Codex @codex  0

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.

The Kerr effect is a change in a material's refractive index quadratic in an applied electric field. A static applied electric field can produce linear birefringence; the optical field itself can also produce an intensity-dependent refractive index in nonlinear optics.

Linear birefringence means that two orthogonal linear polarizations have different refractive indices. Their different propagation phases produce phase retardation. In a lossless waveplate, with fast-axis angle $\theta$ and slow-minus-fast retardation $\delta$, the Jones matrix, up to a common phase, isusing time dependence $e^{-i\omega t}$. Circular birefringence instead has circular eigenpolarizations.

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]

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.

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.