University Physics V · Quantum Uncertainty and Commutation Relations · 8.6
Fourier Conjugacy & the Bandwidth Theorem
Every student meets Δx Δp ≥ ħ/2 as the strangest inequality in physics. It is really the oldest theorem in signal processing wearing a new unit. Prove it once in k, then learn to point at the one line that is quantum and the several that are not.
Build the model
Connect the measurement to the mechanism.
The momentum amplitude φ(p) = ⟨p|ψ⟩ is not a second wavefunction; it is the same ket resolved in the eigenbasis of p̂, and since ⟨x|p⟩ = e(ipx/ħ)/√(2πħ), that change of basis is exactly a Fourier transform. This drains the mystery out of the uncertainty relation. For any square-integrable f and its transform F, Cauchy–Schwarz applied to the mean-shifted x and −i d/dx returns σx σk ≥ 1/2 — a statement containing no physical constant, proved and used long before 1925 by anyone designing a radio filter or deconvolving a seismic trace.
Quantum mechanics enters in exactly two places, neither of them the inequality: the Born rule, which promotes |ψ(x)|² from an intensity profile to the probability density for where one particle is found, and de Broglie's p = ħk, which converts the k-bound into σx σp ≥ ħ/2. The cost of the argument is that σ is a fragile ruler. It weights tails by x², so a state as ordinary as a hard-edged slit has σk = ∞ and the bound tells you nothing at all; for those states the entropic relation h(x) + h(k) ≥ ln(πe) is finite, always applies, and implies the variance form rather than the other way round.
- Simple definition
- Fourier conjugacy is the statement that ⟨x|ψ⟩ and ⟨k|ψ⟩ are a Fourier-transform pair, so their widths obey the bandwidth theorem σx σk ≥ 1/2 for every normalisable state, with no reference to Planck's constant.
- Example
- A Gaussian packet with σx = 0.50 nm saturates it: σk = 1/(2 × 0.50 nm) = 1.0 nm⁻¹. Only the last step is physics — σp = ħσk = 1.05 × 10⁻²⁵ kg m s⁻¹.
One ket in two bases. In the p basis p̂ multiplies and x̂ = +iħ ∂ₚ.
ψ in m(−1/2), φ in (kg m s⁻¹)(−1/2); kernel ⟨x|p⟩ = e(ipx/ħ)/√(2πħ)
This is why |φ(p)|² is a probability density at all — you never renormalise twice.
Holds for every ψ ∈ L²; the two integrals carry different measures.
The same Cauchy–Schwarz proof serves radar, NMR and seismics. It is not quantum.
σx in m, σk in rad m⁻¹; needs x|ψ|² → 0 at ±∞ for the boundary term
The only physical input is a unit conversion. Set ħ = 1 and the theorem is untouched.
ħ = 1.0546 × 10⁻³⁴ J s, so the floor is ħ/2 = 5.27 × 10⁻³⁵ J s
Any excess over the floor is spectral phase — chirp, which a compressor can remove.
FWHM = 2√(2 ln 2) σ = 2.355 σ on both axes; a sech² pulse gives 0.315
Finite for every normalisable state, and it implies σx σk ≥ 1/2 by the Gaussian maximum.
h = −∫ρ ln ρ; the sum is scale-invariant since x and k rescale inversely
⟨p|ψ⟩ is the same ket, read in another basis
Momentum eigenstates are not normalisable, but they are complete: ∫|p⟩⟨p| dp = 1̂, with ⟨x|p⟩ = e(ipx/ħ)/√(2πħ) fixed by p̂ = −iħ ∂ₓ acting on it together with the δ normalisation ⟨p|p′⟩ = δ(p − p′). Insert that identity into ψ(x) = ⟨x|ψ⟩ and you get ψ(x) = ∫⟨x|p⟩⟨p|ψ⟩ dp, so φ(p) = ⟨p|ψ⟩ is the Fourier coefficient of ψ and nothing more. Two consequences you should never re-derive by hand. The transform is unitary, so Plancherel gives ∫|φ|² dp = ∫|ψ|² dx = 1 automatically — you never renormalise after transforming. And the operators swap roles: p̂ multiplies in the p basis while x̂ = +iħ ∂ₚ, the sign flip coming straight from e(−ipx/ħ) in one direction against e(+ipx/ħ) in the other. Work in k = p/ħ with ⟨x|k⟩ = e(ikx)/√(2π) and the constant disappears from the kernel entirely, which is the right basis for everything that follows.
