University Physics V · Finite Potential Wells and Quantum Tunneling · 6.1
Bound States & the Continuum
Every scattering calculation from here on is written in a basis that is half sum and half integral. This lesson sorts the halves: which eigenvectors you can normalise to one, which are distributions you can only normalise to a δ, and how to write a single identity operator that carries both.
Build the model
Connect the measurement to the mechanism.
Take Ĥ = p̂²/2m + V(x̂) on L²(ℝ) with V real and short-range, so V(x) → V∞ faster than 1/|x|. That one asymptotic condition — a floor at infinity — cuts the spectrum in two, and the cut is a statement about which solutions of the same second-order equation survive the boundary condition at |x| = ∞. Below the floor the asymptotic equation is ψ″ = κ²ψ with κ = √(2m(V∞ − E))/ħ, whose two solutions grow and decay; demanding decay at both ends is two conditions on a two-parameter family with one free scale, so it fails at every E except a discrete set.
The survivors are honest vectors of L²(ℝ), non-degenerate in one dimension by a Wronskian argument, and finite in number for a short-range well. Above the floor the asymptotic equation is ψ″ = −k²ψ, both solutions oscillate at constant amplitude, and nothing is rejected: every E > V∞ is in the spectrum, twice over, and no eigenfunction is square-integrable. Those are not Hilbert-space vectors but distributions, normalised as ⟨k|k′⟩ = δ(k − k′), and the price is that the identity becomes a sum plus an integral.
Physical states are still normalisable packets; the continuum labels are only the coordinates you expand them in.
- Simple definition
- A bound state is a square-integrable eigenvector of Ĥ with energy below the potential's value at infinity, normalised by ⟨n|m⟩ = δₙₘ; an unbound state has energy above that floor, is not square-integrable, and is normalised by ⟨k|k′⟩ = δ(k − k′).
- Example
- For V(x) = −αδ(x) with α = 0.3904 eV nm an electron has exactly one bound state, Eb = −1.000 eV with ψb = √κ e(−κ|x|) and κ = 5.12 nm⁻¹; every E > 0 is doubly degenerate, and none of those states can be normalised to 1.
One comparison decides which sector a state belongs to, before any matching or integration.
V∞ in J or eV, usually set to 0. Short-range means |V| falls faster than 1/|x|.
Below the floor one asymptotic solution must be discarded at each end; above it neither can be.
κ and k in m⁻¹; 1/κ is the tail length, 2π/k the far-field wavelength.
Fixes the constant in front of every scattering state, hence every golden-rule matrix element.
ψₙ carries m(−1/2); δ(k − k′) carries metres, so ψₖ is dimensionless, e.g. e(ikx)/√(2π).
Expand any packet as cₙ = ⟨n|ψ⟩ and c(k) = ⟨k⁺|ψ⟩, with Σ|cₙ|² + ∫|c(k)|² dk = 1.
|k⁺⟩ carries incoming e(ikx); k < 0 is incidence from the right, so the integral covers all real k.
Decay at both ends forces W = 0, so bound levels are simple; oscillating tails allow W ≠ 0, so the continuum is twofold.
Two solutions at one real E. W = 0 exactly when they are proportional.
One bound state for any α > 0; above it, cos(k|x| + δ)/√π with tan δ = κ/k and sin(kx)/√π untouched.
α = 0.3904 eV nm gives an electron κ = 5.12 nm⁻¹, Eb = −1.000 eV, tail length 0.195 nm.
The floor at infinity is the whole input
Take Ĥ = p̂²/2m + V(x̂) on L²(ℝ), V real and short-range: V(x) → V∞ as |x| → ∞, falling faster than 1/|x|, with V∞ set to 0 by convention. Far from the well the equation is free, ψ″ = (2m/ħ²)(V∞ − E)ψ, and the sign of V∞ − E decides everything. Below the floor that coefficient is positive and the two asymptotic solutions are e(±κx) with κ = √(2m(V∞ − E))/ħ: one decays, one blows up. Above it the coefficient is negative and both solutions oscillate, e(±ikx) with k = √(2m(E − V∞))/ħ, neither growing nor dying. Notice where the boundary condition has moved. In a box you imposed ψ = 0 on a wall; here you impose behaviour at |x| = ∞, and the entire difference between the two sectors is which asymptotic solutions that condition can reject. For an electron 1.00 eV below the floor, κ = 5.12 nm⁻¹ and the tail dies over 0.195 nm; 1.00 eV above it, k = 5.12 nm⁻¹ and the wave runs on for ever with wavelength 1.23 nm.
