Admissibility and matching conditions

Read −(ħ²/2m)ψ″ + Vψ = Eψ as an ODE and it is entirely undiscriminating. For any V and any E, real or complex, it has a two-dimensional space of solutions: hand it ψ(x₀) and ψ′(x₀) and integration supplies the rest. The equation alone therefore permits every energy, which is to say it selects none. Quantisation is a boundary-value statement, not a differential one. Put an infinite wall at x = 0 and a 10.0 eV well over 0 < x < 1.50 Å, and you can integrate outward at E = −2.4 eV, at −2.540 eV and at −2.6 eV alike. All three solve the ODE exactly. Only the middle one decays as x → ∞; the other two grow exponentially in the classically forbidden region and are not in L², so they are not states. Naming the admissible set — the domain of Ĥ — is precisely what turns a differential equation into a spectrum.

Every matching rule in this topic comes from one manoeuvre. Write the equation as ψ″ = (2m/ħ²)(V − E)ψ and integrate it from x₀ − ε to x₀ + ε, giving ψ′(x₀+ε) − ψ′(x₀−ε) = (2m/ħ²)∫(V − E)ψ dx. Then ask what that integral does as ε → 0. If |V| ≤ Vₘₐₓ on the interval and ψ is continuous there, the integrand is bounded by (Vₘₐₓ + |E|)|ψ|ₘₐₓ and the whole integral is at most 2ε times that, so it vanishes: ψ′ is continuous, and stays continuous across a step of any finite height. If V = −αδ(x₀), the integral picks out −αψ(x₀) exactly, however small ε becomes, so Δψ′ = −(2mα/ħ²)ψ(x₀). If V is infinite over a region, the integral does not tend to zero at all, ψ′ is unconstrained, and the wall constrains ψ instead. Integrate a second time and the same bound gives Δψ = 0 in all three cases: ψ itself never jumps.

Solving piecewise leaves an unknown amplitude on each side, and the instinct is to solve for them. You rarely need to. Divide the ψ′ condition by the ψ condition and the amplitudes cancel, leaving the logarithmic derivative L = ψ′/ψ. Across a bounded interface L is continuous; across V = −αδ(x) it jumps by exactly −2mα/ħ², because ψ is continuous there. For the wall-plus-well below, k cos(ka)/sin(ka) on the inside must equal −κ on the outside — one real equation in one unknown E, with no amplitudes anywhere. This is also what a shooting code compares: integrate in from the left, integrate in from the right, and match ψ′/ψ at a meeting point, since that ratio is invariant under the arbitrary scaling of each piece. In practice one matches the Wronskian ψ_L ψ′_R − ψ′_L ψ_R instead, so that a node sitting at the match point does not send the ratio to infinity.

Take V = −αδ(x) with α > 0. Away from the origin the equation is ψ″ = κ²ψ with κ = √(−2mE)/ħ, so square-integrability leaves A e^(κx) on the left and B e^(−κx) on the right, and continuity at 0 sets B = A: ψ = A e^(−κ|x|), with ψ′(0⁺) = −κA and ψ′(0⁻) = +κA, a delivered kink of −2κA. The demanded kink is −(2mα/ħ²)A. Matching gives κ = mα/ħ², hence E = −ħ²κ²/2m = −mα²/2ħ², and normalising ∫|A|²e^(−2κ|x|)dx = |A|²/κ = 1 fixes A = √κ. For an electron with α = 7.62 eV Å this is κ = 1.000 Å⁻¹ and E = −3.810 eV. Two features deserve names. There is exactly one bound state for every α, because matching produced a single linear equation rather than a transcendental one. And a repulsive δ, V = +βδ(x), demands Δψ′ = +(2mβ/ħ²)ψ(0) > 0, which no decaying e^(−κ|x|) can supply — the sign of the jump condition is the entire reason a well binds and a barrier does not.

An infinite wall is the V₀ → ∞ limit of a step, and taking that limit carefully shows what survives. Beyond a step of height V₀ the decaying solution is ψ ∝ e^(−κx) with κ = √(2m(V₀ − E))/ħ, so for an electron κ ≈ √(V₀ / 3.810 eV Ų): 5.12 Å⁻¹ at V₀ = 100 eV and 51.2 Å⁻¹ at 10 keV. The penetration depth 1/κ collapses from 0.195 Å to 0.0195 Å, and in the limit ψ is pinned to zero at the wall. What does not survive is any statement about ψ′. The logarithmic derivative just beyond the wall is −κ, which diverges, so the matching condition degenerates from “ψ′/ψ is continuous” to “ψ = 0”, leaving ψ′ free to jump. That is the whole content of an infinite wall. Imposing ψ(x_w) = 0 and ψ′(x_w) = 0 together on a second-order ODE forces ψ ≡ 0 by uniqueness: the extra condition does not select a better state, it abolishes every state.

Ĥ is not the formula; it is the formula together with a domain, and the choice changes the answer. On [0, L] the expression −(ħ²/2m)d²/dx² is self-adjoint under Dirichlet ψ(0) = ψ(L) = 0, under Neumann ψ′(0) = ψ′(L) = 0, under the Robin family ψ′(0) = γ₀ψ(0) and ψ′(L) = γ_Lψ(L) with γ real, and under periodic conditions — and these are different operators, not different notations. For an electron on L = 10.0 Å, Dirichlet gives Eₙ = 0.376 n² eV, a ground state at 0.376 eV. Neumann admits the constant function and puts its ground state at exactly 0. Mixing them, ψ(0) = 0 with ψ′(L) = 0, gives kL = (n − ½)π and a ground state at 0.0940 eV. Three spectra, one differential expression. The moral is practical as well as formal: a finite-difference Hamiltonian on a box silently imposes Dirichlet at the outermost grid points, so it answers the boxed question — and the levels nearest the top of a well are the ones that notice.