University Physics IV · Foundations of Quantum Mechanics · 5.7
Free Evolution, Group Velocity & Spreading
A single plane wave is everywhere at once and describes no particle. Add a band of them and you get something localised that moves — but the free-particle dispersion is quadratic, so the components run at different speeds. This lesson settles which speed is the particle's, and how long being localised lasts.
Build the model
Connect the measurement to the mechanism.
A single de Broglie plane wave has the same modulus everywhere, so it picks out no particle at all; localisation has to be bought with a band of wavenumbers. Superpose them — Ψ(x, t) = (2π)(−1/2) ∫ φ(k) e(i[kx − ω(k)t]) dk — and free evolution becomes trivial in k-space, because each component keeps its amplitude and only collects the phase e(−iħk²t/2m). Everything then follows from the one dispersion relation ω = ħk²/2m.
Expand it about the central k₀: the first derivative, dω/dk = ħk₀/m, is the speed at which the whole envelope translates, and it equals p₀/m, the classical velocity — while ω/k₀ = ħk₀/2m, exactly half of it, belongs to the crests inside the envelope and moves no probability anywhere. The second derivative, d²ω/dk² = ħ/m, refuses to vanish, and that is the price of the model: the components dephase, and for a Gaussian the width obeys σₓ(t) = σ₀√(1 + (t/τ)²) with τ = 2mσ₀²/ħ. Nothing is pushing the packet open.
The momentum distribution is frozen, because p̂ commutes with the free Hamiltonian; what you are watching is a fixed spread of speeds σₚ/m flying apart from a common start. Localisation is therefore a loan, and the more tightly you take it out the sooner it is called in — τ goes as σ₀², so an electron pinned to a nanometre is 41% wider after just 17 femtoseconds.
- Simple definition
- A free wave packet is a superposition of plane waves in which time only multiplies each component by e(−iħk²t/2m): the envelope travels at the group velocity ħk₀/m while its width grows from the spread of speeds it already contained.
- Example
- An electron localised to σ₀ = 1.0 nm already carries σₚ = ħ/2σ₀ = 5.3 × 10⁻²⁶ kg m s⁻¹, a velocity spread of 58 km s⁻¹, so it is 1.4 nm wide after τ = 17 fs and about 58 μm wide one nanosecond later.
Free evolution touches only the phase of each component, so the momentum distribution never changes.
φ(k) is fixed once by Ψ(x,0); k in m⁻¹, ω in rad s⁻¹, and |φ(k)|² is the momentum density
Quadratic, not linear — which is the whole reason a matter packet spreads and a light pulse in vacuum does not.
from E = ħω and E = p²/2m with p = ħk; ħ/m = 1.16 × 10⁻⁴ m² s⁻¹ for an electron
vg is the classical particle speed; vₚ shifts when you move the zero of energy, so it can carry nothing.
both in m s⁻¹; vg belongs to the envelope, vₚ to the crests inside it
Below τ the width barely moves; above it σₓ grows linearly. Halving σ₀ quarters τ.
σ₀ the initial rms width in m, τ the spreading time in s; τ ∝ σ₀², so tighter means faster
Late-time spreading is free flight of a fixed velocity spread: σₓ → (σₚ/m) t.
[p̂, Ĥ] = 0 when V = 0, so |φ(k)|² cannot move; σₚ in kg m s⁻¹
A parabola in t whose vertex sits at t* = −mC₀/2σₚ², so a converging packet narrows first and only then spreads.
C₀ = ⟨Δx̂Δp̂ + Δp̂Δx̂⟩₀ in J s; C₀ < 0 is a converging packet, C₀ = 0 a Gaussian at its narrowest
Free evolution is trivial once you work in k
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.
One derivative moves the packet, the next one spreads it
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 − vg t)), which says the envelope has simply been translated bodily by vg t, with vg = (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 vg = 7.27 × 10⁶ m s⁻¹; the same expansion that gives that speed also guarantees the spreading.
The phase velocity is half the particle's, and carries nothing
The crests inside the envelope move at vₚ = ω/k = E/p. Non-relativistically E = p²/2m, so vₚ = p/2m = vg/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.
The Gaussian width law and its single timescale
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.
Nothing pushes it apart — a frozen velocity spread flies free
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.
Spreading is an asymptotic statement, not an instantaneous one
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.
Change one variable at a time
Make the relationship visible.
Drag t and watch the open crest slip back through the packet — one wavelength lost for every two the envelope gains, since vₚ = vg/2. Then cut σ₀ from 3 nm to 1.5 nm: the packet starts taller and narrower but spreads four times sooner, because τ goes as σ₀².
GROUP SPEED vg0.232 nm/fs
PHASE SPEED vₚ0.116 nm/fs
SPREADING TIME τ69.1 fs
RMS WIDTH σ(t)2.12 nm
Live interpretationGROUP SPEED vg: 0.232 nm/fs. PHASE SPEED vₚ: 0.116 nm/fs. SPREADING TIME τ: 69.1 fs. RMS WIDTH σ(t): 2.12 nm
Catch the common trap
Explain before calculating.
A free electron is prepared as a minimum-uncertainty Gaussian with σₓ = 1.0 nm, so its spreading time is τ = 2mσ₀²/ħ = 17 fs. Which statement correctly describes the packet at t = 17 fs?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA free electron packet is centred on the de Broglie wavelength λ₀ = 0.100 nm. Find the central wavenumber, the group velocity, the phase velocity, and the packet's central kinetic energy in eV. Take mₑ = 9.109 × 10⁻³¹ kg and ħ = 1.0546 × 10⁻³⁴ J s.
