Momentum eigenstates and delta normalisation
Every momentum an eigenvalue; probability lives only in bands.
Every real momentum is an eigenvalue of p̂, and not one of its eigenfunctions can be normalised. This lesson keeps them anyway: normalise to a Dirac delta instead of to one, confirm it with a periodic box, keep ħ in the right place when the label changes from k to p, and use plane waves only inside an integral over a weight φ(k).
Build the model
A sharp momentum is a genuine eigenvalue whose eigenfunction no particle can occupy.
1The big idea
The momentum operator p̂ = −iħ d/dx has an eigenfunction for every real p. Solving −iħ f′ = p f gives f = e(ipx/ħ) = e(ikx) with p = ħk, and asking f to stay bounded at both ends rules out complex k. Each of these has |f|² constant, so none can be normalised, and yet none can be thrown away, because every physical ψ is built from them. The repair keeps orthogonality and changes what normalisation means. Integrate two plane waves over a window |x| ≤ X and the overlap is 2 sin((k′ − k)X)/(k′ − k): bounded when k′ ≠ k, equal to 2X when k′ = k, and enclosing area 2π whatever X is. As X grows this becomes 2π δ(k′ − k), so fₖ = e(ikx)/√(2π) obeys ∫ fₖ* fₖ′ dx = δ(k − k′), the continuum version of ∫ψₘ*ψₙ dx = δₘₙ. A periodic box of length L reaches the same place by another road: k is restricted to 2πn/L, the functions e(ikx)/√L are orthonormal in the ordinary sense, and letting L grow turns the sum over n into (L/2π)∫dk. Either way what is physical is an integral: a weight φ(k) with ∫|φ|² dk = 1 builds a normalisable ψ, and |φ(k)|² dk is a probability. The machinery fails the moment a single fₖ is treated as a state. Its |fₖ|² is not a probability density, its self-overlap is δ(0), and any answer that depends on L or contains δ(0) has used a plane wave where only a packet is allowed.
2In one sentence
A momentum eigenfunction is fₖ(x) = e(ikx)/√(2π), on which p̂ = −iħ d/dx returns exactly ħk; since it cannot be normalised to 1, it is normalised to a Dirac delta instead: ∫ fₖ*(x) fₖ′(x) dx = δ(k − k′).
Example. Over a window |x| ≤ X, two functions fₖ whose wavenumbers differ by 1.0 × 10⁹ m⁻¹ overlap by at most 1/(π × 10⁹ m⁻¹) = 3.2 × 10⁻¹⁰ m however long the window is, while fₖ with itself gives X/π = 3.2 × 10⁻⁴ m at X = 1.0 mm and grows without limit.
3Every real momentum is an eigenvalue
The momentum eigenvalue equation is first order: −iħ df/dx = p f. Separating and integrating gives f(x) = A e(ipx/ħ) = A e(ikx) with p = ħk, and no boundary condition has been used yet. The only restriction comes from asking f to stay bounded. For a complex k = a + ib, |f| = |A| e(−bx), which explodes as x → −∞ if b is positive and as x → +∞ if b is negative, so k must be real. Every real k survives, so p̂ has the whole real line as its spectrum and one eigenfunction per value. An electron with k = 1.0 × 10¹⁰ m⁻¹ has p = 1.05 × 10⁻²⁴ kg m s⁻¹, wavelength 2π/k = 0.628 nm and classical speed p/m = 1.16 × 10⁶ m s⁻¹. The sign of k is exactly what p̂ reads: cos(kx) solves the free Schrödinger equation at the same energy, but p̂ cos(kx) = iħk sin(kx), so it is not a momentum eigenfunction. The price of a sharp momentum is the modulus: |f|² = |A|² everywhere, so ∫|f|² dx diverges for every A ≠ 0.
