Wave Packet Spreading

Wave packet spreading is the growth of the spatial width of a normalizable quantum state under free-particle time evolution. A free packet spreads not because probability is lost or because a force acts on it, but because different momentum components move with different group velocities.

This page explains the mechanism. The detailed Gaussian construction is kept at Gaussian Wave Packets, and the Fourier-analysis background is kept at Wave Packets.

The Basic Mechanism

For a one-dimensional free particle,

$$\hat H=\frac{\hat p^2}{2m}, \qquad \omega(k)=\frac{\hbar k^2}{2m}.$$

A normalizable state can be built from plane-wave components:

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

If $a(k)$ is concentrated near $k_0$, expand the dispersion relation:

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

where

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

The constant term gives an overall phase. The linear term translates the packet envelope with velocity $v_g$. The quadratic term changes the shape of the envelope. For the nonrelativistic free particle, $\omega''$ is nonzero, so a generic packet disperses.

Width From Operator Evolution

The same conclusion follows without choosing a particular packet shape. In the Heisenberg picture for a free particle,

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

Define the symmetrized initial covariance

$$C_{xp}(0) =\frac{1}{2} \left\langle \Delta \hat x\,\Delta \hat p +\Delta \hat p\,\Delta \hat x \right\rangle_0,$$

with $\Delta \hat x=\hat x-\langle \hat x\rangle$ and $\Delta \hat p=\hat p-\langle \hat p\rangle$. Then

$$(\Delta x)^2(t) =(\Delta x)^2(0) +\frac{2t}{m}C_{xp}(0) +\frac{t^2}{m^2}(\Delta p)^2(0).$$

This formula is often the cleanest way to see what spreading means. A state with nonzero momentum uncertainty cannot keep a fixed width forever unless its initial position-momentum covariance is arranged to delay spreading over some interval. At sufficiently large $\lvert t\rvert$, the $t^2(\Delta p)^2(0)/m^2$ term dominates.

Gaussian Benchmark

For the minimum-uncertainty Gaussian packet used in Gaussian Wave Packets, the initial covariance vanishes and

$$(\Delta x)_0=\sigma_x, \qquad (\Delta p)_0=\frac{\hbar}{2\sigma_x}.$$

The width evolves as

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

The characteristic spreading time is therefore

$$t_{\mathrm{spread}} =\frac{2m\sigma_x^2}{\hbar}.$$

The dependence is physically important:

• Larger mass gives slower spreading.
• Larger initial width gives much slower spreading.
• Sharper localization gives faster spreading because it requires a broader momentum distribution.

The quadratic dependence on $\sigma_x$ is easy to underestimate. Doubling the initial width increases $t_{\mathrm{spread}}$ by a factor of four.

Energy Dispersion

A free packet also contains a spread of energies because

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

The energy spread is not by itself the whole story: a superposition of energy eigenstates can change phase without necessarily becoming broader in a given coordinate. Spreading occurs because the energy phase depends nonlinearly on momentum, so neighboring momentum components acquire relative phases that change the spatial interference pattern.

Near a central momentum $p_0$, the energy expansion is

$$E(p) =E(p_0) +\frac{p_0}{m}(p-p_0) +\frac{(p-p_0)^2}{2m}.$$

The linear term moves the packet center. The quadratic term is the source of dispersion. This is why a packet with a narrower momentum distribution spreads more slowly.

Classical Limit

The packet center obeys the free classical equation:

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

This agrees with the Ehrenfest Theorem Overview. But the width obeys a separate quantum law. Classical-looking motion requires not only a center that follows the classical trajectory, but also a width that remains small compared with the length scales being measured.

For a broad packet or a massive object, $t_{\mathrm{spread}}$ can be very long. For a sharply localized electron, it can be extremely short. This is one of the practical ways the Correspondence Principle enters wave mechanics: classical behavior is a controlled approximation, not the claim that quantum spreading is absent.