- k₀ = 2π/λ₀ = 2π ÷ (1.00 × 10⁻¹⁰ m) = 6.283 × 10¹⁰ m⁻¹, so the central momentum is p₀ = ħk₀ = 1.0546 × 10⁻³⁴ × 6.283 × 10¹⁰ = 6.626 × 10⁻²⁴ kg m s⁻¹.
- Group velocity is the derivative of ω = ħk²/2m, not the ratio: vg = dω/dk = ħk₀/m = p₀/m = 6.626 × 10⁻²⁴ ÷ 9.109 × 10⁻³¹ = 7.27 × 10⁶ m s⁻¹.
- Phase velocity is the ratio: vₚ = ω₀/k₀ = ħk₀/2m = vg/2 = 3.64 × 10⁶ m s⁻¹. It is smaller than the particle's own speed, and it is not a speed anything travels at.
- Central kinetic energy: E = p₀²/2m = (6.626 × 10⁻²⁴)² ÷ (2 × 9.109 × 10⁻³¹) = 4.390 × 10⁻⁴⁷ ÷ 1.822 × 10⁻³⁰ = 2.410 × 10⁻¹⁷ J, and dividing by 1.602 × 10⁻¹⁹ J eV⁻¹ gives 150 eV.
Answerk₀ = 6.28 × 10¹⁰ m⁻¹, vg = 7.27 × 10⁶ m s⁻¹, vₚ = 3.64 × 10⁶ m s⁻¹ — exactly half — and E = 150 eV. Only vg is the electron's speed.
MediumAn electron is prepared in a minimum-uncertainty Gaussian of rms width σ₀ = 2.0 nm. Find σₚ, the velocity spread σₚ/m, the spreading time τ = 2mσ₀²/ħ, the rms width at t = 50 fs, and the uncertainty product at that moment in units of ħ.
- Minimum uncertainty fixes σₚ = ħ/2σ₀ = 1.0546 × 10⁻³⁴ ÷ (2 × 2.0 × 10⁻⁹) = 2.64 × 10⁻²⁶ kg m s⁻¹.
- That is a velocity spread σᵥ = σₚ/m = 2.64 × 10⁻²⁶ ÷ 9.109 × 10⁻³¹ = 2.89 × 10⁴ m s⁻¹, or 29 km s⁻¹ — and it is frozen, since [p̂, Ĥ] = 0 for a free particle.
- τ = 2mσ₀²/ħ = 2 × 9.109 × 10⁻³¹ × (2.0 × 10⁻⁹)² ÷ 1.0546 × 10⁻³⁴ = 7.287 × 10⁻⁴⁸ ÷ 1.0546 × 10⁻³⁴ = 6.91 × 10⁻¹⁴ s = 69 fs.
- So t/τ = 50/69.1 = 0.724, and σₓ(50 fs) = σ₀√(1 + 0.724²) = 2.0 × √1.524 = 2.0 × 1.234 = 2.47 nm.
- σₚ has not moved, so the product is σₓσₚ = (ħ/2)√(1 + (t/τ)²) = 0.5 × 1.234 ħ = 0.617ħ: the state left the ħ/2 floor the moment it began to evolve.
Answerσₚ = 2.6 × 10⁻²⁶ kg m s⁻¹, σᵥ = 29 km s⁻¹, τ = 69 fs, σₓ(50 fs) = 2.5 nm, and σₓσₚ = 0.62ħ.
HardA free electron packet has σₓ = 5.0 nm and σₚ = 3.0 × 10⁻²⁶ kg m s⁻¹ at t = 0, with a symmetrised correlation C₀ = ⟨Δx̂Δp̂ + Δp̂Δx̂⟩₀ = −2.0 × 10⁻³⁴ J s. Find when the packet is narrowest and how narrow, when it returns to 5.0 nm, and its width at t = 1.0 ps.
- Free flight gives x̂(t) = x̂(0) + p̂t/m, so σₓ²(t) = σₓ²(0) + (C₀/m)t + (σₚ/m)²t² — a parabola in t, with σₚ frozen because V = 0.
- The velocity spread is σᵥ = σₚ/m = 3.0 × 10⁻²⁶ ÷ 9.109 × 10⁻³¹ = 3.293 × 10⁴ m s⁻¹.
- The vertex sits at t* = −mC₀/2σₚ² = (9.109 × 10⁻³¹ × 2.0 × 10⁻³⁴) ÷ (2 × 9.0 × 10⁻⁵²) = 1.822 × 10⁻⁶⁴ ÷ 1.8 × 10⁻⁵¹ = 1.01 × 10⁻¹³ s = 101 fs. C₀ < 0, so the packet is converging and the vertex lies in the future.
- There σₘᵢₙ² = σₓ²(0) − C₀²/4σₚ² = 2.50 × 10⁻¹⁷ − (4.0 × 10⁻⁶⁸ ÷ 3.6 × 10⁻⁵¹) = 2.50 × 10⁻¹⁷ − 1.11 × 10⁻¹⁷ = 1.39 × 10⁻¹⁷ m², so σₘᵢₙ = 3.7 nm.
- A parabola is symmetric about its vertex, so the width is back at 5.0 nm at t = 2t* = 202 fs.
- At t = 1.0 ps: σₓ² = 2.50 × 10⁻¹⁷ − 2.196 × 10⁻¹⁶ + (3.293 × 10⁻⁸)² = 2.50 × 10⁻¹⁷ − 2.196 × 10⁻¹⁶ + 1.085 × 10⁻¹⁵ = 8.90 × 10⁻¹⁶ m², giving σₓ = 29.8 nm, against 32.9 nm for the σᵥ t term alone.
AnswerIt narrows to 3.7 nm at 101 fs, is back to 5.0 nm at 202 fs, and is 29.8 nm wide at 1.0 ps — 3.1 nm short of the σᵥ t asymptote it is approaching from below.