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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.

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.

Put $p=\sqrt{2mE}/\hslash$ and $\gamma =mk/{\hslash }^2$. Integrating the Schrodinger equation across the delta potential gives continuity of $\psi$ andFor incidence from the left, write $\psi =e^{ipx}+r{e}^{-ipx}$ on $x<0$ and $\psi =t{e}^{ipx}$ on $x>0$. Matching gives $t=1+r$ and $ip(t-1+r)=2\gamma t$, henceThe incident and transmitted probability currents have the same velocity factor, so the reflection coefficient and transmission coefficient areFor $k<0$, an even bound state has $\psi =A{e}^{-\kappa |x|}$, $\kappa =\sqrt{-2mE}/\hslash$. The derivative jump now gives $-2\kappa A=2\gamma A$, so $\kappa =-\gamma >0$ andThere is exactly one positive value of $\kappa$, hence exactly one even bound state. Its normalized amplitude is $A=\sqrt{\kappa }$. An odd decaying solution would have to vanish at zero and is therefore trivial.

In the shrinking finite square well, its integrated potential is $k=-2Ul$, not $-Ul$: the full width is $2l$. With $Ul$ fixed and $l\to 0$, $0\le (\alpha l{)}^2\le 2mU{l}^2/{\hslash }^2\to 0$. The matching condition then bounds $\beta$ by $(2mUl/{\hslash }^2)(1+o(1))$, so $E$ remains bounded. Consequently $El\to 0$, and the previous matching condition givesUsing $E=-{\hslash }^2{\beta }^2/(2m)$ recovers the same bound-state energy. All other even levels disappear as the width shrinks.

In Bohmian mechanics the particle has a definite position $\mathbf{X}(t)$ at every time. Its wavefunction obeys the usual autonomous wave equation, while its actual position follows the guidance equationHere $\mathbf{j}$ is the probability current. The Born rule is the quantum-equilibrium choice of initial position distribution $\rho =|\psi {|}^2$; quantum equilibrium equivariance ensures that this distribution persists because it obeys the same probability continuity equation as the wave amplitude. The guidance equation fixes the initial velocity as well as subsequent velocities: the second-order equation below does not permit an independent arbitrary initial velocity.

Define the quantum potentialTaking the gradient of the real Madelung equations givesOn any smooth phase patch $\nabla \times \mathbf{v}=0$, so $\nabla (|\mathbf{v}{|}^2/2)=(\mathbf{v}\cdot \nabla )\mathbf{v}$. Along the actual path, differentiation is the material derivative $D/Dt={\partial }_t+\mathbf{v}\cdot \nabla$. ThereforeThis is the Bohmian mechanics Newton form: the classical force is supplemented by the amplitude-dependent quantum potential. Neither division by $R$ nor a smooth phase is justified at a wavefunction node, so the derivation applies on nonzero-amplitude regions. A nonzero circulation around a node is compatible with the locally curl-free guidance equation.

The agreement is in the energy levels, not in electron trajectories. The Bohr model assumes a particle on a circular orbit, with orbital angular momentum quantized as $N\hslash$. The Coulomb bound states here are spatially distributed wavefunctions with no definite classical orbit. Their probability density is spherically symmetric, and their orbital angular momentum is zero because $\ell =0$. Their radial distributions also depend on $N$ and can have nodes. A real radial wavefunction times a global stationary phase has zero probability current, unlike the circulating point charge in the Bohr picture.

For a particle of mass $m$, the probability density and one-dimensional probability current areThe Schrodinger equation and its complex conjugate, for real $V$, giveThe potential terms cancel precisely because $V$ is real. Thus ${\rho }_t+j_x=0$, the continuity equation for probability. Integrating over an interval says that its probability changes by the incoming current minus the outgoing current.

The global time-dependent phase cancels from the probability current. Put $\chi =e^{ikx}+R{e}^{-ikx}$. Direct multiplication givesThe second term is real, because the expression in parentheses is purely imaginary. HenceThe first plane wave travels to the right with incident current $\hslash k/m$; the second travels to the left with reflected current $-(\hslash k/m)|R{|}^2$. Their wave interference changes the probability density but cancels from the net current. The current vanishes for $|R|=1$, is positive for $|R|<1$, and is negative for $|R|>1$. These plane waves are scattering states, not individually normalizable bound states.

The wavefunction is a plane wave and a simultaneous formal eigenfunction of the momentum operator and the time-translation energy operator:Its momentum and energy are therefore definite, while its probability density is spatially uniform:For a particle of mass $m$, the probability current is $j=(\hslash /m)\mathrm{Im}(\overline{\varphi }{\partial }_x\varphi )=\hslash k{A}^2/m$, directed according to the sign of $k$.

On the whole real line, a nonzero constant-amplitude plane wave is not square-integrable and cannot be normalised to total probability one. It represents an idealised momentum eigenstate, interpreted through generalized eigenfunctions or as a limit of wave packets. Also, arbitrary $k,E$ do not automatically solve the Schrodinger equation for a specified potential energy: for a free particle, $E={\hslash }^2{k}^2/(2m)$, or this value plus the constant potential energy for a constant-potential region. If $A=0$, the zero function is not a physical quantum state.

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 $E/V_0\to \infty$.