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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.

At energy equal to a finite rectangular barrier height, the stationary interior Schrödinger equation has a linear solution $\psi =bx+c$. Its probability current can be nonzero: $J=(\hslash /m)\mathrm{Im}(c^{\ast }b)$. Setting the interior classical kinetic energy to zero does not set the transmitted quantum flux to zero.

If the probability current of a normalizable wavefunction vanishes at spatial infinity, integrating the probability continuity equation over all space shows that the total probability $\int |\psi {|}^2$ is constant in time.

Here $\alpha >0$ and $q,p,\gamma$ are real functions of time. Differentiation in position gives ${\psi }_x=[-\alpha (x-q)+ip/\hslash ]\psi$. The standard probability current is thereforeFor a coherent state moving in a quadratic Hamiltonian, the centre follows the classical equation $\dot{q}=p/m$, so the current is the normalized Gaussian probability density transported at its centre velocity. The factor $\sqrt{\alpha /\pi }$ is essential: omitting it changes the current by a dimensional constant and violates the continuity equation for the normalized density. A spatially constant phase does not affect the current.

For $0<E<V_0$, set $k=\sqrt{2mE}/\hslash$ and $\kappa =\sqrt{2m(V_0-E)}/\hslash$. Continuity at the potential step gives the evanescent amplitude $c=2k/(k+i\kappa )$ for unit incident amplitude. Integrating $|c{|}^2{e}^{-2\kappa x}$ over $x>0$ gives $|c{|}^2/(2\kappa )$. Dividing by the incident probability current $\hslash k/m$ proves the displayed dwell time.

For a stationary one-dimensional scattering state with incident probability current $j_{\mathrm{in}}>0$, the dwell time in a region $\Omega$ is its integrated probability density divided by incident flux. It measures occupancy per incoming rate, not a classical trajectory or a uniquely defined tunnelling traversal time.

For a real positive decay coefficient alpha, a stationary exponential tail has zero probability current because $\mathrm{Im}({\psi }^{\ast }{\partial }_x\psi )=0$. Matching a subthreshold potential step can nevertheless give a nonzero coefficient T with $|T{|}^2>1$ relative to a unit incident plane-wave amplitude. This measures the density just beyond the step, not a flux-transmission probability. It describes penetration without a propagating outgoing channel in the semi-infinite high-potential region.

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.

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 a spinless particle without a magnetic vector potential, Bohmian mechanics sets $\dot{\mathbf{X}}=\hslash \nabla S/m$, where $S$ is the dimensionless quantum phase. Equivalently the velocity is the probability current divided by the probability density. This formula is local to nonzero wavefunction regions.

A hard wall at the origin and an attractive delta potential at distance $a$ form a leaky quantum cavity. For unit incident amplitude from outside, the interior standing wave has coefficient $A=-i\lambda x/[1-e^{2ix}+i\lambda x]$, where $x=ka$ and $\lambda ={\hslash }^2/(mqa)>0$. Its modulus squared is the title expression. A resonance pole $p_n=x_n-i{\gamma }_n$ produces an enhancement with a locally Lorentzian line shape, proportional to $[(x-x_n{)}^2+{\gamma }_n^2{]}^{-1}$. At fixed low $n$ and small $\lambda$, $x_n\simeq n\pi (1+\lambda /2+{\lambda }^2/4)$ and ${\gamma }_n\simeq (\lambda n\pi /2{)}^2$.

The coefficient can exceed one because it measures a standing-wave amplitude, while the net interior probability current is zero. Conservation of probability current gives unit reflected flux: no current passes through the hard wall. Near zero, $|A{|}^2\to {\lambda }^2/(2-\lambda {)}^2$ when $\lambda \ne 2$; the threshold case $\lambda =2$ requires a separate limit.

For the infinite square well of length $a$, the energy eigenvalues are $E_n=n^2{E}_1$, with $E_1={\pi }^2{\hslash }^2/(2m{a}^2)$. At $t_{\ast }=\pi \hslash /E_1$, the dynamical phase is $e^{-i{E}_n{t}_{\ast }/\hslash }=(-1{)}^n$. Since the midpoint parity operator has eigenvalues $(-1{)}^{n+1}$ in the same complete orthonormal basis, the time-evolution operator is exactly $U(t_{\ast })=-R_a$ on every state. At $2t_{\ast }$ every phase is one, giving a full revival.

Thus a state becomes its reflected copy, up to a global phase, at the half revival. Its position probability density is reflected, while its mean momentum changes sign because $P{R}_a=-R_{a}P$. For an initially reflection-symmetric probability density, the initial and half-revival position distributions agree even when their probability currents point in opposite directions.

For the normalized ground state of a reduced quantum harmonic oscillator, ${\varphi }_0(q)={\pi }^{-1/4}{e}^{-q^2/2}$ and $E_0=1/2$. Its classical turning points are $q=\pm 1$. The Born rule assigns probability $P=2{\pi }^{-1/2}{\int }_1^{\infty }{e}^{-q^2}dq=\mathrm{erfc}(1)\simeq 0.157299$ to the classically forbidden region. This is a stationary probability of detection outside the classical interval. It is not a transmitted flux or an escape rate: the real stationary wavefunction has zero probability current.

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.

For the equal superposition of the first two box states with zero initial relative quantum phase, put $\Omega =(E_2-E_1)/\hslash$. The probability in the left half is $P_{\mathrm{L}}=1/2+4\cos (\Omega t)/(3\pi )$. The continuity equation and zero wall probability current implyDirect substitution of $\psi (L/2,t)=L^{-1/2}{e}^{-i{E}_{1}t/\hslash }$ and ${\partial }_x\psi (L/2,t)=-2\pi L^{-3/2}{e}^{-i{E}_{2}t/\hslash }$ into the probability current verifies the sign. Positive current means probability leaves the left half. The quantum interference controls transport even though each component has stationary probability density.

