Spreading of a free Gaussian wave packet
Statement
A free particle initially localised to width σ0 spreads, and the spreading timescale is set by the mass and the initial width alone.
Why it matters
The spreading time separates the quantum world from the classical one. For an electron in an atom a packet delocalises in under a femtosecond; for a macroscopic object the same time exceeds the age of the universe. That single ratio — not the smallness of ħ — is the honest reason classical mechanics works.
Assumptions
Derivation
Result
Reading. Two regimes. For t ≪ τ the width barely changes; for t ≫ τ it grows linearly at rate ħ/2mσ0, which is exactly the velocity spread σp/m implied by the initial momentum uncertainty. The packet spreads because it was made of a range of momenta, and the tighter it was squeezed, the wider that range.
Units check. mσ2/ħ = kg·m2 / (J·s) = s. ✓
Limiting cases
- t ≪ τ — width essentially frozen at σ0; the packet holds its shape.
- t ≫ τ — linear growth at rate ħ/2mσ0 = σp/m, the classical spread in velocities.
- Tighter squeeze (small σ0) — shorter τ ∝ σ02; sharper localisation spreads faster.
Breaks when
- A potential is present — a harmonic well produces non-spreading coherent states; a binding potential produces stationary states that never spread.
- Relativistic energies — the dispersion is no longer quadratic.
- Decoherence from an environment, which in practice localises a macroscopic object far faster than this free evolution would ever delocalise it.
Failure modes
- Believing the particle physically expands. The probability distribution broadens; a position measurement still returns a point.
- Applying the result to a bound particle. Electrons in atoms do not spread away, because they are not free.
Worked number — the scale separation that makes classical physics possible
| e⁻ | Electron, atomic scale m = 9.11 × 10⁻³¹ kg · σ0 = 1 Å | τ = 1.7 × 10⁻¹⁶ s |
| ball | Cricket ball m = 0.16 kg · σ0 = 1 mm | τ = 3.0 × 10²⁷ s |
The electron delocalises across an atom in under a femtosecond. The cricket ball's spreading time exceeds the age of the universe by a factor of roughly seven billion. That single ratio is why classical mechanics works, and it is a far more honest answer than "ħ is small". Run the check →
Discussion
The single line σ(t) = σ0√(1 + (t/τ)²) hides a very specific mechanism. The packet is built (step 2) from a continuum of momentum components, and each one carries the phase e−iℏk²t/2m. Because the frequency ω = ℏk²/2m is quadratic in k, the components advance in phase at different rates and steadily fall out of step. That dephasing is the whole story: in momentum space |φ(k)|² never changes, so spreading is a purely real-space consequence of a curved dispersion relation, not of any force acting on the particle.
The form of the answer follows directly from the complex width σ0² + iℏt/2m produced by the Gaussian integral (step 4): its modulus gives σ(t)² = σ0² + (ℏt/2mσ0)². There is exactly one combination of m, σ0 and ℏ with the dimensions of time, namely τ = 2mσ0²/ℏ, so the result had to be controlled by that timescale alone. The late-time slope ℏ/2mσ0 equals σp/m: the packet flies apart at precisely the velocity spread demanded by its initial momentum uncertainty. Squeeze σ0 smaller and σp = ℏ/2σ0 grows, so the tightest packets spread the fastest — the uncertainty principle turned into kinematics.
The mathematics is not unique to quantum mechanics. The paraxial wave equation of optics is the free Schrödinger equation with the propagation distance z playing the role of t; a laser's Gaussian beam diverges by the identical law, with τ becoming the Rayleigh range and the asymptotic slope becoming the beam divergence angle. Equivalently, the free Schrödinger equation is the diffusion equation with an imaginary diffusion constant D = iℏ/2m — which is why the evolution is dispersive and reversible rather than diffusive and irreversible. The same quadratic-dispersion broadening limits pulse rates in optical fibre (group-velocity dispersion). And the entire result can be obtained without the integral at all: in the Heisenberg picture xH(t) = x(0) + p(0)t/m, so Var(x) grows ballistically as σ0² + (σpt/m)² (see Problem 5). The harmonic oscillator is the pointed exception noted in the assumptions: its restoring force continually re-focuses the momentum-induced spread, so coherent states ride the potential without broadening.
Common misconceptions. First, the particle does not physically swell — only the probability distribution broadens, and any single position measurement still returns a point. Second, spreading is not decoherence or dissipation: the evolution is unitary and time-reversible. Reverse the sign of t and the packet re-converges to σ0 and then expands again; σ0 is the value we happened to start with, not a floor the packet is forced toward. Third, the result is strictly for V = 0: a bound electron sits in a stationary state whose |ψ|² is time-independent and does not spread away.
Worked examples
Example 1 — An electron released at atomic scale. An electron is prepared in a minimum-uncertainty Gaussian of width σ0 = 1.0 Å = 1.0 × 10−10 m. Find its spreading time, its width 1.0 fs later, and the rate at which it delocalises.
