Boundary conditions and the discrete spectrum

Write the eigenvalue problem in the position basis: ψ″(x) = [2m(V(x) − E)/ħ²] ψ(x). It is a second-order linear ODE, so for every complex E its solution set is a two-dimensional vector space — fix ψ(x₀) and ψ′(x₀) at one point and everything else follows. Nothing in that statement singles out any E. A free particle makes it sharp: for every E > 0 both e^(ikx) and e^(−ikx) solve it, and for every E < 0 both e^(κx) and e^(−κx) do. Quantisation is therefore never a property of the differential equation. It is a property of the operator, and an operator is a formula plus a domain: Ĥ acts on the square-integrable functions obeying the stated matching conditions, and E is an eigenvalue only if the two-dimensional solution space meets that domain in something other than the zero vector.

Let V(x) → V∞ at both ends and fix E < V∞. Far out the equation becomes ψ″ = κ²ψ with κ = √(2m(V∞ − E))/ħ, whose branches are e^(+κx) and e^(−κx). Square-integrability as x → −∞ admits only e^(+κx), fixing ψ up to scale: call that one-dimensional set of solutions ψ_L. The same demand as x → +∞ leaves ψ_R. Now count. The solution space is two-dimensional and we have chosen two lines through its origin; generically they are different lines, and the state fails to exist not because the ODE broke down but because the solution decaying on the left grows on the right. Their coincidence is measured by the Wronskian W(E) = ψ_L ψ_R′ − ψ_L′ ψ_R, constant in x because the equation carries no first-derivative term, and analytic in E below V∞ — not entire, because κ carries a square-root branch point at E = V∞. E is an eigenvalue exactly when W(E) = 0, and a function analytic there and not identically zero has isolated zeros, which can accumulate only at the edge V∞. That is the proof of discreteness, and it never mentions a wall.

Turn the Wronskian into an algorithm. Pick a matching point xₘ inside the well, integrate ψ_L rightwards from deep in the left forbidden region starting at ψ = e^(κx), integrate ψ_R leftwards from the right, and form the mismatch Δ(E) = ψ_L′/ψ_L − ψ_R′/ψ_R at xₘ — a log-derivative difference, which is −W(E) divided by ψ_L ψ_R and so shares its zeros. Δ changes sign as E crosses a level, and also through a pole wherever ψ_L or ψ_R vanishes at xₘ, so reject any bracket across which |Δ| blows up; on a genuine bracket brentq converges in a dozen evaluations. Two practical points. Integrate out of each forbidden region toward the well, never into one: integrating outward from the well lets the growing branch swamp the decaying one, and a start accurate to 10⁻¹² is dominated by round-off after twenty decay lengths. And count nodes as you sweep E upward — the nth bound state carries n − 1 interior nodes, so the node count names the root you have found and reveals any you skipped.

That argument needed both asymptotic regions classically forbidden, and this sets the window exactly. For E > V∞ the asymptotic solutions are e^(±ikx): both bounded, neither square-integrable, neither rejected, so every such E survives. The spectrum there is continuous and its kets are δ-normalised rather than members of L². For E < min V the sign flips everywhere: ψ″/ψ = 2m(V − E)/ħ² > 0 at every x, so ψ is convex away from the axis throughout and can never bend back to zero at both ends — the ground state must sit above the bottom of the well. Only min V < E < V∞ admits discrete eigenvalues. Two consequences the wall picture hides. A finite well with no walls at all, open to infinity on both sides, still holds bound states, because E < V∞ = 0 makes the outside forbidden. And an infinite well has V∞ = ∞, which is why every one of its states is discrete and it has no scattering sector at all.

For a symmetric well of half-width a and depth V₀ the matching conditions split by parity: k tan(ka) = κ for even states, −k cot(ka) = κ for odd ones, both subject to k² + κ² = 2mV₀/ħ². Write z = ka and z₀ = (a/ħ)√(2mV₀); every root then satisfies z tan z = √(z₀² − z²) or −z cot z = √(z₀² − z²) with 0 < z < z₀. The tangent branches start at z = 0, π, 2π, … and the cotangent branches at π/2, 3π/2, …, so a new root appears each time z₀ clears a multiple of π/2 and the count is N = ⌈2z₀/π⌉. Depth and width enter only through z₀, so quadrupling V₀ while halving a leaves N unchanged. Because z₀ > 0 always clears the first branch, a one-dimensional well binds at least one state however shallow — a statement that fails in three dimensions, where a shallow well binds nothing. An electron in a well of half-width 0.50 nm and depth 5.0 eV has z₀ = 5.73, so N = ⌈3.65⌉ = 4.

Now send V₀ → ∞ at fixed width. The decay constant κ = √(2m(V∞ − E))/ħ diverges, the penetration depth 1/κ collapses to zero, and k tan(ka) = κ degenerates: tan(ka) → ∞, so ka → (2n − 1)π/2, and combining both parities the condition on the full width L = 2a is simply sin(kL) = 0. It is the same matching condition with the exterior amplitude driven to zero, and it is the only square well whose roots are available in closed form: kₙ = nπ/L and Eₙ = n²π²ħ²/(2mL²) = n²h²/(8mL²). For an electron with L = 1.00 nm, E₁ = 0.376 eV and the levels sit at 1, 4, 9, 16 times it. The closed form is bought with a lie about the wall: continuity of ψ′ is abandoned there because no finite potential is present to enforce it, and a deep but finite well leaks over a decay length 1/κ of 0.195 nm for an electron 1.0 eV below the top.