The WKB approximation, also called JWKB, describes a short local de Broglie wavelength in a slowly varying potential energy. For $p(x)=\sqrt{2m[E-V(x)]}$ in a classically allowed region, its two local waves areThe phase follows the classical momentum; its amplitude varies to conserve probability current. In a forbidden region, putting $\kappa (x)=\sqrt{2m[V(x)-E]}$ gives $\psi \simeq {\kappa }^{-1/2}[C{e}^{{\int }^x\kappa /\hslash }+D{e}^{-{\int }^x\kappa /\hslash }]$, with boundary conditions selecting the physical combination.

The WKB validity condition is $\hslash |p^{\prime }|/p^2\ll 1$, with the analogous condition on $\kappa$. It fails at a turning point where $E=V$. For a simple turning point, locally linearizing V yields an Airy function; matching fixes the relative amplitudes and a $\pi /4$ phase. Two such turning points give the two-turning-point WKB quantization conditionFor an opaque barrier, the WKB tunnelling exponent instead givesThe factor two distinguishes probability from amplitude. Matching determines the prefactor; the exponential alone is not accurate near the barrier top. These formulas connect bound state energy eigenvalues and quantum tunnelling to classical actions. ^[ref]

The WKB approximation describes a wave whose local wavelength changes slowly compared with one wavelength. For the one-dimensional stationary Schrödinger equation, put $p(x)=\sqrt{2m[E-V(x)]}$. In a classically allowed region,The phase is the classical action divided by $\hslash$. The amplitude varies as $p^{-1/2}$ so that a single travelling component has constant probability current. A useful local validity condition is $|\hslash p^{\prime }/p^2|\ll 1$.

For $V>E$, write $\kappa =\sqrt{2m(V-E)}/\hslash$. The wave instead has growing and decaying components proportional to ${\kappa }^{-1/2}\exp (\pm \int \kappa \mathrm{d}x)$. At a simple turning point, $p$ vanishes and neither local WKB expression is valid. Linearizing the potential gives an Airy-function solution that connects the two sides and supplies the turning-point phase shift. It is not legitimate to carry the divergent WKB amplitude straight through the turning point.^[ref]

A useful application is tunnelling through a smooth opaque barrier. With turning points $x_1,x_2$, the leading transmission probability isThe exponent is twice the forbidden-region action in units of $\hslash$, because transmission is a probability rather than a wave amplitude. For an ideal rectangular barrier of height $V_0>E$ and width $d$, the same leading exponential is $\exp [-2d\sqrt{2m(V_0-E)}/\hslash ]$; matching at the abrupt edges supplies an additional prefactor.

For a bound state between two smooth turning points, the related WKB quantization rule is ${\int }_{x_1}^{x_2}p\mathrm{d}x=(n+1/2)\pi \hslash$. In a harmonic potential the integral equals $\pi E/\omega$, recovering $E_n=(n+1/2)\hslash \omega$. This exact agreement of energies for the harmonic oscillator does not make WKB wavefunctions uniformly exact near its turning points.

For the reduced quantum harmonic oscillator, $V(q)=q^2/2$ and the ground-state energy is $E_0=1/2$. Thus the classical turning points satisfy $q^2/2=1/2$, giving $q=\pm 1$. Normalizing the wavefunction gives ${\varphi }_0(q)={\pi }^{-1/4}{e}^{-q^2/2}$, because ${\int }_{-\infty }^{\infty }{e}^{-q^2}dq=\sqrt{\pi }$. The Born rule therefore gives the harmonic-oscillator forbidden-region probabilityThe factor two includes both tails. Using the exact complementary error function gives $\mathrm{erfc}(1)=0.1572992$. A stationary real wavefunction has zero probability current, so this result describes forbidden-region occupancy, not continuous loss of the bound particle.

The probability current isFor the incident and transmitted travelling waves, $j_{\mathrm{in}}=\hslash k|A{|}^2/m$ and $j_{\mathrm{tr}}=\hslash k|F{|}^2/m$. Thus $P(E)=|F/A{|}^2$ here. If the asymptotic wave numbers differed, the ratio of outgoing to incoming velocities would also be required. Current conservation gives $P+|B/A{|}^2=1$ for a real nonabsorbing barrier.

For $E>0$, put $k=\sqrt{2mE}$ and $q=k/\hslash >0$. Outside the support of the potential energy, a generalized eigenfunction of the Hamiltonian operator has the plane wave formHere $R_0$ denotes the support radius, to distinguish it from the quantum reflection probability $R$. The momenta are $\pm k$ and the velocities are $\pm k/m$. Thus $A,D$ describe incoming scattering channels from the left and right, and $B,C$ describe outgoing scattering channels towards the left and right. These energy eigenfunctions are not square-integrable functions; normalizable incoming wave packets are formed by superposition over energy, and their passage through the interaction region gives the physical quantum scattering experiment.

For incidence from the left, normalize $A=1,D=0$ and write $B=r_L,C=t_L$. The quantum reflection amplitude is $r_L$ and the transmission amplitude is $t_L$. The ratios of outgoing to incident probability currents defineNo extra velocity ratio occurs because both exterior regions have the same potential energy and mass. The scattering matrix, with a fixed choice of channel order, isIts first and second columns are the responses to unit incidence from the left and right, respectively. For a reflection-symmetric potential energy, these conventions give $r_L=r_R=r$ and $t_L=t_R=t$.