Answer. After one femtosecond an electron initially localised to a single atom has already spread across about six Ångströms and is delocalising at ≈ 5.8 × 105 m/s. This sub-femtosecond delocalisation is why electrons in atoms must be described by orbitals, not trajectories.
Example 2 — Time-of-flight thermometry of a cold atom cloud. A single 87Rb atom (m = 1.44 × 10−25 kg) is released from a trap in which its ground-state width is σ0 = 1.0 µm. It is imaged after a time-of-flight t = 20 ms. Find the imaged width and use it to read off the momentum spread.
Answer. The cloud grows to 7.4 µm, and because the flight is long the image directly encodes the velocity spread 0.37 mm/s. Converting, σp = ℏ/2σ0 = 5.3 × 10−29 kg·m/s gives an effective temperature kBT = σp²/m, i.e. T ≈ 1.4 nK. Time-of-flight imaging is exactly this measurement: let the momentum width map itself onto a size you can see.
Problems
- (Warm-up.) A proton (m = 1.67 × 10−27 kg) is confined to a nuclear scale σ0 = 1.0 fm = 1.0 × 10−15 m and then released. Find the spreading time τ.
Solution
τ = 2mσ0²/ℏ = 2(1.67×10−27)(1.0×10−15)²/(1.055×10−34) = (3.34×10−57)/(1.055×10−34) = 3.2 × 10−23 s. A proton pinned to nuclear dimensions delocalises in a few tens of zeptoseconds — comparable to nuclear dynamical timescales, which is why a free nucleon cannot be treated as a static point.
- (Onset of spreading.) For the electron of Example 1 (τ = 1.7 × 10−16 s), how long until its width has grown by 1.0%?
Solution
Require σ/σ0 = 1.010, so 1 + (t/τ)² = 1.010² = 1.0201, giving (t/τ)² = 0.0201 and t/τ = 0.142. (Equivalently, the small-time expansion σ ≈ σ0(1 + ½(t/τ)²) gives ½(t/τ)² = 0.01.) Thus t = 0.142(1.73×10−16) = 2.5 × 10−17 s. The width is essentially frozen only for a small fraction of τ.
- (Asymptotic rate = velocity spread.) Show that for t ≫ τ the growth rate dσ/dt approaches σp/m, and evaluate it for the electron of Example 1.
Solution
Write σ(t)² = σ0² + (ℏt/2mσ0)². For t ≫ τ the second term dominates, so σ(t) ≈ ℏt/2mσ0 and dσ/dt → ℏ/2mσ0. From step 2 of the derivation the momentum width is σp = ℏσk = ℏ/2σ0, so dσ/dt = σp/m — the packet flies apart at its own velocity spread. Numerically, dσ/dt = 1.055×10−34/[2(9.11×10−31)(1.0×10−10)] = 5.8 × 105 m/s. Interpretation: the fastest and slowest momentum components simply separate ballistically.
- (Which one wins?) Two electrons are prepared with widths σ0 and 2σ0. The tighter one has a larger momentum spread but starts smaller. At what time do the two packets have equal width, and what is that width? Which is broader before and after?
Solution
Using σ² = (initial)² + (ℏt/2m·initial)²: packet A has σA² = σ0² + (ℏt/2mσ0)² and packet B has σB² = 4σ0² + (ℏt/4mσ0)². Set equal and let b = ℏt/mσ0: (b/2)² − (b/4)² = 3σ0² ⇒ (3/16)b² = 3σ0² ⇒ b = 4σ0, hence t = 4mσ0²/ℏ = 2τA. At that time σA² = σ0² + (2σ0)² = 5σ0², so both have width √5 σ0 ≈ 2.24 σ0. Before the crossover B (wider start) is broader; after it, the tighter packet A overtakes and stays broader forever, because its asymptotic slope ℏ/2mσ0 is twice B's. Tighter preparation loses early but wins late.
- (Derivation without an integral.) Reproduce the result σ(t)² = σ0² + (ℏt/2mσ0)² using the Heisenberg picture, starting from the free Hamiltonian, without ever solving the Schrödinger equation.
Solution
For a free particle H = p²/2m, the Heisenberg equations dp/dt = 0 and dx/dt = p/m integrate exactly to pH(t) = p(0) and xH(t) = x(0) + p(0)t/m — formally identical to classical free motion. Taking the variance in the initial state (with ⟨x⟩ = ⟨p⟩ = 0):
⟨xH²⟩ = ⟨x²⟩0 + (t/m)⟨xp+px⟩0 + (t/m)²⟨p²⟩0.
The initial state is a real, symmetric Gaussian, so the symmetric correlator ⟨xp+px⟩0 = 0 (position and momentum are initially uncorrelated). With ⟨x²⟩0 = σ0² and ⟨p²⟩0 = σp² = (ℏ/2σ0)² (minimum uncertainty),
σ(t)² = σ0² + (σpt/m)² = σ0² + (ℏt/2mσ0)²,
the boxed result exactly. The lesson: the spread grows ballistically as a range of velocities σp/m carries the packet apart. As time runs on, the cross term becomes non-zero — position and momentum become correlated (fast components find themselves out front), which is the physical content of the complex width in step 4.