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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.

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.

A half-wave plate reverses the relative sign of the two components. For initial linear polarization at angle $\alpha$ to the fast axis, the outgoing line is at $-\alpha$ modulo $\pi$. The $45^{\circ }$ input therefore rotates by $90^{\circ }$.

In quantum dense coding, a shared Bell pair lets Alice send two classical bits by transmitting one qubit. Use photon polarization as the computational basis, $|0⟩=|H⟩$, $|1⟩=|V⟩$, and define the Bell statesAlice and Bob initially share $|{\varphi }^{+}{⟩}_{AB}$. Alice encodes a two-bit message $(m,n)$ by applying the Pauli operator $X^n{Z}^m$ to her qubit:The last minus sign is a global phase. Alice transmits her photon to Bob, who now holds both qubits and uses a Bell-basis measurement to identify the message. The four Bell states are orthogonal, so an ideal Bell state analyzer decodes them without error. The resource cost is one previously shared Bell pair and one transmitted qubit per two classical bits; the shared quantum entanglement is consumed. Before transmission Bob's reduced density matrix is $I/2$ for every message, consistent with the no-communication theorem.

The BBO crystal supplies the polarization-entangled photon pair. An ultraviolet laser pumps this nonlinear optical crystal; spontaneous parametric down-conversion produces signal and idler photons with ${\omega }_p={\omega }_s+{\omega }_i$ and phase matching ${\mathbf{k}}_p={\mathbf{k}}_s+{\mathbf{k}}_i$. In a noncollinear type-II phase matching arrangement, collect the intersections of the two emission cones. The alternatives $|H{⟩}_A|V{⟩}_B$ and $|V{⟩}_A|H{⟩}_B$ must be indistinguishable in arrival time, spectrum and spatial mode. Compensation of birefringence and adjustment of the relative quantum phase then give $(|HV⟩+|VH⟩)/\sqrt{2}$; a local Pauli X gate converts it to $|{\varphi }^{+}⟩$. Half-wave plates and other waveplates implement Alice's Pauli gates. The BBO is the pair source; the message is encoded subsequently by the local optical elements.^[ref]

A basic passive linear optics Bell state analyzer with no extra resources identifies only two of the four polarization Bell states unambiguously. Thus the ideal two-bit capacity presupposes complete Bell-basis measurement; that basic optical implementation alone does not attain it deterministically.

The beta barium borate crystal supplies spontaneous parametric down-conversion: a weak ultraviolet pump can produce a pair of lower-frequency photons, with their frequencies and wavevectors selected by conservation of energy and phase matching. A suitably aligned and compensated source makes the two polarization alternatives coherent, giving the stated Bell state. The crystal alone does not guarantee that every collected pair has that state.

The two polarizing beam splitters route horizontal polarization and vertical polarization to different ports. Take $H$ from arm $a$ to the trigger $T$ and $V$ to arm $c$. The ordinary beam splitter in arm $b$ is a lossless, polarization-independent 50:50 splitter, with the unused input in the vacuum state. It splits probability amplitudes, rather than measuring which output a photon takes. The half-wave plate changes the relative phase of its principal polarizations by $\pi$; orient it so that a $V$ photon in $c$ becomes $(H+V)/\sqrt{2}$ in $c^{\prime }$, up to a common phase. Finally, photodetectors absorb arriving photons and supply electrical clicks. They need not resolve photon number for the ideal four-photon, fourfold-coincidence calculation: a click at every detector already requires one photon at each.

Choose the relative wave phases so that the final polarizing beam splitter sends $H_{c^{\prime }}$ to $H_2$, $V_{c^{\prime }}$ to $V_1$, $H_d$ to $H_1$ and $V_d$ to $V_2$. Such phase conventions must be used consistently: an uncompensated relative phase would give a phase-shifted GHZ state.

Use the two-pair component of polarization-entangled down-conversion to account for identical bosons in the same collected modes. Write $a_H^{†}$ for a creation operator in mode $a$ with horizontal polarization, and similarly for the other modes. For the plus Bell state, the pair-creation operator is $K^{†}=a_H^{†}{b}_V^{†}+a_V^{†}{b}_H^{†}$. The normalized four-photon state isThese are normalized Fock states. In particular, $(a_H^{†}{)}^2|0⟩=\sqrt{2}|2H_a⟩$; omitting this factor would misnormalize the four-photon input.

