Quadratic dispersion and why a free packet spreads
Curvature of ω(k), not uncertainty, spreads the packet.
Every free packet in this course widens, and it is tempting to file that under the uncertainty principle. This lesson finds the real cause, the curvature of ω(k), and follows it to an exact law for the width, one that also explains packets that narrow before they spread and light pulses that never spread at all.
Build the model
A free packet widens because its wavenumbers travel at different speeds.
1The big idea
A free packet is a sum of plane waves, and each one advances its phase at its own rate ω(k) = ħk²/2m. Expand about the central wavenumber k₀ and the series ends after three terms: a constant phase, a linear term that carries the envelope rigidly at vg = ħk₀/m, and a quadratic term (ħ/2m)(k − k₀)², the only one able to change the envelope's shape. Physically it says that the group velocity ħk/m depends on k, so the faster components pull ahead of the slower ones. Followed through, this gives an exact law for a packet of any shape: σₓ² is a parabola in t whose t² coefficient is (σₚ/m)², the same law a cloud of classical particles with the same spreads and correlation would obey. Nothing is lost on the way. |φ(k)|² never changes; what spreading builds is a correlation between position and momentum, and a packet prepared with the opposite correlation, its faster part behind, narrows first, reaches a waist no narrower than ħ/(2σₚ), and only then spreads. Light in vacuum, with ω = ck, satisfies the same Fourier inequality and does not spread at all, so the cause is the curvature ω″ = ħ/m, not the uncertainty principle. The model stops where its assumptions do: once the potential has curvature the width is no longer a parabola in t, and at relativistic speeds ω″ falls to ħ/(γ³m).
2In one sentence
Wave-packet spreading is the growth of a free packet's position spread σₓ with time. It happens because ω = ħk²/2m is quadratic, so each wavenumber moves at its own group velocity ħk/m and the packet's components drift apart.
Example. An unchirped Gaussian electron packet 1.0 nm wide has σₖ = 0.50 nm⁻¹ and a velocity spread ħσₖ/m = 5.8 × 10⁴ m s⁻¹, so after 1.0 ps it is about 58 nm wide. A light pulse in vacuum of the same shape is still 1.0 nm wide.
3Expand ω(k) about the centre and the series stops
Write the packet as ψ(x, t) = (1/√(2π))∫φ(k) e(i(kx − ω(k)t)) dk and put k = k₀ + κ. For ω = ħk²/2m the Taylor series ends after three terms, ω = ω₀ + vgκ + (ħ/2m)κ² with vg = ħk₀/m, because ω‴ = 0. Then ψ = e(i(k₀x − ω₀t)) (1/√(2π))∫φ(k₀ + κ) e(iκ(x − vg t)) e(−iħκ²t/2m) dκ. Drop the last factor and the integral depends on x − vg t alone: the envelope slides without changing shape, which is all the previous lesson needed. The last factor is the whole of this one. It is a phase, so it leaves |φ|² alone, but it differs from one κ to the next and grows with t, and components that fall out of step reshape their sum. Across the packet it varies by about (ħ/2m)σₖ²t. For an electron ħ/2m = 0.0579 nm² fs⁻¹, so with σₖ = 0.50 nm⁻¹ it reaches 0.25 rad after 17 fs, the time an unchirped Gaussian with that σₖ takes to widen by √2.
4Each wavenumber travels at its own speed
The quadratic term has a plain meaning. A packet built from wavenumbers near k alone sits where its phase kx − ω(k)t is stationary in k, at x = ω′(k)t = ħkt/m. So the group velocity is a function of k, and a spread σₖ becomes a spread of speeds σᵥ = (ħ/m)σₖ = σₚ/m. For σₖ = 0.50 nm⁻¹ that is 5.8 × 10⁴ m s⁻¹: after 1.0 ps the packet is about 58 nm wide, after 1.0 ns about 58 μm. The growth is linear in t, ballistic rather than diffusive, and at long times the packet becomes a map of its own momentum distribution, |ψ(x, t)|² → (m/ħt)|φ(mx/ħt)|², which is how time-of-flight images of released atoms measure momenta. The classical correspondence is complete here: a cloud of free classical particles with the same spreads of position and velocity, and the same correlation between them, widens by exactly the same amount. What quantum mechanics adds is σₓσₚ ≥ ħ/2, which forbids a narrow packet with a small velocity spread.
