Wave Packets

A wave packet is a superposition of wave modes arranged to produce localization. In quantum mechanics, a square-integrable packet represents a normalizable state, unlike an individual plane wave on the full line.

Three motions should be distinguished:

• the oscillation of the carrier phase;
• the translation of the envelope;
• the change of envelope shape caused by dispersion.

Confusing these scales leads to many incorrect statements about particle velocity and spreading.

Momentum superposition

At time zero, a one-dimensional packet can be written

$$\psi(x,0) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} \phi(p,0)e^{ipx/\hbar}\,dp.$$

For a free or translation-invariant system with dispersion relation $E(p)$, each momentum amplitude acquires a phase:

$$\phi(p,t) = e^{-iE(p)t/\hbar}\phi(p,0).$$

Therefore,

$$\begin{aligned} \psi(x,t) &= \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} \phi(p,0)\\ &\quad\times e^{i[px-E(p)t]/\hbar}\,dp. \end{aligned}$$

The function $\phi(p,0)$ specifies which plane waves are present and with what relative phases. A narrow momentum distribution produces a long spatial envelope; a broad momentum distribution can produce sharper localization.

Normalization and probabilities

The Fourier transform is unitary, so

$$\int_{-\infty}^{\infty} \lvert\psi(x,t)\rvert^2\,dx = \int_{-\infty}^{\infty} \lvert\phi(p,t)\rvert^2\,dp.$$

Since time evolution changes only the momentum-space phase,

$$\lvert\phi(p,t)\rvert^2 = \lvert\phi(p,0)\rvert^2.$$

The momentum probability distribution is constant for a free particle even while the position distribution spreads. Normalization is preserved because the evolution is unitary.

A packet need not have compact support. Gaussian packets have nonzero tails everywhere, and a band-limited packet cannot be exactly localized to a finite interval. Localization means concentration of probability, not a classical point with a sharp edge.

Carrier and envelope

Suppose $\phi(p,0)$ is concentrated near $p_0$. Write

$$p=p_0+q$$

and expand the energy:

$$\begin{aligned} E(p_0+q) &= E_0 +E_0'q +\frac{1}{2}E_0''q^2 +\cdots, \end{aligned}$$

where derivatives are evaluated at $p_0$. Then

$$\begin{aligned} \psi(x,t) &= e^{i(p_0x-E_0t)/\hbar} \frac{1}{\sqrt{2\pi\hbar}}\\ &\quad\times \int dq\, \phi(p_0+q,0) e^{iq(x-E_0't)/\hbar}\\ &\quad\times e^{-iE_0''q^2t/(2\hbar)} \cdots. \end{aligned}$$

The first factor is the rapidly oscillating carrier. The remaining integral is the envelope. The linear term translates the envelope, while the quadratic and higher terms deform it.

This expansion is useful only when the momentum width is narrow enough that the retained derivatives describe $E(p)$ across the occupied band.

Group and phase velocities

The group velocity is

$$v_g = \left. \frac{dE}{dp} \right|_{p_0}.$$

To first order in $q$, the envelope depends on $x-v_gt$ and therefore moves at $v_g$ without changing shape.

The phase velocity of the carrier is

$$v_{\mathrm{ph}} = \frac{E(p_0)}{p_0},$$

when $p_0\ne0$. It tracks surfaces of constant carrier phase, not the center of a localized probability distribution.

For a nonrelativistic free particle,

$$E(p)=\frac{p^2}{2m},$$

so

$$v_g=\frac{p_0}{m}, \qquad v_{\mathrm{ph}}=\frac{p_0}{2m}.$$

The packet center follows the classical velocity $p_0/m$. The carrier crests move at half that speed.

For a relativistic positive-energy particle,

$$E(p) = \sqrt{p^2c^2+m^2c^4},$$

and

$$v_g = \frac{pc^2}{E}.$$

This is less than $c$ for $m>0$. The phase velocity can exceed $c$, but it does not carry a localized signal.

Dispersion and spreading

Dispersion occurs when $E(p)$ is nonlinear across the packet’s momentum support. Different momentum components then have different group velocities. The leading measure is

$$E_0'' = \left. \frac{d^2E}{dp^2} \right|_{p_0}.$$

For a narrow distribution with momentum width $\sigma_p$, the velocity spread is approximately

$$\sigma_v \simeq \lvert E_0''\rvert\sigma_p.$$

After a sufficiently long time, this velocity spread produces a spatial width of order $\sigma_vt$. Higher energy derivatives generate asymmetry and other shape changes.

For the nonrelativistic free particle,

$$E''(p)=\frac{1}{m},$$

so every finite-width packet generally spreads. For the relativistic dispersion,

$$E''(p) = \frac{m^2c^6}{E(p)^3}.$$

An exactly linear dispersion has $E''=0$ and translates a one-directional packet without dispersive broadening. Real systems can still introduce spreading through bandwidth limits, multiple branches, interactions, or inhomogeneity.

