Dispersion and Classical Limit

A free wave packet looks classical when its center follows the classical trajectory and its width remains small compared with the length scale being resolved. Dispersion is the mechanism that can spoil this picture: different momentum components move with different group velocities, so the packet spreads.

The free-particle case is unusually clean because the center motion is exactly classical:

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

The classical-limit question is therefore not whether the center obeys Newton’s law. It does. The question is whether the state remains localized enough for that center to represent the observed motion.

Helpful background. Wave Packet Spreading to track the evolution of position variance.

Free Packet In Momentum Space

With the unitary momentum Fourier convention, a normalizable free packet can be written as

$$\psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} \phi(p) \exp\left[ \frac{i}{\hbar}px -\frac{i}{\hbar}\frac{p^2}{2m}t \right]\,dp.$$

If $\phi(p)$ is concentrated near $p_0$, write $p=p_0+q$. The energy expands as

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

The first term gives an overall phase. The linear term translates the packet with velocity

$$v_0=\frac{p_0}{m}.$$

The quadratic term is dispersion. It changes the relative phases across the momentum distribution and therefore changes the packet shape.

Narrow Momentum Distribution

For a directed beam with nonzero mean momentum, a useful sufficient condition for a well-defined velocity is

$$\Delta p\ll \lvert p_0\rvert.$$

Then the velocity spread

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

is small compared with

$$\lvert v_0\rvert=\frac{\lvert p_0\rvert}{m}.$$

This condition is about relative velocity sharpness. Near rest, $p_0=0$, compare $\Delta p/m$ with an absolute velocity resolution instead; a packet can describe a nearly stationary classical particle. Neither velocity condition is enough by itself. A packet can have a narrow momentum distribution and still be very broad in position. Classical-looking particle motion also requires the spatial width to be small compared with the relevant observation scale.

Short-Time Propagation

Over a time $t$, different velocity components separate by a distance of order

$$\Delta x_{\mathrm{disp}} \sim \Delta v\,t = \frac{\Delta p}{m}t.$$

If this is small compared with the initial width or with the measurement resolution, the packet looks approximately rigid. A practical short-time condition is

$$t\ll t_{\mathrm{disp}},$$

where $t_{\mathrm{disp}}$ is the characteristic time on which the packet width changes appreciably.

For an initially unchirped minimum-uncertainty Gaussian,

$$t_{\mathrm{disp}} \sim \frac{2m\sigma_x^2}{\hbar}.$$

This formula encodes the main scaling:

• larger mass slows dispersion;
• wider initial packets disperse more slowly;
• sharper localization disperses more quickly.

The exact Gaussian formula is derived in Gaussian Wave Packets and interpreted in Wave Packet Spreading.

Large Mass Limit

For fixed initial width $\sigma_x$, the minimum momentum spread is

$$\Delta p_{\min} = \frac{\hbar}{2\sigma_x}.$$

This lower bound does not contain the mass. The velocity spread does:

$$\Delta v_{\min} = \frac{\Delta p_{\min}}{m} = \frac{\hbar}{2m\sigma_x}.$$

Thus large mass suppresses the rate at which different momentum components separate in space. This is one reason macroscopic packets can remain effectively localized over ordinary times and length scales even though the uncertainty relation still holds exactly.

The large-mass limit is not magic. If one waits long enough, or asks for enough spatial resolution, dispersion can still matter. Classical behavior is always relative to scales.

Ehrenfest Center Motion

For a free particle, Ehrenfest theorem gives

$$\frac{d}{dt}\langle x\rangle = \frac{\langle p\rangle}{m}, \qquad \frac{d}{dt}\langle p\rangle=0.$$

Therefore

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

when $x_0=\langle x\rangle(0)$ and $p_0=\langle p\rangle$.

This exact center motion is necessary for a classical-particle approximation, but not sufficient. A broad or split wavefunction may have a mean position that is not where a measurement is likely to find the particle. The Ehrenfest Theorem Overview explains this limitation for general potentials.