5The width follows an exact parabola in time
No Gaussian is needed for the central result. Free evolution multiplies φ(k) by e(−iħk²t/2m), so ⟨p⟩ and σₚ never change. On φ, x acts as i ∂/∂k, and differentiating that phase brings down −iħkt/m: one power of t per derivative, because the phase is quadratic in k. Equivalently, d⟨x²⟩/dt = ⟨xp + px⟩/m and d⟨xp + px⟩/dt = 2⟨p²⟩/m, which is constant, so the ladder stops. Either way σₓ²(t) = σₓ²(0) + (2C₀/m)t + (σₚ/m)²t², with C₀ = ½⟨xp + px⟩ − ⟨x⟩⟨p⟩ at t = 0. Dimensions: (J s)(s)/kg and (kg m s⁻¹)²s²/kg² are both m². For an unchirped Gaussian C₀ = 0 and σₚ = ħ/(2σ₀), so σₓ = σ₀√(1 + (t/τ₀)²) with τ₀ = 2mσ₀²/ħ: 0.17 fs for an electron with σ₀ = 0.10 nm, but 8.7 × 10⁷ years for a 0.145 kg ball with σ₀ = 1.0 nm. Sending m → ∞, or ħ → 0 at fixed σ₀, freezes the width, as the classical limit should.
6Nothing is lost: a chirped packet narrows first
Complete the square: σₓ²(t) = σw² + σᵥ²(t − tw)², with tw = −mC₀/σₚ². C₀ says whether the faster part of the packet leads (C₀ > 0) or trails (C₀ < 0), and free motion raises it steadily, C(t) = C₀ + σₚ²t/m, through zero at the waist. A Gaussian given the phase e(ibx²) has local wavenumber 2bx, so with b < 0 its left side moves right, its right side moves left, and C₀ = 2ħbσ₀² is negative. Take σ₀ = 1.2 nm and b = −0.347 nm⁻² for an electron: it shrinks to σw = 0.54 nm at tw = 9.95 fs, is back to 1.2 nm at 19.9 fs, and spreads from then on. At the waist σwσₚ = ħ/2, a minimum-uncertainty packet. None of this is irreversible. The norm stays one and |φ(k)|² never changes; the packet's narrow past is stored in the phase of φ(k). Complex-conjugate an unchirped packet that has spread for a time T and that phase reverses: the new packet narrows for exactly T, back to its starting width, then spreads again.
7With ω = ck nothing spreads, so the cause is dispersion
Repeat the k-space argument for any ω(k) and the law generalises: σₓ² grows by 2t times the covariance of x with ω′(k), plus t² times the variance of ω′(k) over |φ(k)|². For ω = ħk²/2m, ω′ = ħk/m and that variance is (ħσₖ/m)² whatever the packet's shape. For light in vacuum ω = ck, so ω′ = c for every component, both terms vanish, and ψ(x, t) = f(x − ct) exactly. That pulse obeys σₓσₖ ≥ ½ just as the electron does, so the inequality cannot be what drives spreading. In glass ω(k) is curved and short pulses do broaden; ultrafast optics pre-chirps them so that the glass compresses them instead, the optical twin of the narrowing packet. The relativistic particle supplies a limiting case: ω = √(c²k² + m²c⁴/ħ²) gives ω″ = ħ/(γ³m), which is ħ/m at rest and ħ/(1000m) at γ = 10, so a fast packet spreads along its motion far more slowly and approaches the light-like limit.
Key equations
What each relation says, and the units it is written in.
Free dispersion expanded about k₀
ω(k) = ω₀ + vg (k − k₀) + (ħ/2m)(k − k₀)²
The first term is a global phase and the second slides the envelope at vg. Only the third can change the envelope's shape, and there is no fourth.
ω₀ = ħk₀²/2m in rad s⁻¹; vg = ħk₀/m in m s⁻¹; ħ/2m in m² s⁻¹. Exact, because ω‴ = 0.
Spread of group velocities
σᵥ = (ħ/m) σₖ = σₚ/m
ω″ = dvg/dk = ħ/m turns a spread of wavenumbers into a spread of speeds. It never changes, because free evolution leaves |φ(k)|² alone.
σₖ in m⁻¹, σₚ in kg m s⁻¹, σᵥ in m s⁻¹; ħ/m = 1.158 × 10⁻⁴ m² s⁻¹ for an electron
Exact width of any free packet
σₓ²(t) = σₓ²(0) + (2C₀/m) t + (σₚ/m)² t²
No approximation and no Gaussian needed. A classical cloud with the same spreads obeys the same law, and at long times σₓ ≈ σₚt/m: ballistic, not diffusive.