Exact free Gaussian

At $t=0$, take

$$\psi(x,0) = \frac{ e^{-(x-x_0)^2/(4\sigma_0^2)} e^{ip_0(x-x_0)/\hbar} }{ (2\pi\sigma_0^2)^{1/4} }.$$

The position probability distribution has mean $x_0$ and standard deviation $\sigma_0$. Define

$$\tau = \frac{\hbar t}{2m\sigma_0^2}, \qquad x_c(t) = x_0+\frac{p_0}{m}t.$$

Free evolution gives

$$\begin{aligned} \psi(x,t) &= \frac{(2\pi\sigma_0^2)^{-1/4}} {\sqrt{1+i\tau}}\\ &\quad\times \exp\left[ -\frac{(x-x_c(t))^2} {4\sigma_0^2(1+i\tau)} \right]\\ &\quad\times \exp\left[ \frac{ip_0}{\hbar} \left( x-x_0-\frac{p_0t}{2m} \right) \right]. \end{aligned}$$

Its probability density remains Gaussian:

$$\begin{aligned} \lvert\psi(x,t)\rvert^2 &= \frac{1}{\sqrt{2\pi}\,\sigma_x(t)} \\ &\quad\times \exp\left[ -\frac{(x-x_c(t))^2} {2\sigma_x(t)^2} \right]. \end{aligned}$$

with

$$\sigma_x(t) = \sigma_0 \sqrt{ 1+ \left( \frac{\hbar t}{2m\sigma_0^2} \right)^2 }.$$

The center moves uniformly, while the width grows. The momentum width remains

$$\sigma_p=\frac{\hbar}{2\sigma_0}.$$

At $t=0$, the packet saturates the uncertainty relation. At later times,

$$\sigma_x(t)\sigma_p = \frac{\hbar}{2}\sqrt{1+\tau^2},$$

because position and momentum develop correlations. Detailed phase, current, and propagator interpretations belong in Gaussian Wave Packets.

General free-particle variance

In the Heisenberg picture for a free particle,

$$\hat x(t) = \hat x(0)+\frac{t}{m}\hat p.$$

Define the symmetrized covariance

$$C_{xp} = \frac{1}{2} \left\langle \Delta\hat x\,\Delta\hat p +\Delta\hat p\,\Delta\hat x \right\rangle.$$

Then

$$\sigma_x^2(t) = \sigma_x^2(0) +\frac{2t}{m}C_{xp} +\frac{t^2}{m^2}\sigma_p^2.$$

An unchirped minimum-uncertainty Gaussian has $C_{xp}=0$ at its narrowest time. A correlated packet can initially contract before expanding. Spreading is therefore not always monotonic relative to an arbitrarily chosen time origin.

Uncertainty and localization

Every normalized state in the relevant operator domains satisfies

$$\sigma_x\sigma_p \ge \frac{\hbar}{2}.$$

Fourier scaling makes the tradeoff visible: compressing $\psi(x)$ broadens $\phi(p)$. But not every Gaussian-looking packet saturates the inequality. A quadratic phase, often called a chirp, changes the position–momentum covariance and can increase the product at fixed position density.

The precise inequality, equality condition, and domain assumptions are in Position–Momentum Uncertainty.

Stationary phase and semiclassical rays

At large $x$ and $t$, an oscillatory momentum integral is often dominated by stationary points of

$$\Phi(p) = px-E(p)t.$$

The stationary condition is

$$\frac{d\Phi}{dp}=0 \qquad\Longrightarrow\qquad \frac{x}{t} = \frac{dE}{dp}.$$

Thus the component observed near the ray $x/t$ has group velocity $dE/dp$. This connects packet propagation to semiclassical trajectories. Multiple stationary points can contribute interference, while degenerate stationary points require uniform approximations. See Stationary Phase for the asymptotic method.

Probability transport

For a free Schrödinger packet,

$$j(x,t) = \frac{\hbar}{m} \operatorname{Im} \left[ \psi(x,t)^* \frac{\partial\psi}{\partial x} \right]$$

obeys the continuity equation with $\lvert\psi\rvert^2$. For a narrow packet, the current near its center is approximately the density times $v_g$, but interference can produce local current patterns that are not captured by a single classical velocity.

Ehrenfest’s relation for a free particle is

$$\frac{d}{dt}\langle x\rangle = \frac{\langle p\rangle}{m}.$$

This controls the mean, not the width. A packet can follow the classical trajectory in its center while becoming strongly delocalized.

The current derivation is in Probability Current.

Packets and plane-wave idealizations

Plane waves are useful because momentum and translation-invariant Hamiltonians act diagonally on them. Physical preparation and detection, however, occur over finite regions and times. A packet supplies:

• square-integrable normalization;
• finite momentum resolution;
• an arrival envelope;
• a controlled route to the plane-wave limit.

In scattering theory, one often computes with plane waves and then interprets the result as the broad-packet limit. The packet must be broad enough in position to have narrow momentum support, yet narrow enough relative to the experimental geometry to define incident and outgoing regions. These limits should not be conflated.

