Free evolution, group velocity and spreading

Take whatever state you prepared at t = 0 and resolve it into plane waves: φ(k) = (2π)^(−1/2) ∫ Ψ(x,0) e^(−ikx) dx. Each plane wave is an exact solution of the free time-dependent Schrödinger equation provided it turns at its own rate ω(k) = ħk²/2m, so the state at later times is Ψ(x, t) = (2π)^(−1/2) ∫ φ(k) e^{(i[kx − ω(k)t])} dk. Notice what that does and does not change. Every |φ(k)| is untouched — free evolution multiplies each component by a pure phase — so the momentum probability density, and with it ⟨p⟩ and σₚ, is frozen for all time. All the drama happens in position space, and it is interference: the components start in step at t = 0 and drift out of step at rates set by their own ω. The working method follows. Transform to k, attach one phase per component, transform back.

Suppose φ(k) is concentrated near some k₀. Write ω(k) = ω₀ + (dω/dk)₀(k − k₀) + ½(d²ω/dk²)₀(k − k₀)² and put it back into the integral. The constant term is an overall phase and does nothing observable. The linear term factors into e^{(i(k−k₀)(x − v_g t))}, which says the envelope has simply been translated bodily by v_g t, with v_g = (dω/dk)₀ = ħk₀/m = p₀/m — exactly the classical speed. Truncate there and the packet is rigid forever. The quadratic term is what spoils that: for a free particle d²ω/dk² = ħ/m, which is never zero, so each component's phase runs ahead or behind by an amount growing as t(k − k₀)², and the superposition unravels. Compare light in vacuum, where ω = ck makes the second derivative vanish and a pulse keeps its shape indefinitely. A 150 eV electron has k₀ = 6.28 × 10¹⁰ m⁻¹ and v_g = 7.27 × 10⁶ m s⁻¹; the same expansion that gives that speed also guarantees the spreading.

The crests inside the envelope move at vₚ = ω/k = E/p. Non-relativistically E = p²/2m, so vₚ = p/2m = v_g/2: the crests fall steadily backwards through the packet, losing one wavelength for every two the envelope advances. Two arguments show vₚ cannot be a physical speed. First, it depends on where you put the zero of energy — add a constant V₀ everywhere and every ω shifts by V₀/ħ, so vₚ changes while not one measurable prediction does, whereas dω/dk is untouched. Second, if you keep the rest energy and write E = √((pc)² + (mc²)²), then vₚ = E/p = c²/v, which exceeds c for every massive particle. Nothing is wrong, because a single crest carries no probability anywhere: a pure plane wave has the same |Ψ|² at every point and no feature you could time. What does move is the envelope, and Ehrenfest's theorem confirms it — for V = 0, d⟨x⟩/dt = ⟨p⟩/m and d⟨p⟩/dt = 0, exactly, so the centroid drifts at the classical speed forever.

Take the minimum-uncertainty Gaussian, |Ψ(x,0)|² ∝ e^{(−x²/2σ₀²)}. Its transform is Gaussian too, with σₖ = 1/(2σ₀). Attaching e^{(−iħk²t/2m)} to each component and completing the square turns the real width σ₀² into the complex σ₀²(1 + it/τ), and reading off the modulus gives σₓ(t) = σ₀√(1 + (t/τ)²) with τ = 2mσ₀²/ħ. One number runs the whole story. Below τ the width hardly moves, since σₓ ≈ σ₀(1 + t²/2τ²); above it the root goes linear and σₓ → σ₀t/τ. The peak of |Ψ| falls as σₓ^(−1/2), which is exactly what holds the norm at one. Now feel the σ₀² scaling. An electron localised to 1.0 nm has τ = 2 × 9.11 × 10⁻³¹ × 10⁻¹⁸ ÷ 1.055 × 10⁻³⁴ = 17 fs, and one nanosecond later it is 58 μm across. A 1 g bead localised to 1 mm has τ ≈ 1.9 × 10²⁵ s, some 4 × 10⁷ times the age of the universe. Tighter preparation is not safer; it is the very thing that makes spreading fast.

It is tempting to read spreading as a force, or as the particle dissolving. It is neither. Because p̂ commutes with the free Hamiltonian, σₚ is a constant of the motion — exactly constant, not approximately — so the packet is carrying the same spread of speeds σᵥ = σₚ/m that it had at t = 0, and at late times its width simply grows as σₓ → σᵥ t. A classical cloud of particles released from one point with a spread of velocities does precisely the same thing, and nobody calls that a force. What is quantum about it is the floor underneath: you cannot make σᵥ small while σₓ(0) is also small, because σₓσₚ ≥ ħ/2. One consequence is worth carrying forward. The minimum-uncertainty Gaussian does not stay minimum-uncertainty: with σₚ pinned and σₓ growing, σₓσₚ = (ħ/2)√(1 + (t/τ)²), so the state leaves the floor the instant it starts evolving and never comes back to it.

For any free packet x̂(t) = x̂(0) + p̂t/m, and squaring the deviation gives σₓ²(t) = σₓ²(0) + (C₀/m)t + (σₚ/m)²t², where C₀ = ⟨Δx̂Δp̂ + Δp̂Δx̂⟩₀ is the symmetrised position–momentum correlation at t = 0. That is an upward parabola in t, and what happens at the very first instant is set entirely by the sign of C₀. The minimum-uncertainty Gaussian has C₀ = 0 and starts widening immediately. A packet prepared converging has C₀ < 0: it narrows until t* = −mC₀/2σₚ², bottoming out at σₘᵢₙ² = σₓ²(0) − C₀²/4σₚ², and only then spreads. That is not a curiosity — it is what an electron lens does, imposing exactly such a correlation so the beam comes to a focus downstream. What no preparation escapes is the t² term, which eventually dominates whatever C₀ you chose. Spreading is unconditional in the limit, never at every instant.