Counting conditions: the bound sector is discrete and simple
Below the floor, start at the far left with the only admissible solution — the decaying one, unique up to scale — and integrate rightwards at a trial E. Generically the growing exponential reappears at the far right; killing it is one real condition on one real unknown, so the solutions are isolated points rather than an interval. That is where quantisation comes from: the boundary condition, not a postulate. The levels are also simple in one dimension. If ψ₁ and ψ₂ solve the same E, their Wronskian W = ψ₁ψ₂′ − ψ₂ψ₁′ has W′ = 0, and if both decay then W → 0 far away, so W ≡ 0 everywhere and ψ₂ ∝ ψ₁. How many are there? Finitely many for a short-range well: a finite square well of half-width a holds floor(2z₀/π) + 1 of them with z₀ = a√(2mV₀)/ħ, so V₀ = 10.0 eV and a = 0.200 nm give z₀ = 3.24 and three levels. In one dimension there is always at least one, however shallow the well — false in three dimensions. And a Coulomb tail is not short-range: hydrogen's −13.6/n² eV levels accumulate at the floor from below, infinitely many.
Above the floor: two solutions, both admissible
Above the floor nothing is thrown away, so every E > V∞ is in the spectrum and carries two independent eigenfunctions — the Wronskian is now free to be nonzero, because oscillating tails never vanish. Which two you name is a choice of basis. For scattering, label by what comes in: ψₖ⁺ behaves as e(ikx) + r e(−ikx) on the left and t e(ikx) on the right, and its mirror image is incident from the right. For symmetric V, parity is cheaper. The delta well gives both channels in closed form: the even solution is cos(k|x| + δ)/√π with tan δ = κ/k, so at κ = 5.12 nm⁻¹ and k = 8.00 nm⁻¹ the wave is pulled inwards by δ = 32.6°, while the odd solution sin(kx)/√π vanishes at the well and never feels it, δ = 0. The count is a count of admissible asymptotic solutions, not a slogan. Put a step in the potential so the two floors differ, and for VL < E < VR the continuum is non-degenerate again, because the right-hand growing exponential must still be discarded.
Delta normalisation, and the units that come with it
∫|e(ikx)|² dx diverges, so a scattering eigenfunction is not a vector of L²(ℝ) and cannot be normalised to one. What is finite is the pairing of two different labels: (1/2π)∫e(i(k′−k)x) dx = δ(k − k′), so ψₖ = e(ikx)/√(2π) is δ-normalised. Track the units, because they catch errors: k carries m⁻¹, so δ(k − k′) carries metres, ∫dx supplies one metre, and ψₖ must therefore be dimensionless — where a bound ψₙ carries m(−1/2). Relabelling by energy needs the Jacobian, |E⟩ = |k⟩√(dk/dE) with dk/dE = m/(ħ²k); for an electron at 1.00 eV, k = 5.12 nm⁻¹ and the prefactor is 0.639 nm(−1/2) eV(−1/2). Mathematically these objects sit one level outside the Hilbert space, in the dual of a space of well-behaved test functions, the rigged triple Φ ⊂ H ⊂ Φ×. Physically the message is simpler: nothing prepares |k⟩. States are packets ∫dk c(k)|k⟩ with ∫|c(k)|² dk = 1, and only a packet has a moving ⟨x⟩ and a finite ⟨Ĥ⟩.
One identity, half sum and half integral
Completeness is why the split matters: 1̂ = Σₙ|n⟩⟨n| + ∫dk|k⁺⟩⟨k⁺|, and dropping either piece breaks every expansion. Write ψ = Σₙ cₙ|n⟩ + ∫dk c(k)|k⁺⟩ with cₙ = ⟨n|ψ⟩ and c(k) = ⟨k⁺|ψ⟩; then Σ|cₙ|² + ∫|c(k)|² dk = 1 and ⟨Ĥ⟩ = Σ Eₙ|cₙ|² + ∫dk E(k)|c(k)|². The physical reading is sticking against escape. Halve a delta well suddenly and the old ground state is no longer an eigenstate: its overlap with the new bound state is 2√(κκ₀)/(κ + κ₀), so the electron stays bound with probability 4κκ₀/(κ + κ₀)² = 8/9 and is ionised with probability 1/9 — and that 1/9 is an integral over a continuum, not a sum over levels. Two habits follow. Any sum over states in perturbation theory, in a Green's function, or in a golden-rule rate carries that integral silently. And a bound-state basis alone is never complete, which is why truncating a packet onto the bound levels leaks probability.
