Wave Plates and Retarders

A wave plate is a thin slab of transparent crystal that changes nothing about a beam's brightness, yet completely reshapes its polarization, i.e., introduces a phase shift by delaying one of the perpendicular components relative to another without blocking or absorbing light. Send in light linearly polarized at forty-five degrees, and out the far face comes light whose electric field spirals in a perfect circle. Rotate the same plate a few degrees and a linear beam simply turns — its plane of vibration swung to a new angle without a single photon lost. No absorption, no reflection loss to speak of, no moving parts: just a passive flake of quartz or mica doing to polarization what a lens does to a wavefront. The whole device is captured by one number. Because a birefringent crystal offers two different refractive indices to the two orthogonal linear components of the field, one component crosses the plate slightly slower than the other and emerges behind in phase. That lag — the retardance $\Gamma = (2\pi/\lambda_0)\,\Delta n\,d$ — is set entirely by the birefringence $\Delta n$ and the thickness $d$ , and once you fix it you have fixed everything the plate can do. Choose the thickness so the lag is a quarter cycle and you have a quarter-wave plate, the linear-to-circular converter. Choose half a cycle and you have a half-wave plate, the polarization rotator. Everything below hangs off that single relation. We build the plate from the birefringence of the previous article, tune its thickness to a target retardance, and then watch the two workhorse cases — the quarter- and half-wave plates — emerge as special values of $\Gamma$ , with the polarization-ellipse bookkeeping handled by the two-by-two matrices that the next article, on Jones calculus, makes systematic.

Key takeaways

A wave plate is a birefringent slab cut with its optic axis in the face; it splits the field into fast and slow linear components that accumulate a relative phase $\Gamma = (2\pi/\lambda_0)\,\Delta n\,d$ — and nothing else.
Quarter-wave plate ( $\Gamma = \pi/2$ , i.e. $\Delta n\,d = \lambda_0/4$ ): linear light at 45° to the axes becomes circular, and circular becomes linear.
Half-wave plate ( $\Gamma = \pi$ , i.e. $\Delta n\,d = \lambda_0/2$ ): linear polarization at angle $\alpha$ to the fast axis emerges at an angle of $-\alpha$ relative to that same axis — rotate the plate by $\Delta\theta$ and the polarization turns by $2\Delta\theta$ .
Thin plates need high precision: at 589.3 nm a quartz ( $\Delta n = 0.0091$ ) quarter-wave plate is only 16.2 µm thick, so real plates are multi-order or compound zero-order for robustness.
A retarder is lossless — it redistributes the polarization ellipse and conserves total flux — which is the sharp physical contrast with a polarizer, which projects and dims.
Key idea: a wave plate imposes a single relative phase $\Gamma = (2\pi/\lambda_0)\,\Delta n\,d$ between two orthogonal components and touches nothing else; the quarter- and half-wave plates are just the values $\Gamma = \pi/2$ and $\Gamma = \pi$ .

The core insight

A wave plate does nothing to a beam's intensity — it imposes a single relative phase $\Gamma = (2\pi/\lambda_0)\,\Delta n\,d$ between two orthogonal linear components. Everything the plate can do is fixed by that one number and the orientation of its fast axis; the quarter-wave plate ( $\Gamma = \pi/2$ ) interconverts linear and circular light, and the half-wave plate ( $\Gamma = \pi$ ) rotates linear polarization by twice the plate angle.