The first polarizing beam splitter sends $a_H^{†}$ to $T_H^{†}$ and $a_V^{†}$ to $c_V^{†}$. Only the middle term can give exactly one trigger photon and leave another photon in $c$. In the first term both $a$ photons reach $T$ and none reaches $c$; in the last there is no trigger. Thus the relevant, still unnormalized component is

For the selected paths of the half-wave plate and beam splitter, takeHere $e$ labels the output towards $D_3$. An immaterial common reflection phase has been absorbed in the mode definitions. Keeping one photon at $D_3$ leavesAt the final polarizing beam splitter, the $H_{c^{\prime }}{H}_d$ alternative yields one $H$ at each of $D_1,D_2$, while $V_{c^{\prime }}{V}_d$ yields one $V$ at each. The two opposite-polarization alternatives send both photons to the same detector and fail fourfold coincidence counting. The accepted component is thereforeAfter conditioning and normalization, the three detected output modes haveThe optical mode matching must erase any timing, frequency or other record that distinguishes the two alternatives. Otherwise the same polarization populations could describe an incoherent mixed quantum state, rather than this quantum superposition. The Bouwmeester three-photon GHZ experiment uses this principle. The photonic GHZ preparation by fourfold coincidence is a measurement in the coincidence basis: detecting all four photons projects and consumes them. A click at $T$ by itself does not herald a guaranteed freely available three-photon state.

Encode $|0⟩=|H⟩$ and $|1⟩=|V⟩$ in two orthogonal linear polarizations. Then $|\pm ⟩=(|H⟩\pm |V⟩)/\sqrt{2}$ are linear polarizations at opposite forty-five-degree angles, while $(|H⟩\pm i|V⟩)/\sqrt{2}$ are the two circular polarizations. Their right/left names depend on the viewing convention. A half-wave plate with its axis at $22.5^{\circ }$ to H implements H up to a common phase, since its Jones matrix has entries $\cos 2\theta ,\sin 2\theta$ and $-\cos 2\theta$.

In principle, a nonlinear interaction producing $e^{i\pi n_{1V}{n}_{2V}}$ supplies a deterministic Controlled-Z gate: only the joint VV component acquires a minus sign. A strongly coupled matter mediator can also provide the required conditional interaction. At optical single-photon energies the direct nonlinear interaction is weak; loss, unwanted noise and maintaining indistinguishable spatial and temporal modes make a high-fidelity deterministic gate difficult. Passive linear optics by itself does not supply this deterministic entangling interaction; measurement-based optical gates introduce additional resources and success conditions.

For the printed circuit, the first three H gates create $|+{⟩}^{\otimes 3}$. The two controlled-Z operations from the first qubit to the others giveThe final H gates on the last two qubits therefore produce the GHZ stateDefine $|s_X⟩=(|0⟩+s|1⟩)/\sqrt{2}$ and $|s_Y⟩=(|0⟩+is|1⟩)/\sqrt{2}$ for $s=\pm 1$. In any of the XYY, YXY and YYX bases, direct overlaps givewith the same coefficient formula when the X position is permuted. Hence the three explicit expansions can be written compactly asIn each sum the four sign triples are $(+,+,-),(+,-,+),(-,+,+),(-,-,-)$. They have equal probabilities $1/4$. Thus two Y outcomes determine the X outcome with certainty: their product times the X outcome is $-1$.

In the XXX basis,The four allowed triples are $(+,+,+),(+,-,-),(-,+,-),(-,-,+)$, so the XXX outcome product is $+1$.

The GHZ theorem follows without a statistical inequality. If local predetermined values $X_i,Y_i\in \{\pm 1\}$ reproduced the three mixed-basis correlations, multiplying them would givebecause every Y value is squared. Quantum mechanics instead predicts the XXX product $+1$. This is an all-versus-nothing contradiction with local realism using ideal perfect correlations. Standard two-qubit Bell-state tests compare statistical correlations through a Bell inequality; the GHZ construction makes the inconsistency visible already at the level of these deterministic parity constraints.

Taking the partial trace over either photon of the pure pair state givesThe two Schmidt coefficients are both $1/\sqrt{2}$, so the entanglement entropy is one bit for every $\phi$. A local phase gate on Alice maps $|{\Psi }^0⟩$ to $|{\Psi }^{\phi }⟩$, proving local-unitary equivalence. In strict four-vector Bell-basis terminology only $\phi =0,\pi$ give the standard $|{\Psi }^{+}⟩,|{\Psi }^{-}⟩$; every phase nevertheless gives a maximally entangled Bell-type state.

The eigenpolarizations of the requested observable areA polarizing beam splitter with its transmitted polarization axis at angle $\alpha /2$ to horizontal separates these two orthogonal states. Assign outcome $+1$ to its first detector and $-1$ to the other. Equivalently, use a fixed horizontal/vertical splitter preceded by a half-wave plate whose fast axis is at $\alpha /4$; the diagram shows this equivalent implementation. Bob uses $\beta /2$ for his splitter axis, or $\beta /4$ for the preceding half-wave plate.