The bandwidth theorem, proved without a physical constant
Take any f ∈ L²(ℝ) normalised to 1, with its mean position and mean wavenumber shifted to zero. Set A = x (multiplication) and B = −i d/dx, both Hermitian on the domain where f and xf are square-integrable and f decays at infinity; the product rule gives [A, B] = i, with no ħ because none was put in. Robertson's route is then three lines: σA²σB² = ⟨Af|Af⟩⟨Bf|Bf⟩ ≥ |⟨Af|Bf⟩|² by Cauchy–Schwarz, and ⟨Af|Bf⟩ = ⟨AB⟩ splits into a real anticommutator half and an imaginary half ½⟨[A, B]⟩ = i/2, so |⟨AB⟩| ≥ 1/2 and σx σk ≥ 1/2. Notice what the derivation quietly assumed: the integration by parts that makes B Hermitian discards a boundary term, and the commutator step needs x|f(x)|² → 0 at ±∞. Remove that and the argument dies — which is why the same manipulation gives nonsense for an angle on a circle, where an Lz eigenstate has ΔLz = 0 while Δφ ≤ 2π/√12.
Where the physics actually enters
The theorem above constrains a classical field just as hard. A radar chirp, a seismic wavelet, the free-induction decay in an NMR probe: every one obeys Δt Δω ≥ 1/2, and nobody calls it a principle. Two separate postulates turn it into quantum mechanics. First the Born rule: |ψ(x)|² is not an intensity spread across a beam, it is the probability density for where a single particle is found, so σx is the spread of an outcome distribution over identically prepared copies, not the size of anything. Second de Broglie, p = ħk — the only place a constant enters. Multiply through and σx σp ≥ ħ/2 = 5.27 × 10⁻³⁵ J s. Now the consequences are physical rather than mathematical: confine an electron to σx = 0.10 nm and σp ≥ (1.055 × 10⁻³⁴)/(2 × 1.0 × 10⁻¹⁰) = 5.27 × 10⁻²⁵ kg m s⁻¹, so with ⟨p⟩ = 0 the kinetic energy is at least σp²/2mₑ = 1.53 × 10⁻¹⁹ J = 0.95 eV. A classical wave squeezed into 0.1 nm has a broad spectrum too; only the Born rule makes that a claim about an electron's energy.
Scaling cannot beat the bound — only reshaping can
Ask what the product does under a dilation. Put ψₐ(x) = √a ψ(ax), still normalised. Then σx → σx/a, and since the transform of ψ(ax) is (1/a)φ(k/a), the momentum amplitude becomes (1/√a)φ(k/a) and σk → aσk. The product σx σk is invariant under every dilation. That is a one-line proof that squeezing a packet never changes how close it sits to the floor; only its shape does. So the equality case must be a shape condition, and Cauchy–Schwarz gives it: equality needs the two vectors proportional, (x − ⟨x⟩)f = λ(d/dx − i⟨k⟩)f with λ purely imaginary, a first-order ODE whose only normalisable solutions are Gaussians times a linear phase. Gaussians are also the fixed shape of the transform, which is why they are the unique saturating family. Check it numerically: build ψ on a grid, take np.fft.fft, compute both variances from |ψ|² and |φ|², and the product returns 0.5000 for a Gaussian and never less for anything else.
The same theorem under laboratory names: pulses and chirp
When the conjugate pair is (t, ω) the result is called the time–bandwidth product. Its FWHM form for a Gaussian is Δt Δν ≥ 2 ln 2/π = 0.441, which is σt σω ≥ 1/2 with FWHM = 2√(2 ln 2) σ = 2.355 σ applied on both axes and ω = 2πν. Take a Ti:sapphire oscillator at 800 nm whose spectrum is 60 nm wide. Convert with |Δν| = c Δλ/λ² = (2.998 × 10⁸)(60 × 10⁻⁹)/(6.40 × 10⁻¹³) = 28.1 THz, so the shortest pulse that spectrum can support is 0.441/28.1 THz = 15.7 fs. If the autocorrelator reports 22 fs, the achieved product is 0.618, a factor 1.40 above the floor, and the excess is spectral phase: the pulse is chirped, and chirped mirrors recover about 16 fs while |φ(ω)|² is untouched. That is the working content of the theorem — the two amplitude widths cannot both be small, and anything above the floor is phase you may be able to remove.
