University Physics IV · Photons and Matter Waves · 4.8
Fourier Reciprocity & Uncertainty
Before it is quantum mechanics it is mathematics: no function can be narrow in x and narrow in k at once. Learn to compute both widths properly, find the one shape that reaches the floor, and read the resulting Δx Δp ≥ ħ/2 as a statement about how a state was prepared rather than about how carefully it is measured.
Build the model
Connect the measurement to the mechanism.
Take any normalisable wave packet ψ(x) and its Fourier amplitude φ(k), and define each width as a root-mean-square spread about its own mean — Δx from |ψ|², Δk from |φ|². Then Δx Δk ≥ ½, always, with equality for one shape only. Nothing in that sentence is quantum.
It is a theorem about Fourier pairs, and the same inequality governs a radar chirp, a musical note too short to have a pitch, and the bandwidth an optical fibre has to carry. Squeeze the packet by a factor a and the transform stretches by exactly a, so the product is a pure number attached to the shape and not to the scale; minimising it over all shapes returns ½, reached only by a Gaussian whose exponent carries no imaginary quadratic term. Quantum mechanics then contributes a single line, de Broglie's p = ħk, and the theorem becomes Δx Δp ≥ ħ/2.
What that costs is a habit of thought: because a particle is described by such a packet, no state exists with both a sharp position and a sharp momentum, so a confined electron carries kinetic energy it cannot shed and an atom has a size it cannot fall below. What it does not cost is measurement precision. Δx and Δp are spreads over an ensemble of identically prepared systems, fixed the moment the state is prepared, and no instrument appears anywhere in the derivation.
- Simple definition
- Fourier reciprocity is the rule that a function and its transform cannot both be narrow: the product of their root-mean-square widths, Δx Δk, can never fall below one half.
- Example
- A Gaussian packet with Δx = 0.50 nm has Δk = 1/(2 × 0.50 nm) = 1.0 rad nm⁻¹, so Δp = ħΔk = 1.06 × 10⁻²⁵ kg m s⁻¹ — a velocity spread of 116 km s⁻¹ for an electron.
Standard deviations, not FWHM — and no ħ in sight, because this half is pure Fourier analysis.
⟨x⟩ weighted by |ψ(x)|², ⟨k⟩ by |φ(k)|². Δx in m, Δk in rad m⁻¹; both second moments must exist.
One substitution turns a theorem about functions into a limit on every state a particle can occupy.
ħ = 1.055 × 10⁻³⁴ J s; Δp in kg m s⁻¹, Δx in m. The only physical input is p = ħk.
The unique minimum. Any other envelope, and any chirped Gaussian, sits strictly above ħ/2.
A linear phase e(ik₀x) only moves ⟨k⟩ and still saturates; a chirp e(iβx²) raises Δk while Δx does not move at all.
Converts the inequality into an energy: why a bound electron cannot stop, and why atoms have a size.
Holds in any bound state, where ⟨p⟩ = 0. An electron with Δx = 0.10 nm must carry ⟨T⟩ ≥ 0.95 eV.
The time-axis face of the same theorem: a short pulse is spectrally broad, with no quantum content needed.
Δt the pulse duration in s, Δν the spectral FWHM in Hz. A 12 fs pulse at 800 nm needs 78 nm of spectrum.
The energy-time statement that is actually true, and it is a lineshape result rather than an operator bound.
τ = 16.2 ns for sodium 3p → Γ = 4.06 × 10⁻⁸ eV, Δν = 9.8 MHz. Γ is a FWHM, not an rms width.
