The sine eigenbasis: orthonormality and completeness

Take Ĥ = −(ħ²/2m) d²/dx² on the Dirichlet domain — functions in H²([0, L]) with ψ(0) = ψ(L) = 0. Integrating by parts twice, ⟨m|Ĥψₙ⟩ − ⟨Ĥψₘ|ψₙ⟩ equals the Wronskian boundary term −(ħ²/2m)[ψₘ*ψₙ′ − ψₘ*′ψₙ]₀ᴸ, and the boundary condition kills it at both ends. The left side is (Eₙ − Eₘ)⟨m|n⟩ because every Eₙ is real, and the 1D box has no degeneracy, so E₁, E₂, E₃, … are distinct and every off-diagonal overlap must vanish before you integrate anything. Now integrate anyway, because that is what catches a wrong convention. Product-to-sum gives sin A sin B = ½[cos(A − B) − cos(A + B)], so for m ≠ n the integrand is a difference of cosines at (m − n)π/L and (m + n)π/L, each completing a whole number of half-periods on [0, L]; both integrate to zero. For m = n the first cosine is cos 0 = 1 and contributes L/2, so (2/L)(L/2) = 1.

The eigenvalue problem fixes the shape and not the size: any multiple of sin(nπx/L) still solves it and still vanishes at both walls. Born's rule fixes the size. Since ∫₀ᴸ sin²(nπx/L) dx = L/2 for every n — sin² averages to ½ over a whole number of half-periods, and n only changes how many — the constant is √(2/L), the same for every level. Units follow: |ψ|² is a probability per unit length, so ψ carries m^(−1/2). At L = 1.00 nm that is √(2/L) = 4.47 × 10⁴ m^(−1/2), and every |ψₙ|² peaks at 2/L = 2.00 nm⁻¹ whatever n is; raising n moves the antinodes around without raising the peak. One freedom survives normalisation: e^(iθ)|n⟩ is equally normalised, so the physical state is a ray, not a single ket. Choosing the real sine that is positive just inside x = 0 is a convention, and it is the convention that keeps every cₙ real when ψ is real — say it out loud before comparing signs with anyone else's code.

Orthonormality never implies completeness. Throw away |2⟩: the surviving set still satisfies ⟨m|n⟩ = δₘₙ exactly, while ψ₂ itself now expands to the zero vector and Parseval fails on it. The missing input is spectral. On a bounded interval, −d²/dx² with Dirichlet conditions is a regular Sturm–Liouville operator with compact resolvent, so its eigenvectors form an orthonormal basis of the whole of L²([0, L]). In Dirac notation that is Σₙ|n⟩⟨n| = 1; in the position basis it is the closure relation (2/L)Σₙ sin(nπx/L) sin(nπx′/L) = δ(x − x′), a distribution, so read it only under an integral. Numerically the two halves come apart. Build S with Sₙⱼ = √(2Δx/L) sin(nπxⱼ/L) for n ≤ N on a grid of J points: S @ S.T ≈ I_N tests orthonormality of the states you kept, while S.T @ S ≈ I_J tests completeness and only approaches the identity as N approaches J. Truncation breaks completeness first, and never breaks orthonormality at all.

Because Σₙ|n⟩⟨n| = 1, writing |ψ⟩ = Σcₙ|n⟩ and hitting it with ⟨m| returns cₘ = ⟨m|ψ⟩ at once: orthonormality has already decoupled the equations, so there is no linear system to solve. In the position basis that projection is the Fourier sine coefficient, cₙ = √(2/L)∫₀ᴸ sin(nπx/L) ψ(x) dx. Use symmetry before calculus. The basis has definite parity about the midpoint, ψₙ(L − x) = (−1)ⁿ⁺¹ψₙ(x), so a state symmetric about L/2 has zero overlap with every even n and an antisymmetric one with every odd n. For the normalised parabola ψ = √(30/L⁵) x(L − x) that deletes half the integrals unseen, and the survivors give cₙ = 4√60/(n³π³): c₁ = 0.99928, c₃ = 0.03701, c₅ = 0.00799. The state is 99.86% ground state, which is why a parabola is such a good freehand sketch of ψ₁ — and why c₂ = 0 exactly, not merely small.

Insert the identity into the norm: ⟨ψ|ψ⟩ = ⟨ψ|(Σₙ|n⟩⟨n|)|ψ⟩ = Σₙ⟨ψ|n⟩⟨n|ψ⟩ = Σₙ|cₙ|². Normalisation stops being an integral and becomes a convergent series, and the same move on two states gives ⟨φ|ψ⟩ = Σₙbₙ*cₙ — the change of basis is unitary, so it preserves every bracket, not merely lengths. The physics arrives when the Born rule is read in the energy basis: P(Eₙ) = |cₙ|², so Parseval is literally the statement that the probabilities of the possible energy outcomes sum to one, and ⟨H⟩ = Σₙ|cₙ|²Eₙ. It is also your numerical error bar. Truncating at N discards exactly 1 − Σ(n ≤ N)|cₙ|² of the norm — and therefore of the probability — which for the parabola is 1.4 × 10⁻³ at N = 1 and 7.5 × 10⁻⁵ at N = 3. Print that residual in any box code you write; it is the one convergence diagnostic that costs nothing.

How fast cₙ falls is set by how well ψ imitates the basis at the walls, because every ψₙ vanishes there and so does every ψₙ″. A state matching the first condition but not the second — the parabola, whose second derivative is −2A everywhere — gives cₙ ∝ n⁻³ and |cₙ|² ∝ n⁻⁶. A state failing even the first — the flat ψ = 1/√L, a perfectly good vector of L²([0, L]) though nowhere near the domain of Ĥ — gives only cₙ = 2√2/(nπ) for odd n, so |cₙ|² ∝ n⁻² and eight terms retain just (8/π²)(1 + 1/9 + … + 1/225) = 0.9747 of the norm, against the parabola's 0.99992 with two. The mismatch surfaces as Gibbs. The odd 2L-periodic extension of the flat state is a square wave jumping by 2/√L at each wall, and the partial sum overshoots by 0.0895 of that jump: the eight-term sum reaches 1.1803/√L at x = L/16. Keep sixteen terms and the peak moves to x = L/32 and drops only to 1.1793/√L.