The δ-function well

Write V(x) = −α δ(x). The delta has dimension one over length, so α carries energy × length: it is the area of the well, the product of depth and width of whatever narrow well it stands in for. Squeeze a square well — 3 eV deep over 0.08 nm, then 6 eV over 0.04 nm, then 12 eV over 0.02 nm — and a bound electron whose own size is far larger than the well cannot tell the members of that family apart: it responds to ∫V dx = −α, not to the profile. The delta is the limit that keeps only that number. Notice what the model therefore lacks: any length of its own. Whatever size the bound state has must be built from m, α and ħ, and the only length those three can form is ħ²/mα — that one observation already fixes the whole answer up to numerical factors.

The time-independent equation reads −(ħ²/2m)ψ″ − α δ(x)ψ = Eψ. Integrate every term from −ε to +ε. The right-hand side vanishes as ε → 0, because ψ is finite and the interval shrinks; the delta term yields −αψ(0) whatever ε is; the first term integrates to −(ħ²/2m)[ψ′(ε) − ψ′(−ε)]. So the derivative jumps by Δψ′ = −(2mα/ħ²)ψ(0): a finite kink whose size the strength dictates. ψ itself must stay continuous — a step in ψ would put a δ′ into ψ″ that nothing else in the equation could cancel. Keep the full ladder straight: where V is finite, ψ and ψ′ are both continuous; where V holds a δ, ψ is continuous and ψ′ jumps by exactly this amount; only at a hard infinite wall is ψ pinned to zero with ψ′ unconstrained.

For E < 0, away from the origin the equation is ψ″ = κ²ψ with κ = √(−2mE)/ħ, and normalisability kills one exponential on each side: ψ = A e^(κx) for x < 0 and A e^(−κx) for x > 0, with the same A because ψ is continuous. The bound state is forced to be even — an odd solution has ψ(0) = 0 and never feels the well at all. Now apply the jump: ψ′(0⁺) − ψ′(0⁻) = −κA − κA = −2κA must equal −(2mα/ħ²)A, so κ = mα/ħ². One equation, one root, one bound state: E = −ħ²κ²/2m = −mα²/2ħ². Normalising, ∫|ψ|² dx = A²/κ = 1 gives A = √κ, so the density at the spike is |ψ(0)|² = κ: a stronger well makes a taller, narrower, more deeply bound state, always with unit area under |ψ|².

Work in electron units: κ = mα/ħ² = α⋅mc²/(ħc)², with mc² = 511 keV and ħc = 197.3 eV⋅nm. A strength α = 0.30 eV⋅nm gives κ = 3.94 nm⁻¹, a decay length 1/κ = 0.25 nm, and E = −0.59 eV — a chemistry-scale binding from an ångström-scale spike. The energy splits instructively: ⟨V⟩ = −α|ψ(0)|² = −ακ = 2E, twice the total, while ⟨T⟩ = E − ⟨V⟩ = −E = +0.59 eV. The kink is where that kinetic energy lives: a cusp costs curvature, and curvature is momentum content. Probabilities fall exponentially in units of the decay length, P(|x| > d) = e^(−2κd): 13.5% of the electron sits beyond one decay length, 1.8% beyond two.

Send in a plane wave with E > 0 instead. The same jump condition, applied to e^(ikx) + r e^(−ikx) on the left and t e^(ikx) on the right, fixes r and t in a line each, and the flux ratios come out as T = 1/(1 + |E_b|/E) and R = 1 − T, where |E_b| = mα²/2ħ² is the depth of the bound level. The energy of the state below zero governs the scattering above it: T = ½ exactly at E = |E_b|, transmission climbs towards one at high energy and dies linearly as E → 0. Note that the sign of the spike never mattered — α enters squared, so a delta barrier transmits identically. An attractive well reflecting an electron that classically would simply accelerate over it is as clean a wave effect as this course offers.

The delta stands in for a real well of depth V₀ and width a only while the state it binds is much larger than the well: κa ≪ 1. Check it after solving, since κ = mα/ħ² depends on the answer. A 3.0 eV well 0.08 nm wide has α = 0.24 eV⋅nm and κ = 3.15 nm⁻¹, so κa = 0.25 and the point model is decent; push κa towards one and you must return to the finite well's transcendental matching — which then also holds the excited states a point cannot, since nodes need room. Two warnings travel with the model. It is strictly one-dimensional: in 3D an attractive delta is not a well-defined operator without regularisation, and a shallow 3D well need not bind at all, whereas in 1D any attractive area binds something. And the model has no excited spectrum, so any question about transitions is outside its competence.