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For a finite potential energy discontinuity, the stationary state $\psi$ and its first derivative are continuous at $a$. Integrating the Time-independent Schrodinger equation across a shrinking interval about $a$ gives continuity of ${\psi }^{\prime }$; a jump in $\psi$ would produce a distributional derivative incompatible with a finite potential energy.

For $E>V_0$, put $k=\sqrt{2mE}/\hslash$, $q=\sqrt{2m(E-V_0)}/\hslash$ and use unit incident amplitude:There is no wave incident from the right. The matching equations $1+r=t$, $k(1-r)=qt$ give $r=(k-q)/(k+q)$ and $t=2k/(k+q)$. Comparing reflected and incident probability currents gives the reflection coefficientThe transmitted fraction is $T=(q/k)|t{|}^2=4kq/(k+q{)}^2$, so $R+T=1$.

For $0<E<V_0$, write $\kappa =\sqrt{2m(V_0-E)}/\hslash$ and retain only the decaying right-hand solution $t{e}^{-\kappa (x-a)}$. Matching gives $r=(k-i\kappa )/(k+i\kappa )$, which has modulus one. The evanescent wave carries no transmitted probability current. At $E=V_0$, the bounded right-hand solution is constant and matching again gives $r=1$. ThusClassically, a particle is wholly transmitted for $E>V_0$ and wholly reflected for $E<V_0$; there is no classical partial reflection above the step. At the threshold it has zero right-hand speed and no transmitted flux, a marginal case requiring a convention about motion exactly on the discontinuity. The quantum result approaches $R=0$ as $E/V_0\to \infty$.

The Born rule and probability current in one dimension giveMultiply the Time-dependent Schrodinger equation by ${\Psi }^{\ast }$, subtract its complex conjugate multiplied by $\Psi$, and rearrange to obtain the continuity equation ${\rho }_t+j_x=0$.

For the stated superposition, the interference terms in $j$ are real and cancel from its imaginary part. Henceindependent of $x$ and $t$.

For a wavefunction $\Psi (x,t)$, the probability density and probability current areMultiply the Schrodinger equation by ${\Psi }^{\ast }$, multiply its complex conjugate by $\Psi$, and subtract. The real potential energy terms cancel, leavingwhich is the probability continuity equation ${\rho }_t+j_x=0$.

For a stationary state, ${\rho }_t=0$, so $j_x=0$ and $j$ is constant in space. A normalizable wavefunction has vanishing current at spatial infinity, forcing this constant to be zero. By contrast the nonnormalizable plane wavehas stationary density $|A{|}^2$ and nonzero currentwhen $k\ne 0$.

In the classical experiment, particles whose impact parameter lies between $b$ and $b+db$ occupy an incident annulus of areaFor a spherically symmetric potential, azimuthal symmetry scatters this annulus into the solid-angle stripThus the central-potential classical differential cross-section isIf the relation between $b$ and $\theta$ has several branches, one sums this expression over every branch. The corresponding diagram consists of an incoming line offset by $b$ from the centre, an outgoing asymptote making angle $\theta$ with the incident direction, and the annulus and solid-angle strip just described.

For quantum scattering by a central potential, the far-field wavefunction iswhere $f(\theta )$ is the scattering amplitude. The incident plane wave has probability currentTo leading order in $1/r$, the scattered spherical wave has radial currentThe scattered rate through area $r^{2}d\Omega$ divided by the incident flux is therefore the quantum differential cross-section from a scattering amplitude:

For a one-dimensional wavefunction, the probability current isA plane wave $D{e}^{ikx}$ carries current $(\hslash k/m)|D{|}^2$ to the right, while $D{e}^{-ikx}$ carries the negative of this current. Thus the asymptotic terms are interpreted as an incident wave of amplitude $A$, a reflected wave of amplitude $B$, and a transmitted wave of amplitude $C$. Their currents areCurrent conservation for a real potential gives $j_{\mathrm{in}}+j_{\mathrm{ref}}=j_{\mathrm{tr}}$, and thereforewith $P_{\mathrm{ref}}+P_{\mathrm{tr}}=1$.

A normalizable bound state has $E<0$. Writingits decaying asymptotic behaviour is

For a one-dimensional wavefunction,The probability continuity equation isFor a stationary state $\psi (x,t)=\varphi (x)e^{-iEt/\hslash }$, the probability density is independent of time, so ${\partial }_{x}j=0$ and the probability current is constant in space.

For the plane waveone obtainsThis is a momentum eigenstate with momentum $p=\hslash k$ and, when it satisfies the free Schrodinger equation, energy $E={\hslash }^2{k}^2/(2m)$. Its constant density and current describe a spatially uniform beam carrying probability in the sign of $k$. It is not a normalizable wavefunction, so it represents an idealized state rather than a localized particle.

The one-dimensional Time-dependent Schrodinger equation for a wavefunction in a real potential isMultiplying this equation by ${\Psi }^{\ast }$, multiplying its complex conjugate by $\Psi$, and subtracting gives the probability continuity equationIf $\Psi$ and its first derivative decay sufficiently rapidly as $x\to \pm \infty$, then the probability current $j$ vanishes at both ends. Integrating the continuity equation and using the fundamental theorem of calculus yieldsThis conservation of quantum probability is required by the Born rule: once a normalizable wavefunction has total probability one, its time evolution must preserve that normalization.

For the stated Gaussian wave packet, put $\beta =\hslash t/m$. Sinceits squared modulus isThe Gaussian integral then giveswhich is independent of time, as required.

The probability density and probability current areFor a stationary state, $\psi (x,t)=\chi (x)e^{-iEt/\hslash }$, so $\rho =|\chi {|}^2$ is independent of time. The continuity equation then gives $\partial J/\partial x=0$. Hence

For incidence from the left, normalize the asymptotic wavefunction byFor incidence from the right, define $r^{\prime },t^{\prime }$ analogously with the directions reversed. Conservation of the probability current givesApplying the same conserved sesquilinear current to the two scattering solutions givesThese are exactly the column orthonormality relations $S^{†}S=I$ for the stated One-dimensional S-matrix, so $S$ is unitary.

The Schrodinger equation and its complex conjugate areBecause the potential $U$ is real, its two contributions cancel when differentiating the probability density $\rho ={\psi }^{\ast }\psi$. ThereforeThe expression in square brackets is the probability current $J$, soThis probability continuity equation says that probability can leave a region only through the current across its boundary.

For a plane wave $A{e}^{i(kx-Et/\hslash )}$, the probability current isThe incident amplitude is one and the transmitted amplitude is $T$, soThis is the transmission probability for quantum tunnelling. In the stationary state, $\partial \rho /\partial t=0$, so the continuity equation makes the net current independent of position. The reflected current therefore supplies the remainder:

For real $k$, conservation of the radial probability current, equivalently constancy of the radial Wronskian, equates the incoming and outgoing fluxes and givesFor a real potential, the regular radial solution depends on $k$ through $k^2$. Comparing its asymptotic form at $k$ and $-k$ therefore givesThe first identity places $S_l(k)$ on the unit circle, sofor a real phase shift. Choosing the phase continuously, the second identity givesThese are the unitarity and reflection identities for a partial-wave S-matrix.

Ifthen the ratio of scattered radial probability current through $r^{2}d\Omega$ to the incident probability flux is

The transmission probability is the ratio of transmitted to incident probability current. When the asymptotic wavenumbers agree, it is the squared modulus of the transmission amplitude.