Currents and Continuity (Preview)

The Born rule tells us that $|\psi(x,t)|^2\,dx$ is the probability of finding a particle between $x$ and $x+dx$ at time $t$ . But probability does not just sit still — if the wavefunction changes over time, probability can flow from one region to another, much like charge flowing in a wire. Making that flow mathematically precise leads to the probability current and the continuity equation.

The probability density and its time derivative

Define the probability density

$\rho(x,t) = |\psi(x,t)|^2 = \psi^*(x,t)\,\psi(x,t).$

To see how $\rho$ changes with time, differentiate with respect to $t$ :

$\frac{\partial \rho}{\partial t} = \frac{\partial \psi^*}{\partial t}\,\psi + \psi^*\,\frac{\partial \psi}{\partial t}.$

The Schrödinger equation for a particle of mass $m$ in a real potential $V(x)$ gives

$i\hbar\,\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V\psi,$

$\frac{\partial \psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^2 \psi}{\partial x^2} - \frac{i V}{\hbar}\psi.$

Taking the complex conjugate (remembering $V$ is real),

$\frac{\partial \psi^*}{\partial t} = -\frac{i\hbar}{2m}\frac{\partial^2 \psi^*}{\partial x^2} + \frac{i V}{\hbar}\psi^*.$

Substituting both expressions into $\partial\rho/\partial t$ and noting that the $V$ terms cancel:

$\frac{\partial \rho}{\partial t} = \frac{i\hbar}{2m}\!\left(\psi^*\frac{\partial^2\psi}{\partial x^2} - \psi\frac{\partial^2\psi^*}{\partial x^2}\right).$

The right-hand side is a perfect spatial derivative:

$\frac{\partial \rho}{\partial t} = \frac{\partial}{\partial x}\!\left[\frac{i\hbar}{2m}\!\left(\psi^*\frac{\partial\psi}{\partial x} - \psi\frac{\partial\psi^*}{\partial x}\right)\right].$

Defining the probability current

The expression inside the spatial derivative is the probability current:

$J(x,t) = \frac{\hbar}{2mi}\!\left(\psi^*\frac{\partial\psi}{\partial x} - \psi\frac{\partial\psi^*}{\partial x}\right) = \frac{\hbar}{m}\,\operatorname{Im}\!\left(\psi^*\frac{\partial\psi}{\partial x}\right).$

With this definition, the result above becomes the continuity equation:

$\boxed{\frac{\partial \rho}{\partial t} + \frac{\partial J}{\partial x} = 0.}$

This equation has exactly the same mathematical form as the classical continuity equation for a fluid or for electric charge: a local increase in density must be fed by an inward current.

Global conservation from local conservation

Integrating the continuity equation over all of space,

$\frac{d}{dt}\int_{-\infty}^{\infty}\rho\,dx = -\left[J(x,t)\right]_{-\infty}^{\infty}.$

For any physically reasonable (square-integrable) wavefunction, $J \to 0$ as $x \to \pm\infty$ , so the right-hand side vanishes:

$\frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2\,dx = 0.$

This confirms that if the wavefunction is normalized at $t = 0$ , the Schrödinger equation keeps it normalized for all time. Probability is globally conserved because it is locally conserved.

A quick check: the plane wave

For a free-particle plane wave $\psi = A\,e^{i(kx - \omega t)}$ with $|A|^2 = \rho_0$ (constant), the current is

$J = \frac{\hbar k}{m}\,|A|^2 = \rho_0\,v,$

where $v = \hbar k/m$ is the group velocity. This is exactly what you would write classically for a uniform beam of particles with number density $\rho_0$ all moving at speed $v$ — a satisfying check that the quantum definition reduces to the familiar classical result when the wavefunction is a plane wave.

What comes next

In higher dimensions (and in the intermediate track) $J$ becomes a vector field $\mathbf{J}$ , and the continuity equation reads $\partial\rho/\partial t + \nabla\cdot\mathbf{J} = 0$ . The current also plays a starring role in quantum tunneling (where it lets us define transmission and reflection coefficients) and in the probability-flux interpretation of scattering. For now the key takeaways are: probability has a local current, that current obeys a continuity equation derived directly from the Schrödinger equation, and normalization is preserved automatically.