Two widths, and what counts as a width
Both widths are standard deviations, not eyeball extents: Δx² = ⟨x²⟩ − ⟨x⟩² with |ψ(x)|² as the weight, and Δk² = ⟨k²⟩ − ⟨k⟩² with |φ(k)|². For a Gaussian the full width at half maximum is 2.355 times the rms width, so quoting FWHM for both factors and calling the result Δx Δk overstates it by 2.355², a factor of five and a half. The definition also has teeth, because it requires the second moments to exist. Chop a plane wave into a top hat of length L and Δx = L/√12 = 0.289L looks tidy, but the transform is a sinc function whose |φ|² tails fall only as 1/k², so ∫k²|φ|²dk diverges and Δk is infinite. The theorem is satisfied, uselessly. An rms width punishes tails and sharp edges far more harshly than the eye does, which is exactly why it is the width the theorem is about.
The product belongs to the shape, not the size
Replace ψ(x) by √a ψ(ax), which keeps it normalised. Every length in the packet shrinks by a, so Δx → Δx/a, while the transform stretches, so Δk → aΔk. The product is untouched. Δx Δk is therefore a dimensionless number attached to the shape of the packet and to nothing else: scale is free, shape is not. Minimising it over all shapes is a variational problem, and Cauchy–Schwarz settles it in three lines. Shift so that ⟨x⟩ = ⟨k⟩ = 0, then Δx²Δk² = ∫x²|ψ|²dx · ∫|ψ′|²dx ≥ |∫xψ*ψ′dx|². Integrating by parts gives ∫x(|ψ|²)′dx = −1, so the real part of that last integral is exactly −½ and the right-hand side is at least ¼. Equality in Cauchy–Schwarz needs ψ′ = −λxψ, whose solution is the Gaussian. So the floor exists, it equals ½, and exactly one family of shapes reaches it.
Reading the floor as an energy
In any bound state ⟨p⟩ = 0, so ⟨p²⟩ = (Δp)² and the mean kinetic energy is at least ħ²/(8mΔx²). Confine an electron to Δx = 0.10 nm and it must carry at least 0.95 eV — atomic in scale, from one inequality and no Coulomb physics whatsoever. Push the argument at hydrogen: call the size r, take ⟨p²⟩ ≈ ħ²/r² as an order-of-magnitude reading of the bound, and minimise E(r) = ħ²/2mr² − ke²/r. The turning point is r = ħ²/mke² = 0.0529 nm, the Bohr radius, with E = −13.6 eV, exactly right. The atom does not collapse because shrinking r earns −1/r in potential energy but pays +1/r² in kinetic. Run it on a nucleus and it excludes things: an electron held inside Δx = 5 fm needs pc ≥ ħc/(2Δx) = 197.3/10 = 19.7 MeV, nearly forty times its rest energy, so it would emerge ultrarelativistic at about 20 MeV. Beta electrons arrive with roughly 1 MeV, so they were never sitting in the nucleus.
What the two spreads are spreads of
Δx and Δp describe a preparation, not an apparatus. Make N copies of one state, measure position on half of them and momentum on the other half, and histogram each set. Δx is the standard deviation of the first histogram, Δp of the second, and the theorem says their product cannot fall below ħ/2. Neither number can be read off one system at all: a single electron yields a single position, and a standard deviation needs the whole batch. Nor does either cap what a detector can resolve — locating one electron to 0.01 nm inside a prepared spread of 1.0 nm is no stranger than measuring one adult's height to the millimetre when the population's spread is 70 mm. The single slit is the standard demonstration precisely because it is a preparation and not a measurement: the slit imposes a transverse position spread, and the far-field diffraction pattern is the momentum histogram that spread demands.
Energy and time is a different statement
ΔE Δt ≥ ħ/2 looks like the same theorem with new letters, and it is not, because t labels the evolution rather than being an operator with a spread. Two honest statements replace it. First, Fourier on the time axis: a wave train lasting Δt cannot have a frequency sharper than Δω ≈ 1/(2Δt), so a pulse built from quanta of energy ħω obeys ΔE Δt ≥ ħ/2 with Δt the duration of the pulse. Second, a state decaying exponentially with lifetime τ emits a Lorentzian line of full width at half maximum Γ = ħ/τ, which for sodium's 16.2 ns 3p state is 4.06 × 10⁻⁸ eV, or 9.8 MHz beside an optical frequency of 5.09 × 10¹⁴ Hz. Note the FWHM: a Lorentzian's second moment diverges, exactly as the top hat's did, so the rms bound has nothing to say about it. And the line is quiet but hard to see, because thermal motion in a warm vapour cell buries it under a Doppler width hundreds of times larger — which is why the natural width needs cold atoms or a Doppler-free technique.
