The Gaussian packet, worked exactly
One complex width carries the packet's whole history.
This is the simplest wave packet you can evolve exactly with pencil and paper, so every later argument about spreading leans on it. Do the integral once, carefully, and read three things off the answer: how fast a packet spreads, why ħ/2 is reached only at the start, and why a pebble never notices.
Build the model
Complete the square with a complex coefficient, and the packet's whole history follows.
1The big idea
Choose the momentum amplitude to be a Gaussian, φ(k) ∝ exp[−σ₀²(k − k₀)²], and the free packet can be evolved with no approximation at all. Free evolution multiplies each component by e(−iħk²t/2m), which is itself a Gaussian in k with an imaginary coefficient, so the integrand stays Gaussian and one completion of the square gives ψ(x, t) at every t. All of the time dependence collects in one complex number, the width α = σ₀² + iħt/2m. The real part of 1/α sets the modulus: |ψ|² is a normal distribution centred at ħk₀t/m, with σₓ(t) = σ₀√(1 + (t/τ)²) and τ = 2mσ₀²/ħ. The imaginary part of 1/α sets the phase, which gains a term in x²: the fast components run ahead of the slow ones, exactly as in a classical cloud whose velocities spread by ħ/(2mσ₀). Meanwhile |φ(k)| never changes, so σₚ = ħ/(2σ₀) is frozen and σₓσₚ = ħ/2 holds at t = 0 only. That floor is a theorem about Fourier pairs, reached only by an unchirped Gaussian, with no apparatus anywhere in the calculation. Because τ grows as mσ₀², it is 17 fs for an electron localised to 1 nm and about 6 × 10⁵ years for a microgram grain localised to 1 μm. The width law holds for V = 0 and survives a uniform force unchanged; a step or a barrier gives each k its own reflection amplitude and the integral no longer closes, and a real macroscopic body is never isolated long enough to test the formula.
2In one sentence
A Gaussian wave packet is a free-particle state whose momentum amplitude φ(k) is a Gaussian; it stays Gaussian at every time, with complex width α = σ₀² + iħt/(2m), and its position spread grows as σₓ(t) = σ₀√(1 + (ħt/(2mσ₀²))²).
Example. An electron released with σ₀ = 1.00 nm has σₚ = ħ/(2σ₀) = 5.27 × 10⁻²⁶ kg m s⁻¹ and τ = 2mσ₀²/ħ = 17.3 fs, so 50 fs later its width is 1.00 nm × √(1 + (50/17.3)²) = 3.06 nm, while σₚ is unchanged.
3Pick the amplitude whose integral closes
A packet is ψ(x, 0) = (1/√(2π)) ∫ φ(k) e(ikx) dk, and for almost every φ that integral has no closed form. Take φ(k) = (2σ₀²/π)¹⁄⁴ exp[−σ₀²(k − k₀)²]. Its norm is one, since ∫ exp[−2σ₀²(k − k₀)²] dk = √(π/(2σ₀²)) cancels the prefactor squared, and |φ|² is a normal distribution in k with σₖ = 1/(2σ₀). To transform it, set q = k − k₀, take e(ik₀x) outside, and complete the square: −σ₀²q² + iqx = −σ₀²(q − ix/(2σ₀²))² − x²/(4σ₀²). The shifted Gaussian still integrates to √π/σ₀, because moving a Gaussian's centre off the real axis does not change its integral. The result is ψ(x, 0) = (2πσ₀²)⁻¹⁄⁴ exp[−x²/(4σ₀²)] e(ik₀x), so |ψ|² is a normal distribution with standard deviation σ₀. For σ₀ = 1.00 nm, σₖ = 0.500 nm⁻¹ and σₓσₖ = ½ exactly.
