University Physics V · Foundations of Quantum Mechanics · 2.5
λ = h/p and momentum eigenfunctions
One relation, λ = h/p, takes you from an accelerating voltage to the spacing a beam will diffract from; behind it sits an operator statement, p̂ e^{ikx} = ħk e^{ikx}. Learn to use the first on the bench and to handle the second without ever calling the plane wave a state.
Build the model
Connect the measurement to the mechanism.
De Broglie's step was not to decorate particles with waves but to notice what relativity already permits. The phase of a plane wave, φ = k⋅x − ωt, counts wavefront crossings at an event, and every observer agrees on a count, so (ω/c, k) must transform as a four-vector; a free particle carries exactly one four-vector, pμ, and E = ħω fixes the single constant, giving kμ = pμ/ħ and with it p = ħk, λ = h/p. Read as linear algebra, the same relation says e^{ikx} is an eigenfunction of p̂ = −iħ d/dx with eigenvalue ħk: the matter wavelength is the spectrum of an operator.
The cost is that this eigenfunction is no vector of the state space — |e^{ikx}|² = 1 everywhere, so it is not square-integrable, and no choice of amplitude repairs that. Quantum mechanics keeps it anyway, as an improper basis element normalised to a delta function, ⟨p|p′⟩ = δ(p − p′), so that completeness reads ∫dp |p⟩⟨p| = 1 and every genuine state is a packet ∫dp φ(p)|p⟩. Sharp momentum is therefore never occupied: it is the idealised edge of a basis, and the theory prices it accordingly.
- Simple definition
- λ = h/p assigns every particle of momentum p a wavelength, and in operator language it says e^{ikx} is the eigenfunction of p̂ = −iħ d/dx with eigenvalue ħk = p.
- Example
- An electron accelerated through 54 V has p = √(2mₑ eV) = 3.97×10⁻²⁴ kg m s⁻¹, so λ = h/p = 0.167 nm — comparable to the atomic spacing in a nickel crystal, which is why Davisson and Germer saw diffraction peaks.
One line converts a beam's momentum into the length scale on which it can interfere.
h = 6.626×10⁻³⁴ J s, ħ = h/2π; p in kg m s⁻¹ gives λ in m and k in rad m⁻¹
Grant E = ħω and relativity forces λ = h/p — both relations share the single constant ħ.
φ = k⋅x − ωt is a frame-independent crest count, so (ω/c, k) must ride parallel to (E/c, p)
The matter wavelength becomes an operator statement: measured momenta are the spectrum of p̂.
p̂ = −iħ d/dx generates translations; eigenvalue ħk in kg m s⁻¹, one for every real k
The continuum's replacement for ⟨m|n⟩ = δₘₙ: orthogonality survives, normalisation to 1 does not.
δ(p − p′) carries units 1/[p]; the 1/√(2πħ) makes ∫dp |p⟩⟨p| = 1 come out exact
The bench form: 54 V gives 0.167 nm and 10 kV gives 12.26 pm, straight from the supply dial.
non-relativistic; mₑ = 9.109×10⁻³¹ kg, e = 1.602×10⁻¹⁹ C, V in volts
Use relativistic p once eV stops being tiny beside mₑ c²; the shift grows linearly with V.
2mₑ c² = 1.022 MeV, so the bracket is 1 + V/(1.022×10⁶ V): −0.49% in λ at 10 kV
Counting wavefronts is frame-independent
Attach a phase φ = k⋅x − ωt to a free particle and ask what relativity permits. The phase at an event counts how many crests have swept past it, and a count of discrete events is something all observers agree on: φ must be a Lorentz scalar. Since φ is the contraction of xμ = (ct, x) with the coefficients (ω/c, k) — up to the overall sign the metric signature fixes — those coefficients must themselves form a four-vector. A free particle carries exactly one four-vector, pμ = (E/c, p), so kμ can only be proportional to it, and the Planck–Einstein relation E = ħω — whose constant h the stopping-potential slope of topic 2.3 measured — fixes the constant: kμ = pμ/ħ. The space part is p = ħk, i.e. λ = h/p. The same algebra prices the wave's two speeds: the phase velocity ω/k = E/p exceeds c for a massive particle, but the group velocity dω/dk = dE/dp = v is the particle's own — the crests race ahead carrying nothing while the packet travels with the electron.
Put numbers on the relation
For electrons the bench form is λ = h/√(2mₑ eV) = 1.226 nm/√V. At the 54 V of Davisson and Germer, λ = 1.226/√54 = 0.167 nm — the same few-tenths-of-a-nanometre scale as a crystal lattice, which is why a crystal is the natural diffraction grating and a ruled one is hopeless. At 10 kV, λ = 12.26 pm non-relativistically; the correction of the last formula card pulls it to 12.20 pm. Scale the object up and the relation is just as obedient and far more brutal: a 1 μg dust grain drifting at 1 mm s⁻¹ has p = 10⁻¹² kg m s⁻¹ and λ = 6.6×10⁻²² m, seven orders of magnitude below a nuclear radius of a few femtometres, with no slit structure anywhere to match — nothing diffracts it, and the classical world sits at the small-λ face of the same formula. The practical skill: read the supply voltage, produce λ, and know before the experiment which spacings can Bragg-scatter the beam.
