Piecewise potentials and matching

Start from −(ħ²/2m)ψ″ + V(x)ψ = Eψ and move everything to one side: ψ″ = (2m/ħ²)(V − E)ψ. Nothing is approximate yet. Now suppose V is constant across some interval. The bracket is a fixed number and the equation is a linear ODE with constant coefficients, with three cases. If E > V, write (2m/ħ²)(V − E) = −k², so ψ″ = −k²ψ and ψ = A e^(ikx) + B e^(−ikx), or equivalently A′ sin kx + B′ cos kx. If E < V, write it as +κ², so ψ″ = +κ²ψ and ψ = C e^(κx) + D e^(−κx); the sines are gone. If E = V exactly, ψ″ = 0 and ψ is a straight line, A + Bx. Which pair you write is bookkeeping, not physics: e^(±ikx) for waves arriving and leaving, sin and cos for a standing wave, and cosh and sinh in place of e^(±κx) when a region is closed on both sides. Two constants per region, always — and that count is the currency the rest of the method spends.

Rearranged once more, ψ″/ψ = (2m/ħ²)(V − E): the curvature of ψ measured relative to ψ itself is fixed by how far E sits from the local floor. Where E > V the ratio is negative, so ψ curves back towards the axis — which is what oscillation is. Where E < V it is positive, so ψ curves away from the axis: a positive ψ that is falling either flattens and turns back up or crosses the axis and runs away below it, and only one exact slope lets it sink smoothly towards zero as a pure decaying exponential. That is why nothing wiggles in a classically forbidden region, and why you can never write a sine there. The same expression gives the local wavelength λ = 2πħ/√(2m(E − V)): as the floor rises towards E the wave stretches out, and at a turning point where E = V the curvature vanishes and ψ has an inflection. Sketch the solution before computing it and most sign errors die on the page. A companion rule from the semiclassical limit runs the other way for amplitude — a slower particle lingers, so |ψ|² tends to be larger where E − V is smaller.

The rules are derived, not assumed. Integrate ψ″ = (2m/ħ²)(V − E)ψ from x₀ − ε to x₀ + ε. The left side gives ψ′(x₀ + ε) − ψ′(x₀ − ε); the right side is a bounded function integrated over an interval of width 2ε. If V is finite — however big the jump — the right side vanishes as ε → 0, so ψ′ is continuous. Continuity of ψ itself follows one step earlier: if ψ jumped, ψ′ would contain a delta and ψ″ the derivative of a δ, and no finite (V − E)ψ could match that. So the ordinary rules are ψ continuous everywhere, ψ′ continuous wherever V is bounded. The derivation also tells you when they fail. Against an infinite wall the bound is gone and only ψ = 0 survives, with ψ′ unconstrained. Across V = −α δ(x) the right-hand integral does not vanish; it leaves ψ′(0⁺) − ψ′(0⁻) = −(2mα/ħ²)ψ(0), a fixed kink whose size is the strength of the δ.

Each boundary hands you two homogeneous equations in the region constants. Solving them as a pair drags the overall normalisation through the algebra for no reason, so divide one by the other and match the logarithmic derivative ψ′/ψ instead. It has dimension 1/length, it is unchanged if you rescale ψ, and it turns two equations into one condition plus a leftover amplitude ratio you can fix at the end. Take an infinite wall at x = 0, a flat floor of width a, and a step of height V₀ beyond it. Then ψ_I = A sin kx and ψ_II = B e^{(−κ(x−a))} are already continuous once B = A sin ka, and equating ψ′/ψ at x = a gives k cot ka = −κ, one transcendental equation in E with no amplitudes in sight. Symmetry buys the same saving twice over: if V(−x) = V(x), every non-degenerate eigenstate is even or odd, so you solve on half the axis and match at one boundary instead of two.

With N regions you have 2N constants plus one unknown energy. Each of the N − 1 internal boundaries supplies two equations, so 2(N − 1) in all. For a bound state the two outer regions run to infinity and normalisability kills the growing exponential in each, removing two constants and leaving 2N − 2 unknown coefficients against 2N − 2 homogeneous equations. A homogeneous system has only the trivial solution unless its determinant vanishes, and that determinant depends on E: the discrete spectrum is exactly the set of energies where it does, and the one remaining freedom is fixed by normalisation. A scattering state is counted differently. On one side you set the incoming amplitude by hand — you choose how intense the incident beam is — so no eigenvalue condition survives and E is continuous. A three-region barrier has N = 3: six constants, four matching equations, one incoming amplitude set to 1 and no wave arriving from the right, leaving exactly four unknowns for four equations at every E.

The model is a caricature, and it is worth naming the price. Real potentials are smooth: the image-charge barrier at a metal surface, or the band edge at a semiconductor heterojunction, changes over a length d of a few tenths of a nanometre rather than jumping. Treating it as a sharp step is safe only when d is small compared with the local de Broglie wavelength; for a 1 eV electron λ = 1.23 nm, so d = 0.3 nm is already marginal and the true reflection comes out smaller than the step formula predicts. Push the other way — V varying slowly over many wavelengths — and reflection becomes exponentially weak, which is the WKB limit. The remaining assumptions are structural: one dimension, one particle, a static V with no explicit time dependence, no spin, and non-relativistic kinematics. One rule genuinely changes in a solid. Where the effective mass differs across the junction, matching ψ′ is wrong; continuity of (1/m*)ψ′ is what conserves probability flux.