University Physics V · The Schrödinger Equation · 4.8
Spectra, Degeneracy & Completeness
Once Ĥ is self-adjoint its spectrum splits at the asymptotic potential, and the two halves behave nothing alike. Here you learn to count the bound levels, δ-normalise the ones you cannot count, write closure as a sum plus an integral, and prove that no one-dimensional bound level is ever degenerate.
Build the model
Connect the measurement to the mechanism.
The spectral theorem does not hand you an eigenbasis; it hands you a projection-valued measure, and only part of that measure is built from eigenvectors. Where V(x) tends to finite asymptotes, the equation ψ″ = (2m/ħ²)(V − E)ψ behaves in two incompatible ways. Below the lower asymptote both tails are exponential, admissibility kills the growing one at each end, and the two one-sided conditions can be met at once only at isolated energies: a discrete point spectrum whose eigenvectors are genuine unit-norm elements of L²(ℝ) obeying ⟨m|n⟩ = δₘₙ.
Above the asymptote both tails oscillate, nothing is excluded, and every E belongs to the spectrum — but not one of those solutions is square-integrable, so none of them is a state. They are generalised eigenvectors, distributions living outside the Hilbert space in the dual of a rigged triple, and the best normalisation available is ⟨E|E′⟩ = δ(E − E′). Completeness then cannot be a sum: closure reads 1̂ = Σₙ|n⟩⟨n| + ∫dE Σσ|E, σ⟩⟨E, σ|, the label σ running over one value below the higher asymptote and two above it.
The cost is paid twice. You can never prepare a continuum ket, so every scattering statement is secretly about a normalisable packet; and on a computer the continuum must be caged in a finite box, which fakes it into levels whose spacing you have to watch collapse before you believe anything read off them.
- Simple definition
- The spectrum of Ĥ is every energy at which Ĥ − E fails to have a bounded inverse: a discrete part carrying normalisable eigenvectors, and a continuous part carrying only non-normalisable generalised ones.
- Example
- An electron in a 5.00 eV deep, 0.400 nm wide square well has exactly two bound levels — discrete, non-degenerate, normalisable — while every E above the well top is in the spectrum too, doubly degenerate and normalisable only to a delta.
Insert it anywhere and a normalised state gives Σ|cₙ|² + ∫|c(E)|² dE = 1. Drop the integral and closure fails silently.
n runs over bound levels with Eₙ below the asymptotic V; the integral runs above it, σ over that energy's degeneracy
Fixes the units of the coefficients: |cₙ|² is a probability, |c(E)|² a probability density per joule.
δₘₙ is a dimensionless number; δ(E − E′) carries J⁻¹, so ⟨E|E⟩ = δ(0) diverges
Turns ⟨k|k′⟩ = δ(k − k′) into ⟨E|E′⟩ = δ(E − E′). For an electron at 1.50 eV the prefactor is 0.577 (eV⋅nm)(−1/2).
√(dk/dE) = √(m/ħ²k) is the density-of-states Jacobian; ψE has units (J⋅m)(−1/2)
Bound states force W → 0 in the tails, so W ≡ 0 and ψ₂ ∝ ψ₁: no 1D bound level is ever degenerate.
Needs both ψ to solve the same equation at the same E on one connected interval with V finite
The threshold is energetic, not geometric: it is set by V at infinity, never by how wide or deep the well is.
V± are the limits of V(x) as x → ±∞, and g is the degeneracy of that energy
Spectrum is a bigger word than eigenvalue
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 ∫₀L |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.
The asymptotic potential fixes the threshold
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.
Delta normalisation and the Jacobian everyone drops
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 ψE(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.
One dimension has no degenerate bound levels
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 the fact that a line has two ends; a sphere is not two points, which 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.
The box that fakes a continuum, and how to catch it
Put Ĥ on a grid inside a hard-walled box of length Lbox, 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 Lbox), 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, about k Lbox/π, climbs from 39 to 79. Hence the operational rule: converge every eigenvalue in both grid spacing and box length, and treat any level whose energy tracks Lbox as a discretised slice of the continuum, not as a state of Ĥ.
What the split predicts: part of a state can leave
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⟩, and the two terms behave in opposite ways. The sum is almost periodic: its modulus never decays, and with a single bound level populated it is exactly constant. The integral is the Fourier transform of an integrable function, so by the Riemann–Lebesgue lemma it tends to zero as t → ∞ — that amplitude disperses 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. Clean exponential decay is what you get when the bound projection is zero and the continuum coefficient carries a resonance — but the bookkeeping that predicts it is already this closure relation.
Change one variable at a time
Make the relationship visible.
Drag the box length from 2 to 8: the right-hand bars crowd onto the threshold while the bound bars do not move at all — that immobility is how you tell a real level from a box artefact. Then lower λ and watch levels leave through E = 0 one at a time.
BOUND LEVELS3
GROUND LEVEL E₀-0.694 D
BOX SPACING ΔE₂₁0.375 D
BOX LEVELS BELOW 0.5 D2
Live interpretationBOUND LEVELS: 3. GROUND LEVEL E₀: −0.694 D. BOX SPACING ΔE₂₁: 0.375 D. BOX LEVELS BELOW 0.5 D: 2
Catch the common trap
Explain before calculating.
