Uncertainty Principle & Wave Packets

Choose the shape of a packet and its width. The page draws the packet and the spread of wavenumbers it is made of side by side, and works out Δx·Δk. Give it a particle and a time, and it works out how far the packet has spread once nothing holds it.

Worked example — Uncertainty Principle & Wave Packets

Spread in wavenumber	0.5000 nm⁻¹
Product of the two spreads	0.5000
Spread in momentum	5.273×10⁻²⁶ kg·m/s
Spread in velocity	5.788×10⁴ m/s
Time for the width to double	29.92 fs
Width 100.0 fs after release	5.874 nm

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Uncertainty Principle & Wave Packets

A wave packet is a sum of plane waves e^ikx, with amplitudes φ(k) spread over a range of wavenumbers, that adds up to a wave confined to a region. Its widths Δx and Δk are the standard deviations of |ψ(x)|² and |φ(k)|², and for any such sum Δx·Δk ≥ ½, with equality only for a Gaussian. A matter wave has momentum p = ħk, so the same inequality is Heisenberg's uncertainty principle, Δx·Δp ≥ ħ/2.

What each symbol means

Δx: the standard deviation of position, from |ψ(x)|²
Δk: the standard deviation of wavenumber, from the spectrum |φ(k)|²
Δp: the standard deviation of momentum, ħΔk
ħ: the reduced Planck constant, h/2π = 1.055×10⁻³⁴ J s
t₂: the time a free packet takes to double its Δx

Assumes: A free particle in one dimension, with no force on it after release, and a packet that is real and symmetric when it is let go, so that it is at rest on average and none of its parts has a head start. Non-relativistic energy, E = p²/2m. Widths are standard deviations, which is why the bound is ½: measured as full widths at half maximum, the Gaussian's product is 4 ln 2 ≈ 2.77 instead.
Breaks when: A packet whose phase varies across it, a chirped packet, has the same |ψ(x)|² and a wider spectrum, so it sits above the bound, and if its fast parts start at the back it narrows before it spreads. A force changes the spreading altogether: in a harmonic well a Gaussian of the right width, the oscillator's ground state, keeps that width for ever. And a dust grain is never left alone long enough to spread. Collisions with air molecules and thermal photons record its position far faster than its width could double, which is the subject of decoherence.

Only the Gaussian reaches ½, and a box does not come close 🖖

Every packet here is a sum of plane waves, and Δx·Δk ≥ ½ is a statement about such sums: it holds for sound and for light before it says anything about electrons. Look at the table of five shapes, all with the same Δx. The Gaussian gives 0.5000. A triangle gives 0.5477, a cosine arch 0.5679, an exponential, whose amplitude in k is a Lorentzian, 0.7071. The box, which looks the most tightly confined of the five, gives infinity. Its edges are steps, and a step needs waves of every wavenumber, the reason a <a href="/math/fourier-series">Fourier series</a> for a square wave never runs out of harmonics. Its spectrum falls only as 1/k², and the integral that defines Δk does not converge. Press Box: its spectrum keeps a ripple of side lobes out to both edges of the plot and beyond, where the Gaussian's has long since reached zero. Equality holds for the Gaussian and for nothing else, which is a theorem rather than a quirk of these five. Multiply by ħ, since p = ħk for a matter wave, and it becomes Heisenberg's Δx·Δp ≥ ħ/2.

An electron held to 1 nm doubles its width in 29.92 femtoseconds 🖖

Let a packet go and the waves in it, each carrying its own momentum ħk, drift apart. The spread in velocity is Δv = ħΔk/m, the width grows as √(Δx² + (Δv t)²), and so it has doubled once Δv t = √3 Δx. For a Gaussian that time is 2√3 mΔx²/ħ, proportional to the mass and to the square of the starting width. Press Electron, 1 nm: 29.92 fs. Electron, 0.1 nm: ten times narrower and a hundred times sooner, 0.2992 fs. Then Dust grain. A microgram held to a micrometre doubles its width in 1.041×10⁶ years, and a year after release it is still 1.000 µm wide to four figures. Nothing exempts the grain from spreading. It is 10²¹ times heavier than the electron and starts a thousand times wider, and the doubling time is multiplied by both.