A packet touches the floor once, then leaves it
A Gaussian is the only shape at the floor, and a packet does not stay there. Start a free electron as a minimum-uncertainty Gaussian of width σ₀. Free evolution leaves Δp untouched, since no force acts and the momentum distribution is frozen, while the position width grows as Δx(t) = σ₀√(1 + (ħt/2mσ₀²)²). The product therefore climbs: Δx Δp = (ħ/2)√(1 + (ħt/2mσ₀²)²), equal to ħ/2 only at t = 0. For σ₀ = 1.0 nm the width doubles in t = 2√3 mσ₀²/ħ ≈ 30 fs, at which moment the product is ħ, twice the floor. What free evolution has added is a quadratic phase — a chirp. The density |ψ|² stays Gaussian throughout, only wider, and Δp never moves; what the chirp does is make the packet broader in k than its own width alone would require. Strip that chirp off at time t and Δk would fall back to 1/(2Δx(t)), putting the product on the floor again. A linear phase e(ik₀x) costs nothing by contrast, since it only shifts ⟨k⟩, so a moving Gaussian saturates exactly as a stationary one does. It is the imaginary quadratic term that costs, and free evolution grows one out of nothing.
Change one variable at a time
Make the relationship visible.
Halve Δx from 2.0 to 1.0 nm and watch the k-peak double in width while the product stays pinned at 0.500 ħ. Then raise r, the quadratic phase written across the packet: the left peak is untouched, the right one spreads, and the product climbs off the floor.
Δx2.00 nm
Δk0.25 rad/nm
Δx Δp (in ħ)0.500 ħ
⟨T⟩ (electron)2.4 meV
Live interpretationΔx: 2.00 nm. Δk: 0.25 rad/nm. Δx Δp (in ħ): 0.500 ħ. ⟨T⟩ (electron): 2.4 meV
Catch the common trap
Explain before calculating.
A source emits electrons all prepared in the same state, whose position spread is Δx = 1.0 nm. A detector then locates one of those electrons to within 0.01 nm. Which statement is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA Gaussian electron wave packet is prepared with position spread Δx = 0.50 nm and mean momentum zero. Find the smallest possible momentum spread, the corresponding velocity spread, and the mean kinetic energy the packet forces on the electron. Take ħ = 1.055 × 10⁻³⁴ J s and mₑ = 9.11 × 10⁻³¹ kg.
- The packet is an unchirped Gaussian, so it sits exactly on the floor: Δx Δp = ħ/2.
- Δp = ħ/(2Δx) = 1.055 × 10⁻³⁴ ÷ (2 × 0.50 × 10⁻⁹) = 1.055 × 10⁻²⁵ kg m s⁻¹.
- Δv = Δp/mₑ = 1.055 × 10⁻²⁵ ÷ 9.11 × 10⁻³¹ = 1.16 × 10⁵ m s⁻¹, about 116 km s⁻¹.
- Because ⟨p⟩ = 0, ⟨p²⟩ = (Δp)², so ⟨T⟩ = (Δp)²/2mₑ = (1.055 × 10⁻²⁵)² ÷ (2 × 9.11 × 10⁻³¹) = 6.10 × 10⁻²¹ J = 38 meV.
- Compare with kT at 300 K, 25.9 meV: confining an electron to half a nanometre already costs about one and a half times room-temperature thermal energy.
AnswerΔp = 1.06 × 10⁻²⁵ kg m s⁻¹, Δv ≈ 116 km s⁻¹, and ⟨T⟩ ≈ 38 meV — roughly 1.5 kT at 300 K.
