University Physics IV · Quantum Potentials · 7.6
The Δ-Function Well
The fastest complete bound-state problem in quantum mechanics: one integration across the spike hands you the kink in ψ′, the kink hands you κ, and κ hands you the single allowed energy. Learn it well — every contact interaction you meet later, from crystal impurities to cold-atom models, is this calculation again.
Build the model
Connect the measurement to the mechanism.
Take the finite square well and squeeze it: depth up, width down, the product V₀a held fixed at α. Every well in that family binds an electron; the limit remembers nothing about the shape except the area, and that limit is V(x) = −α δ(x). The model costs one strange feature — a potential that is infinite at a single point — and pays for it immediately: integrating the Schrödinger equation across the spike converts the infinity into a finite, fixed jump in ψ′, and that single condition replaces all the transcendental matching of the finite well.
The jump can be satisfied by exactly one decay constant, κ = mα/ħ², so there is exactly one bound state, at E = −mα²/2ħ² — quadratic in the strength, because the state must build its own size out of m, α and ħ alone; the well supplies no length scale of its own. What you give up is resolution: the delta is honest only while the real well is much narrower than the state it binds, κa ≪ 1, and it can never show you excited states, because a point has no room for nodes.
- Simple definition
- The δ-function well V(x) = −α δ(x) is an infinitely narrow, infinitely deep well of finite area α that binds exactly one state, whose derivative jumps by −(2mα/ħ²)ψ(0) at the spike.
- Example
- For an electron with α = 0.30 eV⋅nm the single level sits at E = −0.59 eV and the wavefunction e-folds every 1/κ = 0.25 nm on either side of the spike.
One integration across the spike replaces both walls' worth of matching conditions from the finite well.
α in eV⋅nm (energy × length) — the area of the well; ψ itself stays continuous through the spike
The jump condition is one equation with one root κ — that uniqueness is why there is exactly one state.
κ in m⁻¹ (nm⁻¹ in practice); 1/κ is the decay length, and ψ(0)² = κ
The only energy m, α and ħ can build — dimensional analysis fixes the form before any solving.
quadratic in the strength: doubling α quadruples the binding; E in J or eV
The bound level's scale controls transmission: T = ½ at E = |Eb|, rising to 1 at high energy, dying as E → 0.
E > 0 incident kinetic energy; |Eb| = mα²/2ħ²; same T for well and barrier, since α enters squared
A well much narrower than the state it binds may as well be a point — only its area survives the squeeze.
V₀ depth (eV) and a width (nm) of the real well; κa compares well width to state size
A spike with area, not a wall with depth
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.
Integrate across the spike
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.
Three lines to the only bound state
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.
Put numbers on the state
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.
The same spike read by a travelling wave
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 + |Eb|/E) and R = 1 − T, where |Eb| = 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 = |Eb|, 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.
Where the point model stops
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.
Change one variable at a time
Make the relationship visible.
Double the strength from 0.15 to 0.30 eV⋅nm: the peak sharpens, the decay length halves, and the level drops fourfold, −0.15 to −0.59 eV. Then slide d outward and watch the tail probability die as e(−2κd) — each extra decay length costs a factor of e² ≈ 7.4.
DECAY CONSTANT κ3.94 nm⁻¹
DECAY LENGTH 1/κ0.254 nm
BOUND ENERGY E-0.59 eV
P(|x| > d)2.0 %
Live interpretationDECAY CONSTANT κ: 3.94 nm⁻¹. DECAY LENGTH 1/κ: 0.254 nm. BOUND ENERGY E: −0.59 eV. P(|x| > d): 2.0 %
Catch the common trap
Explain before calculating.
A delta well of strength α binds an electron at energy −E₀, with decay length ℓ. The strength is doubled to 2α. What happens to the bound spectrum?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron sits in the well V(x) = −α δ(x) with α = 0.30 eV⋅nm. Find the decay constant κ, the decay length, and the bound-state energy. Take mc² = 511 keV and ħc = 197.3 eV⋅nm.
- κ = mα/ħ² = α⋅mc²/(ħc)² = 0.30 × 511000/(197.3)² = 153300/38927 = 3.94 nm⁻¹.
- Decay length 1/κ = 1/3.94 = 0.254 nm — the state e-folds every couple of ångströms on either side of the spike.
- E = −ħ²κ²/2m = −(ħc)²κ²/(2mc²) = −38927 × 15.5/1022000 = −0.59 eV. Equivalently E = −mα²/2ħ² = −(0.30)² × 511000/(2 × 38927) = −0.59 eV — the two routes must agree.
- Check the kink: ψ = √κ e(−κ|x|) has ψ′(0±) = ∓κ√κ, so Δψ′ = −2κ³⁄², while the jump condition demands −(2mα/ħ²)ψ(0) = −2κ⋅√κ. Identical, as it must be.
Answerκ = 3.94 nm⁻¹, decay length 0.25 nm, E = −0.59 eV.
MediumA δ-function well binds an electron at E = −0.50 eV. Find the well strength α, and the probability that a position measurement finds the electron more than 0.50 nm from the spike.
- Invert the energy for the decay constant: κ = √(2m|E|)/ħ = √(2⋅mc²⋅|E|)/ħc = √(2 × 511000 × 0.50)/197.3 = 714.8/197.3 = 3.62 nm⁻¹.
- The strength follows from κ = mα/ħ²: α = ħ²κ/m = (ħc)²κ/mc² = 38927 × 3.62/511000 = 0.276 eV⋅nm.
- The density is |ψ|² = κ e(−2κ|x|); integrating both tails gives P(|x| > d) = 2∫ from d to ∞ of κ e(−2κx) dx = e(−2κd).
- With 2κd = 2 × 3.62 × 0.50 = 3.62: P = e(−3.62) = 0.027 — about 2.7%, even though 0.5 nm is only 1.8 decay lengths out.
Answerα = 0.276 eV⋅nm; P(|x| > 0.5 nm) = e(−3.62) ≈ 0.027, about 2.7%.
HardA nitrogen impurity in a crystal traps an electron; model it first as a square well of depth V₀ = 3.0 eV and width a = 0.080 nm. (i) Replace the well by a delta function and predict the binding energy. (ii) Verify the replacement was legitimate. (iii) Find the probability that a 1.5 eV conduction electron passes the impurity without reflecting.
- Only the area survives the squeeze: α = V₀a = 3.0 eV × 0.080 nm = 0.24 eV⋅nm.
- κ = α⋅mc²/(ħc)² = 0.24 × 511000/38927 = 3.15 nm⁻¹, so Eb = −(ħc)²κ²/(2mc²) = −38927 × 9.92/1022000 = −0.38 eV.
- Legitimacy check, after the fact: κa = 3.15 × 0.080 = 0.25 ≪ 1. The state's decay length 1/κ = 0.32 nm is four times the well's width, so the electron cannot resolve the profile and only the area matters.
- For the travelling electron use the flux ratio T = 1/(1 + |Eb|/E) = 1/(1 + 0.378/1.5) = 1/1.252 = 0.80.
- So R = 0.20: one electron in five reflects from a purely attractive defect — a classical particle would accelerate over it and never turn back.
AnswerEb ≈ −0.38 eV; κa ≈ 0.25, so the delta model is sound; T ≈ 0.80, R ≈ 0.20.