= Lossless two-channel thickness scattering model
{title2=$p_0=\cos^2(\kappa t),\quad p_g=\sin^2(\kappa t)$}
Consider an original illustrative model with exactly two outgoing <scattering channels>: a central beam $0$ and one diffracted beam $g$. Let $t\geq0$ be specimen thickness and $\kappa>0$ a constant coupling with units of inverse length. Normalize the <probability amplitudes> $A_0,A_g$ to incident <probability current>, assume equal longitudinal speeds, and specify
$$
\frac{dA_0}{dt}=i\kappa A_g,\qquad
\frac{dA_g}{dt}=i\kappa A_0,\qquad
A_0(0)=1,\quad A_g(0)=0.
$$
These <ordinary differential equations> define a lossless toy model, not a fit to the <silicon> example or a complete theory of <crystal scattering>. Differentiating once more gives $A_0''=-\kappa^2 A_0$, so
$$
A_0=\cos(\kappa t),\qquad A_g=i\sin(\kappa t),\qquad
p_0=|A_0|^2=\cos^2(\kappa t),\quad p_g=|A_g|^2=\sin^2(\kappa t).
$$
The identity $p_0+p_g=1$ expresses conservation within these two <scattering channels>. Alternatively, substituting the <ordinary differential equations> into $d(|A_0|^2+|A_g|^2)/dt$ makes the two cross terms cancel. Here ideal <bright-field transmission electron microscopy> accepts fraction $p_0$ and rejects fraction $p_g$; ideal <dark-field transmission electron microscopy> accepts precisely the rejected channel. This complementarity requires the declared two-channel completeness and absence of loss.
Even with known $\kappa$, thickness is ambiguous: $\kappa t=\pi/4$ and $3\pi/4$ both give $(p_0,p_g)=(1/2,1/2)$. Near $t=0$, $p_g\simeq(\kappa t)^2$; extending that approximation through a full oscillation would erase the ambiguity incorrectly. The model isolates reversible transfer between <probability amplitudes>; it does not say that every thicker specimen must look darker.
\Image[/tem-lossless-two-channel-thickness.png]
{title=Complementary <scattering channel> fractions and ambiguous thickness in a lossless model}
{description=Original calculations from the stated coupled <probability amplitudes>. Left: accepted central and diffracted fractions versus dimensionless thickness $\kappa t$. Right: their <probability amplitudes> trace a unit circle in the $(A_0,-iA_g)$ plane; the marked distinct thicknesses have identical fractions but different amplitude signs. No measured data or physical thickness calibration is implied.}
{source=tem-lossless-two-channel-thickness.py}
Real <dark-field transmission electron microscopy> usually collects only a subset of the directions rejected in <bright-field transmission electron microscopy>. Other <elastic scattering> directions, <inelastic scattering> and detector losses prevent a general two-image conservation rule. To keep loss and background separate, extend the counting model explicitly. Let $\ell$ be the fraction absent from both recorded beam channels, $p_0+p_g=1$ describe the remaining fraction, $N$ be incident count per exposure, $\eta_j$ the detector efficiency and $b_j$ its mean additive background. Then the <expected value> of recorded count $n_j$ is
$$
\mathbb E[n_j]=\eta_j N(1-\ell)p_j+b_j,\qquad j\in\{0,g\}.
$$
For separately acquired equal-illumination exposures with known $b_j,\eta_j$, the two corrected mean counts sum to $N(1-\ell)$, not necessarily $N$. Removal from these channels is not synonymous with physical absorption: an <Electron> can leave the chosen acceptance or change <kinetic energy> without being stopped in the specimen. Background subtraction removes a mean offset, but its counting <variance> remains.
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