4Plane waves are orthogonal, and their self-overlap is a delta
Normalisation to one fails, but overlaps can still be computed if the integral is first taken over a window |x| ≤ X. With q = k′ − k, the overlap of e(ikx) with e(ik′x) is ∫ e(iqx) dx = 2 sin(qX)/q. At q = 0 it equals 2X, the window's length, and grows without limit. For q ≠ 0 it never exceeds 2/|q|, however large X becomes. Its first zeros sit at q = ±π/X, and its area over all q is exactly 2π for every X. A curve that grows taller and narrower while its area stays fixed is the defining behaviour of a Dirac delta. As X → ∞ the overlap therefore becomes 2π δ(k′ − k), and dividing each plane wave by √(2π) leaves δ(k − k′) alone. At X = 1.0 μm the kernel sin(qX)/(πq) already peaks at X/π = 3.2 × 10⁻⁷ m, with its first zeros at ±3.1 × 10⁶ m⁻¹. One subtlety matters later: at a fixed q ≠ 0 the overlap does not decay as X grows. As a function of q it oscillates ever faster, so its integral against a smooth function of k over any range that excludes q = 0 goes to zero, and only the peak at q = 0 survives.
5The units say which delta you are using
A Dirac delta carries the inverse unit of its argument, because ∫δ(k) dk = 1 is a pure number. So δ(k − k′) is in metres, and the dimensionless fₖ = e(ikx)/√(2π) fits it: its overlap integral over x is in metres too. Label the same functions by p = ħk and the scaling rule δ(u/a) = |a| δ(u) gives δ(k − k′) = δ((p − p′)/ħ) = ħ δ(p − p′). To be normalised to δ(p − p′), which is in (kg m s⁻¹)⁻¹, each function must absorb a factor 1/√ħ: fₚ(x) = e(ipx/ħ)/√(2πħ). Check it: |fₚ|² is in (J s)⁻¹, and integrating over x in metres gives m/(kg m² s⁻¹) = (kg m s⁻¹)⁻¹, as required. Mixing the two conventions is not a small slip. Pairing e(ipx/ħ)/√(2π) with δ(p − p′) mis-states every momentum-space probability by a factor ħ = 1.055 × 10⁻³⁴ J s, and a dimension check exposes it in one line. Pick one label, k or p, and keep its prefactor with it through the whole calculation.
6A periodic box makes the continuum countable
The second repair keeps ordinary normalisation by giving space a finite length. Impose f(x + L) = f(x), as on a ring of circumference L. Then e(ikL) = 1, so k is restricted to kₙ = 2πn/L, and fₙ = e(ikₙx)/√L is normalised to 1 over one period and orthogonal to every other fₙ, because e(i(kₘ − kₙ)x) winds through a whole number of turns. Each allowed k occupies a slot 2π/L wide, so a band dk holds L dk/2π states and sums become integrals: Σₙ g(kₙ) → (L/2π) ∫ g(k) dk. Matching normalisations gives the dictionary fₙ = √(2π/L) fₖ at k = kₙ, box coefficients cₙ → √(2π/L) φ(kₙ), and (L/2π) δₘₙ → δ(k − k′). On a ring of L = 1.0 mm the momentum steps are h/L = 6.6 × 10⁻³¹ kg m s⁻¹, so an electron with p = 1.0 × 10⁻²⁴ kg m s⁻¹ sits near n = 1.5 × 10⁶. The test of a box calculation is that its physical answers stop depending on L once L far exceeds the region the particle explores. The box is a ring, not a well: in a hard-walled well sin(nπx/a) mixes +k with −k and has no sharp momentum.
7Only weighted integrals carry probability
No experiment prepares an fₖ, so what delta normalisation buys is control of superpositions. A state is a weight: ψ(x) = ∫ φ(k) fₖ(x) dk with ∫|φ(k)|² dk = 1. The weight is recovered as φ(k) = ∫ fₖ*(x) ψ(x) dx, because substituting ψ lets δ(k − k′) collapse the k′ integral, and the same substitution gives ∫|ψ|² dx = ∫|φ|² dk, so a normalised weight always builds a normalised ψ. Probabilities come in bands: the chance that ħk lies between ħk₁ and ħk₂ is ∫|φ|² dk over that interval, and a single exact k has probability zero. The mean momentum ∫ ħk |φ(k)|² dk is finite, and for a weight narrow about k₀ it equals ħk₀ = mv, the classical momentum. A plane wave can still answer ratio questions: its probability current divided by its density is ħk/m, the classical velocity, whatever the normalisation. But any result meant to be a probability that comes out containing δ(0), or depending on a box length L, has used a plane wave where only a packet belongs.
Key equations
What each relation says, and the units it is written in.
Momentum eigenvalue equation
p̂ e(ikx) = −iħ d/dx e(ikx) = ħk e(ikx)
Every real k works and no boundary condition picks among them, so the spectrum of p̂ is the whole real line. A complex k would make |e(ikx)| grow exponentially toward one end.
