Two-body operators and the many-electron Hamiltonian

Units are SI throughout. The theory is non-relativistic, so Unit 9's metric diag(1, −1, −1, −1) plays no part and the interaction is the instantaneous Coulomb law e²/(4πε₀|r − r′|). Label an electron's coordinates x = (r, σ) and write ∫dx for Σ_σ∫d³r. In first quantisation a pair interaction is V = Σ v(xₐ, x_b) summed over pairs a < b, equal to ½Σ v(xₐ, x_b) over all a ≠ b, with v(x, y) = v(y, x): a sum over distinct pairs, never a particle with itself. On an orthonormal basis of spin-orbitals φᵢ(x) the numbers it needs are ⟨ij|v|kl⟩ = ∫dx∫dy φᵢ*(x)φⱼ*(y) v(x, y) φₖ(x)φₗ(y). The first label of each side, i and k, belongs to the particle at x; the second, j and l, to the particle at y. Two symmetries follow: ⟨ij|v|kl⟩ = ⟨ji|v|lk⟩ from v(x, y) = v(y, x), and ⟨ij|v|kl⟩ = ⟨kl|v|ij⟩* from hermiticity. Dimensions: φ carries m^(−3/2) and the double integral m⁶, so ⟨ij|v|kl⟩ is an energy in J, while the c and c† of a discrete basis are dimensionless. So V̂ must be ½Σ⟨ij|v|kl⟩c†ᵢc†ⱼ times two annihilators, in an order that is not a convention: the next step derives it.

Act on |kl⟩ = c†ₖc†ₗ|0⟩ with k ≠ l. Anticommuting each annihilator to the right gives cₙcₘc†ₖc†ₗ|0⟩ = (δₘₖδₙₗ − δₘₗδₙₖ)|0⟩. So the trial V̂ = ½Σ⟨ij|v|mn⟩c†ᵢc†ⱼcₙcₘ gives ½Σᵢⱼ(⟨ij|v|kl⟩ − ⟨ij|v|lk⟩)c†ᵢc†ⱼ|0⟩. Swap i and j in the second sum: ⟨ji|v|lk⟩ = ⟨ij|v|kl⟩ and c†ⱼc†ᵢ = −c†ᵢc†ⱼ, so the two halves are equal and V̂|kl⟩ = Σᵢⱼ⟨ij|v|kl⟩c†ᵢc†ⱼ|0⟩. That is first quantisation exactly: v commutes with the antisymmetriser, sends φₖ(x)φₗ(y) to Σ⟨ij|v|kl⟩φᵢ(x)φⱼ(y), and c†ᵢc†ⱼ|0⟩ is the determinant of φᵢ and φⱼ (topic 10.01, quoted). Write the annihilators as cₖcₗ instead and every sign reverses: since cₖcₗ = −cₗcₖ, the mirrored operator is exactly −V̂. For bosons the annihilators commute and the two orders agree; for fermions the order carries the whole sign of the interaction. The one-particle limit is immediate, because cₗcₖc†ₘ|0⟩ = 0: V̂ annihilates every one-electron state, so hydrogen and He⁺ feel no repulsion.

Pass to field operators, ψ(x) = Σφᵢ(x)cᵢ (topic 10.02, quoted): V̂ = ½∫dx∫dy v(x, y) ψ†(x)ψ†(y)ψ(y)ψ(x). Read it from the right: remove an electron at x, count the electrons left at y with n̂(y) = ψ†(y)ψ(y), put the first one back. On ψ†(x₁)…ψ†(x_N)|0⟩ the removal picks each electron a in turn, the count runs over b ≠ a, and the two ordering signs cancel, so V̂ returns ½Σ v(xₐ, x_b) over all a ≠ b times the same state: the first-quantised pair sum for every N, not only N = 2. Now compare the classical energy of a charge cloud, ½∫∫v n̂(x)n̂(y). Moving ψ(x) past ψ†(y) gives n̂(x)n̂(y) = ψ†(x)ψ†(y)ψ(y)ψ(x) + δ(x − y)n̂(x), so the cloud form is V̂ plus the one-body term ½∫dx v(x, x)n̂(x): each electron meeting itself at zero distance, which for e²/(4πε₀|r − r′|) is infinite. Writing every creator to the left of every annihilator deletes that term. The ½ has a different job, undoing the double count of each pair a ≠ b.

Nothing so far used a property of the φᵢ beyond orthonormality, so V̂ has the same form in every basis. Change basis with a unitary U, φ′ₘ = Σᵢφᵢ Uᵢₘ, so that c′†ₘ = Σᵢ c†ᵢ Uᵢₘ (topic 10.02, quoted). Substituting c†ᵢ = Σₘ U*ᵢₘ c′†ₘ and cᵢ = Σₘ Uᵢₘ c′ₘ into V̂ returns the same expression with ⟨mn|v|ps⟩′ = Σ U*ᵢₘ U*ⱼₙ ⟨ij|v|kl⟩ Uₖₚ Uₗₛ: one factor of U per index, conjugated on the bra side, so the coefficients form a four-index tensor. The size of that tensor is the price of the two-body term. M spin-orbitals give M⁴ coefficients. For K real spatial orbitals, the pair symmetry and reality leave P(P + 1)/2 distinct integrals, where P = K(K + 1)/2: 6 for 1s and 2s, 1540 for K = 10, and 1.28 × 10⁷ for K = 100, against K(K + 1)/2 one-body integrals. Transforming them one index at a time, each step summing K terms for each of K⁴ entries, costs about 4K⁵ operations rather than the K⁸ of doing all four at once.

Hold a nucleus of charge Ze fixed at the origin. Then H = Σₐ[pₐ²/2m − Ze²/(4πε₀rₐ)] plus e²/(4πε₀|rₐ − r_b|) summed over pairs a < b, and in Fock space Ĥ = Σ⟨i|h|j⟩c†ᵢcⱼ + ½Σ⟨ij|v|kl⟩c†ᵢc†ⱼcₗcₖ. The operator is the same for N = 1, 2 or 92, because N is only the occupation of the state it acts on. Neither term touches spin, so with spin-orbitals φₐ(r)χ_σ the spin enters only through the labels: ⟨i|h|j⟩ carries δ(σᵢ, σⱼ), and ⟨ij|v|kl⟩ carries δ(σᵢ, σₖ)δ(σⱼ, σₗ). Left out are spin-orbit coupling and the Breit interaction, both of relative order (Zα)² = 2.1 × 10⁻⁴ for helium, nuclear recoil, m/M = 1.37 × 10⁻⁴, and QED. The test is a measured number. The exact ground state of this Ĥ for helium, from large variational calculations (quoted), is −2.903724 e²/(4πε₀a₀) = −79.014 eV. The measured value, minus the sum of helium's two ionisation energies, 24.587 eV and 54.418 eV, is −79.005 eV. The 0.009 eV gap, 1.2 × 10⁻⁴ of the total, is mostly the 0.011 eV reduced-mass shift from the nucleus's finite mass: the model is right to the size of what it leaves out.