Where this lives today

Wave plates are everywhere light must be steered without loss. Every laser laboratory uses a half-wave plate and polarizing cube as a continuously variable, non-absorbing attenuator, and a quarter-wave plate to build an optical isolator that lets light out but blocks its back-reflection. Atomic physics runs on clean circular polarization: a quarter-wave plate after a linear polarizer prepares the $\sigma^{\pm}$ light that drives $\Delta m = \pm 1$ transitions in magneto-optical traps and optical-pumping schemes, and flipping the plate swaps the handedness. The variable retarder has quietly conquered consumer optics. Every liquid-crystal display is a dense array of voltage-controlled retarders sandwiched between polarizers — the pixel is a wave plate you tune electrically to pass or block backlight. Polarized 3-D cinema hands each eye a different circular handedness through quarter-wave films, and camera circular polarizers (a linear polarizer bonded to a quarter-wave plate) cut glare while keeping autofocus beam-splitters happy. In fibre-optic links, in-line polarization controllers built from squeezed-fibre retarders keep coherent receivers locked to the incoming state. Ellipsometry, the workhorse of thin-film metrology, reads a sample's induced retardance to nanometre precision with a rotating compensator that is Soleil and Babinet's instrument in modern dress.

The physical picture

Picture linearly polarized light striking the plate with its electric field tilted at forty-five degrees to the crystal's fast axis. The field splits into two equal pieces, one oscillating along the fast axis, one along the slow. Inside the crystal these two see different refractive indices, so the slow piece falls behind. When both emerge, the slow one lags the fast one by the retardance $\Gamma$ . If that lag is exactly a quarter cycle, the two orthogonal oscillations are ninety degrees out of step — and two equal-amplitude oscillations in quadrature trace a circle. Linear light has become circular.

Figure

A quarter-wave plate: 45° linear light splits into equal fast (x) and slow (y) components, the slow one lags by Γ = π/2, and the recombined field spirals as circular light

A half-wave plate makes the slow component lag by a full half cycle, flipping its sign relative to the fast component. Geometrically the field is reflected across the fast axis, so a linear input at angle $\alpha$ comes out at $-\alpha$ measured from that axis — a net rotation of $2\alpha$ . Nudge the plate itself by a few degrees and the mirror line moves with it, so the output swings twice as fast as the plate. That factor of two is the signature of every half-wave plate, and it makes a plate on a rotation mount into a smooth, lossless polarization knob. How thick must the crystal be? The retardance grows linearly with thickness, $\Gamma = (2\pi/\lambda_0)\,\Delta n\,d$ , so the quarter- and half-wave points sit at fixed thicknesses set by the birefringence. For quartz at the sodium-D wavelength the numbers are startlingly small.

Figure

Retardance Γ grows linearly with plate thickness for quartz at 589.3 nm (Δn = 0.0091); the quarter-wave point is 16.2 µm, the half-wave point 32.4 µm

At sixteen micrometres a quartz quarter-wave plate is thinner than a human hair, and a calcite one — with its far larger birefringence — would be under a micrometre, too fragile to handle. That is why real plates are built thick and clever, as the next sections spell out.

The core equations

Everything starts with the phase each component accumulates crossing the plate. A mode with index $n$ picks up optical phase $k_0 n d$ ; the two orthogonal modes differ only in their index, so their phase difference — the retardance — depends on the birefringence alone.

\Gamma = \frac{2\pi}{\lambda_0}\,(n_{\text{slow}} - n_{\text{fast}})\,d = \frac{2\pi}{\lambda_0}\,\Delta n\,d \qquad (1)

where: $\Gamma$ — relative phase retardation in radians between the slow and fast components; $\lambda_0$ — vacuum wavelength; $\Delta n = n_{\text{slow}} - n_{\text{fast}} = |n_e - n_o|$ — the birefringence (from the previous article); $d$ — physical thickness. The common propagation phase cancels, so only the difference of indices matters — a plate is a pure differential-delay device. Fixing $\Gamma$ to a target value fixes the thickness. The two workhorse retardances are a quarter cycle and a half cycle, giving the two named plates.

d_{\text{QWP}} = \frac{\lambda_0}{4\,\Delta n}\quad(\Gamma = \tfrac{\pi}{2}), \qquad d_{\text{HWP}} = \frac{\lambda_0}{2\,\Delta n}\quad(\Gamma = \pi) \qquad (2)

where: $d_{\text{QWP}}, d_{\text{HWP}}$ — thicknesses giving quarter- and half-wave retardance at $\lambda_0$ . Because $\Delta n$ is small for good plate materials, these thicknesses are tens of micrometres — the origin of the multi-order and compound designs discussed below. To track what the plate does to a polarization state, represent the field by its two complex components and the plate by a two-by-two matrix — the Jones matrix, made systematic in the next article. In its own axis frame (fast axis along $x$ ) a retarder is diagonal, delaying the slow axis relative to the fast.