ħ = 1.055 × 10⁻³⁴ J s; k in m⁻¹ and real; the eigenvalue p = ħk is in kg m s⁻¹
Overlap over a finite window
∫ over |x| ≤ X of e(−ikx) e(ik′x) dx = 2 sin((k′ − k)X)/(k′ − k)
Its area over k′ is 2π for every X, so as X → ∞ it becomes 2π δ(k′ − k). At a fixed k′ ≠ k it stays below 2/|k′ − k| but never decays pointwise.
X in m, k and k′ in m⁻¹, result in m; peak 2X at k′ = k, first zeros at |k′ − k| = π/X
Delta normalisation in k
∫ fₖ*(x) fₖ′(x) dx = δ(k − k′), with fₖ(x) = e(ikx)/√(2π)
The continuum form of ∫ψₘ*ψₙ dx = δₘₙ: zero overlap for k ≠ k′, and an infinite self-overlap δ(0) as the price of a perfectly sharp k.
fₖ is dimensionless; δ(k − k′) is in m, the inverse unit of k; the integral runs over the whole line
The same statement labelled by p
∫ fₚ*(x) fₚ′(x) dx = δ(p − p′), with fₚ(x) = e(ipx/ħ)/√(2πħ)
Since δ(k − k′) = δ((p − p′)/ħ) = ħ δ(p − p′), each function must absorb 1/√ħ. Keep 1/√(2π) in a p calculation and every momentum probability is off by a factor ħ.
fₚ in (J s)⁻¹⁄²; δ(p − p′) in (kg m s⁻¹)⁻¹; fₚ = fₖ/√ħ with p = ħk
Periodic box of length L
fₙ(x) = e(ikₙx)/√L, with kₙ = 2πn/L and n = 0, ±1, ±2, …
Ordinary orthonormality over one period, ∫₀ᴸ fₘ* fₙ dx = δₘₙ. Nothing is infinite any more; the cost is a length L that no physical answer may depend on.
L in m, fₙ in m⁻¹⁄²; allowed k are spaced by 2π/L, allowed momenta by h/L
Weight of a normalisable state
φ(k) = ∫ fₖ*(x) ψ(x) dx
Substitute ψ = ∫ φ(k′) fₖ′ dk′ and δ(k − k′) collapses the k′ integral. Probability lives in bands of k; a single exact k carries none.
ψ in m⁻¹⁄², so φ in m¹⁄²; |φ(k)|² dk is the probability that the momentum lies between ħk and ħ(k + dk)
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At the default gap of 0.5 nm⁻¹, step X from 4 to 9 nm: the dot swings from +0.58 to −0.62 nm inside the fixed ±0.64 nm bound and never settles, while the peak climbs from 1.27 to 2.86 nm and narrows. The area stays 1, so the limit is δ(k′ − k), but only inside an integral over k.
Live results
- PEAK HEIGHT X/π
- 1.27 nm
- FIRST ZERO π/X
- 0.785 nm⁻¹
- OVERLAP AT THE GAP
- 0.58 nm
- BOUND 1/(π × GAP)
- 0.64 nm
Catch the common trap
The mistake this lesson is built to prevent.
Delta normalisation does not turn a plane wave into a state. It is tempting to read fₖ = e(ikx)/√(2π) as a particle spread evenly along the line with probability density |fₖ|² = 1/(2π) = 0.159. That number has no unit, while a probability density on a line must carry m⁻¹, and its integral over the line is infinite rather than 1. The relation ∫ fₖ* fₖ′ dx = δ(k − k′) says nothing about a single fₖ on its own: at k′ = k it returns δ(0), which is infinite. It is a rule for how the fₖ behave inside an integral over k, and what it guarantees is that a packet ψ = ∫ φ(k) fₖ dk has ∫|ψ|² dx = ∫|φ|² dk. So the physical questions are about bands of momentum, never about where a plane-wave particle is or the probability of one exact k.
Test it: what changes?
- Peak height X/π grows without limit
- First zeros at ±π/X move inward
- Overlap at the gap oscillates, never settling
- Bound 1/(π × gap) stays fixed
- Prefactor becomes 1/√(2πħ), not 1/√(2π)
- δ(k − k′) becomes ħ δ(p − p′)
- Weight becomes Φ(p) = φ(k)/√ħ
- Band probabilities unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron in a long, nearly monochromatic beam is idealised by f(x) = e(ikx)/√(2π) with k = 4.0 × 10⁹ m⁻¹. Show that f is a momentum eigenfunction and find p, the de Broglie wavelength and the classical speed. Then apply p̂ to g(x) = cos(kx) and say which momentum values g is built from.
