Bound states and the continuum

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.

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 once the tail falls faster than 1/x², as every finite-range well's does: 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. An attractive tail slower than 1/x² holds infinitely many: hydrogen's Coulomb tail, not even short-range, gives −13.6/n² eV levels that accumulate at the floor from below.

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 V_L < E < V_R the continuum is non-degenerate again, because the right-hand growing exponential must still be discarded.

∫|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⟩.

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.

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.