4Time enters as an imaginary part of the width
Each plane wave evolves by its own phase, φ(k) → φ(k) e(−iħk²t/2m), and that factor is a Gaussian in k with a purely imaginary coefficient. Multiply it in and the exponent, −σ₀²(k − k₀)² − i(ħt/2m)k², is still quadratic in k, so the same square completes. The bookkeeping is shortest with the k² coefficients gathered into one complex number, α(t) = σ₀² + iħt/(2m), whose imaginary part is an area: ħt/m has units J s × s ÷ kg = m². The answer is ψ(x, t) = (2π)⁻¹⁄⁴ (σ₀/α)¹⁄² exp[−(x − vg t)²/(4α)] e(i(k₀x − ω₀t)), with vg = ħk₀/m and ω₀ = ħk₀²/2m. Two checks. At t = 0, α = σ₀² and the starting packet returns. And the equation holds: for k₀ = 0, iħ∂ψ/∂t and −(ħ²/2m)∂²ψ/∂x² both equal −(ħ²/2m)[x²/(4α²) − 1/(2α)]ψ, because all the t-dependence sits in α and dα/dt = iħ/2m. A packet with k₀ ≠ 0 is the same solution carried along at vg with the plane wave e(i(k₀x − ω₀t)) multiplied in.
5Read the modulus: centre, width and norm
|ψ|² needs only Re(1/α) = σ₀²/|α|². It gives |ψ|² = exp[−(x − vg t)²/(2σₓ²)]/(√(2π) σₓ), with σₓ(t) = |α|/σ₀ = σ₀√(1 + (t/τ)²) and τ = 2mσ₀²/ħ. Normalisation survives at every t, because the prefactor 1/σₓ falls exactly as fast as the width grows, and the centre moves at vg = ħk₀/m, the classical velocity. The two limits have different shapes. For t ≪ τ, σₓ ≈ σ₀[1 + ½(t/τ)²], a flat start with no linear term; for t ≫ τ, σₓ ≈ ħt/(2mσ₀), linear in t. For an electron with σ₀ = 1.00 nm, τ = 17.3 fs, and σₓ is 1.41 nm at t = τ, 3.06 nm at 50 fs and 57.9 nm at 1.00 ps. A narrower start spreads sooner: σ₀ = 0.50 nm gives τ = 4.3 fs and a width of 4.66 nm at 40 fs, twice the 2.31 nm reached by a packet that began at 2.00 nm. The two widths are equal at t = 17.3 fs, when ħt/(2m) = 0.50 nm × 2.00 nm; in the figure, compare σ₀ = 0.5 and 2.0 nm at 17 fs and again at 18 fs to see them trade places.
6Read the phase: the chirp is classical sorting
Im(1/α) leaves ψ a phase (x − vg t)²(t/τ)/(4σₓ²), quadratic in the distance from the centre. Its slope is a local wavenumber, kloc = k₀ + (x − vg t)(t/τ)/(2σₓ²), so the leading edge carries more momentum than the trailing edge. For k₀ = 0 the local velocity is ħkloc/m = (x/t)(t/τ)²/(1 + (t/τ)²), which tends to x/t once t ≫ τ: whatever is found at x got there by moving out from near the origin at speed x/t, as in a classical cloud sorted by speed. The width agrees exactly too. Independent classical particles launched with position spread σ₀ and an uncorrelated velocity spread σᵥ spread as σₓ² = σ₀² + σᵥ²t², and σᵥ = ħ/(2mσ₀) reproduces the quantum law term by term. What quantum mechanics adds is that σᵥ cannot be chosen: σ₀ fixes it. Reverse the sign of the chirp and the packet narrows first. The width law is even in t, so this Gaussian is also the focus of a packet that was converging before t = 0.
7Why ħ/2 is reached at one instant only
Free evolution changes only the phase of φ(k), so |φ(k, t)|² = |φ(k)|² and σₚ = ħ/(2σ₀) at every t. The product therefore grows: σₓσₚ = (ħ/2)√(1 + (t/τ)²). What the packet loses after t = 0 is not sharpness but the absence of correlation. Writing ψ = R e(iS), with S′ = ∂S/∂x, the covariance ½⟨xp + px⟩ − ⟨x⟩⟨p⟩ = ħ∫(x − ⟨x⟩)S′R² dx works out to (ħ/2)(t/τ), and Schrödinger's form of the bound from UQMI-04.07, σₓ²σₚ² ≥ cov² + ħ²/4, holds with equality at every t. The floor itself belongs to the Fourier pair: any function and its transform obey σₓσₖ ≥ ½, with equality only for a Gaussian whose phase is linear in x, and ħ enters only through p = ħk. The same theorem makes a laser pulse whose intensity lasts 10 fs rms carry an rms angular-frequency spread of at least 5.0 × 10¹³ s⁻¹. Nothing in the derivation refers to a measurement or an apparatus.
