University Physics IV · The Schrödinger Equation · 6.7
Continuous Spectra & Momentum Space
Momentum has no normalisable eigenstates, so the tidy expansion over a discrete basis breaks down here. Trade the sum for an integral and the Kronecker delta for a Dirac one, and the expansion coefficient becomes a Fourier transform: the same state, written in momentum instead of position.
Build the model
Connect the measurement to the mechanism.
A bound-state problem hands you a countable eigenbasis, orthonormal to a Kronecker δ, with |cₙ|² a probability. Momentum breaks that machinery. Solving −iħ du/dx = pu returns uₚ(x) ∝ e(ipx/ħ) for every real p, so the spectrum is a continuum, and |uₚ|² is the same constant everywhere — the normalisation integral diverges and no prefactor rescues it.
The repair is not to abandon the basis but to weaken what normalisation means: pick the prefactor 1/√(2πħ) so that overlaps come out as δ(p − p′) instead of 1, and the eigenfunction becomes a distribution that is only ever used inside an integral. Everything else follows by replacing Σₙ with ∫dp. The coefficient cₙ becomes a function φ(p), the projection integral becomes the Fourier transform, |φ(p)|² becomes a probability density per unit momentum rather than a probability, and Parseval's theorem guarantees that the single normalisation ∫|ψ|²dx = 1 has already fixed ∫|φ|²dp = 1.
The cost is that no laboratory prepares a state of sharp momentum: the price of a definite p is a state spread over all space, so what physics actually handles is a packet, and the two ways of writing that packet — ψ(x) and φ(p) — are locked together by the transform. That lock is what the uncertainty bound reports.
- Simple definition
- The momentum-space wavefunction φ(p) is the amplitude for momentum p — the coefficient of the δ-normalised momentum eigenfunction in the expansion of ψ, obtained as the Fourier transform of ψ(x), whose modulus squared is a probability per unit momentum.
- Example
- For an electron packet of width σx = 1.0 nm, φ(p) is a Gaussian of spread σp = ħ/2σx = 5.3 × 10⁻²⁶ kg m s⁻¹, so a momentum window of width 1.0 × 10⁻²⁷ kg m s⁻¹ at the peak holds probability |φ(0)|²Δp = 7.6 × 10⁻³.
Every real p is an eigenvalue, so the spectrum is continuous — yet |uₚ|² = 1/2πħ everywhere, so no uₚ is a state.
p in kg m s⁻¹, x in m; the prefactor carries (J s)(−1/2), and 2πħ = h = 6.626 × 10⁻³⁴ J s
The continuum's stand-in for δₘₙ. It fixes the prefactor and turns every later step into a Fourier identity.
from ∫ e(i(p′−p)x/ħ) dx = 2πħ δ(p − p′); δ(p) carries the unit of 1/p, (kg m s⁻¹)⁻¹
Σₙ cₙ ψₙ has become ∫dp φ(p) uₚ — the same expansion with an integral in place of the sum.
[ψ] = m(−1/2), [φ] = (kg m s⁻¹)(−1/2); the 2πħ splits evenly between the pair
No single momentum carries a probability; windows do. Widen the window and the number grows.
|φ|² is a density per unit momentum, unit (kg m s⁻¹)⁻¹; only its integral is a probability
Normalise once, in whichever variable is easier — the other representation comes out normalised free.
any square-integrable ψ; the general form is ∫ψ*χ dx = ∫φ*ξ dp for two states
One transform ties the two widths: squeeze x by three and p spreads by three. That is the uncertainty bound.