The operator behind the wavelength
Now translate the wave statement into the working language. Define p̂ = −iħ d/dx and act on e^{ikx}: the derivative brings down ik, the prefactor −iħ turns that into ħk, and the function returns unchanged — p̂ e^{ikx} = ħk e^{ikx}. That is an eigenvalue equation, and it is why λ = h/p is more than an analogy: measured momenta are eigenvalues of p̂, and the plane wave of wavelength 2π/k is the eigenfunction belonging to ħk. The operator is not guessed. Applying e(−ip̂a/ħ) to ψ(x) reproduces the Taylor series of ψ(x − a), so p̂ is the generator of translations, exactly the role p plays in classical mechanics through the Poisson bracket. Hermiticity comes from integration by parts: ⟨φ|p̂ψ⟩ − ⟨p̂φ|ψ⟩ equals a boundary term −iħ[φ*ψ] that vanishes for normalisable states — a first sight of the domain questions Unit 3 makes precise. One convention to fix: with the time factor e(−iωt), positive k means motion towards +x.
An eigenfunction the state space refuses
Try to admit e^{ikx} as a state and it fails at the first axiom: |e^{ikx}|² = 1 at every x, so ⟨ψ|ψ⟩ = ∫|ψ|² dx diverges, and scaling by an amplitude A only rescales the divergence by |A|². The momentum operator has a perfectly real spectrum yet not one eigenvector in L²(R). The repair is a change of normalisation, not of function. Write ψₚ(x) = e(ipx/ħ)/√(2πħ) and compute the overlap of two of them: ⟨p|p′⟩ = (1/2πħ)∫e(i(p′−p)x/ħ) dx = δ(p − p′), the Fourier representation of the Dirac δ. Orthogonality survives — distinct momenta have zero overlap — but the 'length' of each basis element is a delta spike rather than 1, and the units say so: ψₚ carries (J s)(−1/2), so ⟨p|p′⟩ carries 1/[p], as a density in p must. These improper kets live one rung outside L², in a rigged Hilbert space, and are kept because they make the completeness relation ∫dp |p⟩⟨p| = 1 an exact operator identity.
Physical states are packets over the basis
Insert that identity in front of any state: |ψ⟩ = ∫dp φ(p)|p⟩ with φ(p) = ⟨p|ψ⟩ — the momentum representation, concretely the Fourier transform of ψ(x). Born's rule then reads |φ(p)|² as a probability density: P(p₁ < p < p₂) = ∫|φ(p)|² dp, and the probability of any exact value is an integral over a single point, which is zero. Sharp momentum is never occupied. A beam quoted as 'momentum p₀' is shorthand for a packet whose φ(p) is sharply peaked: a Gaussian of width σₓ = 1 nm around a 54 eV electron has σₚ/p₀ = ħ/(2σₓ p₀) ≈ 1.3% — monochromatic by any experimental standard and still honestly in L². The narrower φ(p), the wider the packet in x; squeeze Δp to zero and the state stretches into the infinite plane wave you were forbidden to occupy. So read λ = h/p as a statement about where φ(p) is centred, not about a wavelength the particle 'has'.
On a grid, the plane wave becomes honest
Numerics never meets a delta function, and the reason is instructive. Put the particle on a ring of circumference L — periodic boundary conditions, the natural home of numpy.fft. The allowed wavenumbers are kₙ = 2πn/L, and e^{ikₙ x}/√L is now genuinely normalised: ⟨kₘ|kₙ⟩ = δₘₙ, Kronecker rather than Dirac. On an N-point grid the momentum operator is a matrix the discrete Fourier transform diagonalises exactly, so numpy.fft.fft(ψ) is literally the change to p̂'s eigenbasis, with eigenvalues ħkₙ up to the Nyquist momentum πħN/L. As L grows, the ladder of allowed momenta packs together with density L/2πħ states per unit momentum, sums over n become (L/2πħ)∫dp, and box normalisation goes over smoothly to delta normalisation. That limit is the honest content of ⟨p|p′⟩ = δ(p − p′): it is what Kronecker orthonormality becomes when the box outgrows every measurement, and every simulation in this course tacitly runs at finite L on the safe side of it.
Change one variable at a time
Make the relationship visible.
Raise V: the carrier shortens as 1/√V while the envelope stands still. Then push σ towards 1 nm and watch σₚ/p fall as the packet flattens towards e(ikx) — only at an unreachable σ = ∞ would it become the δ-normalised eigenfunction itself.
WAVELENGTH λ = h/p0.167 nm
MOMENTUM p = ħk3.97 10⁻²⁴ kg m s⁻¹
SPREAD σₚ/p2.7 %
CARRIER CYCLES PER σ3.0 wavelengths
Live interpretationWAVELENGTH λ = h/p: 0.167 nm. MOMENTUM p = ħk: 3.97 10⁻²⁴ kg m s⁻¹. SPREAD σₚ/p: 2.7 %. CARRIER CYCLES PER σ: 3.0 wavelengths
Catch the common trap
Explain before calculating.