8Mass sets the clock, and where the formula stops
τ = 2mσ₀²/ħ is linear in m and quadratic in σ₀. Localised to 1.00 nm, an electron has τ = 17.3 fs, a proton 31.7 ps and a C₆₀ molecule, of mass 1.197 × 10⁻²⁴ kg, 22.7 ns. A 1.0 μg grain localised to 1.0 μm has τ = 1.9 × 10¹³ s, about 6 × 10⁵ years, and spreads at ħ/(2mσ₀) = 5.3 × 10⁻²⁰ m s⁻¹. As m → ∞ the width law tends to σₓ = σ₀ for all t: the classical particle, which keeps its position. The closed form rests on three assumptions. V = 0: a constant V₀ only multiplies ψ by e(−iV₀t/ħ), and a uniform force moves the centre without changing σₓ(t), but a step or a barrier gives each k its own reflection and transmission amplitudes, and the integral no longer closes. ω = ħk²/2m, the non-relativistic dispersion. And isolation: a real grain is struck by gas molecules and photons long before τ has elapsed, which is physics this model leaves out.
Key equations
What each relation says, and the units it is written in.
Gaussian momentum amplitude
φ(k) = (2σ₀²/π)¹⁄⁴ exp[−σ₀²(k − k₀)²]
The simplest amplitude whose Fourier integral closes in elementary functions, and which still closes once each k carries its phase e(−iħk²t/2m).
σ₀ in m; k and k₀ in m⁻¹; φ in m¹⁄². |φ|² is normal, with mean k₀ and standard deviation σₖ = 1/(2σ₀).
Complex width
α(t) = σ₀² + iħt/(2m)
All of the time evolution in one number: the t = 0 integral goes through unchanged with σ₀² replaced by α(t).
α in m², since ħt/m is J s × s ÷ kg = m². ħ = 1.055 × 10⁻³⁴ J s, m in kg, t in s.
The packet at time t
ψ(x, t) = (2π)⁻¹⁄⁴ (σ₀/α)¹⁄² exp[−(x − vg t)²/(4α)] e(i(k₀x − ω₀t))
Re(1/α) sets |ψ|², and Im(1/α) leaves a phase quadratic in x − vg t. It solves the equation because dα/dt = iħ/2m is exactly the ratio the equation demands.
vg = ħk₀/m in m s⁻¹, ω₀ = ħk₀²/2m in s⁻¹, ψ in m⁻¹⁄²; the root is the branch that is positive at t = 0.
Width at time t
σₓ(t) = σ₀ √(1 + (ħt/(2mσ₀²))²)
Starts flat, σ₀[1 + ½(t/τ)²], and ends linear, ħt/(2mσ₀) = σₚt/m: the width of a classical cloud whose velocities spread by σₚ/m.
σₓ and σ₀ in m, t in s. σₓ is the standard deviation of |ψ|², centred at vg t; the area under |ψ|² stays 1.
Spreading time
τ = 2mσ₀²/ħ
The width reaches √2 σ₀ at t = τ. Linear in m and quadratic in σ₀: halve the initial width and the packet spreads four times sooner.
τ in s (kg m² ÷ J s). Electron, σ₀ = 1.00 nm: 17.3 fs. A 1.0 μg grain, σ₀ = 1.0 μm: 1.9 × 10¹³ s, about 6 × 10⁵ years.
Uncertainty product in time
σₓσₚ = (ħ/2) √(1 + (t/τ)²)
Equal to ħ/2 at t = 0 only. The excess is the x–p covariance (ħ/2)(t/τ), and Schrödinger's bound σₓ²σₚ² ≥ cov² + ħ²/4 stays an equality.
σₚ = ħ/(2σ₀) in kg m s⁻¹, constant: free evolution changes only the phase of φ(k), never |φ(k)|.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At m = 1 and σ₀ = 0.5 nm, drag t to 40 fs: the spike flattens to 4.66 nm and the fringes of Re ψ crowd outward. Set σ₀ = 2.0 nm and it arrives only 2.31 nm wide. Then raise m to 10: the wide packet all but stops spreading.
