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Anderson localization is the confinement of electron wave functions, or other wave modes, by interference between repeated wave scattering paths in static disorder. A localized eigenstate occupies a limited region even in a much larger sample. Disorder can therefore suppress diffusion without absorbing the wave. In 1958, Philip Warren Anderson studied random site energies and intersite coupling, finding a regime without transport. His paper develops a theoretical model; it does not report an Anderson laboratory measurement.^[ref]

The distinction from ordinary multiple scattering is the retention of wave phases: alternative paths contribute probability amplitudes before their squared modulus is taken. Adding path probabilities alone loses the cross terms responsible for interference. This makes Anderson localization useful as a warning about transport models: repeated deflections need not produce unlimited diffusion. It also connects electron transport with wave propagation in a random medium, although the detailed equations and experimental diagnostics differ between electrons, light and acoustic waves.

Philip Warren Anderson's 1977 lecture distinguishes localized magnetic moments from disorder-localized states. It credits Nevill Francis Mott with recognizing the significance of the mobility edge. A Mott insulator instead requires electron interactions and can exist without disorder.^[ref]

John Hasbrouck Van Vleck developed quantum mechanics explanations of magnetic susceptibility, including its temperature dependence; this is a distinct contribution from Anderson localization.^[ref]

Occupying an antibonding molecular orbital opposes a chemical bond. For equivalent $1s$ atomic orbitals of the hydrogen atom, the difference ${\sigma }_u^{\ast }={\sigma }_{-}$ vanishes on the plane halfway between the protons: every point there is equally far from both centres, so the amplitudes cancel. This nodal plane removes the contribution of this molecular orbital to electron density on that plane; it does not make an entire finite region empty. The $1s$-derived antibonding molecular orbital lies above its bonding molecular orbital partner in energy. See Field, MIT 5.61, lecture 25, pp. 1–2.

In the ${\mathrm{He}}_2$ minimal model, $({\sigma }_u^{\ast }{)}^2$ adds $N_a=2$ and cancels the bond order contribution of $({\sigma }_g{)}^2$. Opposite signs on the two lobes describe probability amplitudes, while both lobes have nonnegative probability density after squaring. The star is an orbital label, not an extra electron or an electric charge.

A lossless beam splitter that gives equal output probabilities for a photon incident in one input port. Balance constrains the magnitudes of the probability amplitudes; the beam splitter scattering matrix also specifies their relative wave phases.

The Born rule connects a quantum state to measurement probabilities. Let $\{|j⟩\}$ be a complete orthonormal basis, finite or countably infinite, for a Hilbert space, and letAn ideal measurement distinguishing these basis vectors returns label $j$ with probabilityThis assignment is a physical measurement rule. Its compatibility with total probability one follows from orthonormality and normalization:The probabilities are the squared moduli, not the amplitudes themselves or the squares $a_j^2$. ^[ref].

If the basis is an eigenbasis of an observable, the reported numerical value is its eigenvalue. Several vectors can share one eigenvalue; the probability of that value is then the sum over the corresponding basis labels. More generally, a projective measurement with orthogonal projections $P_r$ assignsfor a pure state and a general density matrix $\rho$, respectively. This also covers mixed states and degenerate eigenspaces. The Born rule determines outcome statistics; the ideal conditional state change is specified separately by the Lüders rule. See projective measurement for that formulation and its assumptions. ^[ref].

For a worked example, take the normalized qubit quantum stateThe squared moduli add to $3/4+1/4=1$. A computational-basis measurement, equivalently a measurement of the $Z$ Pauli matrix, hasThese probabilities do not depend on the relative quantum phase $\varphi$. Now choose the Hadamard basiswhose vectors are eigenstates of the $X$ Pauli matrix, also used by the Pauli X gate. The new probability amplitudes and probabilities areBoth probabilities are nonnegative and their sum is one. The relative quantum phase becomes visible because the two probability amplitudes add before their squared modulus is taken:

A Hadamard gate followed by a computational-basis measurement realizes these $X$ statistics, since $H|\pm ⟩=|0⟩,|1⟩$, respectively. Compare the two bases using fresh, identically prepared copies: a preceding measurement may change the state. The quantum superposition section shows why replacing the preparation by the mixed state $\mathrm{diag}(3/4,1/4)$ keeps the $Z$ statistics but gives $p_{+}=p_{-}=1/2$ for every $\varphi$.