What the computer actually returns
Numerically you truncate: a finite-difference Ĥ on [−X, X] with Dirichlet walls, then numpy.linalg.eigh. The negative eigenvalues are the physical ones, and they converge exponentially because what you discarded is the tail beyond the wall, of relative weight e(−2κX) — for κ = 5.12 nm⁻¹ and X = 2.00 nm that is 1.3 × 10⁻⁹. The positive eigenvalues are not the continuum. They are the box's own modes, kₙ ≈ nπ/(2X), spaced near 1.00 eV by ΔE = (ħ²k/m)Δk = 0.31 eV at 2X = 4.00 nm and 0.15 eV at 8.00 nm — a spacing that halves each time the box doubles, which is the diagnostic: rerun at a different X and keep only the eigenvalues that do not move. What is physical up there is not any single level but the change in the density of states, (1/π)dδ/dE, and the phase shift itself. If you want δ(E) or T(E), solve a boundary-value problem at fixed E instead of an eigenvalue problem.
Change one variable at a time
Make the relationship visible.
Drag the window X from 0.30 to 0.60 nm: the bound weight creeps from 0.954 to 0.998 and stops, while the continuum weight runs 0.071 to 0.154 and keeps climbing. Raising α tightens the bound spike; raising k shrinks the phase shift δ, never the amplitude.
BOUND LEVEL Eb-0.998 eV
CONTINUUM E2.44 eV
BOUND WEIGHT IN ±X0.994
CONTINUUM WEIGHT IN ±X0.147
Live interpretationBOUND LEVEL Eb: −0.998 eV. CONTINUUM E: 2.44 eV. BOUND WEIGHT IN ±X: 0.994. CONTINUUM WEIGHT IN ±X: 0.147
Catch the common trap
Explain before calculating.
An electron meets V(x) = −αδ(x) with α = 0.3904 eV nm, giving κ = mα/ħ² = 5.12 nm⁻¹ and one bound level at E = −1.000 eV. Count the linearly independent eigenfunctions at E = −1.000 eV and at E = +1.000 eV, and say how each is normalised.
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron is bound by V(x) = −αδ(x) with α = 0.300 eV nm. Using ħ²/mₑ = 0.0762 eV nm², find κ, the binding energy and the tail length, then the probability of finding the electron within 0.100 nm of the well. Finally, count the independent eigenfunctions at E = +0.591 eV.
- Integrate the equation across the spike: ψ′(0⁺) − ψ′(0⁻) = −(2mα/ħ²)ψ(0). The only E < 0 solution decaying both ways is ψb = √κ e(−κ|x|), whose kink gives −2κ = −2mα/ħ², so κ = α ÷ (ħ²/m) = 0.300 ÷ 0.0762 = 3.94 nm⁻¹.
- Energy from κ: Eb = −ħ²κ²/(2m) = −(0.0762 × 3.937²)/2 = −0.591 eV. Nothing was quantised by hand; the single demand that ψ decay at both ends picked one E out of a continuum of trial values.
- Tail length 1/κ = 0.254 nm, so the density |ψb|² = κ e(−2κ|x|) falls by a factor e² over each 0.254 nm of travel from the well.
- Probability inside ±0.100 nm: ∫κ e(−2κ|x|) dx = 1 − e(−2κ × 0.100) = 1 − e(−0.787) = 0.545. Widen the window and this saturates at 1 — the mark of a vector of L²(ℝ).
- At E = +0.591 eV, k = √(E ÷ (ħ²/2m)) = √(0.591/0.0381) = 3.94 nm⁻¹, equal to κ because this energy mirrors the bound one. Both asymptotic solutions now oscillate, so there are two eigenfunctions: cos(k|x| + δ)/√π with tan δ = κ/k = 1, that is δ = 45°, and sin(kx)/√π. Neither can be normalised to 1.
Answerκ = 3.94 nm⁻¹, Eb = −0.591 eV, tail length 0.254 nm, P(|x| < 0.100 nm) = 0.545. One state below the floor, two above it, with δ = 45° at this particular energy.
MediumFix the constants in front of a scattering state. (a) Show that ψₖ(x) = e(ikx)/√(2π) satisfies ⟨k|k′⟩ = δ(k − k′), and check its units. (b) Convert to energy normalisation, ⟨E|E′⟩ = δ(E − E′), and evaluate the prefactor for an electron at E = 1.00 eV. (c) Compare with a periodic box of L = 20.0 nm: give the level spacing there and the factor relating the two conventions.