Spatial Resolution And Coarse Graining

Let $L_{\mathrm{obs}}$ be the spatial resolution or length scale relevant to the question. A free packet is particle-like over the time interval of interest if

$$\Delta x(t)\ll L_{\mathrm{obs}}.$$

For scattering, $L_{\mathrm{obs}}$ might be a detector resolution, a barrier width, or the distance over which an asymptotic beam is considered well collimated. For a classical trajectory comparison, it might be the scale on which the trajectory is plotted or measured.

This scale dependence matters. The same packet can be classical-looking for a coarse detector and visibly quantum for a fine interferometric measurement.

De Broglie Wavelength And Envelope Scale

A packet with central momentum $p_0$ has de Broglie wavelength

$$\lambda_{\mathrm{dB}} = \frac{2\pi\hbar}{\lvert p_0\rvert}.$$

For a semiclassical packet, the carrier oscillations are usually much shorter than the envelope scale:

$$\lambda_{\mathrm{dB}}\ll \Delta x.$$

Equivalently, the packet contains many phase oscillations under its envelope. This is compatible with a narrow relative momentum spread:

$$\Delta p\ll \lvert p_0\rvert.$$

For a directed beam, a wavelength comparable to an aperture or obstacle means that diffraction cannot generally be neglected. The carrier-wavelength comparison is not a necessary condition for every classical trajectory approximation: at $p_0=0$, the expression for $\lambda_{\mathrm{dB}}$ is undefined, yet a packet can remain concentrated around a stationary classical trajectory. Use the absolute position and velocity resolutions in that case.

Stationary Phase View

The free packet integral has rapidly oscillating phase

$$\Phi(p) = px-\frac{p^2}{2m}t.$$

The stationary phase condition is

$$\frac{d\Phi}{dp} = x-\frac{p}{m}t =0.$$

Thus the dominant momentum contributing to position $x$ at time $t$ satisfies

$$x=\frac{p}{m}t,$$

which is the classical free trajectory from the origin. This form assumes that the phase of $\phi(p)$ varies slowly near the contributing momentum. A packet translated to $x_0$ has a factor $e^{-ipx_0/\hbar}$; include it in the phase to obtain $x-x_0=pt/m$. Stationary phase is one route from wave mechanics to classical propagation: when phases oscillate rapidly, contributions away from classical stationary paths cancel strongly.

This idea becomes more powerful in WKB and semiclassical propagator methods. The broader map is in Semiclassical Limit Overview, WKB Approximation, and Semiclassical Propagator.

A Practical Classicality Checklist

For a free wave packet over a chosen time interval, check:

• Is $\Delta p/m$ small compared with the relevant velocity resolution (or with $\lvert p_0\rvert/m$ for a directed beam)?
• Does $\Delta x(t)\ll L_{\mathrm{obs}}$ hold throughout that interval?
• Is one localized packet sufficient to describe the measured position distribution, and are interference effects negligible for the observables being used?

The conditions $t\ll t_{\mathrm{disp}}$ and $\lambda_{\mathrm{dB}}\ll\Delta x$ are useful additional checks for an approximately rigid, directed semiclassical packet. They are not required for all motion that looks classical at a specified resolution. A packet can change its width substantially while remaining far narrower than $L_{\mathrm{obs}}$, and a nearly stationary packet need not have a well-defined carrier wavelength.

Preparation correlations also matter. For free evolution over $0\le t\le T$, with $T>0$ and finite position and momentum variances, the smallest achievable value of the largest position variance is

$$\min_{\psi}\ \max_{0\le t\le T}(\Delta x)^2(t) =\frac{\hbar T}{2m}.$$

Exercise 1 derives the bound and a Gaussian that attains it by focusing halfway through the interval. Thus $\hbar T/(2m)\ll L_{\mathrm{obs}}^2$ is the scale separation needed for a freely evolving state to stay narrow throughout that window; velocity resolution remains a separate requirement. This concerns the spreads of one unmeasured, freely evolving state. It does not impose a detector-error bound on a sequence of measurements.

Common Mistakes

• Saying the classical limit is simply $\hbar\to0$ without specifying scales.
• Confusing exact Ehrenfest center motion with full classical particle behavior.
• Assuming a narrower packet is always more classical; narrow packets often spread faster.
• Applying the directed-beam condition $\Delta p\ll\lvert p_0\rvert$ to a packet near rest instead of using an absolute velocity resolution.
• Treating de Broglie wavelength as irrelevant once the packet center moves classically.
• Forgetting that coarse graining and detector resolution are part of practical classicality.