The usefulness of the Born rule is this conversion from linear algebra to experimental predictions: calculate overlaps in the basis actually measured, then take squared moduli. Its appealing geometry is that changing basis redistributes probability while preserving total squared norm. The same rule relates an energy eigenstate expansion to an energy measurement and a position wavefunction to a probability density.

Born's 1926 paper on quantum scattering interpreted the outgoing-wave coefficients in terms of the probabilities of collision outcomes; a corrective footnote specified proportionality to the square of the amplitude. This is a historical scattering formulation, rather than the general density matrix formula above. ^[ref]; ^[ref].

Choose the qubit computational basis $|0⟩,|1⟩$, also called the $Z$ basis, and the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, also called the $X$ basis. Compare a coherent preparation of $|+⟩$ with choosing $|0⟩$ or $|1⟩$ at random with equal probabilities and discarding that label:These are original ideal preparation models. Their $Z$-basis probabilities are both $(1/2,1/2)$, because their quantum state populations agree. Their $X$-basis probabilities are respectively $(1,0)$ and $(1/2,1/2)$. Indeed $⟨+|{\rho }_{+}|+⟩=1$, whereas $⟨+|I/2|+⟩=1/2$. The missing off-diagonal quantum coherence therefore changes a prediction that the common populations cannot determine.

For a relative quantum phase $\varphi$, the coherent family instead hasThe last formula follows by adding or subtracting the two probability amplitudes before taking the squared modulus. At $\varphi =\pi$, the state is $|-⟩$ and the $X$ outcomes switch to $(0,1)$, while the $Z$ outcomes remain $(1/2,1/2)$. In contrast, $-|+⟩$ changes both components by the same global phase and retains the $X$ outcomes $(1,0)$. In photon polarization, the identification $|0⟩=|H⟩$, $|1⟩=|V⟩$ makes $|+⟩$ and $|-⟩$ the two orthogonal diagonal linear polarizations; these labels describe measurement bases, not spatial axes for photon spin.

Even perfect statistics in one basis cannot certify a preparation. The above $Z$ data fit both density matrices. Conversely, distinct preparation histories can yield the same quantum state:The off-diagonal terms cancel in the right-hand ensemble. Every measurement on the qubit alone then has identical probabilities for those two histories. A retained classical label would allow conditioning on different members of the quantum state ensemble; it is extra information beyond the unlabelled density operator.

An output port whose ideal probability amplitude vanishes by destructive interference for a specified coherent input and calibration. Changing a relative wave phase, losing quantum coherence or changing the input can brighten it.

Choose an isolated pair of energy levels, with ground state $|g⟩=(0,1{)}^T$ and excited state $|e⟩=(1,0{)}^T$. Use the Pauli matrices in the ordered basis $(|e⟩,|g⟩)$, so ${\sigma }_z=\mathrm{diag}(1,-1)$. After subtracting the mean energy, prescribe the laboratory Hamiltonian operatorThe unperturbed energy gap is $\hslash {\omega }_0$, the drive has angular frequency $\omega$, and $\varphi$ fixes its oscillation phase at the chosen time origin. All of ${\omega }_0$, $\omega$, the Rabi angular frequency $\Omega$ and the Rabi detuning $\delta$ are in radians per second; ordinary frequency is angular frequency divided by $2\pi$. In particular, $\Omega$ here is the coefficient multiplying the cosine in $H_{\mathrm{lab}}/\hslash$, not twice that coefficient.

Write $|{\psi }_{\mathrm{lab}}⟩=R(t)|{\psi }_{\mathrm{rot}}⟩$ with $R(t)=\exp (-i\omega t{\sigma }_z/2)$. Transforming the Schrödinger equation gives $H_{\mathrm{rot}}=R^{†}{H}_{\mathrm{lab}}R-i\hslash R^{†}\dot{R}$. The rotating-wave approximation drops terms at angular frequency $2\omega$, leavingThis approximation requires $\Omega ,|\delta |\ll {\omega }_0,\omega$, and an interaction duration over which omitted corrections remain small. It also assumes that other energy levels can be neglected. Resonance alone does not justify either assumption. An exactly co-rotating drive has no discarded counter-rotating component; the linearly oscillating laboratory drive above does.

