University Physics IV · Photons and Matter Waves · 4.7
Wave Packets: Phase & Group Velocity
A single de Broglie wave fills all of space and locates nothing. Add a band of them and you get a lump that travels — but at which of two speeds? This lesson separates the envelope from the crests, shows which of them is physics and which is bookkeeping, and prices the localisation you just bought.
Build the model
Connect the measurement to the mechanism.
A plane wave e(i(kx − ωt)) has the same |ψ|² at every point, so it carries a definite momentum and no position whatever. To describe something that is somewhere, superpose a band of them: ψ(x, t) = (1/√(2π)) ∫ φ(k) e(i(kx − ω(k)t)) dk. Everything that follows is squeezed out of the dispersion relation ω(k) that the physics supplies — for a free non-relativistic particle, ħω = ħ²k²/2m makes ω = ħk²/2m, a parabola.
Expand it about the centre of the band. The zeroth term carries the crests along at the phase velocity ω/k = ħk/2m; the first derivative carries the envelope, and with it all the probability, at the group velocity dω/dk = ħk/m = p/m, which is exactly the classical speed of the particle. The second derivative, ħ/m, refuses to vanish, so components of different k run at different speeds and the envelope broadens.
Of the two velocities only the group velocity can be physical: move the zero of energy by V₀ and ψ picks up a global phase e(−iV₀t/ħ) that changes no probability anywhere, yet ω/k shifts by V₀/(ħk) while dω/dk does not stir. That is the price of the model — the band of k that localises the particle is the same band that pulls it apart.
- Simple definition
- A wave packet is a superposition of de Broglie waves spread over a band of wavenumbers; its envelope moves at the group velocity dω/dk, while the crests inside it move at the phase velocity ω/k.
- Example
- For a free electron centred on k₀ = 1.0 × 10¹⁰ m⁻¹, vg = ħk₀/m = 1.16 × 10⁶ m s⁻¹ — the speed a 3.81 eV electron actually has — while the crests inside creep along at half of that, 5.79 × 10⁵ m s⁻¹.
One k has no position at all. A band of them makes a lump about 1/Δk wide.
φ(k) is the amplitude at wavenumber k, k in m⁻¹; |ψ|² is a probability per unit length
The only physics in the integral. Both velocities and the spreading rate are derivatives of this parabola.
from ħω = E = ħ²k²/2m; ħ = 1.055 × 10⁻³⁴ J s, ω in rad s⁻¹, m in kg
Half the particle's speed non-relativistically, c²/v relativistically, and a speed of nothing in either case.
the speed of one crest; it shifts by V₀/(ħk) if the zero of energy moves
The particle's own speed: 1.16 × 10⁶ m s⁻¹ at k₀ = 1.0 × 10¹⁰ m⁻¹, matching √(2E/m) exactly.
evaluated at k₀, the centre of φ(k); vg in m s⁻¹, p in kg m s⁻¹
The whole effect with no calculus: one superposition, two different speeds.
Δk = k₂ − k₁ and k̄ = ½(k₁ + k₂); the envelope moves at Δω/Δk, the crests at ω̄/k̄
τ = 17 fs for a 1.0 nm electron packet but 32 ps for a 1.0 nm proton packet.
σ₀ is the initial rms width in m; Δp = ħ/2σ₀ stays fixed while σₓ grows
One plane wave is everywhere; a particle is not
Take a single de Broglie wave, ψ = A e(i(kx − ωt)). Its modulus squared is |A|² at every x and every t, so the particle is equally likely here, in the next room, and outside the galaxy, and no feature of the wave ever changes that. It is not even normalisable — ∫|A|² dx diverges — so strictly it is not a state, only a convenient limit. The repair is to add a band of them, ψ(x, t) = (1/√(2π)) ∫ φ(k) e(i(kx − ω(k)t)) dk, with φ(k) peaked at some k₀ and of width Δk. The components interfere constructively across a region of width Δx ≈ 1/Δk and cancel outside it, and that lump is the thing you can point at. Nothing has been added to the theory; the rest of this lesson is simply what that integral does as t runs on.
