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Take nonrelativistic quantum mechanics, a short-range interaction, a fixed heavy target, and a common projectile mass $m$. An incoming plane wave in atomic state $a$ has unit coefficient. In an open scattering channel $i$, the target finishes in a normalized atomic state ${\chi }_i$, and the projectile has positive velocity magnitude $v_i=\hslash k_i/m$. At large distance,Here $\Omega$ denotes direction, $d\Omega$ is a solid angle element, and ${\epsilon }_i$ is an internal atomic energy. Different ${\chi }_i$ are orthogonal. The scattering amplitude $f_{ia}$ has units of length. A closed scattering channel has no propagating outgoing probability current at this total energy.

The incoming probability current density is $j_{\mathrm{in}}=v_a$. The outgoing radial probability current density in channel $i$ is $j_{r,i}\sim v_i|f_{ia}{|}^2/r^2$. Multiplying by the area $r^{2}d\Omega$ and dividing by $j_{\mathrm{in}}$ gives the differential scattering cross-section:The $r^{-2}$ dilution cancels the detector's area growth. The velocity ratio remains: a squared wave coefficient alone is not generally a flux fraction. A differential scattering cross-section has units of area per solid angle, rather than being a normalized directional probability density. For selected scattering events in acceptance $D$, an ideal detector with equal efficiency in all retained channels instead givesThe denominator specifies the event selection; it need not equal one square unit of area. Absolute rates require incident flux, target number and detector efficiency in addition to these relative weights. The next finite-mode model normalizes outgoing probability current to one incoming event and predicts event fractions directly.

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.

Consider one chosen model at a fixed total energy, with two open scattering channels, no absorption and no omitted outgoing modes. Channel 1 leaves the target unchanged; channel 2 excites it. Let $v_1=v>0$ and $v_2=v/2$. For the same projectile mass, conservation of energy then requires a target excitation $\Delta =3m{v}^2/8$. These are hypothetical parameters, not a fit to a particular atom. Each incoming or outgoing mode is normalized to unit probability current; the outgoing modes will have the angular profiles specified below.

Choose the S-matrixThis phenomenological unitary matrix specifies the channel conversion; it is not derived here from an interaction potential energy. For incidence in channel 1, its first column gives the outgoing flux amplitudesTo recover ordinary wave coefficients relative to an incident coefficient one, undo the velocity normalization:Thus $|c_2{|}^2=32/25>1$ is allowed: channel 2 travels more slowly. Its outgoing probability current fraction is $16/25$, not $32/25$. Renormalizing the unweighted squares would give $9/41$ and $32/41$, the wrong probabilities. More generally the ordinary coefficient map is $A=V^{-1/2}S{V}^{1/2}$, with $V=\mathrm{diag}(v,v/2)$; it obeys $A^{†}VA=V$. The conserved quantity is the squared coefficient norm weighted by velocity, which represents probability current.

Now specify an original angular realization of these two outgoing modes, with scattering angle $\theta$ and no azimuth dependence:The first normalization follows from $2\pi {\int }_{-1}^1(1+u^2)du=16\pi /3$, where $u=\cos \theta$. These angular profiles define the modes of this model; they are not asserted to follow from Born's approximation or an atomic potential energy. The joint directional probability densities are

Suppose a detector accepts the cone $D:0\le \theta \le \pi /3$, with full azimuth and unit efficiency. Its angular fractions within the channels areMultiplying the angular fractions by the outgoing flux weights givesFor $1600$ independent incident events, the expected values are therefore $171$ and $256$ accepted detections, compared with $576$ and $1024$ over all directions. The remaining expected $1173$ events fall outside the cone. The chosen incident number makes the expectations integers; actual counts fluctuate. Conditional on a detection inside the cone, the channel probabilities are $171/427$ and $256/427$, rather than $9/25$ and $16/25$. Angular acceptance changes the observed channel proportions even with equal detector efficiencies.

The final atomic states are orthogonal, so ignoring the atomic readout adds the two channel intensities; it does not produce $|b_1{g}_1+b_2{g}_2{|}^2$. Coherent paths that reach the same final state can instead interfere, as the existing probability amplitude and quantum superposition examples demonstrate. The distinction is physical distinguishability of final states, not whether the detector operator chooses to display their labels. Flux normalization and angular integration make the scattering interpretation useful: they turn complex outgoing waves into conserved event weights and then into the statistics of a specified detector.