W(\Gamma) = \begin{pmatrix} e^{-i\Gamma/2} & 0 \\ 0 & e^{+i\Gamma/2}\end{pmatrix}, \qquad W_{\text{QWP}} \;\propto\; \begin{pmatrix}1 & 0\\ 0 & i\end{pmatrix}, \qquad W_{\text{HWP}} \;\propto\; \begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix} \qquad (3)

where: $W(\Gamma)$ — retarder matrix in its principal frame (time convention $e^{i(kz-\omega t)}$ , so the slow $y$ -axis gains the extra phase $e^{+i\Gamma}$ relative to the fast $x$ -axis); the quarter-wave and half-wave forms follow by setting $\Gamma = \pi/2$ and $\Gamma = \pi$ and dropping an overall phase. The matrix is unitary — $W^\dagger W = I$ — the algebraic statement that a retarder never removes energy.

The core equations lead directly to the two devices

Derivation APlate thickness from a target retardance

Our aim is to turn the abstract requirement "delay one axis by a quarter cycle" into a concrete number a workshop can cut, and to understand why that number forces a design compromise. The strategy is direct: write the retardance as a linear function of thickness, invert it for the quarter- and half-wave conditions, and then confront the awkward fact that the resulting thicknesses are tiny and wavelength-specific — which is what drives the multi-order and compound-zero-order constructions used in every real plate.

Step 1. Each normal mode crosses the plate accumulating phase $\phi_j = k_0 n_j d = (2\pi/\lambda_0)\,n_j d$ , so the retardance is the difference $\Gamma = \phi_{\text{slow}} - \phi_{\text{fast}} = (2\pi/\lambda_0)\,\Delta n\,d$ . Why: the two modes travel collinearly through the same thickness, so everything except the index difference is common and cancels — the plate is a pure differential delay.

Step 2. Set $\Gamma = \pi/2$ for a quarter-wave plate: $(2\pi/\lambda_0)\,\Delta n\,d = \pi/2$ gives $d_{\text{QWP}} = \lambda_0/(4\,\Delta n)$ . Likewise $\Gamma = \pi$ gives $d_{\text{HWP}} = \lambda_0/(2\,\Delta n)$ , exactly twice as thick. Why: retardance is linear in thickness, so the named plates are equally spaced ladder rungs — doubling the thickness doubles the retardance.

Step 3. Put in quartz at $\lambda_0 = 589.3$ nm, $\Delta n = 0.0091$ : $d_{\text{QWP}} = 589.3\,\text{nm}/(4 \times 0.0091) = 16.2\,\mu\text{m}$ . For calcite, $\Delta n = 0.172$ , the same plate is only $0.86\,\mu\text{m}$ — comparable to a wavelength of light ( $\lambda_0 = 0.58 \mu\text{m}$ ) and impossible to polish and handle. Why: the thickness scales as $1/\Delta n$ , so the very materials that are most strongly birefringent make the most fragile true-zero-order plates.

