Group Velocity and Phase Velocity

Group velocity is the velocity of a narrow wave-packet envelope, while phase velocity is the velocity of surfaces of constant phase for a single wave component. For a nonrelativistic free particle, the group velocity equals the classical particle velocity, but the phase velocity does not.

The distinction matters because plane waves are not localized particles. Localized free-particle states are wave packets, and their observable motion is governed by the motion of the packet envelope.

Plane Wave Phase

A plane-wave component has the form

$$\psi_k(x,t)=A e^{i(kx-\omega t)}.$$

The phase is

$$\Phi(x,t)=kx-\omega t.$$

A surface of constant phase satisfies $d\Phi=0$, so

$$k\,dx-\omega\,dt=0.$$

The phase velocity is therefore

$$v_{\mathrm{ph}} =\frac{dx}{dt} =\frac{\omega}{k},$$

for $k\ne 0$. This is the speed of the oscillatory crests of one Fourier component. It is not, by itself, the velocity of a localized particle.

Packet Envelope

A localized packet is a superposition of nearby wavenumbers:

$$\psi(x,t) =\frac{1}{\sqrt{2\pi}} \int a(k)e^{i(kx-\omega(k)t)}\,dk.$$

Suppose $a(k)$ is sharply concentrated near $k_0$. Expand the dispersion relation:

$$\omega(k) \approx \omega(k_0) +\omega'(k_0)(k-k_0).$$

Keeping only this linear approximation, the packet becomes an envelope moving rigidly at

$$v_g =\omega'(k_0) =\frac{d\omega}{dk}\bigg\rvert_{k_0}.$$

This is the group velocity. It is the velocity of the packet envelope when the packet is narrow in momentum space and spreading is negligible over the time of interest.

The next term,

$$\frac{1}{2}\omega''(k_0)(k-k_0)^2,$$

controls dispersion. Its effect is explained in Wave Packet Spreading.

Free-Particle Formulas

For a nonrelativistic free particle,

$$E=\frac{p^2}{2m}, \qquad p=\hbar k, \qquad E=\hbar\omega.$$

Thus

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

The phase velocity is

$$v_{\mathrm{ph}} =\frac{\omega}{k} =\frac{\hbar k}{2m} =\frac{p}{2m}.$$

The group velocity is

$$v_g =\frac{d\omega}{dk} =\frac{\hbar k}{m} =\frac{p}{m}.$$

The group velocity matches the classical free-particle velocity. This agreement is one concrete form of the Ehrenfest theorem for free motion.

Negative Wavenumbers

The sign matters. If $k_0\gt0$, both $v_g$ and $v_{\mathrm{ph}}$ are positive. If $k_0\lt0$, both are negative. A packet centered at negative momentum moves to the left:

$$v_g=\frac{\hbar k_0}{m}\lt0.$$

The energy is still positive because

$$E=\frac{\hbar^2k_0^2}{2m}.$$

This is why positive energy in one-dimensional free motion corresponds to two directions of propagation.

Why Phase Velocity Is Not Particle Velocity

The probability density of a pure plane wave is spatially constant:

$$\lvert \psi_k(x,t)\rvert^2=\lvert A\rvert^2.$$

There is no localized bump whose position can be tracked. Phase fronts move, but a single plane wave does not describe a localized particle on the full line. A packet has an envelope, and that envelope moves at group velocity.

For the free-particle dispersion relation, the phase velocity is half the group velocity:

$$v_{\mathrm{ph}}=\frac{1}{2}v_g.$$

This factor of one half is not a paradox. It reflects the quadratic relation $E=p^2/(2m)$, not a new particle speed.

Dispersion And Rigid Motion

If the dispersion relation is exactly linear,

$$\omega(k)=\omega_0+v(k-k_0),$$

then $\omega''(k)=0$ and a narrow packet can translate without changing shape. The nonrelativistic free particle is different:

$$\omega''(k)=\frac{\hbar}{m}.$$

Because this is nonzero, free nonrelativistic packets generally spread. The packet center moves with $v_g$, while the width changes according to the momentum spread and covariance of the state.

Worked Example

Consider a Gaussian packet centered at momentum $p_0=\hbar k_0$. Its center moves as

$$\langle x\rangle(t) =x_0+\frac{p_0}{m}t.$$

The group velocity formula gives the same result:

$$v_g=\frac{\hbar k_0}{m}=\frac{p_0}{m}.$$

Meanwhile the phase velocity of the carrier wave is

$$v_{\mathrm{ph}} =\frac{\hbar k_0}{2m} =\frac{p_0}{2m}.$$

The envelope and the phase crests therefore drift through one another. In visualizations of a moving Gaussian packet, the broad probability envelope moves at $v_g$, while the oscillations inside the envelope move at $v_{\mathrm{ph}}$.

Common Mistakes

• Identifying phase velocity with the velocity of a localized particle.
• Forgetting that a plane wave has no localized position to track.
• Computing $v_g$ from $E/p$ instead of from $d\omega/dk$.
• Ignoring the sign of $k$ when describing direction of motion.
• Assuming group velocity alone guarantees rigid motion. Dispersion depends on $\omega''(k)$.
• Treating a broad packet in $k$ as if it had a single exact group velocity.

Where This Is Used

• Free Particle derives the dispersion relation and plane-wave solutions.
• Gaussian Wave Packets shows the envelope motion in the canonical normalizable example.
• Wave Packet Spreading explains why nonzero $\omega''(k)$ changes packet width.
• Wave Packets and Classical Trajectories combines group motion with phase-space localization, external forces, and branch splitting.
• Stationary Phase in Quantum Mechanics derives the long-time map between detection position and the momentum component whose group velocity reaches it.
• Fourier Transform gives the position-momentum transform convention.
• Momentum Representation explains how momentum-space phases generate time evolution.
• Scattering pages use group velocity and probability current to interpret incoming, reflected, and transmitted beams.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.

Exercises

1. Derive $v_{\mathrm{ph}}$ and $v_g$ for the nonrelativistic free-particle dispersion relation.

Solution

For a free particle,

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

The phase velocity is

$$v_{\mathrm{ph}} =\frac{\omega}{k} =\frac{\hbar k}{2m}.$$

The group velocity is

$$v_g =\frac{d\omega}{dk} =\frac{\hbar k}{m}.$$

Since $p=\hbar k$, this gives $v_g=p/m$.

2. A packet is centered at $k_0\lt0$. What is the direction of its group motion?

Solution

The group velocity is

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

For positive mass $m$ and $k_0\lt0$, this is negative. The packet center moves toward decreasing $x$.

3. Explain why a linear dispersion relation does not produce spreading in the linear approximation.

Solution

If

$$\omega(k)=\omega_0+v(k-k_0),$$

then all components in the packet share the same group velocity,

$$\frac{d\omega}{dk}=v,$$

and

$$\frac{d^2\omega}{dk^2}=0.$$

The constant term contributes an overall phase, and the linear term translates the envelope. There is no quadratic phase term to distort the envelope, so the packet does not spread within this approximation.