi /2$ give a noncircular ellipse whose axes coincide with $x,y$. More generally, nonzero wave amplitudes and $\sin \delta \ne 0$ give elliptical polarization, with circular polarization as a special case. If exactly one wave amplitude vanishes, the nonzero wave has linear polarization for every $\delta$; if both vanish, there is no nonzero wave to assign a polarization of waves. ^[ref].

For example, take $\omega =2\pi {\mathrm{s}}^{-1}$, so the wave period is $T=1\mathrm{s}$. The choices $(A_x,A_y,\delta )=(2,1,0)$, $(1,1,\pi /2)$ and $(2,1,\pi /2)$, with electric field wave amplitudes in $V/m$, produce respectively the line segment $E_y=E_x/2$, the circle $E_x^2+E_y^2=1$ and the ellipse $E_x^2/4+E_y^2=1$, when coordinates are measured in $V/m$. At $t=T/4$, their tips are respectively $(0,0)$, $(0,-1)$ and $(0,-1)V/m$. The last two share a point without having the same path.

Resolving the electric field into two transverse components makes polarization of waves useful: the same frequency and wave amplitudes can describe different paths when their phase difference changes. An ideal linear polarizer selects the component along its pass axis, letting its orientation probe the direction of linear polarization through Malus's law. A fixed pair of wave amplitudes and a fixed phase difference describe one stable path; unpolarized light requires a statistical description rather than one fixed ellipse.

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.