The Pauli matrix multiplication law gives $K^2={\Omega }_{\mathrm{eff}}^{2}I$. Separating even and odd terms of the matrix exponential therefore proveswith the continuous limit $U=I$ when $\Omega =\delta =0$. Starting from $|g⟩$, the upper probability amplitude is $-i{e}^{-i\varphi }(\Omega /{\Omega }_{\mathrm{eff}})\sin ({\Omega }_{\mathrm{eff}}t/2)$. The Born rule givesThe rotating transformation only changes the quantum phases of these basis states, so their laboratory populations are the same. Rabi detuning tilts the rotation axis of the Bloch vector: it increases the oscillation rate to ${\Omega }_{\mathrm{eff}}$ but caps the transferred population at ${\Omega }^2/({\Omega }^2+{\delta }^2)$. A fixed drive oscillation phase cancels from this single-pulse population when the initial state is $|g⟩$. It still fixes the relative quantum phase of a resulting quantum superposition and matters when another pulse supplies a phase reference.

Electron diffraction is interference between scattered electron probability amplitudes. It connects the de Broglie wavelength of the incident beam to the arrangement of the scattering material. In elastic scattering from a fixed crystal, $|{\mathbf{k}}_{\mathrm{out}}|=|{\mathbf{k}}_{\mathrm{in}}|=2\pi /\lambda$, and constructive crystal scattering selects wavevector transfers $\mathbf{Q}={\mathbf{k}}_{\mathrm{out}}-{\mathbf{k}}_{\mathrm{in}}$ in the reciprocal lattice. The physical momentum transfer to the electron is $\hslash \mathbf{Q}$ in this convention.

Bragg's law expresses the same condition as $2d\sin {\theta }_B=n\lambda$, where $d$ is the spacing of the chosen lattice planes, $n$ the diffraction order, and ${\theta }_B$ the angle to those planes. The angle between incoming and outgoing propagation directions is $\chi =2{\theta }_B$ for that reflection geometry. In normal-incidence backscattering, a detector's angle $\varphi$ from the outward surface normal instead gives $\chi =\pi -\varphi$; neither detector colatitude nor angle to the surface normal should silently replace ${\theta }_B$. A finite-row electron diffraction model uses a different spacing, the spacing between sites along a surface row, and makes that distinction explicit.

Suppose a constant Hamiltonian operator has a complete discrete orthonormal basis of energy eigenstates, $H|n⟩=E_n|n⟩$. Expanding an initial pure state and applying the time-evolution operator term by term givesA projective measurement resolving these basis vectors therefore assigns the same Born rule probability $|c_n{|}^2$ to each label $n$ at every time. An energy measurement groups labels sharing the same energy eigenvalue: $P(E)={\sum }_{n:E_n=E}|c_n{|}^2$ is also constant. If $c_m,c_n\ne 0$, their probability amplitude ratio instead evolves asOnly energy differences enter this relative quantum phase. Adding a constant $\epsilon I$ to the Hamiltonian operator multiplies the entire pure state by $e^{-i\epsilon t/\hslash }$, a global phase, and leaves the density operator unchanged.

For a fixed observable $A$, the expectation value can nevertheless vary:Terms joining distinct energy eigenvalues can oscillate; diagonal terms cannot. If $[A,H]=0$, these terms between different energies vanish, so the expectation value is constant. This distinguishes conservation of energy from the stronger claim that every observable is stationary.

An energy eigenstate has only a global phase evolution and is a stationary state. A quantum superposition entirely within one degenerate energy eigenspace is also an energy eigenstate and is stationary. For a general density operator in finite dimension, stationarity under a constant Hamiltonian operator means $[H,\rho ]=0$: differentiating $U\rho U^{†}$ proves necessity, and commuting $\rho$ through the exponential proves sufficiency. Merely fixing the diagonal energy populations does not fix the off-diagonal quantum coherence.

A physicist who developed wave mechanics. In 1926, Erwin Schrödinger made the energies of chemical atoms an eigenvalue problem, rather than prescribing a separate quantization rule for each classical orbit. He shared the Nobel Prize in Physics in 1933 with Paul Dirac.^[ref]

The Schrödinger equation turns a proposed Hamiltonian operator into predictions for a wavefunction and its evolution. Boundary conditions can select discrete energy eigenvalues, while quantum superposition makes their relative quantum phases observable through interference. The infinite square well gives a transparent worked example; a hydrogen atom supplies a physical application with a different potential energy.

