Phase velocity, group velocity, and which is the particle
The envelope moves at p/m, the crests at half.
A free packet carries two speeds, and a first course has to settle which one belongs to the particle. The dispersion relation supplies both, stationary phase shows that |ψ|² rides at dω/dk = p/m, a shift of the energy zero exposes ω/k as bookkeeping, and the width of φ(k) decides whether one group velocity exists at all.
Build the model
The crests move at half the particle's speed, and no measurement ever times them.
1The big idea
A plane wave e(ikx − iωt) solves the free Schrödinger equation only if ħω = ħ²k²/2m, and that one relation holds two different speeds. A surface of constant phase kx − ωt moves at ω/k = ħk/2m, the phase velocity. But a plane wave has the same |ψ|² everywhere, so it has no position to carry, and anything a detector registers belongs to a superposition ψ = (1/√(2π)) ∫ φ(k) e(ikx − iωt) dk with ω = ω(k). When φ(k) is peaked at k₀, the components reinforce only where their phases agree to first order in k − k₀, which is where d/dk of kx − ω(k)t vanishes: x = ω′(k₀)t. Expanding ω to first order about k₀ says more: the envelope translates rigidly at vg = dω/dk = ħk₀/m, while the carrier inside it slides at ω₀/k₀ = ħk₀/2m, exactly half as fast. The group velocity is p/m, the classical speed, and it is unchanged when every energy is shifted by a constant; the phase velocity changes under that shift, so it cannot be what a time-of-flight experiment measures. The model stops at the term it dropped. Because ω is quadratic, the components of φ(k) carry group velocities spread over Δv = ħσₖ/m, and the packet keeps one velocity only while Δv t stays small against its width: for a Gaussian, for about k₀/σₖ of its own widths. When φ(k) is broad or split in two there is no single group velocity, and only the mean ⟨x⟩ still moves at ħ⟨k⟩/m, exactly.
2In one sentence
The group velocity vg = dω/dk, evaluated at the peak k₀ of φ(k), is the speed of a narrow-band packet's envelope |ψ|²; for a free particle it is ħk₀/m = p₀/m, twice the phase velocity ω/k at which the individual crests move.
Example. An electron packet centred on k₀ = 1.0 × 10¹⁰ m⁻¹ (λ = 0.628 nm) carries its |ψ|² at ħk₀/m = 1.16 × 10⁶ m s⁻¹, crossing a 1.00 mm gap in 0.864 ns, while its crests advance at only 5.79 × 10⁵ m s⁻¹.
3One plane wave has a speed but carries no particle
Put ψ = e(ikx − iωt) into iħ ∂ψ/∂t = −(ħ²/2m) ∂²ψ/∂x². The left side returns ħωψ and the right side (ħ²k²/2m)ψ, so every real k works provided ω = ħk²/2m: the dispersion relation, which is E = ħω with E = p²/2m and p = ħk. The crests of Re ψ = cos(kx − ωt) sit where kx − ωt = 2πn and move at ω/k = ħk/2m, the phase velocity. For an electron with k = 1.0 × 10¹⁰ m⁻¹ (λ = 0.628 nm) that is 5.79 × 10⁵ m s⁻¹, yet classical mechanics gives an electron of momentum ħk = 1.05 × 10⁻²⁴ kg m s⁻¹ the speed p/m = 1.16 × 10⁶ m s⁻¹, twice as fast. There is no contradiction yet, because the plane wave describes no motion: |ψ|² = 1 at every x and every t. Its crests move while its probability stays put, so any speed a detector could measure must belong to a superposition of plane waves, whose |ψ|² is not uniform.
4Chord and tangent on the ω(k) curve
Plot ω against k. The phase velocity ω₀/k₀ is the slope of the chord from the origin to the point (k₀, ω₀); the group velocity is the slope of the tangent there. On the parabola ω = ħk²/2m the tangent has slope ħk₀/m and the chord ħk₀/2m, so their ratio is exactly 2 at every k₀. That is a property of curves with ω ∝ k², not of electrons, and other waves show it by contrast. Light in vacuum has ω = ck, a straight line through the origin, so chord and tangent coincide and vₚ = vg = c: a light pulse in vacuum keeps its shape. Deep-water waves have ω = √(gk), whose tangent is half the chord, so their crests outrun the group, appearing at its back and vanishing at its front. A matter wave, whose ω curves upward, does the reverse: its crests fall behind the envelope. Chord and tangent answer different questions about one curve, and only the local slope says how a localised disturbance travels.