MediumA mode-locked Ti:sapphire laser centred at 800 nm emits Gaussian pulses of 12 fs FWHM duration. Find the minimum spectral width the pulse must carry, in Hz and in nm, then find the shortest pulse that could survive a 10 nm-wide bandpass filter placed in the beam.
- For Gaussian envelopes the time-bandwidth product in FWHM measure is Δν Δt ≥ 0.441, with equality for a transform-limited pulse.
- Δν = 0.441 ÷ (12 × 10⁻¹⁵ s) = 3.675 × 10¹³ Hz.
- Convert to wavelength: Δλ = λ²Δν/c = (800 × 10⁻⁹)² × 3.675 × 10¹³ ÷ 2.998 × 10⁸ = 7.85 × 10⁻⁸ m = 78 nm.
- That is 9.8% of the centre wavelength — a 12 fs pulse is not monochromatic in any useful sense, and every optic it passes must be broadband.
- Run the relation backwards for the filter: Δν = cΔλ/λ² = 2.998 × 10⁸ × 10 × 10⁻⁹ ÷ (800 × 10⁻⁹)² = 4.68 × 10¹² Hz, so Δt ≥ 0.441 ÷ 4.68 × 10¹² = 9.4 × 10⁻¹⁴ s.
AnswerΔν ≈ 3.7 × 10¹³ Hz and Δλ ≈ 78 nm. A 10 nm filter caps the bandwidth at 4.7 × 10¹² Hz and so stretches the pulse to at least 94 fs, eight times longer.
HardThe sodium 3p state decays to 3s with lifetime τ = 16.2 ns, emitting at 589.0 nm. Find (a) the natural linewidth as an energy and as a frequency FWHM, (b) the corresponding wavelength spread, and (c) how it compares with Doppler broadening in a vapour cell at 500 K. Take m(Na) = 3.82 × 10⁻²⁶ kg and k = 1.381 × 10⁻²³ J K⁻¹.
- Exponential decay of the amplitude gives a Lorentzian line of FWHM Γ = ħ/τ = 1.055 × 10⁻³⁴ ÷ 1.62 × 10⁻⁸ = 6.51 × 10⁻²⁷ J = 4.06 × 10⁻⁸ eV.
- In frequency, Δν = Γ/h = 1/(2πτ) = 1 ÷ (2π × 16.2 × 10⁻⁹) = 9.82 × 10⁶ Hz, about 10 MHz.
- Wavelength spread: Δλ = λ²Δν/c = (589.0 × 10⁻⁹)² × 9.82 × 10⁶ ÷ 2.998 × 10⁸ = 1.14 × 10⁻¹⁴ m, roughly 11 fm — a relative width of 1.9 × 10⁻⁸.
- Doppler FWHM is ΔνD = ν₀√(8kT ln2 / mc²), with ν₀ = c/λ = 5.09 × 10¹⁴ Hz. The bracket is 8 × 1.381 × 10⁻²³ × 500 × 0.6931 ÷ (3.82 × 10⁻²⁶ × 8.988 × 10¹⁶) = 1.115 × 10⁻¹¹, whose square root is 3.34 × 10⁻⁶.
- So ΔνD = 5.09 × 10¹⁴ × 3.34 × 10⁻⁶ = 1.70 × 10⁹ Hz, which is 1.70 × 10⁹ ÷ 9.82 × 10⁶ = 173 times the natural width.
- The natural width is a floor set by the lifetime and cannot be reduced; the Doppler width can, by cooling the atoms, using a transverse beam, or using Doppler-free saturated-absorption spectroscopy — which is how the 10 MHz figure is actually measured.
AnswerΓ = 4.06 × 10⁻⁸ eV, Δν ≈ 9.8 MHz, Δλ ≈ 11 fm. At 500 K the Doppler width is 1.70 GHz, 173 times larger, so the natural width is invisible in a hot cell.