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For fixed circular eigenvectors ${\mathbf{u}}_{\pm }=(1,\mp i{)}^{\mathsf{T}}/\sqrt{2}$, let $P_{\pm }={\mathbf{u}}_{\pm }{\mathbf{u}}_{\pm }^{†}$. A uniform optical medium with amplitude propagation coefficients ${\gamma }^{\pm }$ has the displayed Jones matrix. Put $\gamma =({\gamma }^{+}+{\gamma }^{-})/2$ and $\delta =({\gamma }^{+}-{\gamma }^{-})/2$. ThenThe sum of the two modal exponentials supplies a hyperbolic cosine, not its reciprocal. The determinant is $e^{-2\gamma z}$, and $\mathcal{J}(z_1+z_2)=\mathcal{J}(z_1)\mathcal{J}(z_2)$. Real modal coefficients give circular dichroism; purely imaginary coefficients give lossless circular birefringence. Complex modal labels do not have an intrinsic numerical ordering.

The optical coherency matrix records the second moments of transverse complex amplitudes. Normalize its trace to light intensity. Under a deterministic Jones matrix $M$, $C\mapsto MC{M}^{†}$. Unpolarized light has $C=(I_0/2)I_2$, so its output intensity is $(I_0/2)\mathrm{tr}(M{M}^{†})$; it cannot be represented by one fixed Jones vector.

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.

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

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.

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.

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.

The normalized circular eigenvectors from part (a) define the orthogonal projection matricesThus the spectral decomposition of the Jones matrix isThis simply substitutes $a=({\lambda }_{+}+{\lambda }_{-})/2$ and $b=({\lambda }_{+}-{\lambda }_{-})/2$ and makes each eigenmode's wave amplitude transmission explicit.

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.

Choose the first polarizer's pass axis as ${\mathbf{e}}_1$ and write the second as $\mathbf{a}=({\mathbf{e}}_1+{\mathbf{e}}_2)/\sqrt{2}$. The positive signed rotation takes ${\mathbf{e}}_1$ toward $\mathbf{a}$ by the short $\pi /4$ angle. Let $B_{\parallel }$ be signed according to $\theta ={\rm Y}L{B}_{\parallel }$. The Jones matrices of ideal polarizers are their orthogonal projection matrices,Removing the unimportant common propagation phase, the lossless glass block has the rotation matrixThe elements act from right to left, soWith the common optical phase retained, multiply this by $e^{i\bar{\varphi }}$. Extra scalar attenuation would also multiply the result, but none is supplied. A single Jones vector does not represent unpolarized light; the device's Jones matrix is nevertheless well defined for each incident component and can be applied to their statistical mixture.

Keep source, observer and magnetic field in their original places, but interchange the physical polarizers, including their pass-axis orientations. The forward order is then $P_2$, glass, $P_1$, so the new Jones matrix is again $P_{1}R(\theta )P_2$. ConsequentlyThis is the same intensity change as exchanging source and observer. At the field that maximizes the original forward transmission, the swapped arrangement blocks the beam. If both operations were performed together, the original order relative to propagation would be restored and so would the original transmission. The answer concerns exchanging the two differently oriented pass axes; simply relabelling two identical polarizers while leaving their axes fixed would not change the apparatus.