University Physics V · The Quantum Wavefunction · 3.6
Position & Momentum Representations
Students meet ψ(x) first and mistake it for the state. This topic breaks the habit: the same ket has momentum components φ(p) = ⟨p|ψ⟩, the Fourier transform shuttles between the two component lists, and unitarity makes one normalisation serve both. Afterwards, ‘which basis?’ is a question you ask before every calculation.
Build the model
Connect the measurement to the mechanism.
Nothing in the postulates privileges position. A state is a ket |ψ⟩ in L²(ℝ); ψ(x) = ⟨x|ψ⟩ is merely its component list along one improper basis, and any other self-adjoint operator's eigenfunctions serve as well. Momentum earns its basis by generating translations: the unitary shift exp(−ip̂a/ħ) sends ψ(x) to ψ(x−a), and expanding to first order in a forces p̂ = −iħ d/dx.
Its eigenvalue equation returns plane waves e(ipx/ħ)/√(2πħ) — δ-normalised, not square-integrable — and inserting their resolution of the identity turns the expansion coefficient ⟨p|ψ⟩ into the Fourier transform in the symmetric 2πħ convention. That map is unitary: Plancherel's theorem makes ∫|ψ|²dx = ∫|φ|²dp, so one normalisation serves both bases and |φ(p)|² is the momentum probability density with no further factor. The price of the continuum is real.
The kets |x⟩ and |p⟩ lie outside the Hilbert space, orthonormality means δ(p−p′) rather than a Kronecker δ, and the commutator [x̂, p̂] = iħ that encodes the whole structure is impossible for finite matrices — take the trace: Tr(x̂p̂ − p̂x̂) = 0 while Tr(iħ1) = iħN. Every grid calculation therefore approximates this pair of bases, and knowing where the approximation fails is part of owning the representation.
- Simple definition
- The momentum representation φ(p) = ⟨p|ψ⟩ lists the same state's components along δ-normalised plane waves; it is the Fourier transform of ψ(x), and |φ(p)|² is the probability density for momentum.
- Example
- A Gaussian packet with position spread σx = 0.50 Å has momentum representation of spread σp = ħ/(2σx) = 1.1 × 10⁻²⁴ kg m s⁻¹ — squeeze one component list and the other broadens in exact reciprocity.
The matrix element of the change of basis — and the prefactor 1/√(2πħ) is exactly what makes the transform unitary.
δ-orthonormal, ⟨p|p′⟩ = δ(p − p′); |⟨x|p⟩|² = 1/2πħ is constant, so its norm integral diverges
Same ket, new components: insert 1 = ∫|p⟩⟨p| dp and read off ⟨p|ψ⟩. No new state is created.
symmetric 2πħ convention; x in m, p in kg m s⁻¹; the inverse differs only in the sign of the exponent
Normalise once, in either basis; |φ(p)|² is then the momentum density with no compensating factor.
the Fourier map is unitary on L²(ℝ): every norm and every inner product survives the basis change
Where the derivative comes from: momentum is defined by what it does — slide the state — not by decree.
a is the shift in m; expand both sides to first order in a to extract the generator
Work where your operator is diagonal: kinetic energy is a painless multiplication by p²/2m in the p-basis.
the sign flips between representations so that [x̂, p̂] = iħ holds in both
The continuum is compulsory: any grid version of x̂ and p̂ gets the commutator wrong somewhere.
X, P any N×N matrices — the trace of every finite commutator vanishes
Momentum is what shifts you
Define the translation operator by what it does: (T(a)ψ)(x) = ψ(x − a). It preserves every inner product — shift both functions and the overlap integral is unchanged — so it is unitary, and T(a)T(b) = T(a+b) makes the family a one-parameter group. Stone's theorem then guarantees a self-adjoint generator: T(a) = exp(−ip̂a/ħ), with the ħ inserted so the generator carries momentum's units. Expand both sides for a small shift: ψ(x − a) ≈ ψ(x) − aψ′(x), while (1 − iap̂/ħ)ψ = ψ − iap̂ψ/ħ. Matching the first-order terms forces p̂ψ = −iħψ′. The differential operator is a consequence, not an axiom: momentum is the observable whose exponential slides wavefunctions along the line, exactly as classical momentum generates displacement through Poisson brackets.
The eigenfunctions escape the space
Solve the eigenvalue equation −iħu′ = pu: uₚ(x) = C e(ipx/ħ), one solution for every real p, oscillating forever with |uₚ|² = |C|² constant. The norm integral diverges, so no momentum eigenfunction lives in L²(ℝ): the spectrum is purely continuous and the 'basis' is improper. The workable substitute is δ-orthonormality. Using ∫e(i(p−p′)x/ħ)dx = 2πħ δ(p − p′), the choice C = 1/√(2πħ) gives ⟨p|p′⟩ = δ(p − p′) — a Dirac δ where a discrete basis would have a Kronecker δ — and a resolution of the identity ∫|p⟩⟨p| dp = 1 that works inside every integral. The kets |p⟩ are calculational scaffolding: legitimate to expand across, impossible to occupy. A physical state is always a square-integrable superposition of them — a wave packet.