The dispersion relation carries all the physics
The integral is identical for light, sound, ripples and electrons. Only ω(k) differs, and ω(k) is where the physics enters. For a free non-relativistic particle, E = p²/2m with E = ħω and p = ħk gives ω(k) = ħk²/2m, a parabola through the origin. For light in vacuum, ω = ck, a straight line. That difference decides everything. On a straight line through the origin, ω/k and dω/dk are the same number, so light's crests and envelope both travel at c and a vacuum pulse keeps its shape forever. On a parabola, ω/k = ħk/2m, dω/dk = ħk/m, and the second derivative ħ/m is not zero — so a matter packet has two different speeds and cannot keep its shape. At k₀ = 1.0 × 10¹⁰ m⁻¹: ω = 5.79 × 10¹⁵ rad s⁻¹, ω/k = 5.79 × 10⁵ m s⁻¹, dω/dk = 1.16 × 10⁶ m s⁻¹.
Stationary phase puts the lump at dω/dk
Expand about the centre of the band: ω(k) ≈ ω₀ + ω′(k₀)(k − k₀) + ½ω″(k₀)(k − k₀)². Pull out the carrier e(i(k₀x − ω₀t)) and what remains, to first order, depends on x and t only through the combination x − ω′(k₀)t: the envelope keeps its shape and slides at vg = ω′(k₀). Equivalently, the components add up only where their phases are stationary in k, and that condition is x = ω′(k₀)t. Two waves show it without calculus. Superpose k₁ = 0.95 × 10¹⁰ m⁻¹ and k₂ = 1.05 × 10¹⁰ m⁻¹ for a free electron: ω₁ = 5.226 × 10¹⁵ and ω₂ = 6.385 × 10¹⁵ rad s⁻¹, so the envelope runs at Δω/Δk = 1.16 × 10⁶ m s⁻¹ while the carrier runs at ω̄/k̄ = 5.81 × 10⁵ m s⁻¹ — a ratio of 1.995, heading for exactly 2 as the band narrows.
Only the group velocity survives a change of energy zero
Potential energy has no absolute zero: shift it by V₀ and no force and no measurement changes. In the wave, that shift multiplies ψ by e(−iV₀t/ħ), a global phase, so |ψ|² is untouched at every x and t. But it also sends ω(k) → ω(k) + V₀/ħ, which leaves dω/dk exactly alone and moves ω/k by V₀/(ħk). Adding just 10 eV to the zero lifts the phase velocity of that 3.81 eV electron from 5.79 × 10⁵ to 2.10 × 10⁶ m s⁻¹, while its group velocity sits still. A quantity a bookkeeping choice can rescale is not the speed of anything. Relativity says it louder: with E = γmc², ω/k = E/p = c²/v exceeds c for every massive particle, while dω/dk = dE/dp = pc²/E = v is the speed you measure.
The curvature of ω(k) spreads the packet
Keep the second-order term and the envelope stops moving rigidly. For a Gaussian packet of initial rms width σ₀, the exact free solution gives σₓ(t) = σ₀√(1 + (t/τ)²) with τ = 2mσ₀²/ħ, while the momentum spread Δp = ħ/2σ₀ never changes — the components that were there from the start simply run apart at Δv = ħ/2mσ₀. For an electron with σ₀ = 1.0 nm, τ = 17 fs and Δv = 5.79 × 10⁴ m s⁻¹, so after 1.0 ns the packet is 58 µm across. Narrow packets die fastest, since τ goes as σ₀²; heavy ones barely spread, since τ goes as m, and a 1.0 nm proton packet needs 32 ps to reach √2 σ₀. That mass scaling is a large part of why the classical limit looks classical.