The Davisson–Germer experiment gives a complementary experimental route into matter waves: an accelerated electron beam scattered from a nickel crystal produces direction-dependent collected electric current. Its apparatus and angular scans make the wave prediction testable. Interpreting a wavefunction as a probability amplitude also requires the Born rule; that statistical interpretation deserves its own history alongside Erwin Schrödinger's dynamics.

A nonzero probability density inside a barrier does not by itself imply transport. In one dimension, the probability current of a wavefunction is $J=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$. A single evanescent wave $C{e}^{-\kappa x}$ has ${\psi }^{\ast }{\psi }^{\prime }=-\kappa |C{|}^2{e}^{-2\kappa x}$, a real quantity, hence $J=0$. This is what happens on the inaccessible side of a potential barrier extending to infinity with $E<V_0$.^[ref]

For the finite-barrier wavefunction $A{e}^{\kappa x}+B{e}^{-\kappa x}$, the cross terms instead giveThe relative quantum phase of the two probability amplitudes permits a nonzero, spatially constant probability current, even though neither exponential function separately carries it. The probability continuity equation gives ${\partial }_{x}J=0$ for this stationary state. Matching the wavefunction and its derivative also makes $J$ continuous at both faces, soHere $R=|r{|}^2$ is the quantum reflection probability, and $T=J_{\mathrm{tr}}/J_{\mathrm{inc}}$ is the transmission probability. The net exterior probability current equals the interior probability current; it is not lost as the wavefunction decreases. A physical incident wave packet supplies finite total probability, while the stationary state describes the energy component used to compute its transmitted flux.

In a two-level system, let the Hamiltonian operator have matrix $H=\mathrm{diag}(0,2)$ in the orthonormal basis $|0⟩,|1⟩$. The normalized real trial quantum states and their Rayleigh quotient areThe excited-state probability amplitude is $\sin \theta$, so its probability is ${\sin }^2\theta$. Thus $E(\theta )\ge 0=E_0$, with equality if and only if the excited-state probability amplitude vanishes: $\theta \in \pi \mathbb{Z}$. The remaining quantum state is $|0⟩$ up to a global phase. At $\theta =\pi /4$ the trial energy is $1$; at $\theta =\pi /2$ it is $2$. These are upper bounds on the ground-state energy, even though neither is a minimizing trial.

The real trial family contains the ground state, so it attains the exact minimum despite not covering every complex quantum state. More generally, a family covering all normalized vectors in a two-dimensional Hilbert space gives the exact lowest eigenvalue of any Hermitian matrix by minimizing its Rayleigh quotient. For a complex trial $c_0|0⟩+c_1|1⟩$ with $|c_0{|}^2+|c_1{|}^2=1$, the same calculation gives $E=2|c_1{|}^2$: a relative quantum phase cannot change this diagonal Hamiltonian energy. This finite example is the weighted mean proof of the variational method with just two energy eigenstates.

Take a nonrelativistic particle of constant mass $m$ with energy $E>0$, and choose the same zero of potential energy on both sides of a rectangular potential barrier:The equality of the exterior potential energies matters: it gives the same exterior wavenumber $k=\sqrt{2mE}/\hslash$ and makes $T=|t{|}^2$. Here $\hslash =h/(2\pi )$ is the reduced Planck constant. This model has a finite jump in potential energy, no delta potential, and no absorbing term.

For $0<E<V_0$, the time-independent Schrödinger equation gives the spatial wavefunctionThe full stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. We set the incident probability amplitude to one and impose no incoming wave from the right; this scattering wavefunction is not a normalizable wavefunction. The tunnelling decay constant $\kappa$ describes exponential decay, rather than an oscillatory wavenumber in the barrier. Both exponential functions are needed on the finite interval: the growing term is not excluded by a condition at infinity.

The wavefunction and its derivative are continuous at each face, giving four linear matching conditions:Their solution supplies the transmission amplitudeA short derivation is to use the two right-face equations to express $A{e}^{\kappa a}$ and $B{e}^{-\kappa a}$ in terms of $t$, then substitute into the left-face equations. Taking the squared modulus gives the rectangular barrier transmission probability.

