Hilbert space and operator domain

States in the box are vectors in L²([0, L]): square-integrable functions on the interval, identified when they agree almost everywhere, paired by ⟨f|g⟩ = ∫₀ᴸ f*(x)g(x) dx. It is a separable infinite-dimensional Hilbert space, and ψ carries units of m^(−1/2) so that ⟨ψ|ψ⟩ = 1 is a pure number. One thing is easier here than on the whole line: the interval is bounded, so every bounded function is automatically square-integrable and e^(ikx) is an honest vector of this space for every real k, with no delta normalisation needed. The functions √(2/L) sin(nπx/L) are one orthonormal basis of the space, not the space itself, and they arrive only after a particular Hamiltonian has been chosen. At this stage the space knows nothing about walls.

Write τ = −(ħ²/2m) d²/dx². By itself τ is a recipe, and it cannot be an operator on all of L²: by the Hellinger–Toeplitz theorem an everywhere-defined symmetric operator is bounded, and τ is not bounded. So an operator is a pair (τ, D) with D a dense subspace, and you must say what D is. Symmetric means ⟨f|τg⟩ = ⟨τf|g⟩ for every f, g ∈ D — this is what physicists call Hermitian. Self-adjoint is strictly stronger. The adjoint τ† is defined on every f for which g ↦ ⟨f|τg⟩ is continuous on D, and self-adjointness demands D(τ†) = D(τ), not merely D(τ†) ⊇ D(τ). Only the equality buys you the spectral theorem: a real spectrum, not merely real eigenvalues, which symmetry alone already guarantees; a complete spectral resolution, which on the box is an orthonormal eigenbasis; and a unitary e^(−iĤt/ħ) that conserves the norm. In finite dimensions the distinction is empty, because every domain is the whole space. In L²([0, L]) it is the entire subject.

Integrate by parts twice and everything collapses to the walls: ⟨f|τg⟩ − ⟨τf|g⟩ = −(ħ²/2m)[f*g′ − f*′g]₀ᴸ. Symmetry is exactly the vanishing of that Wronskian-type bracket, and many domains achieve it — Dirichlet ψ(0) = ψ(L) = 0; Neumann ψ′(0) = ψ′(L) = 0; Robin ψ′(0) = ψ(0)/a₀ and ψ′(L) = −ψ(L)/a_L for real lengths a; periodic ψ(0) = ψ(L) with ψ′(0) = ψ′(L); antiperiodic, with a sign. Counting the family is the deficiency-index calculation: τ†ψ = ±iλψ is a second-order ODE whose two solutions are both square-integrable on a bounded interval, so n₊ = n₋ = 2 and the self-adjoint extensions form a U(2), four real parameters wide. These are different operators with different physics. For an electron on [0, 1.00 nm], Dirichlet gives 0.376, 1.504, 3.384 eV; Neumann gives 0, 0.376, 1.504 eV with a constant ground state; periodic gives 0, then 1.504 and 6.016 eV each twofold degenerate. The spectrum belongs to the domain.

The choice among that family is made by physics, not by taste. In a finite well of depth V₀ the exterior solution decays as e^(−κ|x|) with κ = √(2m(V₀ − E))/ħ, and matching ψ and ψ′ at the left wall gives ψ′(0)/ψ(0) = κ — a Robin condition whose parameter is the penetration depth a = 1/κ. Put numbers on it: an electron near 0.376 eV in a 10.0 eV well has κ = 1.59 × 10¹⁰ m⁻¹, so a = 0.063 nm, more than 6% of a 1.00 nm box, and the honest operator there is Robin, not Dirichlet. Deepening the wall helps only as V₀^(−1/2): at 100 eV, a = 0.020 nm. Sending V₀ → ∞ drives a → 0 and delivers ψ(0) = ψ(L) = 0 in the limit. Dirichlet is therefore a limit, and the kink in ψ at each wall is the fingerprint of that limit — continuity of ψ survives it, continuity of ψ′ does not.

One integration by parts gives ⟨f|p̂g⟩ − ⟨p̂f|g⟩ = −iħ[f*g]₀ᴸ. On the Dirichlet domain g(0) = g(L) = 0, the bracket vanishes, and p̂ = −iħ d/dx is symmetric. But notice what that argument does to the adjoint: the bracket vanishes for every f whatsoever, because g is already zero at both walls, so D(p̂†) is the whole of H¹([0, L]) with no wall condition — strictly larger than D(p̂). Self-adjointness needs equality, so p̂ fails. The deficiency count says the same thing arithmetically. Solving p̂†ψ = ±i(ħ/L)ψ gives ψ± = e^(∓x/L), both square-integrable because the interval is bounded, so n₊ = n₋ = 1 and the extensions form a U(1) circle: ψ(L) = e^(iθ)ψ(0), with eigenvalues pₙ = ħ(θ + 2πn)/L and eigenfunctions e^(ipₙx/ħ)/√L. Every one of them has |ψ| = 1/√L at both walls, so none can satisfy ψ(0) = 0. No self-adjoint momentum operator lives on the Dirichlet domain.

Run the same deficiency calculation on three geometries and the pattern is clear. On the whole line e^(∓x/L) is not square-integrable, the indices are (0,0), and p̂ is already self-adjoint with spectrum ℝ. On the half-line [0, ∞) only e^(−x/L) survives, the indices are (1,0), and because they are unequal there is no self-adjoint extension at all — a radial momentum operator simply does not exist. On [0, L] both survive, the indices are (1,1), and you inherit a circle of choices. What does survive on the Dirichlet domain is 2mĤ = p̂†p̂, the expression p̂² with Dirichlet walls, which is self-adjoint even though p̂ is not, so ⟨p²⟩ = (nπħ/L)² and the product Δx Δp are legitimate — x̂ maps the Dirichlet domain into itself, so ⟨[x̂, p̂]⟩ = iħ picks up no boundary term. What fails is applying p̂² twice. For ψ ∝ x(L − x), p̂²ψ is a nonzero constant that does not vanish at the walls, so p̂⁴ψ is undefined and the naive answer ⟨p⁴⟩ = ħ⁴∫ψ*ψ⁗ dx = 0 is meaningless; the spectral sum gives 120ħ⁴/L⁴.