The pair correlations are $⟨Z\otimes Z⟩=-1$, $⟨X\otimes X⟩=\cos \phi$, and $⟨Z\otimes X⟩=⟨X\otimes Z⟩=0$. Expanding the two observables therefore givesThe half-angle between the Bloch observable parameter and the physical linear-polarization analyser angle is essential.

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.

On Bob's polarization qubit, takeSuitable conditional unitary operators areIndeed $(I\otimes XZ)|{\Phi }^{+}⟩=|{\Psi }^{-}⟩$, $(I\otimes X)|{\Phi }^{-}⟩=|{\Psi }^{-}⟩$, and $(I\otimes Z)|{\Psi }^{+}⟩=-|{\Psi }^{-}⟩$. The last minus sign is a harmless global phase.

For an ideal half-wave plate whose fast axis makes angle $\theta$ with horizontal polarization, discard the common phase and use the Jones matrixIt implements $Z$ at $0^{\circ }$ and $X$ at $45^{\circ }$. Thus for outcome ${\Phi }^{+}$ use two half-wave plates, first $0^{\circ }$ then $45^{\circ }$ in the propagation direction; for ${\Phi }^{-}$ use one at $45^{\circ }$; for ${\Psi }^{+}$ one at $0^{\circ }$; for ${\Psi }^{-}$ use no plate. Ideal retardance is assumed at the photon wavelength, and the setting is chosen after receiving the detector's classical result.

An ideal cross-Kerr interaction between two single modes is $H_{\mathrm{int}}=\hslash \chi n_1{n}_2$, with $n_j=a_j^{†}{a}_j$. The number operators commute with the interaction, so this coupling changes phases without transferring photons between the modes. To implement the specified positive conditional phase, usewhere $n_{2V}=|V⟩⟨V|$ within the one-photon polarization subspace. The horizontal branch has zero interacting occupation. Since $e^{i\Theta n}|\alpha ⟩=|e^{i\Theta }\alpha ⟩$, the specified conditional rotation follows. This is an ideal lossless single-mode model; an actual travelling-wave medium's bandwidth, loss and noise are not established by writing this Hamiltonian.

Use $X_1=(a+a^{†})/2$ and $X_2=(a-a^{†})/(2i)$. A coherent state has centre $(\mathrm{Re}\alpha ,\mathrm{Im}\alpha )$ and variance $1/4$ in either optical quadrature. Multiplication of $\alpha$ by $e^{i\Theta }$ rotates that centre anticlockwise through $\Theta$, keeping the coherent state noise disk unchanged. For real positive $\alpha$ the two centres are $(\alpha ,0)$ and $(\alpha \cos \Theta ,\alpha \sin \Theta )$.

Normalize the initial polarization of an electromagnetic wave qubit. The output of the medium isThe joint state is generally an entangled state. The overlap magnitude is $|⟨\alpha |\beta ⟩|=e^{-2|\alpha {|}^2{\sin }^2(\Theta /2)}$. The state factorizes when the amplitudes coincide, while nearly distinguishable amplitudes make the polarization nearly maximally entangled with the optical mode. Before any polarization conditioning, tracing out mode 2 gives the mixture $(|\alpha ⟩⟨\alpha |+|\beta ⟩⟨\beta |)/2$, with no interference between branches.

Include the normalization factors in the half-wave plate map: $|H⟩\mapsto (|H⟩+|V⟩)/\sqrt{2}$ and $|V⟩\mapsto (-|H⟩+|V⟩)/\sqrt{2}$. The transformed joint state isConditioning on a vertical result and setting $\Theta =\pi$ therefore preparesThis pure optical cat state is the conditional output; it is not the unconditioned reduced state immediately after the Kerr medium. The measurement erases the polarization label, allowing the two amplitudes to interfere. The calculated optical Wigner functions below display the resulting fringes; their negative regions are quasiprobability, not negative detection probabilities. These illustrative plots choose $\alpha =1.15$: the rotation panel uses $\Theta =2\pi /3$, while the cat panels use the conditional Kerr rotation $\Theta =\pi$ and three different relative phases $\phi$.

The coherent-state overlap follows by expanding in the Fock basis:Thus $s=⟨\alpha |-\alpha ⟩=e^{-2|\alpha {|}^2}$ andThe postselection probability for Kerr-cat preparation includes both the normalized input qubit and the wave plate. At $\alpha =0,\varphi =\pi$ the vertical branch has zero probability and no normalized conditional state. For the remaining calculation take $\alpha$ real, as required.