Exercises

1. A free particle of mass $m>0$ evolves without intervening measurements for $0\le t\le T$, where $T>0$. Among normalized full-line states with finite position and momentum variances, find the smallest possible value of $\max_{0\le t\le T}(\Delta x)^2(t)$. First apply the uncertainty relation to $x(0)$ and $x(T)$, then test a Gaussian that is unchirped at $t=T/2$. Can the optimal state have zero mean momentum?

Solution

The free Heisenberg evolution gives

$$x(T)=x(0)+\frac{T}{m}p(0), \qquad [x(0),x(T)]=i\hbar\frac{T}{m}.$$

The uncertainty relation, interpreted in quadratic-form sense when operator products are not defined, implies

$$\Delta x(0)\,\Delta x(T)\ge\frac{\hbar T}{2m}.$$

If $M=\max_{0\le t\le T}(\Delta x)^2(t)$, both endpoint standard deviations are at most $\sqrt M$. Consequently $M\ge\hbar T/(2m)$, independently of the initial covariance.

To attain the bound, arrange an unchirped minimum-uncertainty Gaussian with position variance $s^2$ at $t=T/2$. Propagating it forwards and backwards gives

$$(\Delta x)^2(t) =s^2+\frac{\hbar^2(t-T/2)^2}{4m^2s^2}.$$

Its largest variance occurs at the two endpoints:

$$M=s^2+\frac{\hbar^2T^2}{16m^2s^2} \ge\frac{\hbar T}{2m}, \qquad s^2=\frac{\hbar T}{4m} \quad\text{at equality}.$$

The initial state has $C_{xp}(0)=-\hbar/2$ and $(\Delta p)^2=m\hbar/T$. It contracts until the midpoint, where its variance is half the endpoint value, and then expands symmetrically. The covariance mechanism and contractive Gaussian states are discussed by Roy (2018, Eqs. 8–19).

The mean momentum is arbitrary, including zero: translation of the mean trajectory does not change any variance. When $L_{\mathrm{obs}}^2\gg\hbar T/(2m)$, this zero-mean packet stays localized around its stationary center even though it has no nonzero carrier momentum and its width is not constant. Its velocity spread must still be compared with the velocity resolution. These are preparation spreads of the same freely evolved state; they do not include the disturbance or errors of sequential position measurements.

2. A packet has mean momentum $p_0$ and spread $\Delta p=0.01\lvert p_0\rvert$. What does this say about its velocity spread?

Solution

The mean speed scale is

$$\lvert v_0\rvert=\frac{\lvert p_0\rvert}{m},$$

and the velocity spread is

$$\Delta v=\frac{\Delta p}{m} = 0.01\frac{\lvert p_0\rvert}{m} = 0.01\lvert v_0\rvert.$$

The packet has a one-percent relative velocity spread.

3. Use stationary phase to recover the classical relation between $x$, $p$, and $t$ for a free particle.

Solution

The free-particle phase in the momentum integral is

$$\Phi(p)=px-\frac{p^2}{2m}t.$$

Stationary phase requires

$$\frac{d\Phi}{dp} = x-\frac{p}{m}t =0.$$

Thus

$$x=\frac{p}{m}t,$$

which is the classical free-particle trajectory from the origin with velocity $p/m$.

4. Why is a narrow spatial packet not automatically more classical?

Solution

A narrow spatial packet has a large momentum spread by the uncertainty relation. The velocity spread is $\Delta v=\Delta p/m$, so its components separate more quickly. Unless the mass is large or the time interval is short, sharper localization can make dispersion more important, not less.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• M. V. Berry and K. E. Mount, “Semiclassical approximations in wave mechanics,” Reports on Progress in Physics 35, 315-397, 1972.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• S. M. Roy, “Rigorous quantum limits on monitoring free masses and harmonic oscillators,” Physical Review A 97, 032108 (2018), doi:10.1103/PhysRevA.97.032108, arXiv:1711.10261.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.