Live results
- SPREADING TIME
- 6.2 fs
- WIDTH NOW
- 2.96 nm
- UNCERTAINTY PRODUCT
- 4.93 × ħ/2
- VELOCITY SPREAD
- 96.5 km s⁻¹
Catch the common trap
The mistake this lesson is built to prevent.
A Gaussian is the minimum-uncertainty shape, so the tempting conclusion is that a spreading Gaussian stays at σₓσₚ = ħ/2 and pays for its growing σₓ with a sharpening σₚ. Neither half is true. Free evolution multiplies each φ(k) by e(−iħk²t/2m), a phase of modulus one, so |φ(k)|², and with it σₚ = ħ/(2σ₀), is frozen from the moment of release: a momentum measurement at 1 ps gives the same histogram as one at t = 0. The product therefore grows as (ħ/2)√(1 + (t/τ)²). A Gaussian |ψ|² is not enough to reach ħ/2; the equality needs ψ itself to be a Gaussian times e(ik₀x), with a phase linear in x, and after t = 0 the phase of ψ carries a term in x², the signature of fast parts running ahead of slow ones. Nothing disturbed the particle. The momenta it had from the start are carrying its parts apart.
Test it: what changes?
- Spreading time τ falls fourfold
- Momentum spread σₚ = ħ/(2σ₀) doubles
- Velocity spread ħ/(2mσ₀) doubles
- Late-time width ħt/(2mσ₀) doubles
- Spreading time τ = 2mσ₀²/ħ grows in proportion
- Width at t falls toward σ₀
- Velocity spread ħ/(2mσ₀) falls
- Momentum spread σₚ unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron and a 1.0 μg dust grain are each prepared as Gaussian packets at rest on average, with σ₀ = 10 nm. Find each spreading time τ and each width 1.0 ns later, check the units of τ, and say what the grain shows about the classical limit.
View solution →
- τ = 2mσ₀²/ħ. Units: kg m² ÷ J s = kg m² ÷ (kg m² s⁻¹) = s, as a time must be.
- Electron: τ = 2 × 9.11 × 10⁻³¹ kg × (1.0 × 10⁻⁸ m)² ÷ 1.055 × 10⁻³⁴ J s = 1.73 × 10⁻¹² s = 1.73 ps.
- At t = 1.0 ns, t/τ = 579, deep in the linear regime: σₓ = 10 nm × √(1 + 579²) = 5.79 μm. The far-field form agrees: ħ/(2mσ₀) × t = 5.79 × 10³ m s⁻¹ × 1.0 × 10⁻⁹ s = 5.79 μm.
- Grain: τ = 2 × 1.0 × 10⁻⁹ kg × (1.0 × 10⁻⁸ m)² ÷ 1.055 × 10⁻³⁴ J s = 1.90 × 10⁹ s, about 60 years. At 1.0 ns, t/τ = 5.3 × 10⁻¹⁹, and the width grows by ½σ₀(t/τ)² = 1.4 × 10⁻⁴⁵ m; even after a full year it grows by only 1.4 pm on 10 nm.
- Limit: τ ∝ m, so as m → ∞ at fixed σ₀ the law tends to σₓ = σ₀ for every t. A heavy body keeps its position spread, which is the classical behaviour.
Answer. Electron: τ = 1.73 ps and σₓ = 5.8 μm after 1.0 ns. Grain: τ = 1.9 × 10⁹ s (about 60 years), and its width grows by about 10⁻⁴⁵ m in the same nanosecond.
MediumExample 2
An electron packet with k₀ = 0 is released at t = 0 as a real Gaussian with σ₀ = 1.00 nm. At t = 3τ, find t, σₓ, and the local wavenumber and local velocity at x = +σₓ, and compare that velocity with x/t. Then find the x–p covariance and show that it accounts for all of σₓσₚ above ħ/2.
View solution →
- τ = 2mσ₀²/ħ = 17.3 fs, so t = 3τ = 51.8 fs, and σₓ = σ₀√(1 + 3²) = √10 × 1.00 nm = 3.16 nm.
- The phase of ψ is x²(t/τ)/(4σₓ²) plus a constant, so kloc = ∂(phase)/∂x = x(t/τ)/(2σₓ²). At x = σₓ this is 3/(2 × 3.162 nm) = 0.474 nm⁻¹.
