Energy budget of a rotated anisotropic slab

A lossless slab cannot absorb light, so reflectance and transmittance must add to one: $R + T = 1$. Build a uniaxial crystal whose optic axis is tilted out of the plane of incidence, though, and the obvious bookkeeping seems to break.

The naive budget comes up short

Take a calcite-like slab ($n_o = 1.658$, $n_e = 1.486$, lossless) in air and orient its optic axis out of plane with euler = (π/6, π/4, 0):

using TransferMatrix
air  = Layer(λ -> 1.0, 1.0)
slab = Layer(λ -> 1.658, λ -> 1.658, λ -> 1.486, 0.5; euler = (π/6, π/4, 0.0))

r = transfer(1.0, [air, slab, air])          # normal incidence
r.Rpp + r.Tpp        # ≈ 0.9873  — looks like 1.3% has vanished

The slab is transparent, yet Rpp + Tpp ≈ 0.987. No energy is actually lost — the budget is simply incomplete.

The missing pieces are the converted channels

A rotated anisotropic crystal does something an isotropic film never does: it converts polarization. Send in a purely p‑polarized wave and a small s‑polarized component appears in both directions — the cross‑polarized reflectance Rps and the cross‑polarized transmittance Tps. That converted light is real, carries energy, and is exactly what the naive sum omits. Adding both channels closes the budget to machine precision:

r.Rpp + r.Rps + r.Tpp + r.Tps        # = 1.0   (to ~1e-15)

A few conventions make this add up cleanly:

• The coefficient is indexed r_{in,out}, so the p‑input cross‑reflection is Rps (not Rsp, which is the s-input term — the two coincide only at normal incidence).
• Every transmittance is a per‑mode Poynting flux: Tpp is the power carried by the p‑like substrate eigenmode alone, Tps the power carried by the s‑like eigenmode, each evaluated with its own wavevector. The four channels therefore partition the transmitted energy exactly — nothing is counted twice.

The per‑input‑polarization statements of energy conservation are therefore

\[R_{pp} + R_{ps} + T_{pp} + T_{ps} = 1, \qquad R_{ss} + R_{sp} + T_{ss} + T_{sp} = 1 .\]

Panel (a) sweeps the incidence angle with the optic axis held out of plane. The naive Rpp + Tpp (blue) dips below one at every angle; the complete Rpp + Rps + Tpp + Tps (black) is flat at one. The shaded gap between them is the converted power Rps + Tps.

Panel (b) explains when the conversion happens. Holding the incidence angle fixed and rotating the optic axis in azimuth, the converted power is exactly zero when the axis lies in the plane of incidence (α = 0): there the mirror symmetry of the problem keeps p and s decoupled, just like an isotropic medium. Tilt the axis out of that plane and the conversion — and the apparent energy "leak" — switches on, mostly in transmission (Tps) with a smaller reflected share (Rps).

The lesson generalizes: for any anisotropic stack, always account for the cross‑polarized channels — both reflected and transmitted — before concluding that energy is or isn't conserved.

The full runnable script is examples/anisotropic_energy_budget.jl.