5Stationary phase finds the packet
Write the packet as ψ(x, t) = (1/√(2π)) ∫ φ(k) e(iθ) dk with θ(k) = kx − ω(k)t, and take φ(k) real, peaked at k₀ with width σₖ. The size of each contribution is set by φ; whether the contributions add or cancel is decided by θ. Across the peak θ changes by about θ′(k₀)σₖ. If that is many radians, the integrand turns through many cycles while φ barely changes, the contributions cancel, and ψ is small. They add only where θ is stationary at the peak, θ′(k₀) = x − ω′(k₀)t = 0. So the peak of |ψ| sits at x = ω′(k₀)t and moves at vg = ω′(k₀) = ħk₀/m, and |ψ| falls away once |x − vg t| exceeds roughly 1/σₖ. A phase in the amplitude moves the start: φ(k) = |φ(k)| e(−ikx₀) adds −x₀ to θ′, so the peak starts at x₀. The dimensions check: ħk/m carries (J s)(m⁻¹)/kg = m s⁻¹.
6Linearise ω and the crests slip backward
Because ω is quadratic, its Taylor series about k₀ ends after three terms: ω(k) = ω₀ + vg(k − k₀) + (ħ/2m)(k − k₀)². Drop the last term and the integral can be done for any φ: ψ(x, t) ≈ e(iω₀t) ψ(x − vg t, 0), the starting packet translated by vg t and multiplied by a phase. So |ψ(x, t)|² = |ψ(x − vg t, 0)|², and every feature of the probability density moves at vg. Write the start as a real envelope times a carrier, ψ(x, 0) = A(x) e(ik₀x), and the result is A(x − vg t) e(ik₀x − iω₀t): the envelope moves at vg and the crests at ω₀/k₀ = vₚ. For a free particle vₚ = vg/2, so relative to the envelope the crests drift backward at vg/2: each appears at the leading edge, crosses the packet and fades at the trailing edge. For the electron with λ = 0.628 nm, a new crest passes the peak every λ/(vg − vₚ) = 1.09 fs, exactly one period 2π/ω₀.
7Why the detector reads vg
Three checks pick out vg. Classical correspondence: vg = ħk₀/m = p₀/m, the speed Newton gives a particle of momentum p₀, or √(2E/m) for kinetic energy E. Ehrenfest's theorem: d⟨x⟩/dt = ⟨p⟩/m, which is ħk₀/m for a narrow φ(k). And, decisively, the energy zero. Adding a constant V₀ to the Hamiltonian multiplies every solution by e(−iV₀t/ħ), which changes no probability density and no expectation value of x or p; only ⟨H⟩ moves, by exactly V₀. It turns ω into ħk²/2m + V₀/ħ, so vₚ becomes ħk/2m + V₀/(ħk) while dω/dk is untouched. Counting the electron's rest energy, V₀ = mc² = 511 keV, lifts vₚ at k = 1.0 × 10¹⁰ m⁻¹ to 7.76 × 10¹⁰ m s⁻¹, which agrees with c²/vg, the phase velocity the relativistic dispersion relation gives. A speed that bookkeeping can set anywhere, even above c, is not what a stopwatch records. A time-of-flight measurement registers the arrival of |ψ|², and |ψ|² moves at vg.
8When there is no single group velocity
Linearising dropped (ħ/2m)(k − k₀)². Restored, it says each part of φ(k) moves at its own ħk/m, so a packet of k-width σₖ holds velocities spread over Δv = ħσₖ/m, and it moves as one object only while Δv t ≪ σₓ. For a Gaussian, σₓσₖ = ½, which makes this t ≪ T = 2mσₓ²/ħ, a time in which the packet travels vg T = (k₀/σₖ)σₓ: k₀/σₖ of its own widths. An electron with k₀ = 1.0 × 10¹⁰ m⁻¹ and σₓ = 1.5 nm has T = 38.9 fs and covers 45 nm, thirty widths, as a recognisable packet. At σₓ = 0.15 nm, σₖ is a third of k₀, T = 0.389 fs, and the packet blurs after three widths. Split φ(k) into two peaks and ψ splits into two packets. Yet ⟨x⟩ always moves at exactly ħ⟨k⟩/m, because ω′ is linear in k and so averages to ω′(⟨k⟩); for a split packet that is a mean, not a place the particle is found.