- Local velocity: ħkloc/m = 1.158 × 10⁻⁴ m² s⁻¹ × 4.74 × 10⁸ m⁻¹ = 5.49 × 10⁴ m s⁻¹. Ballistic estimate: x/t = 3.162 nm ÷ 51.8 fs = 6.10 × 10⁴ m s⁻¹. The ratio is 0.900 = (t/τ)²/(1 + (t/τ)²) = 9/10, and it approaches 1 as t/τ grows.
- σₚ = ħ/(2σ₀) = 5.27 × 10⁻²⁶ kg m s⁻¹, unchanged since release, so σₓσₚ = (ħ/2)√10 = 1.58ħ.
- Covariance: with ψ = R e(iS) and ⟨x⟩ = ⟨p⟩ = 0, cov = ħ∫xS′R² dx = ħ(t/τ)/(2σₓ²) × σₓ² = (ħ/2)(t/τ) = 1.50ħ. Then cov² + ħ²/4 = 2.25ħ² + 0.25ħ² = 2.50ħ², and √2.50 ħ = 1.58ħ: Schrödinger's bound holds with equality.
Answer. t = 51.8 fs and σₓ = 3.16 nm; at x = σₓ, kloc = 0.474 nm⁻¹ and ħkloc/m = 5.49 × 10⁴ m s⁻¹, 90% of x/t = 6.10 × 10⁴ m s⁻¹; σₓσₚ = 1.58ħ, exactly √(cov² + ħ²/4) with cov = 1.50ħ.
HardExample 3
A single ⁸⁷Rb atom, m = 1.443 × 10⁻²⁵ kg, is released from a trap as an unchirped Gaussian packet and imaged after 25.0 ms of flight. Along the horizontal axis, where gravity does not act, it moves freely. Which σ₀ makes the width at the camera smallest, what is that width, and how wide would a packet released at σ₀ = 1.00 μm arrive?
View solution →
- Write σₓ² = σ₀² + (ħt/(2mσ₀))². The first term rewards a tight release and the second punishes it, because a tight release carries a broad σₚ = ħ/(2σ₀).
- Take u = σ₀² as the variable: σₓ² = u + (ħt/2m)²/u, and d(σₓ²)/du = 1 − (ħt/2m)²/u² = 0 at u = ħt/(2m). Numerically, with ħ = 1.0546 × 10⁻³⁴ J s to match the four-figure mass, ħt/(2m) = 1.0546 × 10⁻³⁴ J s × 0.0250 s ÷ (2 × 1.443 × 10⁻²⁵ kg) = 9.14 × 10⁻¹² m², so σ₀ = 3.02 μm.
- At the optimum the two terms are equal: σₓ² = 2u = ħt/m = 1.827 × 10⁻¹¹ m², so σₓ = √(ħt/m) = 4.27 μm, √2 times the release width. Its spreading time 2mσ₀²/ħ equals the flight time, 25.0 ms: the best release is the one whose τ matches t.
- Release at σ₀ = 1.00 μm instead: τ = 2 × 1.443 × 10⁻²⁵ kg × (1.00 × 10⁻⁶ m)² ÷ 1.0546 × 10⁻³⁴ J s = 2.74 ms, t/τ = 9.14, and σₓ = 1.00 μm × √(1 + 83.5) = 9.19 μm, 2.15 times the optimum.
- Checks: ħt/m has units J s × s ÷ kg = m². As m → ∞, √(ħt/m) → 0, the classical particle that can be both localised and slow. And the best width grows only as √t: doubling the flight time widens it by √2, not by 2.
Answer. σ₀ = 3.02 μm gives the narrowest arrival, σₓ = 4.27 μm, the release width whose τ equals the 25.0 ms flight; a 1.00 μm release arrives 9.19 μm wide.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
σₓ = σ₀√(1 + (t/τ)²), with τ = 2mσ₀²/ħ.
Replace σ₀² by α = σ₀² + iħt/(2m) to evolve the Gaussian exactly. σₚ = ħ/(2σ₀) never changes, so σₓσₚ = ħ/2 only at t = 0. The law needs V = 0 or a uniform force and isolation; a step or barrier breaks the closed form.
Next: The well, the boundary conditions, and what they cost →