The infinite square well

Do not start from "the particle cannot be there". Start from a finite well of depth V₀ and take the limit. Outside, ψ ∝ e^(−κ|x|) with κ = √(2m(V₀ − E))/ħ, so as V₀ → ∞ the decay length 1/κ shrinks to nothing and the outside amplitude collapses. Continuity of ψ then hands you ψ(0) = ψ(L) = 0 — the boundary conditions are inherited, not assumed. What does not survive the limit is continuity of ψ′. Integrating the Schrödinger equation across a wall gives Δψ′ = (2m/ħ²)∫Vψ dx, which vanishes for finite V but not for infinite V, so the slope is allowed to jump at each edge. For ψ₁ that jump runs from 0 outside to √(2/L)(π/L) just inside. It is the first sign that this model is a limit, not a potential anyone could build.

Inside, V = 0 and ψ″ = −k²ψ with k = √(2mE)/ħ, whose general solution is ψ = A sin kx + B cos kx. Impose ψ(0) = 0: cos 0 = 1, so B = 0. Impose ψ(L) = 0 with A ≠ 0: sin kL = 0, so kL = nπ, and E = ħ²k²/2m follows immediately. Now handle the integer carefully. n = 0 gives ψ ≡ 0 everywhere — a function you cannot normalise, describing a particle that is nowhere — so it is not a state and E = 0 is not in the spectrum. Negative n returns sin(−nπx/L) = −sin(nπx/L), the same function up to an overall sign, and a global phase is physically invisible, so n = −3 and n = 3 are one state, not two. The spectrum is therefore n = 1, 2, 3, … with no repeats and no rung at zero energy.

Born's rule only means something once ∫|ψ|² dx = 1. With ψ = A sin(nπx/L), use sin²θ = (1 − cos 2θ)/2: ∫₀^L A² sin²(nπx/L) dx = A²[x/2 − (L/4nπ) sin(2nπx/L)]₀^L = A²L/2, since the sine vanishes at both limits. So A = √(2/L) for every n — the constant does not grow with n, even though higher states wiggle faster, because sin² averages to exactly ½ whatever the frequency. Two checks. The units are right: in one dimension |ψ|² is a probability per unit length, so ψ carries m^(−1/2), and √(2/L) does. And the shape is right: |ψₙ|² peaks at 2/L, exactly twice the classical density 1/L of a particle bouncing at constant speed, because the quantum particle piles probability into n humps and empties it at the nodes between them.

Compute ∫₀^L ψₘψₙ dx with sin a sin b = ½[cos(a − b) − cos(a + b)]. Both cosines integrate to zero across the well when m ≠ n, and the first becomes cos 0 = 1 when m = n, returning δₘₙ. The orthogonality is not luck: Ĥ is Hermitian, so eigenfunctions belonging to different eigenvalues are automatically orthogonal, and the integral merely confirms a theorem you already had. What you gain is a basis. These sines are exactly the Fourier sine series on [0, L], which is complete, so any state of the well — a Gaussian squeezed in at t = 0, a triangle, a single half-lobe — can be written as ψ = Σ cₙψₙ with cₙ = ∫ψₙ*ψ dx, and Σ|cₙ|² = 1 makes |cₙ|² the probability of measuring Eₙ. Orthonormality is precisely what lets one overlap integral extract one coefficient.

Eₙ = n²h²/(8mL²) carries two scalings. Up the ladder E ∝ n², so gaps widen as Eₙ₊₁ − Eₙ = (2n + 1)E₁ — unlike the harmonic oscillator's even rungs or hydrogen's converging ones, and a fingerprint you can test against a measured spectrum. Across boxes E ∝ 1/L², the price of confinement: halve a quantum dot from 5.0 nm to 2.5 nm and every level quadruples. Numbers make it concrete. For an electron Eₙ = n² × 0.376 eV⋅nm² / L², so a 0.50 nm well gives E₁ = 1.50 eV and a first gap of 4.51 eV — already a 275 nm ultraviolet photon, and every higher gap lies deeper in the ultraviolet. Put a proton in the same box and E₁ falls by the mass ratio to 0.82 meV. And E₁ can never be zero, because a ψ with no curvature is a straight line, and no straight line except ψ ≡ 0 vanishes at both walls: curvature is forced, and curvature is kinetic energy.

The buy is real: an exact spectrum, an exact basis, and a first model of any tightly confined electron — a quantum dot, a nanowire segment, a π-electron on a conjugated chain. The costs are worth naming out loud. ψₙ is not a momentum eigenstate: p̂ψₙ ∝ cos(nπx/L), not sin, because a standing wave is an equal superposition of +ħkₙ and −ħkₙ. So ⟨p⟩ = 0 while ⟨p²⟩ = ħ²kₙ² = 2mEₙ, and for n = 1 the uncertainty product is Δx Δp = 0.568ħ, comfortably above the ħ/2 floor. The walls are the bigger fiction. A real well has finite depth, so the wavefunction penetrates and the effective width exceeds L — which is why the infinite well always overestimates the levels — and it holds only finitely many bound states rather than an unbounded ladder.