Worked Estimate

Take an electron initially localized to $\sigma_x=1\,\mathrm{nm}$. With $m_e=9.11\times 10^{-31}\,\mathrm{kg}$,

$$t_{\mathrm{spread}} =\frac{2m_e\sigma_x^2}{\hbar} \approx 1.7\times 10^{-14}\,\mathrm{s}.$$

A nanometer-scale electron packet spreads on a femtosecond scale. A packet with the same initial width but a proton mass spreads about $1836$ times more slowly. A packet a thousand times wider spreads a million times more slowly. These estimates explain why wave packet spreading is obvious in microscopic examples but often negligible in coarse macroscopic descriptions.

Common Mistakes

• Thinking a free quantum particle must move as a rigid localized packet.
• Treating spreading as a loss of normalization. Unitary evolution preserves total probability.
• Confusing motion of the packet center with growth of the packet width.
• Assuming that a narrow initial packet is more classical. Narrow localization usually means faster spreading.
• Ignoring the position-momentum covariance term, especially for initially chirped or squeezed packets.
• Treating phase velocity as the particle velocity. The envelope moves with group velocity.

Where This Is Used

• Free Particle gives the plane-wave basis and dispersion relation.
• Gaussian Wave Packets gives the exact solvable packet used as the standard benchmark.
• Dispersion and Classical Limit explains when spreading remains negligible for classical-looking motion.
• Wave Packets and Classical Trajectories places this spreading law inside the full trajectory-validity checklist.
• Wave Packets explains the Fourier superposition behind localization.
• Position-Momentum Uncertainty explains why narrowing a packet broadens its momentum distribution.
• Unitary Time Evolution explains why the norm is preserved while the spatial density changes.
• Wave Packets and Scattering uses spreading packets to interpret reflection, transmission, and tunneling as time-dependent processes rather than only stationary-wave calculations.

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. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.

Exercises

1. Use $\hat x(t)=\hat x(0)+\hat p(0)t/m$ to derive the general width formula for a free packet.

Solution

Let $\Delta \hat x(t)=\hat x(t)-\langle \hat x(t)\rangle$ and $\Delta \hat p=\hat p-\langle \hat p\rangle$. Since $\hat p$ is constant for a free particle,

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

Squaring and taking the expectation value gives

$$(\Delta x)^2(t) =(\Delta x)^2(0) +\frac{t}{m} \left\langle \Delta \hat x\,\Delta \hat p +\Delta \hat p\,\Delta \hat x \right\rangle_0 +\frac{t^2}{m^2}(\Delta p)^2(0).$$

With

$$C_{xp}(0) =\frac{1}{2} \left\langle \Delta \hat x\,\Delta \hat p +\Delta \hat p\,\Delta \hat x \right\rangle_0,$$

this is the stated formula.

2. A minimum-uncertainty Gaussian has $\sigma_x(t)=\sigma_x\sqrt{1+(t/t_{\mathrm{spread}})^2}$. At what time has its width doubled?

Solution

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

$$2=\sqrt{1+\left(\frac{t}{t_{\mathrm{spread}}}\right)^2}.$$

Squaring gives

$$4=1+\left(\frac{t}{t_{\mathrm{spread}}}\right)^2,$$

so

$$t=\sqrt{3}\,t_{\mathrm{spread}}.$$

3. Explain why increasing the mass makes a packet more classical-looking without changing the uncertainty principle.

Solution

The uncertainty principle constrains the product $(\Delta x)(\Delta p)$ and does not disappear for large mass. However, the free spreading rate depends on velocity spread:

$$\Delta v=\frac{\Delta p}{m}.$$

For the same momentum uncertainty, a larger mass gives a smaller velocity spread. For a Gaussian packet,

$$t_{\mathrm{spread}} =\frac{2m\sigma_x^2}{\hbar},$$

so increasing $m$ increases the spreading time. The packet can then remain narrow relative to the measured length scales for a longer time.