For a free particle of mass $m>0$ in nonrelativistic quantum mechanics, choose the potential energy zero and writeApplying the momentum operator $-i\hslash \nabla$ returns $\mathbf{p}\psi$; this fixes the sign convention. The angular frequency is measured relative to this chosen energy zero. Adding a constant energy changes only a time-dependent global phase, leaving the spatial de Broglie wavelength unchanged. No rest-energy term is included in this nonrelativistic quantum mechanics convention.

The plane wave has $|\psi {|}^2=|A{|}^2$ and probability current $\mathbf{j}=(\mathbf{p}/m)|A{|}^2$. Its real part can be drawn as a sine or cosine, but that graph is a component of a probability amplitude, not a transverse displacement. On infinite space this is a generalized momentum eigenstate, not a state satisfying wavefunction normalization. A narrow wave packet instead moves with group velocity ${\nabla }_{\mathbf{k}}\omega =\mathbf{p}/m$; the phase velocity $\omega /|\mathbf{k}|=|\mathbf{p}|/(2m)$ is not its transport speed.

The equality of canonical momentum and kinetic momentum here assumes a field-free region with the magnetic vector potential chosen zero. For a charged particle in a magnetic field, the kinetic momentum is $\mathbf{\Pi }=\mathbf{p}-q\mathbf{A}$. An arbitrary canonical phase gradient must not then be substituted into $m\mathbf{v}$; the free particle formula needs its stated assumptions.

For an infinite square well initially on $[-a,a]$, a normalized ground state is $a^{-1/2}\cos (\pi x/(2a))$. If the endpoints change instantaneously to $\pm \eta a$, with $\eta >1$, the sudden approximation leaves the wave function unchanged, extended by zero outside the old well. Its overlap with the new normalized ground state $(\eta a{)}^{-1/2}\cos (\pi x/(2\eta a))$ is $4{\eta }^{3/2}\cos (\pi /(2\eta ))/[\pi ({\eta }^2-1)]$. The Born rule squares this probability amplitude; for $\eta =2$ the result is $64/(9{\pi }^2)$, and the limit as $\eta \downarrow 1$ is $1$.

The Hadamard gate is the unitary operatorwhere $I$ is the identity operator. It maps the computational basis to the Hadamard basis:

These outputs have equal computational-basis measurement probabilities, but their relative quantum phase differs. Applying another Hadamard gate sends them to different definite outputs. Calling this gate a classical coin toss would discard precisely the information that makes this possible. ^[ref].

Starting from $|0⟩$, compare two quantum circuits, reading matrix products from right to left:Here $Z$ is the Pauli Z gate. After the first Hadamard gate, the probability amplitudes are $(1/\sqrt{2},1/\sqrt{2})$. Inserting $Z$ changes them to $(1/\sqrt{2},-1/\sqrt{2})$ without changing their squared magnitudes. At the final gate, the contributions to output zero cancel and those to output one add. Without $Z$, cancellation occurs at output one instead. This is controlled interference, with the quantum phase determining which answer emerges.

The same calculation works for any phase gate $P_{\varphi }=\mathrm{diag}(1,e^{i\varphi })$:For example, $\varphi =\pi /3$ gives $P(1)=1/4$, while $\varphi =\pi /2$ gives $P(1)=1/2$. The output-one amplitude at $\varphi =\pi /2$ is $(1-i)/2$: its two complex contributions add before the Born rule is applied. A phase gate alone leaves the populations unchanged; recombination converts its relative quantum phase into observable probabilities.

If quantum decoherence completely removes the computational basis coherence between the gates, the intermediate density matrix becomes $I/2$. Both the phase gate and final Hadamard gate leave $I/2$ unchanged, so $P(1)=1/2$ for every $\varphi$. The disappearing fringe explains why preserving quantum coherence matters: it enables the gate sequence to carry phase information into the final measurement.

The normalized Hadamard transform maps a vector indexed by binary strings through entries $2^{-n/2}(-1{)}^{x\cdot y}$, where the dot product is evaluated modulo two. It is its own inverse and an orthogonal transformation, hence also a unitary operator on complex input vectors. The Hadamard gate is the $n=1$ quantum case, and the Hadamard basis is the image of the computational basis. The Walsh-Hadamard transform gives the same tensor product construction.