- (a) ⟨k|k′⟩ = (1/2π)∫e(i(k′−k)x) dx = δ(k′ − k), the Fourier representation of the δ. Units: k carries m⁻¹, so δ(k − k′) carries metres; ∫dx supplies one metre, so ψₖ is dimensionless — unlike a bound ψₙ, which carries m(−1/2).
- (b) Relabel by energy with the Jacobian |E⟩ = |k⟩√(dk/dE). Since E = ħ²k²/2m, dE/dk = ħ²k/m and dk/dE = m/(ħ²k), so ψE(x) = e(ikx)√(m/(2πħ²k)).
- Numbers at E = 1.00 eV: k = √(1.00 ÷ 0.0381) = 5.12 nm⁻¹, then dk/dE = 1 ÷ (0.0762 × 5.123) = 2.56 nm⁻¹ eV⁻¹, and the prefactor is √(2.562/2π) = 0.639 nm(−1/2) eV(−1/2) — the units m(−1/2) E(−1/2) that δ(E − E′) demands.
- (c) In a periodic box the allowed k are 2πn/L, spaced Δk = 2π ÷ 20.0 nm = 0.314 nm⁻¹, so near 1.00 eV the levels sit ΔE = (ħ²k/m)Δk = 0.0762 × 5.123 × 0.314 = 0.123 eV apart — a spacing that vanishes as 1/L.
- The two conventions differ by the density of states: e(ikx)/√L = √(2π/L) × e(ikx)/√(2π), and Σₙ → (L/2π)∫dk. Every physical rate multiplies one factor by the other, which is why L cancels out of the answer.
AnswerψE = 0.639 e(ikx) nm(−1/2) eV(−1/2) at k = 5.12 nm⁻¹; a 20.0 nm box gives ΔE = 0.123 eV, and box and delta conventions differ by √(2π/L), the factor the density of states L/2π undoes.
HardAn electron occupies the ground state of V = −α₀δ(x) with α₀ = 0.3904 eV nm, so κ₀ = 5.123 nm⁻¹ and E₀ = −1.000 eV. The strength is switched suddenly to α = α₀/2, too fast for the state to respond. Find the probability the electron is still bound, the weight thrown into the continuum, and the mean energy of that continuum part.
- A sudden switch leaves ψ alone and changes the basis. The new well has κ = κ₀/2 = 2.562 nm⁻¹ and one bound state ψb = √κ e(−κ|x|) at Eb = −ħ²κ²/2m = −0.250 eV, a quarter of E₀ because E ∝ α².
- Overlap of two cusped exponentials: ⟨ψb|ψ⟩ = √(κκ₀)∫e(−(κ+κ₀)|x|) dx = 2√(κκ₀)/(κ + κ₀). So Pbound = 4κκ₀/(κ + κ₀)², which at κ = κ₀/2 is 4(1/2)/(3/2)² = 8/9 = 0.889.
- By completeness the rest is not lost, it is ionised: Σ + ∫ = 1 forces ∫dk|c(k)|² = 1 − 8/9 = 1/9 = 0.111. The initial state is even, so its overlap with every odd channel sin(kx)/√π is exactly zero, and all of that weight sits in the even channel cos(k|x| + δ)/√π.
- Check the energy in position space: ⟨T⟩ = (ħ²/2m)∫|ψ′|² dx = ħ²κ₀²/2m = 1.000 eV, and ⟨V⟩ = −α|ψ(0)|² = −ακ₀ = −0.1952 × 5.123 = −1.000 eV, so ⟨Ĥ⟩ = 0.000 eV — an exact cancellation whenever κ = κ₀/2.
- Now read the same number spectrally: ⟨Ĥ⟩ = (8/9)(−0.250 eV) + ∫dk E(k)|c(k)|² = 0, so the continuum carries +0.222 eV and its mean energy is 0.222 ÷ 0.111 = 2.00 eV, twice the original binding energy. Keeping only the bound term would have given −0.222 eV, wrong by the whole continuum contribution.
AnswerPbound = 8/9 = 0.889 at Eb = −0.250 eV; the continuum takes 1/9 = 0.111, all of it even-parity, with mean energy 2.00 eV. ⟨Ĥ⟩ = 0 only when both sectors are counted.