Group and Phase Velocity

A quantum matter wave is rarely a single pure sinusoid. Instead, a localized particle is described by a wave packet: a superposition of sinusoidal components with wave-numbers spread around some central value $k_0$ . Two distinct speeds characterize how this packet moves.

Phase velocity

A single plane-wave component $e^{i(kx - \omega t)}$ has a fixed phase on the surface $kx - \omega t = \text{const}$ . Solving for $x$ gives the phase velocity

$v_p = \frac{\omega}{k}.$

For a free non-relativistic particle, the quantum mechanical dispersion relation follows directly from substituting the de Broglie relations $E = \hbar\omega$ and $p = \hbar k$ into the classical kinetic energy $E = p^2/(2m)$ :

$\hbar\omega = \frac{(\hbar k)^2}{2m} \implies \omega = \frac{\hbar k^2}{2m}.$

The phase velocity of each component is therefore

$v_p = \frac{\omega}{k} = \frac{\hbar k}{2m} = \frac{p}{2m}.$

Notice that this equals half the classical particle speed $v_{\text{cl}} = p/m$ . Individual plane-wave crests do not travel at the physically meaningful speed of the particle.

Group velocity

The group velocity is the speed at which the envelope — the overall shape of the wave packet — propagates. For a narrow packet peaked at $k_0$ it is

$v_g = \left.\frac{d\omega}{dk}\right|_{k=k_0}.$

Differentiating the free-particle dispersion relation $\omega = \hbar k^2 / (2m)$ with respect to $k$ gives

$v_g = \frac{\hbar k_0}{m} = \frac{p}{m}.$

This is precisely the classical particle velocity. The group velocity, not the phase velocity, carries the particle's energy and matches what a detector would record as the arrival time of the particle.

Dispersion

When $v_g$ depends on $k$ — equivalently, when $d^2\omega/dk^2 \neq 0$ — different frequency components travel at different speeds and the packet spreads over time. For the free-particle dispersion relation $\omega \propto k^2$ we have $d^2\omega/dk^2 = \hbar/m \neq 0$ , so a free quantum wave packet always disperses (broadens) as it propagates. This spreading is a direct precursor to the Heisenberg uncertainty principle: a perfectly sharp position at $t = 0$ requires a very broad superposition of momenta, and those momenta soon pull the packet apart.

Try it

This is a numerical exercise — your code should return a number. Use the free-particle dispersion relation $\omega = \hbar k^2 / (2m)$ to compute the group velocity of an electron with wave-number $k = 1.0 \times 10^{10}\ \text{m}^{-1}$ . The group velocity formula gives $v_g = \hbar k / m$ ; with $\hbar = 1.0546 \times 10^{-34}\ \text{J\,s}$ and $m = 9.109 \times 10^{-31}\ \text{kg}$ this evaluates to approximately $1.158 \times 10^{6}\ \text{m/s}$ .

Run your code to see the quantum state.