 Lossless two-channel thickness scattering model

Consider an original illustrative model with exactly two outgoing scattering channels: a central beam $0$ and one diffracted beam $g$. Let $t\ge 0$ 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 specifyThese 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 $A_0^{\prime \prime }=-{\kappa }^2{A}_0$, soThe 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.

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 $n_j$ isFor 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.