Probability Current

Probability current describes the local transport of quantum probability. For one spinless nonrelativistic particle with constant mass, the standard kinetic term, and a local real scalar potential,

$$\rho=|\psi|^2, \qquad \mathbf j = \frac{\hbar}{2mi} \left( \psi^*\nabla\psi-\psi\nabla\psi^* \right) = \frac{\hbar}{m}\operatorname{Im}(\psi^*\nabla\psi).$$

The density tells how much probability is present per coordinate volume; $\mathbf j$ tells the signed rate and direction at which it crosses a surface. The formula is Hamiltonian dependent, not a universal current for every quantum model.

Required background. Wavefunctions and Probability Density supplies $\rho$. Time-Dependent Schrödinger Equation in Coordinate Space supplies the evolution law from which the current is identified.

Helpful background. Normalization Conventions explains why continuum and unit-flux wave amplitudes have different meanings.

Current is fixed by the Hamiltonian

For the scalar Schrödinger Hamiltonian, the TDSE implies

$$\partial_t\rho+\nabla\cdot\mathbf j=0.$$

The Continuity Equation owns the full local and integral derivation. The present page owns the physical current that appears in that law and how it is used.

Continuity alone fixes only $\nabla\cdot\mathbf j$. In more than one spatial dimension, adding a divergence-free field leaves the same continuity equation. The conventional current is selected by the Hamiltonian’s local coupling, symmetries, boundary flux, or an independently defined current operator. This ambiguity matters when comparing effective, spinful, nonlocal, or lattice models.

Phase and velocity interpretation

Away from nodes, write

$$\psi=R e^{iS/\hbar}, \qquad \rho=R^2.$$

Then the scalar-potential current is

$$\mathbf j=\rho\frac{\nabla S}{m}.$$

The ratio $\mathbf v_{\mathrm{flow}}=\mathbf j/\rho$ is a useful local flow field where $\rho\ne0$. It is not a classical particle trajectory, and $S$ is undefined at a node. A spatially constant phase produces no current, while a phase gradient can produce flow even when the density is stationary.

With electromagnetic vector potential $\mathbf A$,

$$\mathbf j = \rho\frac{\nabla S-q\mathbf A}{m}.$$

The gradient of $S$ alone is gauge dependent; the mechanical combination $\nabla S-q\mathbf A$ is physical.

Plane waves and normalization

For a one-dimensional traveling wave

$$\psi(x)=Ae^{ikx},$$

the current is

$$j=\frac{\hbar k}{m}|A|^2.$$

The sign of $k$ fixes the direction. An exact plane wave on the full line is a generalized state, not a square-normalizable probability distribution. The units and numerical value of $A$ depend on whether the state is box normalized, $k$ normalized, $p$ normalized, energy normalized, or flux normalized.

For counterpropagating waves of the same $|k|$,

$$\psi=Ae^{ikx}+Be^{-ikx},$$

direct substitution gives

$$j=\frac{\hbar k}{m}\left(|A|^2-|B|^2\right).$$

The interference term modulates the density but cancels from the net current for this equal-energy pair. A real standing wave therefore has zero current, although it is not a classical particle at rest.

Scattering flux

For a stationary one-dimensional state incident from the left, use the signed asymptotic currents

$$j_{\mathrm{inc}}>0, \qquad j_{\mathrm{ref}}<0, \qquad j_{\mathrm{trans}}>0.$$

Conservative single-channel scattering obeys

$$j_{\mathrm{inc}}+j_{\mathrm{ref}} =j_{\mathrm{trans}}.$$

Equivalently, in positive magnitudes, $j_{\mathrm{inc}}=|j_{\mathrm{ref}}|+j_{\mathrm{trans}}$. Define

$$R=\frac{|j_{\mathrm{ref}}|}{j_{\mathrm{inc}}}, \qquad T=\frac{j_{\mathrm{trans}}}{j_{\mathrm{inc}}}, \qquad R+T=1.$$

If the asymptotic waves use incident amplitude one and transmission amplitude $t$, then

$$T=\frac{v_R}{v_L}|t|^2,$$

where $v_L$ and $v_R$ are the positive channel velocities. For quadratic dispersion with the same mass, $v=\hbar k/m$; more generally the group velocity is $v=(1/\hbar)dE/dk$ on the selected branch. An asymptotically evanescent channel carries no transmitted flux. Inside a finite forbidden region, however, a complex mixture of growing and decaying exponentials can carry the constant current required by tunneling.

The identity $R+T=1$ fails when the model contains absorption, gain, time-dependent driving, inelastic channels, or untracked degrees of freedom.