Step 4. Escape the fragility two ways. A multi-order plate is cut thick, with $\Gamma = 2\pi m + \pi/2$ for some integer $m$ : robust, but the large $2\pi m$ background drifts violently with wavelength and temperature. A compound zero-order plate stacks two thick plates with crossed axes so their retardances nearly cancel, $\Gamma = (2\pi/\lambda_0)\,\Delta n\,(d_1 - d_2)$ , leaving a small, stable net quarter wave from two rugged crystals. Why: only the difference $d_1 - d_2$ enters, so you get low-order behaviour with high-order mechanical strength. *These results confirm: $d_{\text{QWP}} = \lambda_0/(4\Delta n)$ and $d_{\text{HWP}} = \lambda_0/(2\Delta n)$ ; for quartz at 589.3 nm this gives 16.2 µm and 32.4 µm respectively, and for calcite 0.86 µm.*

Go deeper · Derivation B · 2 minWhat the quarter- and half-wave plates do to a state

Derivation BWhat the quarter- and half-wave plates do to a state

Here we prove the two headline actions — the quarter-wave plate makes circular light and the half-wave plate rotates linear light by twice its angle — directly from the retarder matrix, so the geometric claims of the physical picture become algebra a reader can check line by line. The route is to apply the diagonal plate matrix to a definite input Jones vector and read off the output, twice: once for a quarter-wave plate on 45° linear light, once for a half-wave plate on general linear light at an arbitrary plate angle.

Step 1. Take a quarter-wave plate, fast axis along $x$ , so $W_{\text{QWP}} \propto \mathrm{diag}(1, i)$ , and feed it 45° linear light $\mathbf{J}_{\text{in}} = \tfrac{1}{\sqrt2}(1, 1)^{T}$ . The product is $\mathbf{J}_{\text{out}} = \tfrac{1}{\sqrt2}(1, i)^{T}$ . Why: the slow ( $y$ ) component is multiplied by $i$ , a quarter-cycle lag, while the fast component is untouched — the input had to be at 45° so the two components were equal, the condition a circle needs.

Step 2. Read that output as a field in time: $E_x = \cos\omega t$ , $E_y = \cos(\omega t - \pi/2) = \sin\omega t$ , whose tip traces $(\cos\omega t, \sin\omega t)$ — a circle, turning from $\hat{x}$ toward $\hat{y}$ . Feeding $-45^\circ$ instead (or flipping the fast axis) reverses the handedness. Why: equal amplitudes ninety degrees out of phase is the definition of circular polarization; the sign of the phase sets left versus right.

Step 3. Now a half-wave plate with fast axis at angle $\theta$ : rotate the frame, apply $\mathrm{diag}(1, -1)$ , rotate back, giving $W_{\text{HWP}}(\theta) = \begin{pmatrix}\cos2\theta & \sin2\theta\\ \sin2\theta & -\cos2\theta\end{pmatrix}$ . This is a reflection about the line at angle $\theta$ . Why: $\Gamma = \pi$ flips the sign of the slow component, which is exactly a mirror reflection across the fast axis; conjugating a mirror by rotations moves the mirror line.

Step 4. Apply it to linear light at angle $\alpha$ , $\mathbf{J}_{\text{in}} = (\cos\alpha, \sin\alpha)^{T}$ . The product simplifies to $(\cos(2\theta - \alpha),\,\sin(2\theta - \alpha))^{T}$ — still linear, now at $2\theta - \alpha$ . Why: reflecting a direction $\alpha$ across a line at $\theta$ lands it at $2\theta - \alpha$ , so rotating the plate by $\Delta\theta$ swings the output by $2\Delta\theta$ — the factor of two. *These matrix results can be confirmed directly: $W_{\text{QWP}}\,(1,1)^{T}/\sqrt2 = (1,i)^{T}/\sqrt2$ (circular), and $W_{\text{HWP}}(\theta)\,(\cos\alpha,\sin\alpha)^{T} = (\cos(2\theta-\alpha), \sin(2\theta-\alpha))^{T}$ .*

Limiting cases

The retardance $\Gamma$ is the single dial; its special values name the devices, and its wavelength dependence names the failure modes.