Squeeze it ten times tighter and it ends up nearly ten times wider 🖖

A narrower start means a wider spread of velocities, and given long enough the velocities win. Press Electron, 1 nm and read the width row of the results: 100 fs after release the packet is 5.874 nm wide. Press Electron, 0.1 nm. It started ten times narrower and is now 57.88 nm. For any time t there is a best start, √(ħt/2m), which for 100 fs is 2.406 nm and ends at √2 times that, 3.402 nm; start narrower or wider and the packet is wider at 100 fs. The ladder of starting widths shows the turn, and on the spreading plot the purple line, the start ten times narrower, crosses the blue one within the first few femtoseconds and stays above it. Holding an electron tighter now buys a vaguer electron later.

Problem solved in full

Why an atom is the size it is 5 steps
An electron is held near a proton by their attraction. If the uncertainty principle sets its least momentum, how big is the atom, and how tightly is the electron bound? Press Electron, 0.1 nm for the first step.

Open this problem in the tool
1. Held to Δx = 0.1 nm, the least momentum spread the bound allows is ħ/2Δx, which the page prints for the Gaussian:
  
  $Δ p = \frac{ℏ}{2 Δ x} = \frac{1 . 0546 \times 1 0 ^{- 34} J s}{2 \times 1 0 ^{- 10} m} = 5.273 \times 1 0^{- 25} kg m/s$
2. That spread alone costs kinetic energy:
  
  $E_{k} = \frac{Δ p ^{2}}{2 m _{e}} = \frac{( 5 . 273 \times 1 0 ^{- 25} ) ^{2}}{2 \times 9 . 109 \times 1 0 ^{- 31} kg} = 1.526 \times 1 0^{- 19} J = 0.9525 eV$
3. Now call the confinement r, take the momentum as about ħ/r, and add the proton's pull. The total energy is
  
  $E (r) = \frac{ℏ ^{2}}{2 m _{e} r ^{2}} - \frac{e ^{2}}{4 π ε _{0} r}$
4. It is least where its slope is zero:
  
  $\frac{d E}{d r} = - \frac{ℏ ^{2}}{m _{e} r ^{3}} + \frac{e ^{2}}{4 π ε _{0} r ^{2}} = 0 \Rightarrow r = \frac{4 π ε _{0} ℏ ^{2}}{m _{e} e ^{2}} = 5.292 \times 1 0^{- 11} m$
5. and the least energy is
  
  $E (r) = - \frac{m _{e} e ^{4}}{2 ( 4 π ε _{0} ) ^{2} ℏ ^{2}} = - 2.180 \times 1 0^{- 18} J = - 13.61 eV$
Answer

The estimate puts the electron 0.05292 nm from the proton, bound by 13.61 eV: the Bohr radius, and within 0.1% the ionisation energy of hydrogen. The agreement is partly luck, because p ≈ ħ/r is a choice, and with ħ/2r the radius comes out four times smaller. The reason an atom has a size at all is not luck. Squeezing the electron closer raises its kinetic energy as 1/r² while the attraction gains only as 1/r, so below some radius squeezing costs more than it gains. A classical electron has no such floor and would fall in.

$r = 0.05292 nm = a_{0}, E = - 13.61 eV (hydrogen: - 13.598 eV)$

Learning path

Quantum mechanics: matter as waves

References (3)

A packet built from plane waves, how its width grows once released, σ²(t) = Δx² + α²t²/(4Δx²) with α = ħ/m, and Δx·Δk ≳ ½ from the same analysis: R. Fitzpatrick, "Evolution of Wave Packets" and "Heisenberg's Uncertainty Principle", in Quantum Mechanics (lecture notes), Department of Physics, University of Texas at Austin, 2010.
The mathematics of the inequality: Heisenberg's Δx·Δk ≥ ½ for any square-integrable function, with equality only for Gaussians, and its many variants: G. B. Folland and A. Sitaram, "The uncertainty principle: A mathematical survey", Journal of Fourier Analysis and Applications 3, 207–238 (1997).
The constants, ħ = h/2π, the electron and proton masses and ε₀, and the Bohr radius the worked problem lands on: E. Tiesinga, P. J. Mohr, D. B. Newell and B. N. Taylor, "CODATA recommended values of the fundamental physical constants: 2018", Reviews of Modern Physics 93, 025010 (2021); values as printed on NIST's CODATA 2018 wall chart.

By the 42Math team · Reviewed against published sources. Corrections welcome.