Electromagnetic coupling and gauge invariance

For the minimally coupled kinetic momentum

$$\boldsymbol\pi=-i\hbar\nabla-q\mathbf A,$$

the probability current is

$$\mathbf j = \frac{1}{m}\operatorname{Re}(\psi^*\boldsymbol\pi\psi) = \frac{\hbar}{m}\operatorname{Im}(\psi^*\nabla\psi) -\frac{q}{m}\mathbf A|\psi|^2.$$

Under

$$\mathbf A' = \mathbf A+\nabla\chi, \qquad \Phi'=\Phi-\partial_t\chi, \qquad \psi'=e^{iq\chi/\hbar}\psi,$$

$\boldsymbol\pi\psi$ transforms covariantly with the same phase as $\psi$, while the bilinear probability current itself is gauge invariant. Omitting the $-q\mathbf A$ term destroys that invariance. Minimal Coupling in Wave Mechanics owns the full gauge and operator construction.

Dimensions and related currents

In three dimensions, $[\rho]=L^{-3}$ and $[\mathbf j]=L^{-2}T^{-1}$; surface integration gives probability per unit time. In one dimension, $[j]=T^{-1}$ because a point is the boundary of an interval.

For the stated spinless model, a particle of charge $q$ has charge density $q\rho$ and charge current $q\mathbf j$. Pauli and spin–orbit Hamiltonians can also contain magnetization or other Hamiltonian-specific current terms, so the simple multiplication by $q$ must not be exported without checking the model.

Validity and calculation checks

The displayed scalar formula must be reconsidered for position-dependent mass, nonlocal kernels, curved measures, spin–orbit or Pauli terms, lattice Hamiltonians, and many-particle reduced densities. A Hermitian nonlocal Hamiltonian may conserve the global norm without admitting this simple local current. On a lattice, probability transfer is naturally assigned to directed bonds rather than to a continuum vector field.

A current calculation should pass these checks:

• $j$ reverses sign when a traveling-wave momentum reverses;
• a real stationary wavefunction in a real scalar potential gives $j=0$;
• the signed incident, reflected, and transmitted currents balance;
• the velocity or Jacobian factor matches the continuum normalization label;
• the electromagnetic result is unchanged by a gauge transformation;
• boundary flux agrees with the rate of probability change in the region.

Common pitfalls

Treating density as direction. Two waves can have the same $|\psi|^2$ and opposite currents.

Using amplitude ratios as probabilities. Unequal channel velocities require a flux factor.

Calling the electromagnetic current gauge covariant. The kinetic derivative of the wavefunction is covariant; the physical bilinear current is invariant.

Ignoring generalized-state normalization. A plane-wave amplitude has no probabilistic meaning until its convention is stated.

Exercises

1. A wave has $\psi=Ae^{ikx}+Be^{-ikx}$ with $k>0$. Derive its current.

Solution

Insert the wave and its derivative into $j=(\hbar/m)\operatorname{Im}(\psi^*\psi')$. The two interference terms are real and do not contribute to the imaginary part, so

$$j=\frac{\hbar k}{m}\left(|A|^2-|B|^2\right).$$

2. Prove gauge invariance of the minimally coupled current.

Solution

With $\psi'=e^{iq\chi/\hbar}\psi$ and $\mathbf A'=\mathbf A+\nabla\chi$,

$$(-i\hbar\nabla-q\mathbf A')\psi' =e^{iq\chi/\hbar} (-i\hbar\nabla-q\mathbf A)\psi.$$

Multiplication by $\psi'^*=e^{-iq\chi/\hbar}\psi^*$ cancels the phase. Taking the real part and dividing by $m$ therefore leaves $\mathbf j$ unchanged.

3. A transmitted channel has twice the incident group velocity and amplitude $t=1/2$. Find $T$.

Solution

The flux ratio is

$$T=\frac{v_R}{v_L}|t|^2 =2\left(\frac12\right)^2 =\frac12.$$

Using $|t|^2$ alone would miss the velocity factor.

4. In three dimensions let $\mathbf j'=\mathbf j+\nabla\times\mathbf F$ for a smooth field $\mathbf F$. Show that $\mathbf j'$ obeys the same continuity equation.

Solution

Because the divergence of a curl vanishes,

$$\nabla\cdot\mathbf j' =\nabla\cdot\mathbf j +\nabla\cdot(\nabla\times\mathbf F) =\nabla\cdot\mathbf j.$$

This demonstrates why continuity alone does not select a unique local current in more than one dimension.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.