Case	Result	Meaning
Quarter-wave plate	$\Gamma = \pi/2$ ( $\Delta n\,d = \lambda_0/4$ )	45° linear ↔ circular; a general linear input becomes elliptical with axes along the plate
Half-wave plate	$\Gamma = \pi$ ( $\Delta n\,d = \lambda_0/2$ )	rotates linear light by $2\theta$ ; flips the handedness of circular light
Full-wave plate	$\Gamma = 2\pi$ ( $\Delta n\,d = \lambda_0$ )	identity on polarization at $\lambda_0$ only; used as a compensator or a sharp chromatic filter
Input along an axis	any $\Gamma$	unchanged (up to phase): the input is an eigen-polarization, so nothing splits
Multi-order plate	$\Gamma = 2\pi m + \pi/2$	robustly thick but strongly wavelength- and temperature-sensitive through the large $2\pi m$

Common trap“A quarter-wave plate is a quarter-wave plate at every wavelength”

The retardance is $\Gamma = (2\pi/\lambda_0)\,\Delta n\,d$ , and both $\lambda_0$ in the denominator and $\Delta n$ itself depend on colour — so a plate cut to give exactly $\pi/2$ at 589 nm gives something else at 633 nm. A cheap multi-order plate is worse: its retardance is $2\pi m + \pi/2$ with $m$ perhaps twenty, so the whole $2\pi m$ background sweeps through many full turns across the visible, and it behaves as a true quarter-wave plate only at a comb of discrete wavelengths and only near its design temperature. Treat the " $\lambda/4$ " label as a promise about one color at one temperature at normal incidence. For broadband work, you need an achromatic plate (two materials with opposing $\Delta n$ dispersion) or a variable Soleil–Babinet compensator; and always check whether the quoted plate is zero-order, true-zero-order, or multi-order before trusting it in a new band.

Worked examples

1 · Cutting a quartz quarter-wave plate (confidence builder)

Quartz has birefringence $\Delta n = 0.0091$ at the sodium-D line, $\lambda_0 = 589.3$ nm. Find the thickness of a true-zero-order quarter-wave plate, compare it to calcite, and explain why one of these is used commercially and the other is not.

Step 1. Quarter-wave condition: $d_{\text{QWP}} = \lambda_0/(4\,\Delta n) = 589.3\,\text{nm}/(4 \times 0.0091) = 589.3\,\text{nm}/0.0364 = \mathbf{16.2\ \mu m}$ . The half-wave plate is twice this, $32.4\ \mu\text{m}$ .

Step 2. Repeat for calcite, $\Delta n = 0.172$ : $d_{\text{QWP}} = 589.3\,\text{nm}/(4 \times 0.172) = \mathbf{0.86\ \mu m}$ — a single-crystal quarter-wave plate would be less than one wavelength thick.

Step 3. The design consequence: 16 µm of quartz is already delicate, and 0.86 µm of calcite is unworkable. Commercial true-zero-order quartz plates are cemented between glass windows for support, or the crystal is made multi-order (say $\Delta n\,d = 10.25\lambda_0$ , hundreds of micrometres) for handling and then paired with a second crossed plate so the net retardance is a stable quarter wave — the compound zero-order design.

Step 4. Why quartz, not calcite: the smaller birefringence of quartz is a feature, because it makes the plate thick enough to fabricate as a robust single crystal, while its low dispersion keeps $\Gamma$ near $\pi/2$ over a usefully wide band. Calcite's huge $\Delta n$ is superb for beam-splitting polarizers but wrong for retarders. Quartz QWP = 16.2 µm, HWP = 32.4 µm; calcite QWP = 0.86 µm (impractically thin).

2 · A half-wave plate as a lossless polarization rotator (exam level)

A vertically polarized beam ( $\alpha = 90^\circ$ from the horizontal $x$ -axis) passes through a half-wave plate whose fast axis is set at $\theta = 20^\circ$ . Find the output polarization direction and the net rotation, then explain how the same plate gives continuous, lossless attenuation when followed by a fixed polarizer.

Step 1. Apply the half-wave rule, output angle $= 2\theta - \alpha = 2(20^\circ) - 90^\circ = 40^\circ - 90^\circ = \mathbf{-50^\circ}$ , i.e. polarization now $50^\circ$ below the horizontal. The net rotation of the plane is $2(\theta - \alpha) = 2(20^\circ - 90^\circ) = -140^\circ$ (equivalently $+40^\circ$ modulo $180^\circ$ for a line).

