Position and momentum representations

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.

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.

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.

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 — a derivative to take in the left form, only a multiplication by p in the right.

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.

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.