How to Calculate an Electron’s Expectation Value of Velocity

To calculate an electron’s expectation value of velocity, you need its quantum state and the appropriate velocity operator. In the usual nonrelativistic case with no electromagnetic vector potential, the operator is the momentum operator divided by the electron’s mass. Applying it to the wavefunction and evaluating the expectation integral gives the average velocity in that state.

The velocity operator

In one dimension, the momentum operator in position space is $\hat p=-i\hbar\,\partial/\partial x$. For an electron of mass $m_e$, the nonrelativistic velocity operator is therefore

$$\hat v=\frac{\hat p}{m_e}=-\frac{i\hbar}{m_e}\frac{\partial}{\partial x}.$$

For a normalized wavefunction $\psi(x,t)$, its expectation value is

$$\langle v\rangle=\int_{-\infty}^{\infty}\psi^*(x,t)\left(-\frac{i\hbar}{m_e}\frac{\partial}{\partial x}\right)\psi(x,t)\,dx.$$

In three dimensions, replace the derivative with the gradient:

$$\langle\mathbf v\rangle=\frac{1}{m_e}\int \psi^*(\mathbf r,t)(-i\hbar\nabla)\psi(\mathbf r,t)\,d^3r.$$

This is a vector expectation value. The state must be normalized, or the integral must be divided by $\int |\psi|^2\,d^3r$ if it is not. The expectation can depend on time, so use the wavefunction at the time you want to describe.

Worked example: a Gaussian wave packet

Consider a one-dimensional wave packet with a Gaussian envelope and a plane-wave phase:

$$\psi(x)=\frac{1}{(2\pi\sigma^2)^{1/4}}\exp\left[-\frac{(x-x_0)^2}{4\sigma^2}\right]e^{ik_0x}.$$

Here $\sigma$ sets the spread of the packet, $x_0$ is its center, and $k_0$ is its central wave number. The Gaussian factor is real, while the phase factor carries the average momentum. Differentiate the wavefunction:

$$\frac{\partial\psi}{\partial x}=\left[-\frac{x-x_0}{2\sigma^2}+ik_0\right]\psi.$$

Substituting into the expectation integral gives

$$\begin{aligned} \langle v\rangle &=\frac{-i\hbar}{m_e}\int \psi^*\frac{\partial\psi}{\partial x}\,dx\\ &=\frac{-i\hbar}{m_e}\left[-\frac{1}{2\sigma^2}\int (x-x_0)|\psi|^2\,dx+ik_0\int |\psi|^2\,dx\right]. \end{aligned}$$

The packet is normalized, so the second integral is $1$. Its probability density is symmetric about $x_0$, so the first integral is $0$. Thus

$$\langle v\rangle=\frac{\hbar k_0}{m_e}.$$

For example, take $k_0=1.00\times10^{10}\,\mathrm{m^{-1}}$. Using $\hbar\approx1.0546\times10^{-34}\,\mathrm{J\,s}$ and $m_e\approx9.1094\times10^{-31}\,\mathrm{kg}$ gives

$$\langle v\rangle\approx1.16\times10^6\,\mathrm{m/s}.$$

The width and center of this particular packet do not change the mean velocity; the phase gradient $k_0$ sets it. This is a nonrelativistic result, appropriate when the speed is much less than the speed of light.

What the expectation value means

An expectation value is the average result of many measurements on identically prepared systems. It does not generally mean that one electron has a definite velocity equal to $\langle v\rangle$. Also, the magnitude of the average velocity, $|\langle\mathbf v\rangle|$, is not generally the average speed, $\langle|\mathbf v|\rangle$.

For a real, stationary bound-state wavefunction in a one-dimensional potential, the expectation of momentum—and therefore velocity under these assumptions—is usually zero. The probability density can still be spread over a range of positions, and the electron’s momentum measurements need not all give zero. A zero mean says that positive and negative momentum contributions cancel in the average.

The same result has a useful dynamical interpretation. For a nonrelativistic particle with no vector potential, Ehrenfest’s theorem gives $d\langle x\rangle/dt=\langle p\rangle/m_e=\langle v\rangle$. Thus the expectation value of velocity is the rate of change of the expected position, not the derivative of a particle’s measured position along a classical path.

Magnetic vector potential and other caveats

If a magnetic vector potential $\mathbf A$ is present, the mechanical velocity operator is not simply $\hat{\mathbf p}/m_e$. For an electron with charge $-e$, minimal coupling gives

$$\hat{\mathbf v}=\frac{\hat{\mathbf p}+e\mathbf A}{m_e},$$

where $\hat{\mathbf p}=-i\hbar\nabla$ and $e$ is the positive elementary charge. Use this form when the Hamiltonian includes a vector potential. In relativistic quantum theory, velocity also requires a relativistic treatment; $\mathbf p/m_e$ is the nonrelativistic operator.

Common mistakes

• Using momentum as velocity: Convert using the mass: $\langle v\rangle=\langle p\rangle/m_e$ only in the nonrelativistic case without a vector potential.
• Leaving out the complex conjugate: The expectation integral contains $\psi^*$, not just $\psi$.
• Confusing average velocity with speed: A state can have zero average velocity but nonzero average speed.
• Using a state that is not normalized: Normalize it first, or divide the expectation integral by the state’s norm.
• Treating the result as a trajectory: The expectation is a statistical average, not a guarantee about a single measurement.