An electron beam is prepared with mean momentum ħk₀, and the plane wave ψ(x) = e(ik₀x) is written down to describe it. Which statement survives scrutiny?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyElectrons are accelerated from rest through 54 V, the Davisson–Germer setting. Find the momentum p, the de Broglie wavelength λ, and the wavenumber k, and verify that e^{ikx} returns ħk under −iħ d/dx.
- Energy from the supply: E = eV = 54 × 1.602×10⁻¹⁹ J = 8.65×10⁻¹⁸ J — only about 10⁻⁴ of mₑ c² = 511 keV, so the non-relativistic p = √(2mₑ E) is safe.
- p = √(2 × 9.109×10⁻³¹ × 8.65×10⁻¹⁸) = √(1.576×10⁻⁴⁷) = 3.97×10⁻²⁴ kg m s⁻¹.
- λ = h/p = 6.626×10⁻³⁴ / 3.97×10⁻²⁴ = 1.67×10⁻¹⁰ m = 0.167 nm, matching the shortcut 1.226 nm/√54.
- k = 2π/λ = 3.76×10¹⁰ m⁻¹, and −iħ (d/dx) e^{ikx} = (−iħ)(ik) e^{ikx} = ħk e^{ikx}, with ħk = 1.055×10⁻³⁴ × 3.76×10¹⁰ = 3.97×10⁻²⁴ kg m s⁻¹ — the eigenvalue reproduces p.
Answerp = 3.97×10⁻²⁴ kg m s⁻¹, λ = 0.167 nm, k = 3.76×10¹⁰ m⁻¹; the plane wave is an eigenfunction of p̂ with eigenvalue ħk = p.
MediumAn electron column runs at 10.0 kV. Find the non-relativistic de Broglie wavelength, apply the relativistic correction, and state the fractional shift — is the unit's claim of a half-percent correction at 10 kV right?
- Non-relativistic: λₙᵣ = 1.2264 nm/√10000 = 0.012264 nm = 12.264 pm — the 1.226 nm/√V card kept to one more figure.
- Relativistic momentum from (K + mec²)² = (pc)² + (mec²)²: pc = √(K² + 2K mec²) = √(100 + 2×10×511) keV = √(10320 keV²) = 101.59 keV, against the non-relativistic pc = √(2K mec²) = √10220 = 101.09 keV.
- λᵣₑₗ = hc/pc = 1239.84 eV nm / 101 587 eV = 0.0122047 nm = 12.2047 pm, against λₙᵣ = 1239.84 eV nm / 101 094 eV = 12.2642 pm. Equivalently λᵣₑₗ = λₙᵣ/√(1 + K/2mec²) = 12.2642 pm/√1.00978 = 12.2047 pm.
- Fractional shift: (12.2642 − 12.2047)/12.2642 = 0.0595/12.2642 = 0.49%, the same as 1 − 1/√1.00978 = 0.49%. The field delivered 0.49% more momentum than the non-relativistic estimate, so the wavelength is 0.49% shorter.
Answerλₙᵣ = 12.26 pm, λᵣₑₗ = 12.20 pm, a shift of ≈ 0.49% — relativity is a genuine half-percent correction at 10 kV.
HardA 54 eV electron beam is really a minimum-uncertainty Gaussian packet with position spread σₓ = 1.0 nm. Find the momentum spread σₚ and the ratio σₚ/p₀, the probability that a momentum measurement lands within ±σₚ of ħk₀, and the probability of obtaining exactly ħk₀.
- The Fourier transform of a Gaussian is a Gaussian, so φ(p) = ⟨p|ψ⟩ is centred on ħk₀; minimum uncertainty means σₓ σₚ = ħ/2, so σₚ = ħ/(2σₓ) = 1.055×10⁻³⁴/(2 × 1.0×10⁻⁹) = 5.27×10⁻²⁶ kg m s⁻¹.
- With p₀ = 3.97×10⁻²⁴ kg m s⁻¹ from the 54 eV beam, σₚ/p₀ = 5.27×10⁻²⁶/3.97×10⁻²⁴ = 1.3% — the envelope spans σₓ/λ ≈ 1.0/0.167 ≈ 6 carrier wavelengths: quasi-monochromatic yet honestly normalisable.
- |φ(p)|² is a Gaussian density of standard deviation σₚ, so P(|p − ħk₀| ≤ σₚ) is the one-σ Gaussian mass, 68.3%.
- P(p = ħk₀ exactly) is a finite density integrated over a single point: zero. Sharp momentum belongs only to the improper ket |p₀⟩, δ-normalised and outside L² — no preparation puts a particle in it.
Answerσₚ = 5.27×10⁻²⁶ kg m s⁻¹, σₚ/p₀ ≈ 1.3%; P(within ±σₚ) = 68.3%; P(exactly ħk₀) = 0.