What localisation costs — and what it does not
The band is not optional. Δx Δk ≥ ½ holds for any square-integrable packet, so pinning an electron to σ₀ = 1.0 nm demands Δk ≥ 5 × 10⁸ m⁻¹ and therefore Δp ≥ ħ/2σ₀ = 5.3 × 10⁻²⁶ kg m s⁻¹. Those are the very components that make it spread, so the sharper the packet the shorter it lasts: the two statements are one statement. What the packet does not do is swell as a lump of charge. |ψ|² is a probability density for one electron, and a screen 1.0 ns downstream still records one point-like arrival — only its position is now less predictable. Nor is the growth unconditional: a packet prepared converging narrows first, reaches a minimum, and only then spreads without bound.
Change one variable at a time
Make the relationship visible.
Push t to 1.40 with k₀ = 3.0: the open crest has slipped a full wavelength back through the packet, because vₚ is exactly half vg. Then move V₀ on its own — the crests speed up, stall, or run backwards while the envelope never budges.
GROUP vg = dω/dk3.00
PHASE vₚ = ω/k1.50
vₚ / vg0.500
WIDTH σ(t)/σ₀1.009
Live interpretationGROUP vg = dω/dk: 3.00. PHASE vₚ = ω/k: 1.50. vₚ / vg: 0.500. WIDTH σ(t)/σ₀: 1.009
Catch the common trap
Explain before calculating.
A free-electron packet is centred on k₀ = 1.0 × 10¹⁰ m⁻¹, where ħk₀/m = 1.16 × 10⁶ m s⁻¹ and ω/k₀ = 5.79 × 10⁵ m s⁻¹. Someone re-zeroes the energy scale, adding V₀ = 10 eV to every energy, so ω(k) → ω(k) + V₀/ħ. What happens to the two velocities?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA free electron is described by a narrow packet centred on k₀ = 1.00 × 10¹⁰ m⁻¹. Find ω, the phase velocity and the group velocity, then check which of the two is the speed the electron actually has. Take ħ = 1.055 × 10⁻³⁴ J s and m = 9.109 × 10⁻³¹ kg.
- Dispersion relation: ω = ħk₀²/2m = (1.055 × 10⁻³⁴)(1.00 × 10¹⁰)² / (2 × 9.109 × 10⁻³¹) = 1.055 × 10⁻¹⁴ / 1.822 × 10⁻³⁰ = 5.79 × 10¹⁵ rad s⁻¹.
- Phase velocity: vₚ = ω/k₀ = 5.79 × 10¹⁵ / 1.00 × 10¹⁰ = 5.79 × 10⁵ m s⁻¹.
- Group velocity: vg = dω/dk = ħk₀/m = (1.055 × 10⁻³⁴)(1.00 × 10¹⁰)/(9.109 × 10⁻³¹) = 1.16 × 10⁶ m s⁻¹, exactly twice vₚ.
- Test them: E = ħω = 6.11 × 10⁻¹⁹ J = 3.81 eV, so the classical speed is √(2E/m) = √(1.34 × 10¹²) = 1.16 × 10⁶ m s⁻¹. That is vg, not vₚ.
Answerω = 5.79 × 10¹⁵ rad s⁻¹, vₚ = 5.79 × 10⁵ m s⁻¹, vg = 1.16 × 10⁶ m s⁻¹. The group velocity reproduces √(2E/m); the phase velocity, at half of it, matches nothing measurable.
MediumSuperpose two equal-amplitude free-electron waves with k₁ = 0.95 × 10¹⁰ m⁻¹ and k₂ = 1.05 × 10¹⁰ m⁻¹. Find the speed of the envelope and the speed of the carrier, and work out how many carrier wavelengths fit inside one beat lobe.
- With ħ/2m = 1.055 × 10⁻³⁴ / 1.822 × 10⁻³⁰ = 5.791 × 10⁻⁵ m² s⁻¹: ω₁ = 5.791 × 10⁻⁵ × 9.025 × 10¹⁹ = 5.226 × 10¹⁵ rad s⁻¹ and ω₂ = 5.791 × 10⁻⁵ × 1.1025 × 10²⁰ = 6.385 × 10¹⁵ rad s⁻¹.
