Spectra, degeneracy and completeness

The spectrum of Ĥ is the set of real E for which Ĥ − E has no bounded inverse. For a self-adjoint operator it is real and closed, but nothing in that definition promises eigenvectors. Take the free particle, Ĥ = p̂²/2m on L²(ℝ): its spectrum is the whole half-line [0, ∞), and it has no eigenvectors at all. At E = 1.50 eV the equation Ĥψ = Eψ is solved perfectly smoothly by e^{ikx} with k = 6.275 nm⁻¹, yet ∫₀^L |e^{ikx}|² dx = L diverges, so that solution is not a vector in the space and E is not an eigenvalue. At the other extreme, the infinite well has nothing but eigenvalues. Most Hamiltonians in this unit carry both kinds at once. So "the differential equation has a solution at E" is never the test. The test is whether a square-integrable solution exists, and that is a statement about how ψ behaves at the ends of the interval, not about the ODE.

Write the equation as ψ″ = (2m/ħ²)(V − E)ψ and look at the tails. Far to the right, where V → V₊, an energy below V₊ gives ψ″ = κ²ψ with κ = √(2m(V₊ − E))/ħ, whose two solutions are e^(+κx) and e^(−κx); square-integrability throws one away. The same happens on the left. Integrate inward from each end and you reach the midpoint with two solutions, each fixed up to a scale; matching ψ and ψ′ there is two conditions against one free ratio, leaving one real equation in E. It fails at a generic energy and holds only at isolated ones — the discreteness comes from the boundary condition, not from the shape of the well. Raise E above V₊ and κ becomes ik: both e^{+ikx} and e^(−ikx) stay bounded, nothing is rejected, every energy is admitted. For an electron bound by 2.00 eV, κ = 7.25 nm⁻¹ and the tail decays with length 1/κ = 0.138 nm — a level is either localised on that scale or not localised at all.

Continuum solutions are not in the Hilbert space, so they need a home: the rigged triple Φ ⊂ ℋ ⊂ Φ′, where Φ holds well-behaved test states and the generalised eigenvectors live in its dual. The strongest statement available is distributional, ∫ e^{(i(k−k′)x)} dx = 2π δ(k − k′). Label by k and ψₖ = e^{ikx}/√(2π) gives ⟨k|k′⟩ = δ(k − k′). But closure integrates over energy, and changing the label changes the delta: δ(k − k′) = (dE/dk) δ(E − E′), so |E⟩ = √(dk/dE)|k⟩ with dk/dE = m/ħ²k. That Jacobian is the density of states, and dropping it is the commonest slip here. The result ψ_E(x) = √(m/2πħ²k) e^{ikx} carries units (J⋅m)^(−1/2): for an electron at 1.50 eV, with ħ²/2m = 0.03810 eV nm², k = 6.275 nm⁻¹ and the prefactor is 0.577 (eV⋅nm)^(−1/2). So |⟨E|ψ⟩|² is a probability per joule. A value of 5.0 eV⁻¹ is perfectly legal; only its integral over a band is a probability, and over a 20 meV window that one gives 0.10.

Let ψ₁ and ψ₂ solve the same equation at the same E and build W = ψ₁ψ₂′ − ψ₂ψ₁′. Then W′ = ψ₁ψ₂″ − ψ₂ψ₁″ = (2m/ħ²)(V − E)(ψ₁ψ₂ − ψ₂ψ₁) = 0, so W is one constant everywhere. If both states are bound, ψ and ψ′ vanish as |x| → ∞, so that constant is zero; W = 0 means ψ₂′/ψ₂ = ψ₁′/ψ₁, and integrating gives ψ₂ ∝ ψ₁ — the same ray, not a second state. Notice what the proof used, because each assumption is a loophole. It used vanishing tails, which the continuum has not: above both asymptotes W ≠ 0 and the degeneracy is exactly two, one left-incident solution and one right-incident. It used one connected interval with V finite, so an infinite barrier splitting ℝ into two identical wells produces exactly degenerate pairs. And it used a single coordinate: at fixed E a second-order ODE has a two-dimensional solution space, and W tests dependence inside it. In three dimensions Ĥψ = Eψ is a partial differential equation; the argument survives only inside each radial equation, whose levels are indeed non-degenerate, while the 2l + 1 values of m, and for hydrogen and the oscillator several values of l, share one energy. That is why hydrogen's n² and the 3D oscillator's (n+1)(n+2)/2 are no counterexample. A corollary comes free: if V is even, ψ(−x) solves the same equation at the same E, so it is a multiple of ψ(x) whose square is 1, and every 1D bound state has definite parity.

Put Ĥ on a grid inside a hard-walled box of length L_box, hand it to scipy.linalg.eigh, and everything comes back discrete — a finite box has no continuum at all. The way to separate the two parts is to move the wall. A genuine bound level barely notices: for the weakly bound level of the 5.00 eV well, κ = 5.199 nm⁻¹, and with the wall d = 4.80 nm beyond the well edge the shift scales as e^(−2κd) = e^(−49.9) ≈ 2 × 10⁻²². The faked levels behave like box levels instead. Near energy E their spacing is ΔE = πħ²k/(m L_box), which for an electron at 6.00 eV in a 10.0 nm box is 0.300 eV and drops to 0.150 eV when the box is doubled, while the count below 6.00 eV climbs from 40 to 80: the two bound levels plus a share that grows as k L_box/π. Hence the operational rule: converge every eigenvalue in both grid spacing and box length, and treat any level whose energy tracks L_box as a discretised slice of the continuum, not as a state of Ĥ.

Feed closure into the propagator. For a static Ĥ, |ψ(t)⟩ = Σₙ e^{(−iEₙ t/ħ)} cₙ |n⟩ + ∫ dE e^(−iEt/ħ) c(E) |E⟩. Unitarity keeps the norm of each term fixed, Σ|cₙ|² and ∫|c(E)|² dE, so the two behave in opposite ways only once you project onto a fixed normalisable state |φ⟩. The bound term gives Σₙ e^{(−iEₙ t/ħ)} cₙ⟨φ|n⟩, an almost periodic function that never dies away. The continuum term gives ∫ dE e^(−iEt/ħ) c(E)⟨φ|E⟩, the Fourier transform of an integrable function, so by the Riemann–Lebesgue lemma it tends to zero as t → ∞ — that part of the state leaves every bounded region and never returns. The survival amplitude ⟨ψ(0)|ψ(t)⟩ = Σₙ |cₙ|² e^{(−iEₙ t/ħ)} + ∫ dE |c(E)|² e^(−iEt/ħ) therefore does not decay to zero. Park a packet in a well holding one bound level and give that level 64% of the norm, |c₁|² = 0.64: the survival probability falls from 1 towards |c₁|⁴ = 0.41 and stays there, while the other 36% leaves for good. Near-exponential decay is what you get when the bound projection is zero and the continuum coefficient carries a narrow resonance, never exactly exponential because the spectrum is bounded below — but the bookkeeping that predicts it is already this closure relation.