Step 2. Check the factor of two: turn the plate to $\theta = 25^\circ$ and the output moves to $2(25^\circ) - 90^\circ = -40^\circ$ , a $10^\circ$ swing for a $5^\circ$ plate rotation. The output always turns twice as fast as the plate — the half-wave signature.

Step 3. Now follow the plate with a fixed horizontal polarizer. The transmitted fraction is Malus's law on the rotated linear state, $\cos^2(\text{output angle})$ ; sweeping the plate sweeps this smoothly between full transmission and full extinction. Because the plate itself is lossless, the rejected light is not absorbed but rotated into the polarizer's blocked axis — and with a polarizing beam splitter it emerges out a side port rather than heating anything.

Step 4. That is exactly how high-power lasers are attenuated: a motorized half-wave plate plus a polarizing cube gives hysteresis-free, continuously variable power without an absorptive element that would burn. Output at $-50^\circ$ ; a half-wave-plate-plus-polarizer is a lossless variable attenuator via $\cos^2$ of the $2\theta$ -rotated angle.

Physics of the portrait

Retarders were not invented; they were found, in flakes of mica and crystal that painted colour onto polarized light long before anyone could explain the phase they imposed.

Origins · 1 minretarder and how this was discovered

retarder

the slab that delays one axis · Arago → Fresnel → Soleil–Babinet

The retarder began as a puzzle of colour. In 1811 François Arago, examining thin flakes of mica and quartz between crossed polarizers, saw them blaze with wavelength-dependent tints — the first chromatic polarization, which is nothing but a wave plate whose retardance $\Gamma = (2\pi/\lambda_0)\,\Delta n\,d$ runs through different fractions of a wave for different colours, so each wavelength emerges with a different polarization and is transmitted differently by the analyzer. Augustin-Jean Fresnel, building the wave theory that made sense of it, saw in 1817 that two total-internal-reflection bounces in a glass prism could impose a quarter-wave retardance without any crystal at all — the Fresnel rhomb, an achromatic quarter-wave retarder still used where broadband circular polarization is needed. The controllable version came from the French instrument-makers: Jacques Babinet's compensator of the 1830s used two quartz wedges to give a position-dependent retardance, and Jean-Baptiste Soleil's modification around 1845 added a third plate so the retardance could be dialled uniformly across the beam — the Soleil–Babinet compensator that still measures sample birefringence in laboratories today. The modern liquid-crystal cell is the same idea with an electric knob: a voltage tilts the molecules, changing $\Delta n$ and hence $\Gamma$ in microseconds.

History

1811

— Arago discovers chromatic polarization: thin crystal flakes colour polarized light between crossed polarizers

1816

— Arago and Fresnel establish that two perpendicular polarizations do not interfere (the Fresnel–Arago laws)

1817

— Fresnel introduces the rhomb, an achromatic quarter-wave retarder using total internal reflection

1837

— Babinet builds a wedge compensator giving a variable retardance across the beam

1845

— Soleil modifies it into the uniformly tunable Soleil–Babinet compensator

Going further

Texts. Hecht, Optics, 5th ed., §8.7 (Retarders) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., §6.1 (polarization devices) · Born & Wolf, Principles of Optics, §15.4 (crystal plates and compensators) · Collett, Field Guide to Polarization (retarder catalogue). The retardance relation $\Gamma = (2\pi/\lambda_0)\,\Delta n\,d$ and the quarter/half-wave thicknesses $\lambda_0/(4\Delta n)$ , $\lambda_0/(2\Delta n)$ are standard results. The quartz and calcite numbers use $\Delta n = 0.0091$ (quartz) and $0.172$ (calcite) at $\lambda_0 = 589.3$ nm. The QWP and HWP Jones matrices and the $2\theta$ rotation property follow from direct symbolic calculation; the time convention throughout is $e^{i(kz-\omega t)}$ , matching the companion Jones-calculus and Stokes articles. No material value is asserted beyond these tabulated birefringences.