- Sum to product: cos(k₁x − ω₁t) + cos(k₂x − ω₂t) = 2 cos(½(Δk x − Δω t)) cos(k̄x − ω̄t), with Δk = 1.00 × 10⁹ m⁻¹, Δω = 1.158 × 10¹⁵ rad s⁻¹, k̄ = 1.00 × 10¹⁰ m⁻¹ and ω̄ = 5.805 × 10¹⁵ rad s⁻¹.
- Envelope speed: Δω/Δk = 1.158 × 10¹⁵ / 1.00 × 10⁹ = 1.16 × 10⁶ m s⁻¹, which is precisely ħk̄/m — the group velocity at the centre of the band.
- Carrier speed: ω̄/k̄ = 5.81 × 10⁵ m s⁻¹, a ratio of 1.995 to the envelope. It sits 0.25% above ω(k̄)/k̄ = 5.79 × 10⁵ m s⁻¹ only because ω(k) is curved, so the mean of ω is not ω of the mean.
- Lobe length: the envelope argument advances by π between zeros, so Δx = 2π/Δk = 6.28 nm, against a carrier wavelength 2π/k̄ = 0.628 nm — ten crests to a lobe.
AnswerEnvelope 1.16 × 10⁶ m s⁻¹, carrier 5.81 × 10⁵ m s⁻¹, ratio 1.995 heading for 2; each 6.28 nm lobe holds ten carrier wavelengths.
HardAn electron is prepared as a minimum-uncertainty Gaussian packet of rms width σ₀ = 1.0 nm centred on k₀ = 1.0 × 10¹⁰ m⁻¹. (a) After how long is it √2 times as wide? (b) How wide is it after 1.0 ns, and how far has its centre gone? (c) Repeat (b) for a proton (m = 1.673 × 10⁻²⁷ kg) prepared identically.
- Momentum spread: Δp = ħ/2σ₀ = 1.055 × 10⁻³⁴ / 2.0 × 10⁻⁹ = 5.28 × 10⁻²⁶ kg m s⁻¹, so the components differ in speed by Δv = Δp/m = 5.79 × 10⁴ m s⁻¹. Δp is fixed for all time; only σₓ grows.
- (a) σₓ(t) = σ₀√(1 + (t/τ)²) with τ = 2mσ₀²/ħ = 2(9.109 × 10⁻³¹)(1.0 × 10⁻⁹)² / 1.055 × 10⁻³⁴ = 1.73 × 10⁻¹⁴ s. The width is √2 σ₀ at t = τ = 17 fs.
- (b) At t = 1.0 ns, t/τ = 5.79 × 10⁴ ≫ 1, so σₓ ≈ σ₀ t/τ = Δv t = (5.79 × 10⁴)(1.0 × 10⁻⁹) = 5.79 × 10⁻⁵ m = 58 µm — 58 000 times its starting width. The centre has moved vg t = (1.16 × 10⁶)(1.0 × 10⁻⁹) = 1.16 mm.
- (c) τ is proportional to m, so τₚ = 1836 × 1.73 × 10⁻¹⁴ = 3.17 × 10⁻¹¹ s = 32 ps. At 1.0 ns, t/τₚ = 31.5, giving σₓ = 1.0 × √(1 + 31.5²) = 32 nm, while the centre has moved (ħk₀/mₚ)t = (631 m s⁻¹)(1.0 ns) = 0.63 µm.
- Read the scaling: spreading at late times is Δv t = ħt/2mσ₀, so it falls off with both mass and initial width. The proton is 1836 times heavier and its packet ends up 1834 times narrower.
Answerτ = 17 fs. After 1.0 ns the electron packet is 58 µm wide having travelled 1.16 mm; the identically prepared proton packet is only 32 nm wide having travelled 0.63 µm.