Key equations
What each relation says, and the units it is written in.
Free-particle dispersion relation
ω(k) = ħk²/(2m)
Substituting e(ikx − iωt) into the V = 0 Schrödinger equation forces it. Because ω is quadratic in k rather than linear, crests and envelope cannot share one speed.
ω in rad s⁻¹, k in m⁻¹, m in kg, ħ = 1.055 × 10⁻³⁴ J s; it is E = ħω with E = p²/2m and p = ħk
Phase velocity
vₚ = ω/k = ħk/(2m)
Half the classical speed p/m. It is the only speed a single plane wave has, and it moves no probability, because |e(ikx − iωt)|² = 1 everywhere.
m s⁻¹; the speed of a surface of constant phase kx − ωt, such as one crest of Re ψ
Group velocity by stationary phase
vg = ω′(k₀) = ħk₀/m
The phase kx − ω(k)t is stationary at k₀ only where x = vg t, so that is where |ψ|² peaks. It equals the classical p₀/m, the speed a time-of-flight detector reads.
m s⁻¹; ω′ = dω/dk, k₀ in m⁻¹ is the peak of φ(k), and ħk₀ = p₀ in kg m s⁻¹
Linearised packet: envelope and carrier
ψ(x, t) ≈ e(iω₀t) ψ(x − vg t, 0)
|ψ|² is its t = 0 shape moved by vg t. Writing ψ(x, 0) = A(x) e(ik₀x) gives A(x − vg t) e(ik₀x − iω₀t): envelope at vg, crests at ω₀/k₀ = vg/2.
ω₀ = ω(k₀) in rad s⁻¹; the general phase is i(k₀vg − ω₀)t, which is iω₀t here because k₀vg = 2ω₀
Phase velocity after shifting the energy zero
vₚ = ħk/(2m) + V₀/(ħk)
dω/dk and |ψ|² are untouched, so vg stays ħk/m, while vₚ goes wherever V₀ puts it: counting the rest energy mc² makes it about c²/vg, faster than light.
V₀ in J is a constant added to every energy, so ω becomes ω + V₀/ħ and ψ gains only the factor e(−iV₀t/ħ)
Widths travelled before the shape is lost
vg T/σₓ = k₀/σₖ
Group velocities across φ(k) spread over ħσₖ/m. With σₖ ≪ k₀ the packet crosses many widths as one object; with σₖ near k₀ it smears out as fast as it moves.
Gaussian with σₓσₖ = ½; σₓ in m, σₖ in m⁻¹; T = 2mσₓ²/ħ in s is the time for σₓ to grow by √2
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Step t from 0 to 8 fs: the open dot moves 4.63 nm, but the marked crest drifts back and is replaced. Set V₀ = 0.95 eV and the crests ride with the peak while |ψ| does not change at all. Then drop the initial width to 0.4 nm: the width grows by √2 in 2.8 fs, while the packet crosses only four widths.
Live results
- ENVELOPE PEAK POSITION
- 3.47 nm
- CREST SPEED (PHASE)
- 0.289 nm fs⁻¹
- SPREADING TIME
- 24.9 fs
- WIDTHS BEFORE SPREADING
- 12
Catch the common trap
The mistake this lesson is built to prevent.
The shortcut goes: a de Broglie wave has frequency f = E/h and wavelength λ = h/p, so the particle moves at fλ = E/p. For a free electron E = p²/2m, and the shortcut returns p/2m, half the speed the electron actually has; include the rest energy and it returns more than c. The quantum mechanics is fine; the error is calling fλ the particle's speed. fλ = ω/k is the phase velocity, the speed of a point of constant phase on one plane wave, and one plane wave has the same |ψ|² everywhere, so it describes no motion of probability at all. Its value also depends on where the zero of energy sits, so it changes when nothing physical does. The particle is found where the envelope of a superposition is, and the envelope moves at dω/dk = dE/dp = p/m, whatever constant is added to E.