C₀ = ½⟨xp + px⟩ − ⟨x⟩⟨p⟩ at t = 0, in J s; t in s; every term in m²
Time of the narrowest width
tw = −m C₀/σₚ²
Completing the square in the width law. C₀ < 0, faster part behind, puts the waist in the future, so the packet narrows before it spreads.
tw in s, m in kg, C₀ in J s, σₚ in kg m s⁻¹; tw < 0 means the waist is already past
Local velocity of a chirped packet
v(x) = 2ħbx/m, for ψ ∝ e(−x²/4σ₀²) e(ibx²)
With b < 0 both edges move inward, and averaging m x v(x) gives C₀ = 2ħbσ₀² < 0. Free spreading turns b positive by itself.
b in m⁻², x in m, v in m s⁻¹; the phase bx² gives local wavenumber 2bx
Linear dispersion, the control case
ω = ck gives ψ(x, t) = f(x − ct) for any f
Same Fourier widths and the same σₓσₖ ≥ ½ as an electron packet, yet no spreading at all. The width grows only when ω″ ≠ 0.
c = 3.00 × 10⁸ m s⁻¹; ω″ = 0, so every component has vg = c and σᵥ = 0
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Drag t from 0 to 30 fs: this converging packet shrinks to its waist near 10 fs, then spreads past its starting width. Set β = +2 and it only spreads; set β = 0 and try a smaller σ₀, which spreads faster because σᵥ = ħ/(2mσ₀).
Live results
- WIDTH σₓ AT t
- 0.54 nm
- WAIST TIME tw
- 10.0 fs
- WAIST WIDTH σw
- 0.54 nm
- VELOCITY SPREAD σᵥ
- 108 km s⁻¹
Catch the common trap
The mistake this lesson is built to prevent.
The usual story is that a packet spreads because the uncertainty principle pushes it apart. It does not survive the numbers. σₓσₚ ≥ ħ/2 constrains each instant separately and is already met at t = 0, and σₚ is fixed for all time, so as σₓ grows the product climbs far above ħ/2, which the inequality allows but does not demand. A light pulse in vacuum obeys the same Fourier inequality and keeps its width, and a chirped electron packet gets narrower for a while, which no push from the uncertainty principle could do. The cause is ω″ = ħ/m, which makes speed depend on wavenumber. The inequality's real role is smaller: it sets the least velocity spread a packet of a given width can have, which is why narrow packets spread fast.
Test it: what changes?
- Velocity spread σᵥ = ħ/(2mσ₀) shrinks
- Spreading time τ₀ = 2mσ₀²/ħ lengthens
- Waist stays at t = 0
- Faster part trails, so C₀ < 0
- Packet narrows to a waist, then spreads
- At the waist, σwσₚ = ħ/2
- Velocity spread σᵥ rises with |β|
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron packet is prepared as an unchirped Gaussian of width σ₀ = 2.0 nm centred on k₀ = 10 nm⁻¹. Find its spread of group velocities, its width after 200 fs, and how far its centre moves. How wide would a light pulse in vacuum with the same envelope be after the same time?
View solution →
- An unchirped Gaussian is a minimum-uncertainty packet, so σₖ = 1/(2σ₀) = 0.25 nm⁻¹ = 2.5 × 10⁸ m⁻¹.
- Quadratic dispersion turns that into a speed spread: σᵥ = (ħ/m)σₖ = (1.158 × 10⁻⁴ m² s⁻¹)(2.5 × 10⁸ m⁻¹) = 2.89 × 10⁴ m s⁻¹, or 0.0289 nm fs⁻¹.
- Unchirped means C₀ = 0, so the width law is σₓ² = σ₀² + σᵥ²t². At 200 fs, σᵥt = 5.79 nm and σₓ = √(2.0² + 5.79²) nm = √37.5 nm = 6.1 nm.
- The centre moves at vg = ħk₀/m = 1.158 × 10⁶ m s⁻¹, covering 232 nm in 200 fs. At long times the width grows by 2.5 nm for every 100 nm travelled, since σᵥ/vg = σₖ/k₀ = 0.025.
- Checks: σᵥt is (m s⁻¹)(s) = m, and t → 0 returns 2.0 nm. A light pulse in vacuum has ω = ck, every component moves at c, and it is still 2.0 nm wide.
Answer. σᵥ = 2.9 × 10⁴ m s⁻¹; after 200 fs the electron packet is about 6.1 nm wide and its centre has moved 232 nm. The light pulse is still 2.0 nm wide.