An electron moves in a one-dimensional potential equal to −V₀ on [0, L] and zero outside, deep enough to hold exactly three bound levels. Which statement about the spectrum of Ĥ is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron sits in a one-dimensional square well of depth V₀ = 5.00 eV and width L = 0.400 nm. Using ħ²/2m = 0.03810 eV nm², find how many bound levels it holds, state the degeneracy of each part of the spectrum, and write down the closure relation for this Hamiltonian.
- The well parameter is z₀ = (L/2)√(2mV₀)/ħ = (L/2)√(V₀ ÷ (ħ²/2m)) = 0.200 nm × √(5.00/0.03810) nm⁻¹ = 0.200 × 11.456 = 2.291.
- Bound levels alternate even, odd, even… and one appears per interval of π/2 in z₀, so the count is ⌈2z₀/π⌉ = ⌈4.582/3.1416⌉ = ⌈1.459⌉ = 2.
- Both lie below the asymptotic value V(±∞) = 0 and decay at both ends, so the Wronskian argument makes each non-degenerate. Above E = 0 the two asymptotes are equal, both e^{+ikx} and e(−ikx) are admissible, and every energy is doubly degenerate.
- Closure therefore reads 1̂ = |1⟩⟨1| + |2⟩⟨2| + ∫₀^∞ dE (|E,+⟩⟨E,+| + |E,−⟩⟨E,−|). The two bound projectors alone form a rank-2 operator — nowhere near the identity on L²(ℝ).
AnswerTwo bound levels, each non-degenerate; a continuum at every E > 0 with degeneracy 2; closure is a two-term sum plus an integral over the continuum.
MediumWrite the free-electron scattering state as ψE(x) = A e^{ikx}. Fix A so that ⟨E|E′⟩ = δ(E − E′), evaluate it at E = 1.50 eV with ħ²/2m = 0.03810 eV nm², then find the probability that an energy measurement on a normalised packet with |⟨E|ψ⟩|² = 5.0 eV⁻¹ near 1.50 eV lands in the 20 meV band centred there.
- In the k label, ∫ e(i(k−k′)x) dx = 2π δ(k − k′), so ψₖ = e^{ikx}/√(2π) gives ⟨k|k′⟩ = δ(k − k′). Each sign of k is its own branch: ⟨E, σ|E′, σ′⟩ = δσσ′ δ(E − E′).
- Changing the label changes the delta: δ(k − k′) = (dE/dk) δ(E − E′). So set |E⟩ = √(dk/dE)|k⟩; with E = ħ²k²/2m, dE/dk = ħ²k/m, giving A = √(m / 2πħ²k).
- Numbers: k = √(E ÷ (ħ²/2m)) = √(1.50/0.03810) = √39.37 = 6.275 nm⁻¹, and m/(2πħ²k) = 1/[4π (ħ²/2m) k] = 1/(4π × 0.03810 × 6.275) = 1/3.004 = 0.3329 eV⁻¹ nm⁻¹.
- So A = √0.3329 = 0.577 (eV⋅nm)(−1/2), which is 4.56 × 10¹³ (J⋅m)(−1/2). It is not dimensionless: ψE is normalised per unit energy, not to unit probability.
- |⟨E|ψ⟩|² = 5.0 eV⁻¹ is a density, so exceeding 1 is no contradiction. The probability is its integral over the band: P ≈ 5.0 eV⁻¹ × 0.020 eV = 0.10.
AnswerA = √(m/2πħ²k) = 0.577 (eV⋅nm)(−1/2) at 1.50 eV, i.e. 4.56 × 10¹³ (J⋅m)(−1/2); the 20 meV band carries P ≈ 0.10.
HardYou diagonalise a finite-difference Ĥ for that 5.00 eV, 0.400 nm well inside a hard-walled box of length Lbox = 10.0 nm, then repeat at 20.0 nm. The weakly bound level has binding energy 1.03 eV. Predict what happens to (i) the two bound eigenvalues, (ii) the level spacing near E = +6.00 eV, and (iii) the number of eigenvalues below 6.00 eV. Use ħ²/2m = 0.03810 eV nm².
- A hard box has no continuum, so eigh returns nothing but discrete numbers. The diagnostic is never what the spectrum looks like at one box size — it is how each level moves when the wall moves.
- Bound levels: κ = √(1.03/0.03810) = √27.03 = 5.199 nm⁻¹, and the wall sits d = (10.0 − 0.400)/2 = 4.80 nm beyond the well edge. The wall's effect scales as e(−2κd) = e(−49.9) ≈ 2 × 10⁻²², so both bound eigenvalues are unmoved to every digit eigh prints.
- Faked levels: for a box kₘ = mπ/Lbox, so near energy E the spacing is ΔE = (dE/dk)(π/Lbox) = 2(ħ²/2m)k × π/Lbox. At E = 6.00 eV, k = √(6.00/0.03810) = √157.5 = 12.549 nm⁻¹, giving ΔE = 2 × 0.03810 × 12.549 × π/10.0 = 0.300 eV.
- Doubling the box to 20.0 nm halves it, ΔE = 0.150 eV. No physics changed; the spacing is an artefact of the wall.
- Counting: eigenvalues below 6.00 eV number about k Lbox/π = 12.549 × 10.0/π = 39.9, so 39 at 10.0 nm and 79 at 20.0 nm. The bound count stays at 2 while the continuum count scales with the box — that contrast is the whole test.
AnswerBound levels frozen (shift ~10⁻²² eV); spacing near 6.00 eV halves from 0.300 eV to 0.150 eV; the count below 6.00 eV goes 39 → 79. Only the box-independent levels belong to the discrete spectrum of Ĥ.