Test it: what changes?
- Crest speed ħk₀/2m + V₀/(ħk₀) rises
- Envelope speed ħk₀/m unchanged
- |ψ| and |ψ|² unchanged everywhere
- vg = 2vₚ no longer holds
- Larger σₖ: wider spread of group velocities
- Spreading time 2mσₓ²/ħ shortens
- Fewer widths travelled before spreading, k₀/σₖ
- Envelope speed ħk₀/m unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
A chopper releases a short burst of neutrons whose wave packets are narrowly peaked at de Broglie wavelength λ₀ = 0.180 nm. Find the phase and group velocities, check the group velocity against classical mechanics, and predict when the burst reaches a detector 2.00 m downstream. Use mₙ = 1.675 × 10⁻²⁷ kg and ħ = 1.0546 × 10⁻³⁴ J s.
View solution →
- The peak wavenumber is k₀ = 2π/λ₀ = 2π/(1.80 × 10⁻¹⁰ m) = 3.49 × 10¹⁰ m⁻¹, so the peak momentum is p₀ = ħk₀ = 3.68 × 10⁻²⁴ kg m s⁻¹.
- With ω = ħk²/2m the two speeds are vg = ħk₀/m = (3.68 × 10⁻²⁴ kg m s⁻¹)/(1.675 × 10⁻²⁷ kg) = 2.20 × 10³ m s⁻¹ and vₚ = ħk₀/2m = 1.10 × 10³ m s⁻¹. The units check: kg m s⁻¹ divided by kg is m s⁻¹.
- Classical check: the kinetic energy is E = p₀²/2m = 4.05 × 10⁻²¹ J = 25.2 meV, about the thermal energy at room temperature, and √(2E/m) = 2.20 × 10³ m s⁻¹, the same as vg. The phase velocity is not the classical speed.
- The detector registers |ψ|², which travels at vg, so the burst arrives after t = L/vg = (2.00 m)/(2.20 × 10³ m s⁻¹) = 9.10 × 10⁻⁴ s. Timing it with vₚ would predict 1.82 ms, when nothing arrives.
Answer. vₚ = 1.10 × 10³ m s⁻¹ and vg = 2.20 × 10³ m s⁻¹, the classical speed; the burst reaches the detector 0.910 ms after release.
MediumExample 2
An electron packet has φ(k) = N exp(−(k − k₀)²/(4σₖ²)) e(−ikx₀), with k₀ = 8.0 × 10⁹ m⁻¹, σₖ = 2.0 × 10⁸ m⁻¹ and x₀ = −12.0 nm. Use stationary phase to locate the peak of |ψ|² at t = 20 fs, count the crests that slip backward past that peak meanwhile, and decide whether the linearised picture can still be trusted then. Take m = 9.109 × 10⁻³¹ kg and ħ = 1.0546 × 10⁻³⁴ J s.
View solution →
- The phase of the integrand is θ(k) = k(x − x₀) − ħk²t/2m. Stationarity at the peak, θ′(k₀) = x − x₀ − ħk₀t/m = 0, puts the peak of |ψ|² at x = x₀ + vg t with vg = ħk₀/m.
- vg = (1.0546 × 10⁻³⁴ J s)(8.0 × 10⁹ m⁻¹)/(9.109 × 10⁻³¹ kg) = 9.26 × 10⁵ m s⁻¹. In 20 fs the peak moves 18.5 nm, so it sits at x = −12.0 nm + 18.5 nm = 6.5 nm.
- Crests move at vₚ = vg/2 = 4.63 × 10⁵ m s⁻¹, so relative to the envelope they drift backward by (vg − vₚ)t = 9.26 nm. With λ₀ = 2π/k₀ = 0.785 nm that is 11.8 crests, the same as ω₀t/2π with ω₀ = ħk₀²/2m = 3.70 × 10¹⁵ rad s⁻¹.