Insert the identity and the Fourier transform appears
The change of basis is one line of Dirac algebra. Start from φ(p) = ⟨p|ψ⟩ and insert the position resolution of the identity: φ(p) = ∫⟨p|x⟩⟨x|ψ⟩ dx = (2πħ)(−1/2) ∫ψ(x) e(−ipx/ħ) dx. Run the same trick the other way with ∫|p⟩⟨p| dp = 1 to invert: ψ(x) = (2πħ)(−1/2) ∫φ(p) e(+ipx/ħ) dp. This is the Fourier transform in the symmetric convention, the 2πħ split evenly between the two directions — and the split matters, because an asymmetric convention would smuggle a constant into one of the norms and Born's rule would then need a compensating factor in one basis. Note what did not happen: no new state was produced. ψ(x) and φ(p) are two component lists of one ket, the way one arrow has different components in rotated axes — except that the rotation here connects two continuous bases, and it is the momentum list that makes every ⟨p̂ⁿ⟩ trivial.
Plancherel: one normalisation serves both bases
A unitary map preserves inner products, so ⟨ψ|ψ⟩ evaluated as ∫|ψ(x)|²dx and as ∫|φ(p)|²dp is the same number — Plancherel's theorem. Normalise once, in whichever basis is easier, and Born's rule is ready in both: P(a < p < b) = ∫ₐᵇ|φ(p)|²dp, with φ carrying dimensions of (momentum)(−1/2) just as ψ carries (length)(−1/2). The Gaussian pair makes it concrete: ψ(x) = (2πσ²)(−1/4) e(−x²/4σ²), with position spread σ, transforms to another Gaussian with momentum spread ħ/2σ, and both densities integrate to exactly 1. What narrows in x broadens in p, the product held at σx⋅σp = ħ/2 — the minimum the uncertainty relation permits. Expectation values can now be taken wherever they are cheap: ⟨p⟩ = ∫ψ*(−iħψ′)dx = ∫p|φ(p)|²dp — one integration by parts in the left form, none in the right.
Work in the basis that diagonalises your operator
In its own basis every operator is multiplication: ⟨p|p̂|ψ⟩ = p φ(p). The position operator moves off-diagonal instead — writing x e(−ipx/ħ) = iħ (d/dp) e(−ipx/ħ) inside the transform gives ⟨p|x̂|ψ⟩ = +iħ dφ/dp, the sign flipped relative to p̂ = −iħ d/dx precisely so that [x̂, p̂] = iħ holds in both representations. The practical rule: kinetic energy p̂²/2m is multiplication by p²/2m in the momentum basis, a potential V(x̂) is multiplication in the position basis, and neither is pleasant in the other's home. Numerics exploits this daily: the split-step Fourier method applies V in the x-basis, FFTs to the p-basis, applies the kinetic phase, and FFTs back — numpy.fft doing the unitary shuttle, provided the grid's momentum axis is built as pₖ = 2πħk/(NΔx) and the 1/√(2πħ) bookkeeping matches the discrete transform's own convention.
The cost: no finite matrices, ever
Everything above leans on [x̂, p̂] = iħ, and that relation cannot survive discretisation intact. Take any two N×N matrices X and P: Tr(XP) = Tr(PX), so the trace of their commutator is exactly zero — while Tr(iħ1) = iħN is not. No finite matrices satisfy [X, P] = iħ, whatever their size, so a grid version of this topic is always an approximation and must break somewhere. Where it breaks is instructive. Put ψ on N points of spacing Δx, take x̂ diagonal and p̂ from the FFT: the commutator applied to a smooth packet in the middle of the grid returns iħψ to many digits, but for a state pressed against the box edge, or oscillating near the Nyquist momentum πħ/Δx, it fails completely. The continuum bases are not a convenience — they carry content a finite model provably cannot, which is why the infinite-dimensionality of quantum mechanics on a line is a theorem rather than a preference.
Change one variable at a time
Make the relationship visible.
Narrow σx and the right-hand curve widens in exact reciprocity. Then drag x₀: the packet slides while the momentum density stays put — a shift only multiplies φ(p) by a phase the modulus cannot see. The boost p₀ plays the same trick in mirror image.
σp = 1/(2σx)0.71
σx⋅σp (= ħ/2)0.50
PEAK |ψ|²0.57
PEAK |φ|²0.56
Live interpretationσp = 1/(2σx): 0.71. σx⋅σp (= ħ/2): 0.50. PEAK |ψ|²: 0.57. PEAK |φ|²: 0.56
Catch the common trap
Explain before calculating.
