Probability density, nodes and intervals

⟨x|n⟩ is not the observable. The Born rule says the observable object is ρₙ(x) = |⟨x|n⟩|², whose dimension in one dimension is 1/length, so a value of ρ is never a probability — only ρ dx is. Normalisation is the statement ⟨n|n⟩ = 1 read in the position basis: ∫₀ᴸ (2/L)sin²(nπx/L) dx = 1, which works for every n because the interval holds exactly n half-periods of sin(nπx/L) and sin² averages to ½ over any whole number of them. Outside [0, L] the density is identically zero, and that is not an approximation but the domain: Ĥ is defined on H²([0, L]) with ψ(0) = ψ(L) = 0, so the walls are imposed before any integral is attempted. Note what the modulus discards. ψ₂ changes sign at L/2; ρ₂ does not. Every fact about relative phase — the thing that makes superpositions interfere — is invisible in the density of a single eigenstate.

“The particle is in [a, b]” is a yes/no observable, and its operator is the projector Π̂ = ∫ₐᵇ|x⟩⟨x|dx. It is Hermitian, satisfies Π̂² = Π̂, and has spectrum {0, 1}, so its expectation value in any normalised state lies between 0 and 1 — exactly what a probability must do. In the position basis ⟨n|Π̂|n⟩ collapses to ∫ₐᵇ ρₙ dx, and with sin²θ = (1 − cos 2θ)/2 the integral is elementary: P = (b−a)/L − [sin(2πnb/L) − sin(2πna/L)]/(2πn). Two properties come free. Disjoint intervals add, because Π̂₁Π̂₂ = 0; and the whole box gives Π̂ = 1̂, so P = 1 without further work. For n = 1 and [0, L/4] the formula returns 0.25 − 1/(2π) = 0.0908 — nine per cent of the hits where a classical bouncer would spend twenty-five per cent of its time, because the ground-state density is still climbing out of the wall there.

Zeros of ρₙ are zeros of sin(nπx/L), so they sit at x = jL/n. Of the values j = 0…n, two are the walls, which are boundary conditions rather than nodes; the n − 1 interior ones are the nodes, evenly spaced by L/n. Between them sit n antinodes at (2j−1)L/(2n), each rising to the same 2/L, so no lobe is favoured over another within a level. The count is not an accident of the box: Sturm's oscillation theorem says that for a real Sturm–Liouville problem with Dirichlet conditions, the eigenfunctions ordered by increasing eigenvalue carry 0, 1, 2, … interior zeros. That makes the node count a label. Given eigenvectors from numpy.linalg.eigh on a grid, counting sign changes identifies each level without comparing eigenvalues, and it keeps working when V is not flat and no closed form exists. Physically nodes are curvature, and curvature is kinetic energy: ⟨p̂²⟩ = (nπħ/L)² rises as the square of the number of half-waves squeezed between the walls.

Because ρ is a density, the probability of any single point is zero — at a node and at an antinode alike. What separates them is how fast a shrinking window loses probability. At a generic x₀ the density is locally flat, so P ≈ ρ(x₀)ε for a window of width ε. At a node ψ crosses zero linearly with slope √(2/L)(nπ/L), so ρ ≈ (2/L)(nπ/L)²(x − x₀)², and integrating gives P ≈ (nπ)²ε³/(6L³): a cubic vanishing, not a linear one. Put numbers on it with n = 3 and ε = 0.02L. On the node at L/3 the exact formula returns 1.18 × 10⁻⁴; on the antinode at L/6 it returns 3.99 × 10⁻², a factor of about 340. The node is a deep suppression of the count, not a wall inside the box.

Attach the time factor and nothing happens: Ψ(x, t) = ⟨x|n⟩e^{(−iEₙt/ħ)} has |Ψ|² = ρₙ(x), because the phase has modulus one. The stronger statement is that nothing is flowing either. The probability current j = (ħ/m) Im(ψ*∂ψ/∂x) vanishes identically for ψₙ, since ψₙ is real up to a constant phase, and continuity ∂ρ/∂t + ∂j/∂x = 0 then reads 0 = 0. So ρₙ is the frequency histogram of position measurements made on many identically prepared boxes, not the trace of one particle's motion. And that histogram is a one-shot object: under the projection postulate, an ideal measurement that finds the particle in [a, b] leaves the state Π̂|n⟩ / ‖Π̂|n⟩‖, which is no longer ψₙ and no longer an energy eigenstate, so a second measurement on the same box is drawn from a different density altogether.

A classical particle bouncing at constant speed spends equal time in equal lengths, giving a flat 1/L. ρₙ never approaches that pointwise: at every n, however large, it still swings between 0 and 2/L. What converges is the local average. Over a window of width Δ centred on x, ⟨ρₙ⟩ = 1/L − sin(nπΔ/L)cos(2πnx/L)/(nπΔ), a correction bounded by L/(nπΔ) relative to 1/L, and exactly zero whenever Δ is a whole number of periods L/n. Take L = 5.00 nm and n = 20, so the period is 0.250 nm. A window of 0.600 nm centred at 1.00 nm averages 0.1748 nm⁻¹, 13% below 1/L = 0.200 nm⁻¹, because 0.600 nm is 2.4 periods. Change it to 0.250 nm — a narrower window, not a wider one — and the average is exactly 0.200 nm⁻¹. Correspondence here is a statement about resolution matched to L/n, not about ħ becoming small.