What orbital energies mean: Koopmans and Brillouin

Work in SI units with energies in eV, using the hartree Eₕ = 27.211 eV for atomic values; nothing here is relativistic, so the course metric diag(1, −1, −1, −1) never enters. Take topic 10.06's converged determinant |Φ₀⟩ with canonical orbitals, f|i⟩ = εᵢ|i⟩ and εᵢ = hᵢᵢ + Σⱼ (Jᵢⱼ − Kᵢⱼ), quoted. Annihilate orbital k and keep every other orbital as it was: |Φₖ⟩ = cₖ|Φ₀⟩. Topic 10.05's contractions give its energy with k simply absent. hₖₖ leaves the one-body sum, and the pairs (k, j) and (j, k), each weighted ½, leave the two-body sum. What leaves is hₖₖ + Σⱼ (Jₖⱼ − Kₖⱼ) = εₖ, so E(N − 1; k) = E_HF − εₖ, derived. The same contraction with c†ₐ on a virtual gives E(N + 1; a) = E_HF + εₐ, the affinity version. Dimensions: every term is an energy. Limit: a single electron has nobody to repel, so ε = h = E and the theorem is exact; hydrogen's −ε = 13.606 eV becomes the measured 13.598 eV once the reduced mass is put in.

Add the orbital energies over occupied spin-orbitals: Σεᵢ = Σhᵢᵢ + Σᵢⱼ (Jᵢⱼ − Kᵢⱼ). The determinant's energy carries the same pair sum with a factor ½, so E_HF = Σεᵢ − ½Σᵢⱼ (Jᵢⱼ − Kᵢⱼ) = ½Σᵢ (hᵢᵢ + εᵢ), derived. The cause is bookkeeping, not approximation: εᵢ prices electron i in the field of all the others, so the repulsion between i and j sits in εᵢ and again in εⱼ. Closed-shell helium makes it concrete. Its two spin-orbitals share ε = −0.91796 Eₕ = −24.98 eV and E_HF = −2.86168 Eₕ = −77.87 eV, values topic 10.08 computes; 2ε = −49.96 eV lies J = 2ε − E_HF = 1.02577 Eₕ = 27.91 eV above it, the one 1s-1s repulsion that both orbital energies contain. For one electron the pair sum is empty and E = ε, as it should be. The same arithmetic forbids adding ionisation energies read from one calculation. The first electron costs about 25 eV, the second 54.42 eV from He⁺, whose orbital has contracted, and only their measured sum, 79.0 eV, is the atom's binding energy.

E_HF does not care which orthonormal basis of the occupied space you use: a unitary mixing U of occupied orbitals multiplies the determinant by det U, a phase. Orbital energies do care. Remove instead a normalised mixture u = Σₖ Uₖ φₖ, freezing the rest, and the same contraction raises the energy by −⟨u|f|u⟩ = −Σ Uₖ* Fₖₗ Uₗ, minus a Rayleigh quotient of the occupied block of the Fock matrix. It is stationary under every small change of U exactly when F U = ε U, that is, when u is a canonical orbital, derived. That was Koopmans' own 1934 result. Canonical orbitals are therefore not one choice among equals: they are the orbitals whose removal energy is stationary, and the ones a photoelectron spectrum sees. Methane shows it. Four equivalent C-H bond orbitals describe its ground state exactly as well as the canonical a₁ and t₂ orbitals, yet the measured spectrum has two valence bands, near 23 eV and 14 eV, in a one-to-three pattern, and no band for a single bond.

Put occupied orbital i into virtual a: |Φᵢᵃ⟩ = c†ₐcᵢ|Φ₀⟩. Contracting H between it and Φ₀ leaves hₐᵢ + Σⱼ ⟨aj||ij⟩ = ⟨a|f|i⟩, derived. Now rotate i towards a through an angle θ. A determinant is linear in each orbital, so Φ(θ) = cos θ Φ₀ + sin θ Φᵢᵃ exactly, and for real orbitals E(θ) = E₀ + Δ sin²θ + ⟨Φᵢᵃ|H|Φ₀⟩ sin 2θ with Δ = ⟨Φᵢᵃ|H|Φᵢᵃ⟩ − E₀. The slope at θ = 0 is 2⟨Φᵢᵃ|H|Φ₀⟩. Topic 10.06 made E stationary under every such rotation, so every single-excitation coupling vanishes: ⟨Φᵢᵃ|H|Φ₀⟩ = ⟨a|f|i⟩ = 0. Brillouin's theorem is not an extra fact about Hartree-Fock; it is Hartree-Fock, written as a statement about matrix elements. It needs stationarity only, not canonical orbitals, and it fails for any determinant short of convergence, where the slope 2⟨a|f|i⟩ tells the next iteration which way to rotate.

Expand the exact ground state in determinants built from the Hartree-Fock orbitals, Ψ = c₀Φ₀ + Σ cᵢᵃΦᵢᵃ + Σ cᵢⱼᵃᵇΦᵢⱼᵃᵇ + …, which is configuration interaction, named here. H is a two-body operator, so it links Φ₀ to nothing beyond doubles, and Brillouin removes the singles. The first-order wavefunction of Rayleigh-Schrödinger theory, quoted from Quantum Mechanics II, therefore contains doubles only, and the leading energy correction is a sum over them. With H₀ = Σ f(n), whose determinant energies are sums of εᵢ, it is Møller-Plesset second order, E⁽²⁾ = −Σ |⟨ab||ij⟩|²/(εₐ + ε_b − εᵢ − εⱼ), quoted and not pursued. Every term is negative, and one electron has no pairs, so E⁽²⁾ = 0 and Hartree-Fock is exact there, as it must be. Singles are not absent from Ψ: they enter at second order through their coupling to the doubles. What Brillouin forbids is the direct step. The orbital energies also earn a second job here: their differences are the denominators.

Koopmans' ion is not stationary. Its own Fock operator lacks the departed electron, f − Jₖ + Kₖ, so its Brillouin element for j → a is −⟨ak||jk⟩, not zero: the hole pulls the remaining orbitals inward, and letting them rotate lowers the ion. Relaxation therefore always makes −εₖ too large. Correlation pushes the other way, because the N-electron atom has more pairs to correlate than its ion and is lowered more. For helium, quoted from topic 10.08: −ε = 24.98 eV, relaxation 1.53 eV, correlation 1.14 eV, leaving Koopmans 0.39 eV above the measured 24.587 eV. Neon's 2p gives −ε = 23.14 eV against the measured 21.56 eV. A core hole relaxes far more: neon's −ε₁ₛ = 891.8 eV overshoots the measured 870.2 eV by 21.6 eV, as eight outer electrons contract onto a hole that acts on them like an extra nuclear charge. In a large system a delocalised hole spreads over N sites and the relaxation falls away as N grows, which is why band theory may read Hartree-Fock bands as removal energies. For affinities both errors raise the true value, so −εₐ fails worst.