A state is handed to you in the momentum representation, φ(p) = ⟨p|ψ⟩. How does the position operator x̂ act on φ(p)?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA particle is prepared in the top-hat state ψ(x) = 1/√L for |x| ≤ L/2 and zero outside, with L = 1.0 nm. Find its momentum representation φ(p) and the smallest |p| at which the momentum density vanishes.
- Write the transform in the symmetric convention: φ(p) = (2πħ)(−1/2) ∫ from −L/2 to L/2 of L(−1/2) e(−ipx/ħ) dx.
- The integral is elementary: ∫e(−ipx/ħ)dx = (iħ/p)[e(−ipL/2ħ) − e(+ipL/2ħ)] = (2ħ/p) sin(pL/2ħ).
- So φ(p) = √(2ħ/πL) · sin(pL/2ħ)/p. As p → 0, sin(pL/2ħ) ≈ pL/2ħ, giving the finite peak φ(0) = √(L/2πħ) — the division never blows up.
- The density |φ(p)|² first vanishes when pL/2ħ = π, i.e. |p| = 2πħ/L = h/L. For L = 1.0 nm: p = 6.626 × 10⁻³⁴ ÷ 1.0 × 10⁻⁹ = 6.6 × 10⁻²⁵ kg m s⁻¹.
Answerφ(p) = √(2ħ/πL) sin(pL/2ħ)/p, a sinc centred on p = 0, with its first zero at |p| = h/L = 6.6 × 10⁻²⁵ kg m s⁻¹. Halve the box and that momentum scale doubles.
MediumTransform the normalised Gaussian ψ(x) = (2πσ²)(−1/4) e(−x²/4σ²) with σ = 0.50 Å to the momentum basis, verify Plancherel's theorem on the result, and evaluate the momentum spread σp.
- Use ∫e(−ax²+bx)dx = √(π/a) e(b²/4a) with a = 1/4σ² and b = −ip/ħ: the transform integral equals 2σ√π · exp(−p²σ²/ħ²).
- Attach the prefactors: φ(p) = (2πħ)(−1/2)(2πσ²)(−1/4) · 2σ√π e(−p²σ²/ħ²) = (2πσp²)(−1/4) e(−p²/4σp²) with σp = ħ/2σ — a Gaussian again.
- Plancherel check: |φ(p)|² is a normal density of standard deviation σp, so ∫|φ|²dp = 1 automatically — the normalisation bought once in the x-basis, with no new factor.
- Numbers: σp = ħ/2σ = 1.055 × 10⁻³⁴ ÷ (2 × 5.0 × 10⁻¹¹) = 1.1 × 10⁻²⁴ kg m s⁻¹, and σx⋅σp = ħ/2 = 5.3 × 10⁻³⁵ J s — the minimum-uncertainty product.
Answerφ(p) = (2πσp²)(−1/4) e(−p²/4σp²) with σp = ħ/2σ = 1.1 × 10⁻²⁴ kg m s⁻¹; both densities integrate to 1, and σx⋅σp = ħ/2 exactly.
HardThe infinite-well ground state on [0, L] is ψ(x) = √(2/L) sin(πx/L). A common claim reads it as an equal mix of momenta ±πħ/L. Compute the momentum density |φ(p)|² and test the claim at p = 0 and p = ±πħ/L.
- Set k = π/L and q = p/ħ. Then φ(p) = (2πħ)(−1/2) √(2/L) ∫ from 0 to L of sin(kx) e(−iqx) dx; split the sine into exponentials and use e(±ikL) = −1.
- Both exponential integrals produce the factor (1 + e(−iqL)), and the sum collapses to k(1 + e(−iqL))/(k² − q²).
- Take the modulus squared with |1 + e(−iqL)|² = 4cos²(qL/2): |φ(p)|² = (4k²/πħL) · cos²(qL/2)/(k² − q²)².
- At p = 0: |φ(0)|² = 4/(πħLk²) = 4L/π³ħ ≈ 0.129 L/ħ. At q = ±k the 0/0 limit gives cos²(qL/2)/(k² − q²)² → L²/16k², so |φ(±πħ/L)|² = L/4πħ ≈ 0.080 L/ħ.
- The ratio is (4L/π³ħ)/(L/4πħ) = 16/π² ≈ 1.62: the density is single-peaked at p = 0, not twin-peaked. The two-plane-wave reading fails because sin(kx) is truncated to the box — the sharp edges feed every momentum.
Answer|φ(p)|² = (4k²/πħL) cos²(pL/2ħ)/(k² − p²/ħ²)² with k = π/L. It is single-peaked at p = 0, where the density is 16/π² ≈ 1.62 times its value at p = ±πħ/L — not an equal mix of two sharp momenta.