σ = 1.0 nm gives σₚ = 5.3 × 10⁻²⁶ kg m s⁻¹, and σ σₚ = ħ/2 exactly
Every real p is an eigenvalue, and none of them is a state
Momentum in one dimension is the operator −iħ d/dx, so the eigenvalue problem −iħ du/dx = pu integrates at once to uₚ(x) = A e(ipx/ħ). Nothing restricts p: on the whole line there is no boundary condition to impose, so the spectrum is the entire real axis — a continuum, not a list. The trouble arrives at the next step. |uₚ(x)|² = |A|² is the same number at every point, so ∫|uₚ|²dx over the line diverges for every A ≠ 0. Read physically, the state claims the particle is exactly as likely to be found in a 1 nm window here as in a 1 nm window a light-year away, and no rescaling turns that into a total of one. This is not a defect peculiar to momentum. The same happens to the energy eigenfunctions of any unbound problem — a free particle, or a scattering state above a barrier. A continuous spectrum simply does not come with normalisable eigenfunctions.
Delta normalisation, and the box that does the same job
Keep the eigenfunctions and weaken the condition they must satisfy. Take the overlap of two of them: ∫uₚ* uₚ′ dx = |A|²∫e(i(p′−p)x/ħ)dx = |A|² 2πħ δ(p′ − p), using the Fourier representation of the Dirac δ. Choosing |A|² = 1/2πħ makes that overlap exactly δ(p′ − p), the continuum's replacement for δₘₙ: zero for different labels, and for the same label a spike whose integral is one. The delta is a distribution, meaningful only inside an integral, and it carries the unit of 1/p, which is what forces uₚ to carry (J s)(−1/2). If distributions grate, put the system in a box of length L with periodic ends instead: momenta are then discrete, pₙ = 2πnħ/L, spaced by 2πħ/L, every eigenfunction is normalisable, and sums pass to integrals through Σₙ → (L/2πħ)∫dp. Every physical answer must come out independent of L, and letting L → ∞ returns the delta convention.
The expansion becomes a Fourier transform
With a discrete basis you write ψ = Σₙ cₙ ψₙ and recover the coefficients by projection, cₙ = ∫ψₙ*ψ dx. Do the same with a continuous label and the sum becomes an integral: expanding, ψ(x) = ∫φ(p) uₚ(x) dp; projecting with the delta orthonormality, φ(p) = ∫uₚ*(x) ψ(x) dx. Written out, φ(p) = (2πħ)(−1/2)∫ψ(x)e(−ipx/ħ)dx and ψ(x) = (2πħ)(−1/2)∫φ(p)e(ipx/ħ)dp: the Fourier transform pair, in the symmetric convention where the 2πħ splits evenly and the transform variable is p = ħk rather than k. Nothing new has been assumed. φ is the same object as cₙ, indexed continuously, and the pair says only that these are two ways of writing one state. Which you use is a question of which integral is easier — a free particle is trivial in p and awkward in x, a delta potential the other way round.
Read |φ(p)|² as a density, and let the units enforce it
Because ∫|φ|²dp = 1 and p carries a unit, |φ(p)|² must carry the unit of 1/p, inverse kg m s⁻¹. It is a probability per unit momentum, and it becomes a probability only after an integral: P(p₁ ≤ p ≤ p₂) = ∫|φ|²dp. That the density can exceed one is no paradox but a consequence of that unit, exactly as |ψ(x)|² for a tightly bound state runs to 10¹⁸ m⁻¹. Take the 1 nm electron packet: σₚ = ħ/2σₓ = 5.3 × 10⁻²⁶ kg m s⁻¹, so the peak density is 1/(σₚ√(2π)) = 7.6 × 10²⁴ (kg m s⁻¹)⁻¹. Ask for the probability of a momentum inside a window of width 1.0 × 10⁻²⁷ kg m s⁻¹ at that peak and the answer is 7.6 × 10²⁴ × 1.0 × 10⁻²⁷ = 7.6 × 10⁻³. Ask for the probability of exactly p = 0 and the answer is zero, for every state — which is why a continuum is described by a density and never by a list of |cₙ|².