- Validity: σₓ = 1/(2σₖ) = 2.5 nm, and the group velocities inside φ(k) spread over Δv = ħσₖ/m = 2.32 × 10⁴ m s⁻¹. In 20 fs that spread separates the fast and slow parts by only Δv t = 0.46 nm, under a fifth of σₓ, so the envelope still translates almost rigidly.
- The picture fails as t approaches T = 2mσₓ²/ħ = 108 fs, by which time the packet has travelled vg T = 100 nm, which is k₀/σₖ = 40 of its own widths.
Answer. The peak is at x ≈ 6.5 nm, having moved 18.5 nm at 9.26 × 10⁵ m s⁻¹; about 12 crests (11.8) have slipped back through it; Δv t = 0.46 nm ≪ σₓ = 2.5 nm, so the linearised picture holds, until t nears 108 fs.
HardExample 3
An electron starts in ψ(x, 0) = N exp(−x²/(4σ²)) [e(ik₁x) + e(ik₂x)] with σ = 2.5 nm, k₁ = 4.0 × 10⁹ m⁻¹ and k₂ = 8.0 × 10⁹ m⁻¹. Decide whether this state has a group velocity. Then find where the probability is at t = 0.60 ps, at what times a detector 1.00 μm away can register the electron, and where ⟨x⟩ is at 0.60 ps. Take m = 9.109 × 10⁻³¹ kg and ħ = 1.0546 × 10⁻³⁴ J s.
View solution →
- Each term is a Gaussian times a carrier, so φ(k) is two Gaussians centred on k₁ and k₂, each of width σₖ = 1/(2σ) = 2.0 × 10⁸ m⁻¹. They sit 4.0 × 10⁹ m⁻¹ = 20σₖ apart and do not overlap, so each carries probability ½, N² = 1/(2√(2π)σ), and ⟨k⟩ = 6.0 × 10⁹ m⁻¹.
- Linearising about ⟨k⟩ fails almost at once: at each peak the dropped term (ħ/2m)(k − ⟨k⟩)²t, with k − ⟨k⟩ = ±2.0 × 10⁹ m⁻¹, reaches 1 rad after only 4.3 fs. Across φ, ω′(k) = ħk/m takes the values 4.63 × 10⁵ and 9.26 × 10⁵ m s⁻¹, a spread as large as the smaller speed, so the state has no single group velocity.
- Each peak is narrow on its own (σₖ/k = 0.05 and 0.025), so stationary phase applies to each: ψ splits into two packets moving at v₁ = ħk₁/m = 4.63 × 10⁵ m s⁻¹ and v₂ = ħk₂/m = 9.26 × 10⁵ m s⁻¹. At 0.60 ps they are centred at 278 nm and 556 nm, 278 nm apart.
- Each packet has spread too. Its time scale T = 2mσ²/ħ = 108 fs is long past, so its width is close to Δv t = (ħσₖ/m)t = 13.9 nm: about 14 nm against a 278 nm gap, two cleanly separated lumps.
- A detector at L = 1.00 μm clicks at L/v₂ = 1.08 ps or at L/v₁ = 2.16 ps, each with probability ½. Nothing arrives at L/(ħ⟨k⟩/m) = 1.44 ps, the time a single mean velocity would predict.
- Yet d⟨x⟩/dt = ∫ |φ(k)|² (ħk/m) dk = ħ⟨k⟩/m = 6.95 × 10⁵ m s⁻¹ exactly, so ⟨x⟩ = 417 nm at 0.60 ps: midway between the lumps, almost ten widths from each, where |ψ|² is about 10⁻²¹ of its peak value.
Answer. No single group velocity: the state splits into packets at 4.63 × 10⁵ and 9.26 × 10⁵ m s⁻¹, centred at 278 nm and 556 nm at 0.60 ps; clicks come at 1.08 ps or 2.16 ps, never 1.44 ps; ⟨x⟩ = 417 nm, where there is almost no probability.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Time of flight measures vg = dω/dk = ħk₀/m.
Locate the peak of |ψ|² by stationary phase, x = ω′(k₀)t: it moves at ħk₀/m, while crests move at ħk₀/2m. One group velocity exists only for a narrow φ(k), and only while ħσₖ t/m stays small against the width; a split φ(k) gives separate packets.