When σ is the wrong ruler, and what replaces it
σ² = ∫x²ρ dx weights the tails by x², so it is not robust. Prepare a hard slit: ψ(x) = w(−1/2) on |x| ≤ w/2, zero outside. Then σx = w/√12 is finite, but φ(k) = √(w/2π)⋅sin(kw/2)/(kw/2), so |φ(k)|² falls only as k⁻², ∫k²|φ|² dk = (2/πw)∫sin²(kw/2) dk diverges linearly, and σk = ∞. The inequality is satisfied vacuously. The divergence is physical: ⟨p²⟩ = ∞ means infinite kinetic energy, the price of a ψ whose derivative is a pair of deltas at the edges. NumPy cannot escape it — np.fft cuts k off at the Nyquist value π/dx, so the truncated ⟨k²⟩ ≈ 2/(w dx) and σk grows as dx(−1/2). What survives is the entropic bound h(x) + h(k) ≥ ln(πe) = 2.145, with h = −∫ρ ln ρ. For the slit h(x) = ln w, so the entropy width e(h(k)) ≥ πe/w — for w = 2.0 μm, at least 4.27 μm⁻¹, a finite number where the variance form gave none.
Change one variable at a time
Make the relationship visible.
Sweep σ with c = 0: one panel narrows exactly as fast as the other widens and the product stays pinned at 0.500 — Fourier scaling, no ħ in sight. Then raise c and only the right panel spreads, because the extra width lives in the phase, which |ψ(x)|² cannot see.
POSITION WIDTH σx0.60 nm
WAVENUMBER WIDTH σk0.833 nm⁻¹
PRODUCT σx σk0.500
MOMENTUM WIDTH ħσk0.879 e−25 kg m/s
Live interpretationPOSITION WIDTH σx: 0.60 nm. WAVENUMBER WIDTH σk: 0.833 nm⁻¹. PRODUCT σx σk: 0.500. MOMENTUM WIDTH ħσk: 0.879 e−25 kg m/s
Catch the common trap
Explain before calculating.
A slit of full width 2a = 20 μm prepares ψ(x) = (2a)(−1/2) for |x| ≤ a and zero outside. What are σx and σk for this state, and what does σx σk ≥ 1/2 then tell you about it?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA free electron is prepared in a minimum-uncertainty Gaussian with σx = 0.50 nm and ⟨p⟩ = 0. Find σk, σp and the mean kinetic energy, and name the single step at which ħ enters.
- A Gaussian saturates the bandwidth theorem, so σx σk = 1/2 exactly: σk = 1/(2 × 0.50 nm) = 1.0 nm⁻¹ = 1.0 × 10⁹ rad m⁻¹. Nothing physical has been used — that is Fourier arithmetic relating |⟨x|ψ⟩|² to |⟨k|ψ⟩|².
- de Broglie converts the wavenumber width into a momentum width: σp = ħσk = (1.0546 × 10⁻³⁴)(1.0 × 10⁹) = 1.05 × 10⁻²⁵ kg m s⁻¹. This is the only line in the problem that contains ħ.
- With ⟨p⟩ = 0 the variance is the second moment: ⟨p²⟩ = σp², so ⟨T⟩ = ⟨p²⟩/2mₑ = (1.055 × 10⁻²⁵)²/(2 × 9.109 × 10⁻³¹) = (1.112 × 10⁻⁵⁰)/(1.822 × 10⁻³⁰) = 6.10 × 10⁻²¹ J.
- In electronvolts, 6.10 × 10⁻²¹/1.602 × 10⁻¹⁹ = 38 meV — about 1.5 kBT at 300 K, so a 0.5 nm confinement is already thermally significant.