For two qubits, applying one Hadamard gate to each prepares the uniform quantum superpositionEach probability amplitude is $1/2$, so each computational-basis measurement outcome has probability $1/4$. Applying the Hadamard transform again reconstructs $|00⟩$: signed contributions cancel at the other three outputs. Its usefulness lies in combining simple local quantum gates into an operation on exponentially many amplitudes. Extracting useful information still requires an appropriate subsequent transformation and measurement.

The Heisenberg uncertainty relation constrains the standard deviations of position and momentum in one preparation:On the real line, the position operator $x$ acts by multiplication, and the momentum operator is $p=-i\hslash d/dx$, with $\hslash$ the reduced Planck constant. Their canonical commutation relation $[x,p]=i\hslash I$ and the Robertson uncertainty principle give the bound on normalized quantum states satisfying the moment and product-domain assumptions above; Schwartz functions are a convenient sufficient class. Thus making a preparation more sharply localized in position forces a wider momentum distribution, even for ideal measurement in quantum mechanics.

An explicit saturating example is the normalized Gaussian wave packet, for $\sigma >0$,The wavefunction is a probability amplitude; its squared magnitude is the probability density function, by the Born rule. That density is a normal distribution of mean zero and standard deviation $\Delta x=\sigma$, rather than a density whose width is read directly from the amplitude's exponent. With the unitary, negative-exponent convention for the Fourier transform, its momentum amplitude isThe Gaussian integral evaluates this Fourier transform. Consequently $|{\tilde{\psi }}_{\sigma }(p){|}^2$ is another normal distribution, with mean zero and standard deviationFor example, in dimensionless units with $\hslash =1$, $\sigma =1/2,1,2$ give $(\Delta x,\Delta p)=(1/2,1),(1,1/2),(2,1/4)$. The Gaussian wave packets all saturate the same bound: narrowing one density broadens the other. More precisely, choose a reference length $L$ and set $X=x/L$, $P=pL/\hslash$; then $[X,P]=iI$, and the plotted widths are $\sigma /L$ and $L\Delta p/\hslash$.

The Heisenberg uncertainty relation connects localization to the cost in kinetic energy. In a quantum harmonic oscillator with mass $m>0$ and angular frequency $\omega >0$, the variance contribution to the expectation value of the energy is bounded using the arithmetic-geometric mean inequality:The mean-dependent contributions are nonnegative. The Gaussian eigenstate for a quadratic potential with ${\sigma }^2=\hslash /(2m\omega )$ and zero means reaches this ground state energy. This is why the tradeoff is more than a limitation on preparation: it predicts an unavoidable positive minimum energy.

Heisenberg's original article appeared in March 1927 as "Über den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik", Zeitschrift für Physik 43, pp. 172–198. Its abstract discusses simultaneous imprecision of canonically conjugate quantities and explanatory thought experiments. The precise standard deviation formulation here distinguishes preparation spreads from a slogan about measurement disturbance. ^[ref].

High harmonic generation converts an intense driving electric field into radiation at high integer multiples of its frequency. High harmonic generation in a gas is a process in nonlinear optics: an electron can leave a neutral target by tunnel ionization, gain kinetic energy in the driving electric field, and return to radiate upon recombination. This three-step picture is an approximate account of a probability amplitude for emission in quantum mechanics, not a measured path of one electron.^[ref]

Why odd multiples? In an ideal centrosymmetric gas driven by a single frequency, assume the radiating electric dipole moment satisfies $d(t+T/2)=-d(t)$, where $T$ is the driving period. Its Fourier series coefficient obeys $d_q=-(-1{)}^q{d}_q$, so every even coefficient vanishes. A finite envelope, an oriented target or an additional driving frequency can invalidate that symmetry argument. Producing many odd components also does not prove that their spectral phases form a short optical pulse.

The 1988 report involving Anne L'Huillier observed odd multiples of a $1064\mathrm{nm}$ driving wavelength in rare gases, reaching order 33 in argon. The unexpectedly slow fall of yield beyond the fifth order supplied a broad spectral plateau. A spectral observation alone did not establish the relative spectral phases needed for an attosecond pulse train.^[ref]

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.