Using $a|\pm \alpha ⟩=\pm \alpha |\pm \alpha ⟩$, put $D=1+s\cos \varphi$. Directly evaluating diagonal and cross terms givesThe last quantity is the mean photon number in the conditional cat, which differs from the input coherent state mean ${\alpha }^2$. Since $a{a}^{†}=n+1$,However, its first moment iswhich is generally nonzero. The original PDF's assertion of a zero mean for arbitrary $\varphi$ is inconsistent with its stated pure state. Consequently its displayed expression is the second moment, not the general variance. The phase-dependent cat-state quadrature variance isThe printed result does become the variance for $\sin \varphi =0$, notably the even and odd cat states. It could also describe an ensemble mixing equal weights of $+\varphi$ and $-\varphi$: that different state's mean vanishes while its second moment is unchanged. Such phase averaging is not part of the specified pure-state preparation.

For nonzero finite $\alpha$, $0<s<1$ and $D>0$. The correct quadrature squeezing condition in this quadrature is thereforeFor a representative phase in $-\pi <\varphi \le \pi$, this is $|\varphi |<\arccos (-e^{-2{\alpha }^2})$. If instead one applies the printed second moment as a variance, the resulting condition is $\cos \varphi >0$, or $|\varphi |<\pi /2$ modulo $2\pi$. That is sufficient, but not necessary, for the actual pure state's squeezing. For example, at $\varphi =\pi /2$, $⟨X_2^2⟩=1/4$ while $\mathrm{Var}{X}_2=1/4-{\alpha }^2{e}^{-4{\alpha }^2}<1/4$. This explicitly tests the difference.

For a fixed phase other than $\pi$ modulo $2\pi$, expandAs the input mean number ${\alpha }^2\to 0$, the normalized state tends to vacuum, $⟨X_2⟩\sim -\alpha \tan (\varphi /2)$ andThis expansion is nonuniform near the odd-cat phase. At exactly $\varphi =\pi$ the vacuum terms cancel and the normalized odd coherent-state superposition tends to $|1⟩$, giving $\mathrm{Var}{X}_2\to 3/4$, while its postselection probability tends to zero. The even coherent-state superposition at $\varphi =0$ instead contains only even photon numbers and tends smoothly to vacuum.

At large input mean number, $s\to 0$, the two coherent state branches become nearly orthogonal, and $\mathrm{Var}{X}_2\to 1/4$ for every fixed phase. The mean tends to zero as well. This does not make the state an incoherent mixture: phase-space interference still distinguishes the ideal quantum superposition, although its contribution to this low-order moment is exponentially small. The loss of coherence in an optical cat state arises when ordinary loss reveals the branch to the environment and suppresses that interference, with factor $e^{-2(1-\eta ){\alpha }^2}$ for transmissivity $\eta$.

For the especially simple even cat,It starts at $1/4$, falls to about $0.1108$ at $|\alpha |\simeq 0.7995$, and returns towards $1/4$ from below. The odd-cat curve decreases from $3/4$ towards $1/4$ from above. Other fixed-phase curves have the small- and large-amplitude limits derived above; for negative $\cos \varphi$, squeezing ends when $e^{-2{\alpha }^2}=-\cos \varphi$. In balanced homodyne detection, centre each quadrature distribution on its measured mean before inferring its variance.

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.

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.

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.

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.

A postselection scheme in which two polarization-entangled photon pairs pass through polarizing beam splitters, a half-wave plate and a nonpolarizing beam splitter. One photon at each of a trigger and three output detectors selects a GHZ state in the three output polarizations, provided optical mode matching removes distinguishable pair labels. All four detections are needed for the selected event; the trigger alone does not guarantee the three other output occupations.

A 2018 photonic experiment makes the distinction between populations and quantum coherence controllable. A 400-nm laser pumped a crystal to generate paired photons, filtered near 800 nm. One photon triggered collection; the other entered the apparatus. A half-wave plate and polarizing beam splitter selected horizontal photon polarization; another half-wave plate prepared balanced $H,V$ quantum superpositions. Separated polarization paths and diffraction gratings let a spatial light modulator control their frequency-dependent quantum phases and spectral weights. Recombined paths passed through quartz plates whose adjustable thickness controlled polarization–frequency coupling. Quarter- and half-wave plates plus a polarizing beam splitter selected tomography bases; jointly registered photodetector events supplied the readout. Quantum state tomography inferred polarization density matrices at each thickness. The dephasing model preserves diagonal populations while changing off-diagonal quantum coherence. Here the unobserved photon frequency acts as the environment for the polarization qubit. Finite modulator resolution and imperfect spectral shaping limited agreement with the intended dynamics; this experiment does not establish exact $I/2$ preparation or the ideal probabilities above. ^[ref]

The relative phase accumulated by orthogonal eigenpolarizations is $\Delta =2\pi L(n_2-n_1)/{\lambda }_0$. A quarter-wave plate has relative phase $\pi /2$ modulo $2\pi$, while a half-wave plate has $\pi$ modulo $2\pi$.