View solution →
- p̂f = −iħ d/dx [e(ikx)/√(2π)] = −iħ(ik) f = ħk f, since (−i)(i) = 1. The result is a constant times f, so f is an eigenfunction with p = ħk; the prefactor 1/√(2π) plays no part in the eigenvalue.
- p = ħk = (1.055 × 10⁻³⁴ J s)(4.0 × 10⁹ m⁻¹) = 4.2 × 10⁻²⁵ kg m s⁻¹. Dimensions: J s m⁻¹ = kg m² s⁻¹ m⁻¹ = kg m s⁻¹, a momentum.
- λ = 2π/k = 2π/(4.0 × 10⁹ m⁻¹) = 1.6 × 10⁻⁹ m = 1.6 nm, the same as h/p. Classically v = p/m = (4.22 × 10⁻²⁵)/(9.11 × 10⁻³¹) = 4.6 × 10⁵ m s⁻¹, only 0.15% of c, so p = mv needs no relativistic correction.
- p̂g = −iħ d/dx cos(kx) = iħk sin(kx). That is not a constant times cos(kx), so g is not a momentum eigenfunction, even though it solves the free Schrödinger equation with the same energy ħ²k²/2m = 0.61 eV as f.
- Writing cos(kx) = ½e(ikx) + ½e(−ikx) shows g is built from p = +ħk and p = −ħk with equal weights. Neither f nor g is a state, but a long packet with this cos(kx) profile gives momenta in narrow bands about +4.2 × 10⁻²⁵ and −4.2 × 10⁻²⁵ kg m s⁻¹ with equal probability, never near zero, although the mean is zero.
Answer. p = 4.2 × 10⁻²⁵ kg m s⁻¹, λ = 1.6 nm, v = 4.6 × 10⁵ m s⁻¹; cos(kx) is not a momentum eigenfunction and is built equally from +ħk and −ħk.
MediumExample 2
A conduction electron moves on a thin metal ring of circumference L = 1.5 μm, modelled as a periodic box. Find the allowed wavenumbers and the spacing of allowed momenta; the quantum number n whose momentum lies closest to 3.0 × 10⁻²⁵ kg m s⁻¹, with the electron's speed; the normalised eigenfunctions and their units; and how many allowed states lie in a wavenumber band 1.0 × 10⁸ m⁻¹ wide.
View solution →
- Periodicity f(x + L) = f(x) with f = e(ikx) demands e(ikL) = 1, so kₙ = 2πn/L with n = 0, ±1, ±2, … The spacing is 2π/L = 2π/(1.5 × 10⁻⁶ m) = 4.19 × 10⁶ m⁻¹.
- The momentum spacing is ħ × 2π/L = h/L = (6.626 × 10⁻³⁴ J s)/(1.5 × 10⁻⁶ m) = 4.42 × 10⁻²⁸ kg m s⁻¹.
- n = p/(h/L) = (3.0 × 10⁻²⁵)/(4.417 × 10⁻²⁸) = 679.1, so n = 679, with pₙ = 679 × 4.417 × 10⁻²⁸ = 3.00 × 10⁻²⁵ kg m s⁻¹ and speed pₙ/m = 3.29 × 10⁵ m s⁻¹. Neighbouring states differ by 1/679 = 0.15%, far finer than any momentum measurement resolves.
- Normalise over one circumference: ∫₀ᴸ |A|² dx = |A|² L = 1, so fₙ = e(ikₙx)/√L with 1/√L = 816 m⁻¹⁄². Its density |fₙ|² = 1/L = 6.7 × 10⁵ m⁻¹ carries m⁻¹, a genuine probability density, and different n are orthogonal because e(i(kₘ − kₙ)x) winds m − n whole turns round the ring.
- Each allowed k occupies 2π/L, so a band 1.0 × 10⁸ m⁻¹ wide holds (1.0 × 10⁸)(1.5 × 10⁻⁶)/2π = 23.9, about 24 states. Doubling L doubles the count: L/2π states per unit k is the factor that turns Σₙ into (L/2π) ∫ dk as L → ∞.
