Wave packets and the uncertainty principle

A plane wave ψ = e^{(i(kx − ωt))} has a single wavenumber, so through p = ħk a single momentum, but |ψ|² = 1 at every x: it describes a particle that could be anywhere. Add a second wave and a location starts to appear. Waves with k₁ = 9.5 nm⁻¹ and k₂ = 10.5 nm⁻¹ sum to a carrier at 10 nm⁻¹ inside an envelope 2 cos(Δk x/2) whose beats repeat every 2π/Δk = 6.3 nm. Fill the band between them with many waves and the repeats disappear: the sum stays large only near x = 0. Treat each component as a phasor. At x = 0 they all point the same way; at distance x the phase of component k is kx, so across the band the phasors fan out by x Δk. When the fan has opened into a full circle, at x Δk = 2π for a flat band, they sum to zero, which is the packet's first null: 6.3 nm for a band 1.0 nm⁻¹ wide. A narrower band needs a larger x to close the fan, so a narrow spread of k can only build a wide packet.

To turn “narrow needs broad” into a law, measure both widths the same way: Δx is the standard deviation of |ψ(x)|² and Δk that of |φ(k)|². A Gaussian packet ψ ∝ e^{(−x²/4σ²)} e^(ik₀x) transforms into φ ∝ e^{(−σ²(k − k₀)²)}; squaring each gives Δx = σ and Δk = 1/(2σ), so Δx Δk = ½ whatever σ is. No shape does better, which is the Fourier uncertainty theorem, and most do worse. Give the Gaussian a chirp, a phase e^{(iCx²/4σ²)} that makes the local wavenumber drift across it: |ψ|² is untouched, but Δk grows by √(1 + C²), so C = 1 costs 41% more bandwidth and buys no localisation at all. The constant also depends on which width you quote. The full widths at half maximum of the same Gaussian pair multiply to 8 ln 2 × ½ = 2.77, and a flat band measured to its first null gives 2π, so order-of-magnitude statements say only that Δx Δk is of order 1. The floor of exactly ½ belongs to rms widths.

De Broglie's p = ħk carries the theorem straight into mechanics: Δx Δp ≥ ħ/2, with ħ = 1.055 × 10⁻³⁴ J s. Nothing in the derivation mentions a measurement. Δx and Δp are the spreads of the results that ideal position and momentum measurements would return on many identically prepared particles, and a state narrow in both does not exist to be prepared. The size of ħ decides where this matters. An electron confined to Δx = 0.10 nm has Δp ≥ 5.3 × 10⁻²⁵ kg m s⁻¹, and because ⟨p²⟩ ≥ Δp², its mean kinetic energy is at least Δp²/2m = 0.95 eV: atomic confinement costs electron-volts, the energy scale of chemistry. A 1 μg dust grain held to 1 μm has Δv ≥ 5.3 × 10⁻²⁰ m s⁻¹, far below any thermal or instrumental jitter. That is the correspondence check: as m grows at fixed Δx, or as ħ → 0, the bound fades and a sharp classical trajectory is recovered.

Let the packet evolve. Each component carries its own time factor e^(−iω(k)t), and for a narrow band ω(k) ≈ ω₀ + v_g(k − k₀) with v_g = [dω/dk]₀. Pulling out e^{(i(k₀x − ω₀t))} leaves an integral that depends on x and t only through x − v_g t, so the envelope translates at v_g while the crests move at the phase velocity ω₀/k₀. For a free electron ω = E/ħ = ħk²/2m, so v_g = ħk₀/m = p/m, the classical velocity, while vₚ = ħk₀/2m is half of it and the crests slide backwards through the envelope. At k₀ = 8.0 nm⁻¹ (λ = 0.79 nm, 2.4 eV) the packet moves at 9.3 × 10⁵ m s⁻¹ and its crests at 4.6 × 10⁵ m s⁻¹. The relativistic check agrees: with ω² = c²k² + (mc²/ħ)², v_g = c²k/ω = pc²/E, again the particle velocity and always below c, even though vₚ = c²/v_g exceeds c. Energy and information ride on the envelope.

The next term in the expansion, ½(ħ/m)(k − k₀)², is where the band of k matters: components faster than v_g run ahead, slower ones fall behind, and the leading edge collects the short wavelengths. For a packet that starts as an unchirped Gaussian the result is exact: Δx(t)² = Δx₀² + (Δv t)², with Δv = Δp/m = ħ/(2mΔx₀). Initial size and velocity spread add in quadrature, and tₛ = 2mΔx₀²/ħ marks where spreading takes over. An electron packet with Δx₀ = 0.50 nm has tₛ = 4.3 fs and doubles its width after √3 tₛ = 7.5 fs; a 1 μg grain held to 1 μm has tₛ = 1.9 × 10¹³ s, some 600 000 years. Throughout, |φ(k)|² never changes, because a free particle's momentum distribution is conserved. So Δp stays fixed while Δx grows, and Δx Δp = (ħ/2)√(1 + (t/tₛ)²) climbs off the floor: the packet has acquired exactly the chirp that spoils minimum uncertainty.

Position and momentum are both observables, and the bound on their spreads follows from their operators. Time is not an observable here; it is the parameter the state evolves in, so ΔE Δt cannot be read the same way. Its precise content concerns change. A stationary state goes as e^(−iEt/ħ) forever and has a sharp energy. An excited state that decays with mean lifetime τ has an amplitude e^{(−iE₀t/ħ)} e^(−t/2τ), and a wave train that dies away is a packet in time: its Fourier transform is a Lorentzian in energy with full width at half maximum Γ = ħ/τ. A level with τ = 20 ns is therefore blurred by Γ = 3.3 × 10⁻⁸ eV, a natural linewidth of 1/(2πτ) = 8.0 MHz, set by the decay rather than by any spectrometer. Particle physics reads the relation backwards: the Z boson's measured width of 2.50 GeV gives τ = ħ/Γ = 2.6 × 10⁻²⁵ s, a lifetime no clock could time.