One normalisation, and operators that swap roles
Parseval's theorem, ∫|ψ|²dx = ∫|φ|²dp, follows from substituting both transforms and collapsing the resulting ∫e(i(p−p′)x/ħ)dx into 2πħ δ(p − p′). The state is therefore normalised in both pictures at once; there is never a second normalisation to perform. The same substitution carries operators across. Momentum, which acts as −iħ ∂/∂x on ψ, becomes plain multiplication by p in the momentum picture, so ⟨p⟩ = ∫ψ*(−iħ ∂ψ/∂x)dx = ∫p|φ(p)|²dp. Position turns into a derivative with the opposite sign, ⟨x⟩ = ∫φ*(iħ ∂φ/∂p)dp. The two are mirror images, which is the structural fact behind the canonical commutator. For the 1 nm electron with ⟨p⟩ = 0 the momentum picture gives the kinetic energy in one line: ⟨T⟩ = ⟨p²⟩/2m = σₚ²/2m = (5.27 × 10⁻²⁶)²/(2 × 9.11 × 10⁻³¹) = 1.5 × 10⁻²¹ J, some 9.5 meV that confinement alone supplies.
Sharp edges in x buy a heavy tail in p
The transform punishes discontinuity. A state that is flat inside a region of half-width a and cut to zero at its edges has φ(p) = √(ħ/πa)⋅sin(pa/ħ)/p, so |φ|² falls only as 1/p² far out. Parseval still holds — the total is exactly one, and 90.3% of it sits inside the first zeros at |p| = h/2a, which for a = 0.50 nm is 6.6 × 10⁻²⁵ kg m s⁻¹ — but ⟨p²⟩ = ∫p²|φ|²dp has an integrand that never decays, so the mean square momentum, and with it the kinetic energy, is infinite. That is the momentum-space reading of the admissibility rules: ψ must be continuous everywhere, and ψ′ continuous wherever V is finite, because a step in ψ, or a kink where the potential is finite, loads the momentum amplitude with a tail too heavy to carry a finite energy. Smoothness in x is the same statement as decay in p.
Change one variable at a time
Make the relationship visible.
Drag σx from 1.50 down to 0.50 nm: the position curve triples in height, the momentum curve drops to a third of its height and spreads three times as wide, and σx σp stays at ħ/2. Then move ⟨p⟩ — the momentum peak slides while |ψ(x)|² does not stir, because the phase carries the momentum.
MOMENTUM SPREAD σp0.659 ×10⁻²⁵ kg m/s
PEAK DENSITY |φ|²0.61 per ×10⁻²⁵
σx σp IN UNITS OF ħ0.50
MEAN KE, ELECTRON64 meV
Live interpretationMOMENTUM SPREAD σp: 0.659 ×10⁻²⁵ kg m/s. PEAK DENSITY |φ|²: 0.61 per ×10⁻²⁵. σx σp IN UNITS OF ħ: 0.50. MEAN KE, ELECTRON: 64 meV
Catch the common trap
Explain before calculating.
A one-dimensional state has momentum amplitude φ(p). Near p₀ = 2.0 × 10⁻²⁴ kg m s⁻¹ its momentum density is |φ(p)|² = 4.0 × 10²³ (kg m s⁻¹)⁻¹, effectively flat across a narrow window of width Δp = 1.0 × 10⁻²⁶ kg m s⁻¹ centred on p₀. What is the probability that a momentum measurement returns a value inside that window?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA momentum eigenfunction in one dimension satisfies −iħ du/dx = p u. Solve it, show that no constant normalises it to one, and find the constant that normalises it to a Dirac delta instead.
- Separate and integrate: du/u = (ip/ħ)dx, so uₚ(x) = A e(ipx/ħ), and this solves the equation for every real p. The momentum spectrum is the whole real line.
- Try the usual normalisation. |uₚ(x)|² = |A|² at every x, so integrating from −L/2 to L/2 gives |A|²L, which grows without bound as L → ∞. No nonzero A can make it 1.
- Ask for delta normalisation instead: ∫uₚ*(x)uₚ′(x)dx = |A|²∫e(i(p′−p)x/ħ)dx = |A|² 2πħ δ(p′ − p).