MediumExample 2
An electron packet with zero mean momentum is prepared as ψ(x, 0) ∝ e(−x²/4σ₀²) e(ibx²), with σ₀ = 1.50 nm and b = −0.20 nm⁻². Show that it narrows first, find when it is narrowest and how narrow it gets, and check the waist against σₓσₚ ≥ ħ/2.
View solution →
- The phase bx² gives local wavenumber 2bx: −0.60 nm⁻¹ at x = +1.50 nm and +0.60 nm⁻¹ at x = −1.50 nm. Both edges move toward the centre, at ħ(0.60 nm⁻¹)/m = 6.9 × 10⁴ m s⁻¹.
- Momentum spread from ħ²∫|∂ψ/∂x|² dx: σₖ² = 1/(4σ₀²) + 4b²σ₀² = 0.111 + 0.360 = 0.471 nm⁻², so σₖ = 0.686 nm⁻¹. The chirp has multiplied σₖ² by 4.24.
- Covariance: C₀/ħ is the average of x times the local wavenumber 2bx, so C₀/ħ = 2b⟨x²⟩ = 2bσ₀² = −0.900. It is negative: the faster part of the packet trails.
- Waist time: tw = −mC₀/σₚ² = −(C₀/ħ)/((ħ/m)σₖ²) = 0.900 ÷ (0.1158 nm² fs⁻¹ × 0.471 nm⁻²) = 16.5 fs.
- Waist width: σw² = σ₀² − (C₀/ħ)²/σₖ² = 2.25 − 0.81/0.471 = 0.531 nm², so σw = 0.728 nm. Then σwσₖ = 0.728 × 0.686 = 0.500, so σwσₚ = ħ/2 exactly. The parabola is symmetric about tw, so the width is 1.50 nm again at 33.0 fs.
Answer. Narrowest at tw ≈ 16.5 fs, with σw ≈ 0.73 nm and σwσₚ = ħ/2; back to 1.50 nm at 33 fs, and spreading without limit after that.
HardExample 3
A free atom is released many times over and its horizontal position spread is measured at three times: σₓ = 0.885 μm at t = 0, 0.885 μm at t = 2.00 ms and 2.25 μm at t = 4.00 ms. Use the exact width law to find when the packet was narrowest, how narrow, and its velocity spread. If the packet is Gaussian, what is the atom's mass? What can you say if it is not?
View solution →
- The law is exact for σₓ², so square the data: 0.7832, 0.7832 and 5.0625 μm² at 0, 2.00 and 4.00 ms. Write σₓ² = A + Bt + Ct².
- Equal values at 0 and 2.00 ms put the vertex midway, tw = 1.00 ms, so B = −2C. Then 5.0625 = 0.7832 + 8C gives C = 0.5349 μm² ms⁻².
- σᵥ = √C = 0.731 μm ms⁻¹ = 7.31 × 10⁻⁴ m s⁻¹, and σw² = A − Ctw² = 0.2483 μm², so σw = 0.498 μm. B < 0 means C₀ < 0: the packet was released converging.
- At the waist the covariance is zero and a Gaussian is minimum-uncertainty, σw × mσᵥ = ħ/2. So m = ħ/(2σwσᵥ) = 1.055 × 10⁻³⁴ J s ÷ (2 × 4.98 × 10⁻⁷ m × 7.31 × 10⁻⁴ m s⁻¹) = 1.45 × 10⁻²⁵ kg. Units: J s ÷ (m × m s⁻¹) = kg.
- In atomic mass units, 1.45 × 10⁻²⁵ kg ÷ 1.661 × 10⁻²⁷ kg = 87 u, which is rubidium-87.
- For any other shape σwσₚ > ħ/2, so the same data give only m ≥ 1.45 × 10⁻²⁵ kg. A classical cloud would fit the same parabola and say nothing about m at all: the mass enters only through ħ.
Answer. Waist at tw = 1.00 ms with σw = 0.498 μm, and σᵥ = 7.31 × 10⁻⁴ m s⁻¹. A Gaussian packet needs m = 1.45 × 10⁻²⁵ kg ≈ 87 u, rubidium-87; any other shape only gives m ≥ 1.45 × 10⁻²⁵ kg.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Spreading comes from ω″ = ħ/m, not the uncertainty principle.
Use σₓ²(t) = σₓ²(0) + (2C₀/m)t + (σₚ/m)²t² for any free packet: σₚ never changes, and C₀ < 0 means it narrows first, to a waist at tw = −mC₀/σₚ². It fails once the potential is curved or motion relativistic; with ω = ck nothing spreads.