Fourier conjugacy and the bandwidth theorem

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.

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 L_z eigenstate has ΔL_z = 0 while Δφ = 2π/√12.

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 there it is read as a property of the transform, not a law of nature. 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 ≥ (1.0546 × 10⁻³⁴)/(2 × 1.0 × 10⁻¹⁰) = 5.27 × 10⁻²⁵ kg m s⁻¹, so with ⟨p⟩ = 0 the mean 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.

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: the previous topic shows that Cauchy–Schwarz equality with zero covariance forces (x − ⟨x⟩)f = λ(d/dx − i⟨k⟩)f with λ real and negative, whose only normalisable solutions are Gaussians times a linear phase. That the Gaussian is its own Fourier transform is not the reason: every Hermite function shares that property, yet the nth has σx σk = n + 1/2. 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.

When the conjugate pair is (t, ω) the result is called the time–bandwidth product. Converting both standard deviations to FWHM with the factor 2√(2 ln 2) = 2.355, and using ω = 2πν, turns σt σω ≥ 1/2 into Δt Δν ≥ (2.355)² × (1/2)/(2π) = 2 ln 2/π = 0.441 for a Gaussian. Run it backwards to size a laser: a 10 fs Gaussian pulse needs Δν ≥ 0.441/(10 fs) = 44.1 THz, which at 800 nm is Δλ = λ²Δν/c = (6.40 × 10⁻¹³)(4.41 × 10¹³)/(2.998 × 10⁸) = 94 nm of spectrum, one reason few-cycle oscillators are built on a gain medium as broad as Ti:sapphire. A measured pulse longer than its transform limit carries the same spectrum with a non-flat spectral phase: it is chirped, and chirped mirrors or a prism pair can strip the quadratic part 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.

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⟩ grows as 1/(w dx) and σk 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.