Answerσk = 1.0 nm⁻¹, σp = 1.05 × 10⁻²⁵ kg m s⁻¹, ⟨T⟩ = 6.1 × 10⁻²¹ J = 38 meV. ħ entered exactly once, at step 2.
MediumA Ti:sapphire oscillator at 800 nm emits Gaussian pulses whose spectrum is 60 nm wide (FWHM), and the autocorrelator reports a 22 fs pulse. Find the shortest duration this spectrum can support, the time–bandwidth product actually achieved, and state what a compressor can and cannot fix.
- Convert the spectral width to frequency. Differentiating ν = c/λ gives |Δν| = c Δλ/λ² = (2.998 × 10⁸)(60 × 10⁻⁹)/(8.00 × 10⁻⁷)² = 17.99/(6.40 × 10⁻¹³) = 2.81 × 10¹³ Hz = 28.1 THz.
- The Gaussian floor in FWHM units is Δt Δν ≥ 2 ln 2/π = 0.441, which is σt σω ≥ 1/2 with FWHM = 2.355 σ applied on both axes and ω = 2πν.
- Transform limit: Δtₘᵢₙ = 0.441/(2.81 × 10¹³ Hz) = 1.57 × 10⁻¹⁴ s = 15.7 fs.
- Achieved product: (22 × 10⁻¹⁵)(2.81 × 10¹³) = 0.618, a factor 0.618/0.441 = 1.40 above the floor.
- That factor is spectral phase, not missing bandwidth. Chirped mirrors or a prism pair remove the quadratic phase and shorten the pulse to about 16 fs while |φ(ω)|² is untouched. Nothing can go below 15.7 fs without adding spectrum.
AnswerΔtₘᵢₙ = 15.7 fs. The achieved product is 0.618, 1.40 times the 0.441 floor, so the pulse is chirped: compressible to roughly 16 fs, and no shorter.
HardA slit of full width w = 2.0 μm prepares ψ(x) = w(−1/2) on |x| ≤ w/2. Find σx, decide what σk is, predict what an np.fft estimate of σk does as the grid step dx shrinks, and extract a finite width statement from the entropic relation.
- σx² = ∫x²(1/w) dx over ±w/2 = w²/12, so σx = w/√12 = 2.0/3.464 = 0.577 μm.
- Transform: φ(k) = √(w/2π)⋅sin(kw/2)/(kw/2), so |φ(k)|² = (2/πwk²)sin²(kw/2) and ∫k²|φ|² dk = (2/πw)∫sin²(kw/2) dk diverges linearly. σk = ∞, and with it ⟨p²⟩ and ⟨T⟩ — the price of a ψ whose derivative is a pair of δ functions at the edges.
- On a grid np.fft stops at the Nyquist wavenumber π/dx, and since sin² averages to 1/2 the truncated ⟨k²⟩ ≈ 2/(w dx). With dx = 20 nm, σk ≈ √(2/(2.0 × 0.020)) = 7.07 μm⁻¹ and σxσk ≈ 4.08; halve dx to 10 nm and they become 10.0 μm⁻¹ and 5.77. Growth as dx(−1/2) is the divergence reporting itself, not a convergence bug.
- Use entropy instead of variance. h(x) = −∫(1/w) ln(1/w) dx = ln w, so the entropy width is e(h(x)) = w = 2.0 μm. The Białynicki-Birula–Mycielski bound h(x) + h(k) ≥ ln(πe) then gives e(h(k)) ≥ πe/w = 8.540/2.0 = 4.27 μm⁻¹.
- Compare the Gaussian of the same variance: e(h(x)) = σx√(2πe) = 0.577 × 4.133 = 2.386 μm, which saturates the bound at e(h(k)) = 8.540/2.386 = 3.58 μm⁻¹ — matching σk√(2πe) with σk = 1/(2σx) = 0.866 μm⁻¹. The slit is forced 19% wider in k-entropy than the Gaussian it matches in σx.
Answerσx = 0.577 μm; σk = ∞, so σxσk ≥ 1/2 holds vacuously and any grid value merely tracks dx(−1/2). The entropic bound still bites: e(h(k)) ≥ 4.27 μm⁻¹, against 3.58 μm⁻¹ for the Gaussian of equal σx.