Answer. kₙ = 2πn/L, spaced by 4.19 × 10⁶ m⁻¹, with momentum steps of 4.42 × 10⁻²⁸ kg m s⁻¹; n = 679 and v = 3.29 × 10⁵ m s⁻¹; fₙ = e(ikₙx)/√L with 1/√L = 816 m⁻¹⁄²; about 24 states in the band.
HardExample 3
An electron's momentum-space weight is flat across a band: φ(k) = 1/√Δ for |k − k₀| ≤ Δ/2 and zero outside, with k₀ = 5.0 × 10⁹ m⁻¹ and Δ = 2.0 × 10⁸ m⁻¹. Using fₖ = e(ikx)/√(2π), build ψ(x) and check its norm; find the probability that p lies within ħΔ/4 of ħk₀; rewrite the weight in the p convention and check its norm; and repeat the norm check in a periodic box of L = 10 μm.
View solution →
- ψ(x) = ∫ φ(k) fₖ(x) dk = (1/√(2πΔ)) ∫ e(ikx) dk over the band = (1/√(2πΔ)) e(ik₀x) × 2 sin(Δx/2)/x. So |ψ|² = 2 sin²(Δx/2)/(πΔx²), which at x = 0 equals Δ/2π = 3.18 × 10⁷ m⁻¹: a density in m⁻¹, as it must be.
- Norm: with ∫ sin²(ax)/x² dx = πa and a = Δ/2, ∫|ψ|² dx = (2/(πΔ))(πΔ/2) = 1, matching ∫|φ|² dk = (1/Δ) × Δ = 1. That the two agree is exactly what the delta normalisation of fₖ guarantees.
- Band probability: ∫|φ|² dk from k₀ − Δ/4 to k₀ + Δ/4 = (Δ/2)(1/Δ) = 0.50. In momentum that band is ħk₀ ± ħΔ/4 = 5.27 × 10⁻²⁵ ± 5.3 × 10⁻²⁷ kg m s⁻¹. The probability of p = ħk₀ exactly is zero, because a band of zero width holds none.
- In the p convention Φ(p) = φ(k)/√ħ, since fₖ = √ħ fₚ and dk = dp/ħ. So Φ = 1/√(ħΔ) = 6.89 × 10¹² (kg m s⁻¹)⁻¹⁄² across a band ħΔ = 2.11 × 10⁻²⁶ kg m s⁻¹ wide, and ∫|Φ|² dp = ħΔ/(ħΔ) = 1. Keeping 1/√Δ would give ∫(1/Δ) dp = ħ = 1.05 × 10⁻³⁴ J s, not even a pure number.
- Box check with L = 10 μm: allowed k are 2π/L = 6.28 × 10⁵ m⁻¹ apart, so the band holds LΔ/2π = 318.3 slots, that is 318 or 319 states depending on where its edges fall. Each has cₙ = √(2π/L) φ(kₙ), so |cₙ|² = 2π/(LΔ) = 3.14 × 10⁻³, and Σ|cₙ|² = 0.999 or 1.002: 1 within 0.3%, and exactly 1 as L → ∞.
- Limits: as Δ → 0 the band probability stays 0.50, but |ψ(0)|² = Δ/2π → 0 and the central lobe between x = ±2π/Δ (±31 nm here) spreads without bound, so sharp momentum is approached only by losing localisation. The mean momentum is ħk₀ = mv with v = 5.79 × 10⁵ m s⁻¹, the classical value.
Answer. ψ = 2e(ik₀x) sin(Δx/2)/(x√(2πΔ)), normalised; P = 0.50 for p within 5.3 × 10⁻²⁷ kg m s⁻¹ of 5.27 × 10⁻²⁵ kg m s⁻¹; Φ = 6.89 × 10¹² (kg m s⁻¹)⁻¹⁄² across 2.11 × 10⁻²⁶ kg m s⁻¹; the box sum is 1 within 0.3%.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Plane waves normalise to a delta; only bands of k carry probability.
Use fₖ = e(ikx)/√(2π) with δ(k − k′), or fₚ = e(ipx/ħ)/√(2πħ) with δ(p − p′), never a mixture. Build states as ψ = ∫ φ(k) fₖ dk with ∫|φ|² dk = 1; a probability containing δ(0) or depending on a box length L has misused a plane wave.