Prerequisites. Birefringence · Uniaxial and Biaxial Crystals · Polarization of Light · Linear Polarization · Circular Polarization. Leads to. Jones Calculus · Stokes Parameters and the Poincaré Sphere · Mueller Calculus · Elliptical Polarization.

Which is the fast axis, and does it matter which way I orient the plate?

The fast axis is the one with the smaller refractive index, so light polarized along it travels quickest and picks up the least phase. For most linear operations only the angle of the fast axis matters, and swapping fast for slow (rotating the plate by 90°) changes the retardance by $2\pi$ minus the original — for a quarter-wave plate it flips circular handedness, and for a half-wave plate it makes no difference to a linear rotation. When you need a definite handedness of circular output, the fast-axis direction is what sets it, so plates are marked and the mark matters.

If a wave plate doesn't absorb any light, where does the "lost" power go when I use one to dim a beam?

The plate itself dims nothing — it only reshapes polarization. The dimming happens at the polarizer you place after it. A half-wave plate rotates the linear polarization, and the following polarizer then transmits only the projected component by Malus's law; the rest is either absorbed by the polarizer or, if you use a polarizing beam splitter, sent cleanly out a second port. So the "lost" power is redirected or absorbed downstream, never by the retarder, which is exactly why this scheme survives at kilowatt laser powers where an absorptive attenuator would melt.

What actually happens if my input isn't at 45° to a quarter-wave plate?

You get elliptical light rather than circular. The quarter-wave plate always delays the slow component by 90°, but a circle needs the two components to be equal as well as in quadrature, and only a 45° input splits equally. At 0° or 90° the input lies along an eigen-axis and passes unchanged; in between you get an ellipse whose principal axes line up with the plate's axes and whose ellipticity grows as you approach 45°. This is often useful: a quarter-wave plate plus a rotation mount is a compact way to dial any ellipticity you like.

Why do I sometimes see the retarder matrix written as diag(1, i) and sometimes diag(1, −i)?

Both appear because two things vary between textbooks: whether the time dependence is written $e^{-i\omega t}$ or $e^{+i\omega t}$ , and which axis (fast or slow) is taken as the phase reference. Flipping either one conjugates every retardance phase and swaps which handedness the circular states denote. Nothing physical changes — the plate still turns 45° linear light into circular light — but the label right versus left circular flips. This article fixes $e^{i(kz-\omega t)}$ and puts the extra phase on the slow axis, giving $\mathrm{diag}(1,i)$ ; if you compare with another source, first reconcile its two conventions before worrying about a disagreement.

How is a half-wave plate different from an optically active medium that also rotates polarization?

Both rotate linear light, but differently. A half-wave plate reflects the polarization across its fast axis, so the sense of rotation depends on which side of the axis the input lies, and reversing the beam through the plate undoes the rotation. An optically active medium (sugar solution, quartz along its axis) rotates by a fixed angle in a fixed sense regardless of input orientation, and the rotation is reciprocal. The Faraday effect is a third case — a magnetically induced rotation that is non-reciprocal, adding up on a return pass, which is what makes optical isolators possible. Only the half-wave plate's rotation is really a reflection in disguise.

Can I stack two quarter-wave plates to make a half-wave plate?

Yes, if their fast axes are aligned: two quarter waves in series give $\Gamma = \pi/2 + \pi/2 = \pi$ , a half-wave plate. But align the second plate's fast axis perpendicular to the first and the retardances subtract to zero — the pair becomes a (near) full-wave plate that does nothing, which is precisely the compound-zero-order trick for making a robust low-order plate from two thick crystals. Cascading retarders just adds their retardances vectorially on the Poincaré sphere, so the geometry of the stack, not just the count, sets the result.