- Setting that equal to δ(p′ − p) gives |A|² = 1/2πħ = 1/h, so A = 1/√h = 1/√(6.626 × 10⁻³⁴ J s) = 3.9 × 10¹⁶ (J s)(−1/2).
Answeruₚ(x) = e(ipx/ħ)/√(2πħ), prefactor 3.9 × 10¹⁶ (J s)(−1/2). No choice of A normalises it to one: momentum eigenfunctions are distributions, not states.
MediumAn electron is prepared in the minimum-uncertainty packet ψ(x) = (2πσ²)(−1/4) e(−x²/4σ²) with σ = 1.0 nm and ⟨p⟩ = 0. Find the momentum spread σₚ, check the uncertainty product, and find the mean kinetic energy. Take ħ = 1.055 × 10⁻³⁴ J s and mₑ = 9.11 × 10⁻³¹ kg.
- |ψ|² = (2πσ²)(−1/2) e(−x²/2σ²) is a normalised Gaussian density of standard deviation σₓ = σ = 1.0 nm.
- Transforming with the e(−ipx/ħ) kernel returns another Gaussian: |φ(p)|² = (2πσₚ²)(−1/2) e(−p²/2σₚ²) with σₚ = ħ/2σ. The Gaussian is the one shape the transform preserves.
- σₚ = 1.055 × 10⁻³⁴ ÷ (2 × 1.0 × 10⁻⁹) = 5.27 × 10⁻²⁶ kg m s⁻¹.
- Product: σₓ σₚ = 1.0 × 10⁻⁹ × 5.27 × 10⁻²⁶ = 5.27 × 10⁻³⁵ J s, which is ħ/2 exactly — the Gaussian saturates the bound.
- With ⟨p⟩ = 0, ⟨p²⟩ = σₚ² = 2.78 × 10⁻⁵¹ kg² m² s⁻², so ⟨T⟩ = ⟨p²⟩/2mₑ = 2.78 × 10⁻⁵¹ ÷ (1.82 × 10⁻³⁰) = 1.53 × 10⁻²¹ J.
Answerσₚ = 5.3 × 10⁻²⁶ kg m s⁻¹, σₓ σₚ = ħ/2, and ⟨T⟩ = 1.5 × 10⁻²¹ J ≈ 9.5 meV — energy that confinement alone supplies, with no force acting.
HardA state is flat inside a region and cut off at its edges: ψ(x) = 1/√(2a) for |x| ≤ a and zero outside, with a = 0.50 nm. Find φ(p), verify Parseval's theorem, find the probability that |p| < h/2a, and evaluate ⟨p²⟩.
- φ(p) = (2πħ)(−1/2)(2a)(−1/2) ∫e(−ipx/ħ)dx from −a to a = (4πħa)(−1/2) × 2ħ sin(pa/ħ)/p = √(ħ/πa) · sin(pa/ħ)/p.
- Parseval: put u = pa/ħ, so dp = (ħ/a)du and ∫|φ|²dp = (1/π)∫sin²u/u² du = (1/π)(π) = 1. The transform carried the normalisation across untouched.
- First zeros: sin(pa/ħ) = 0 at p = ±πħ/a = ±h/2a = 6.626 × 10⁻³⁴ ÷ (1.0 × 10⁻⁹) = 6.6 × 10⁻²⁵ kg m s⁻¹.
- Central lobe: (1/π)∫sin²u/u² du from −π to π = (2/π)Si(2π) = 2(1.4181)/π = 0.903, so 90.3% of the momentum probability lies inside those zeros.
- ⟨p²⟩ = ∫p²|φ|²dp = (ħ/πa)∫sin²(pa/ħ)dp. The integrand never decays, so the integral diverges: the step in ψ costs infinite kinetic energy.
Answerφ(p) = √(ħ/πa)⋅sin(pa/ħ)/p; ∫|φ|²dp = 1; 90.3% of the probability lies within |p| < 6.6 × 10⁻²⁵ kg m s⁻¹; ⟨p²⟩ diverges, which is why an admissible ψ must be continuous.