The infinite-wall limit

Integrate the time-independent equation ψ″ = (2m/ħ²)(V − E)ψ across a slice [a, b]: ψ′(b) − ψ′(a) = (2m/ħ²)∫ₐᵇ (V − E)ψ dx. If |V| ≤ Vₘₐₓ on that slice, the right side is bounded by (2m/ħ²)(Vₘₐₓ + |E|)⋅max|ψ|⋅(b − a), which vanishes as b − a → 0. That is the entire proof that ψ′ is continuous, and it can fail in two ways. A delta well V = λδ(x − c) is unbounded but integrable, and leaves a prescribed jump Δψ′ = (2mλ/ħ²)ψ(c). An infinite wall is unbounded on every slice that touches it, so no bound survives at all and the jump is whatever the solution needs. Continuity of ψ itself rests on something quite different: ⟨T⟩ = (ħ²/2m)∫|ψ′|² dx must be finite, and a step in ψ makes ψ′ a δ whose square is not integrable. A step in ψ′ only makes ψ′ a bounded step function, which integrates perfectly well. Hence the hierarchy you should carry: ψ continuous always, ψ′ guaranteed continuous only where V is bounded.

Put the wall at x = 0, with V = V₀ for x < 0 and V = 0 inside. The exterior solution that stays normalisable is ψ = B e^(κx) with κ = √(2m(V₀ − E))/ħ, so ψ′(0⁻)/ψ(0⁻) = κ. Matching that ratio, rather than ψ and ψ′ separately, removes both unknown amplitudes and leaves ψ′(0⁺) = κ ψ(0⁺): a Robin boundary condition whose parameter is κ. Neumann is κ = 0, Dirichlet is κ = ∞, and everything between is a partly transparent wall. Now raise V₀ at fixed n. The interior slope cannot run away with κ — it is at most kA with k = √(2mE)/ħ — so the only way the ratio can track a diverging κ is ψ(0) = ψ′(0)/κ → 0. Put numbers on it. An electron in a 1.0 nm well with V₀ = 10 eV has k = 2.79 nm⁻¹ and κ = 15.96 nm⁻¹, so ψ(0)/ψₘₐₓ = sin(kδₑ) = 0.17: the state is still 17% of its peak height at the wall. Deepen to V₀ = 100 eV and that falls to 0.059.

In the box, ψₙ = √(2/L) sin(nπx/L), so ψₙ(0) = 0 while ψₙ′(0⁺) = √(2/L)(nπ/L); with ψ ≡ 0 outside, ψ′(0⁻) = 0. The derivative therefore jumps by 4.44 nm^(−3/2) at each wall for n = 1 and L = 1.0 nm. Nothing breaks. The kink costs no energy in the form ⟨T⟩ = (ħ²/2m)∫₀^L |ψ′|² dx, because ψ′ is a bounded function on the interval; and that integral really is ⟨ψ|p̂²|ψ⟩/2m here, since integrating by parts leaves the surface term ψ*ψ′|₀^L, which vanishes precisely because ψ(0) = ψ(L) = 0. The Dirichlet condition is what makes the two expressions for kinetic energy agree. Where the kink does bite is anywhere you differentiate twice at a wall: ψ″ carries a delta there, so the momentum-space wavefunction falls off only as 1/p² instead of the finite well's 1/p³, a plane-wave basis that ignores the walls converges slowly at the edges, and ⟨p̂⁴⟩ diverges for every box eigenstate even though ⟨p̂²⟩ is perfectly finite.

Inside the finite well the solution is ψ ∝ sin(k(x + δₑ)), and the Robin condition k cot(kδₑ) = κ fixes the offset through tan(kδₑ) = k/κ. The interior sine reaches zero a distance δₑ outside each wall, so the ground state fits half a wavelength across L + 2δₑ rather than across L, giving k = π/(L + 2δₑ). Run the electron example. L = 1.0 nm and V₀ = 10 eV give z₀ = 8.100, and the lowest root of z tan z = √(z₀² − z²) is z₁ = 1.3974, so k = 2.795 nm⁻¹ and E₁ = 0.298 eV. Then κ = 15.96 nm⁻¹, k/κ = 0.1751, δₑ = 0.0620 nm — about 1% below the crude 1/κ = 0.0627 nm — and L_eff = 1.124 nm, whose π/L_eff reproduces k exactly. The Dirichlet box, using L = 1.0 nm, returns 0.376 eV: 26% high, as 4/(κL) = 0.25 predicted. The sign is structural. Leaking into the wall lengthens the wavelength, and a longer wavelength means less energy. And because κₙ falls as Eₙ rises, the fractional error 4/(κₙL) grows with n: the top of the ladder is the worst part of it.

The convergence is genuine but partial. As V₀ → ∞ the finite-well eigenfunctions converge to the box ones in the L² norm and uniformly on any closed interval inside (0, L). They do not converge in derivative at the walls: every finite well has ψ′ continuous there and the box never does, so the limit of the derivatives is not the derivative of the limit. The spectrum breaks a second way. A well of depth V₀ holds only ⌈(L/π)√(2mV₀)/ħ⌉ bound states — six for L = 1.0 nm and V₀ = 10 eV, seventeen at V₀ = 100 eV — while the box has an infinite ladder. So the two limits do not commute. Fix n and send V₀ → ∞, and the finite-well level converges to n²π²ħ²/(2mL²); fix V₀ and send n → ∞, and the state leaves the well for the scattering continuum instead. Whenever you quote a box result, quote the n you trust with it: the condition is n ≪ (L/π)√(2mV₀)/ħ, not merely V₀ ≫ E₁.

The decision is one ratio, δ/L, and it takes a line to compute. A conduction electron held in by a work function Φ ≈ 4.5 eV has δ = √(0.0381/4.5) nm = 0.092 nm; in a 10 nm metal grain that is 0.9% per wall and a 3.7% shift in energy, so a hard wall is honest there. The π-electrons of a cyanine dye sit in a box about 1 nm long behind a barrier of a few eV, so δ/L is nearer 0.1 — which is why a box fit to the absorption maxima always returns a length longer than the bond geometry allows: the extrapolation length is hiding inside the fitted L. A GaAs/AlGaAs quantum well is the outright failure case. With a barrier ΔE_c = 0.30 eV and m* = 0.067 mₑ, so that ħ²/2m* = 0.569 eV nm², the extrapolation length is δₑ = 1.40 nm, and a 10 nm well leaks 14% of its width into each barrier. The box predicts 56.1 meV where the true ground state is 34.2 meV, and the well holds three bound states, not infinitely many. Device modelling never uses a hard wall.