Periodic and finite-volume packets

In a periodic box, momentum is discrete and the integral becomes a sum:

$$\psi(x,t) = \frac{1}{\sqrt L} \sum_{n\in\mathbb Z} c_n e^{i[p_nx-E(p_n)t]/\hbar},$$

where

$$p_n=\frac{2\pi\hbar n}{L}.$$

Finite-volume packets can wrap around the boundary and exhibit recurrences. Before recurrence, a sufficiently localized packet can approximate full-line motion if its tails remain negligible near the identified boundary.

Numerical propagation

Momentum-space evolution is especially efficient when $E(p)$ is diagonal:

1. Fourier transform $\psi(x,0)$ to $\phi(p,0)$.
2. Multiply by $e^{-iE(p)\Delta t/\hbar}$.
3. Inverse transform to position space when observables or potentials require it.

For split-operator evolution, kinetic and potential phases are alternated between momentum and position representations. Numerical reliability requires:

• enough spatial extent to prevent wraparound;
• small enough $\Delta x$ to resolve the momentum tail;
• small enough time step for the chosen splitting order;
• norm, mean, width, and energy checks;
• convergence under independent changes of box size and grid spacing.

Absorbing layers intentionally make the finite-grid evolution nonunitary and must be distinguished from physical packet loss.

Common mistakes

• Treating the carrier crest as the packet center.
• Using $v_{\mathrm{ph}}$ as the particle or signal velocity.
• Assuming every packet moves rigidly.
• Ignoring $E''(p)$ across a packet with broad momentum support.
• Calling a plane wave a localized normalized state.
• Assuming all Gaussian densities are minimum-uncertainty states.
• Forgetting that a chirped packet can initially contract.
• Interpreting motion of $\langle x\rangle$ as proof that the entire density follows a classical trajectory.
• Using the narrow-band group velocity formula near a cusp or band edge without checking the expansion.
• Allowing a numerical packet to wrap around a periodic FFT grid unnoticed.
• Checking grid spacing but not the finite simulation window.

Exercises

1. Compute the group and phase velocities for a nonrelativistic free particle. Which one equals the classical velocity?

Solution

With $E(p)=p^2/(2m)$,

$$v_g = \frac{dE}{dp} = \frac{p}{m}.$$

The phase velocity is

$$v_{\mathrm{ph}} = \frac{E}{p} = \frac{p}{2m}.$$

The group velocity equals the classical velocity. The phase velocity is half as large and describes the motion of carrier phase, not the packet center.

2. Suppose the dispersion relation is exactly linear over the occupied momenta:

   $$E(p)=E_0+v(p-p_0).$$

   Show that the packet translates without changing shape.

Solution

Insert the dispersion relation into the phase:

$$px-E(p)t = p(x-vt) -\left(E_0-vp_0\right)t.$$

The second term is independent of $p$ and becomes a global phase. The first changes $x$ to $x-vt$ in the initial packet integral. Hence

$$\psi(x,t) = e^{-i(E_0-vp_0)t/\hbar} \psi(x-vt,0).$$

The prefactor is a global phase. Therefore,

$$\lvert\psi(x,t)\rvert^2 = \lvert\psi(x-vt,0)\rvert^2,$$

which is rigid translation at speed $v$.

3. For the free Gaussian above, find the positive time at which the position standard deviation has doubled.

Solution

Set $\sigma_x(t)=2\sigma_0$:

$$2 = \sqrt{ 1+ \left( \frac{\hbar t}{2m\sigma_0^2} \right)^2 }.$$

Squaring gives

$$\left( \frac{\hbar t}{2m\sigma_0^2} \right)^2 =3.$$

Hence

$$t_{\mathrm{double}} = \frac{2\sqrt{3}\,m\sigma_0^2}{\hbar}.$$

Heavier particles and initially broader packets spread more slowly.

4. Derive the general free-particle variance formula from $\hat x(t)=\hat x(0)+t\hat p/m$.

Solution

The centered operator evolves as

$$\Delta\hat x(t) = \Delta\hat x(0) +\frac{t}{m}\Delta\hat p.$$

Squaring while preserving operator order gives

$$\begin{aligned} \Delta\hat x(t)^2 &= \Delta\hat x(0)^2\\ &\quad+ \frac{t}{m} \left( \Delta\hat x\,\Delta\hat p +\Delta\hat p\,\Delta\hat x \right)\\ &\quad+ \frac{t^2}{m^2}\Delta\hat p^2. \end{aligned}$$

Taking expectation values and using

$$C_{xp} = \frac{1}{2} \left\langle \Delta\hat x\,\Delta\hat p +\Delta\hat p\,\Delta\hat x \right\rangle$$

yields

$$\sigma_x^2(t) = \sigma_x^2(0) +\frac{2t}{m}C_{xp} +\frac{t^2}{m^2}\sigma_p^2.$$

A negative initial covariance can make the width decrease temporarily, but the positive quadratic term dominates at sufficiently large $\lvert t\rvert$.

References

• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, World Scientific, 1998.
• M. V. Berry, ‘Waves and Thom’s theorem,’ Advances in Physics 25, 1–26 (1976).
• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992.
• E. J. Heller, ‘Time-dependent approach to semiclassical dynamics,’ Journal of Chemical Physics 62, 1544–1555 (1975).