Energy of a single determinant: direct and exchange terms

SI units throughout, with atomic integrals quoted in hartrees, Eₕ = 27.211 eV; nothing here is relativistic, so the course metric (+, −, −, −) never enters. Take orthonormal spin-orbitals and the determinant |Φ⟩ = c₁† c₂† … c_N† |0⟩, with nᵢ = 1 for i ≤ N and 0 otherwise. All we use is cᵢ cⱼ† + cⱼ† cᵢ = δᵢⱼ, cᵢ cⱼ = −cⱼ cᵢ and cᵢ|0⟩ = 0. Move cₖ to the right: each creator with a different label it passes costs a factor −1; at cₖ† it leaves cₖ cₖ† = 1 − cₖ† cₖ, and the second piece runs on to the vacuum and dies. So cₖ|Φ⟩ is zero if k is empty, and otherwise ± the determinant with k removed, the sign set by how many occupied orbitals are listed before k. The one-body contraction follows: ⟨Φ|cᵢ† cⱼ|Φ⟩ is the overlap of the hole states cᵢ|Φ⟩ and cⱼ|Φ⟩, which vanishes unless the same orbital was removed. Hence ⟨cᵢ† cⱼ⟩ = δᵢⱼ nᵢ and the one-body energy is Σᵢ ⟨i|h|i⟩ over occupied orbitals, derived. For helium's 1s² with hydrogenic Z = 2 orbitals that is 2 × (−2 Eₕ) = −108.8 eV.

Because ⟨Φ|cᵢ† cⱼ† is the bra of cⱼ cᵢ|Φ⟩, the expectation ⟨cᵢ† cⱼ† cₗ cₖ⟩ is the overlap of two states with two holes each. Determinants with different holes are orthogonal, so it vanishes unless k and l are i and j in some order, both occupied, with k ≠ l. If i = k and j = l the two hole states are identical and the overlap is +1; if i = l and j = k they differ only in the order of the annihilators, cₖ cₗ = −cₗ cₖ, and it is −1. So ⟨cᵢ† cⱼ† cₗ cₖ⟩ = ⟨cᵢ† cₖ⟩⟨cⱼ† cₗ⟩ − ⟨cᵢ† cₗ⟩⟨cⱼ† cₖ⟩, the product of contractions minus the exchanged product, derived. Check it on Φ = c₁† c₂† c₃† |0⟩: c₃ c₁|Φ⟩ = −c₂†|0⟩ while c₁ c₃|Φ⟩ = +c₂†|0⟩, so ⟨c₃† c₁† c₃ c₁⟩ = −1, exactly the formula's second term. The general statement, quoted and not proved here, is that p creators and p annihilators in a determinant give the p × p determinant of one-body contractions: every complete pairing, signed by its permutation.

Insert that result into the interaction quoted from topic 10.04, V = ½ Σ ⟨ij|v|kl⟩ cᵢ† cⱼ† cₗ cₖ, whose reversed order cₗ cₖ is what makes the first pairing the direct one. The Kronecker deltas collapse four sums to two: ⟨Φ|V|Φ⟩ = ½ Σᵢⱼ (⟨ij|v|ij⟩ − ⟨ij|v|ji⟩) over occupied i and j, derived. In coordinates x = (r, σ) the direct term Jᵢⱼ = ∫∫ |ψᵢ(x₁)|² v(r₁₂) |ψⱼ(x₂)|² is the classical Coulomb energy of two charge clouds. The exchange term ∫∫ ψᵢ*(x₁) ψⱼ*(x₂) v(r₁₂) ψⱼ(x₁) ψᵢ(x₂) has no classical reading: it is the Coulomb energy of the overlap density ψᵢ*ψⱼ with its own complex conjugate, never negative for a Coulomb v, and it exists only because of the previous section's minus sign. The ½ survives because Σᵢⱼ runs over ordered pairs and meets each pair of electrons twice. Dimensions check: v = e²/(4πε₀ r₁₂), and each |ψ|² carries m⁻³ that its own volume element cancels, so J and K are energies on the scale e²/(4πε₀ a₀) = 27.2 eV.

Write each spin-orbital as ψᵢ = φᵢ(r) χᵢ, a spatial orbital times a two-component spinor. H contains no spin operator, so the spin sums factor out of both integrals: the direct term carries ⟨χᵢ|χᵢ⟩⟨χⱼ|χⱼ⟩ = 1, while the exchange term carries |⟨χᵢ|χⱼ⟩|², which is 1 for parallel spins and 0 for antiparallel. So ⟨ij|v|ji⟩ = |⟨χᵢ|χⱼ⟩|² Kᵢⱼ with a purely spatial Kᵢⱼ ≥ 0, derived: only parallel spins exchange, and when they do the energy falls, the seed of Hund's first rule. The figure's model makes this vivid. For two atoms in a one-dimensional box of length L with a contact interaction v = g δ(x₁ − x₂), the direct and exchange integrands coincide on the line x₁ = x₂, so K₁₂ = J₁₂ = g/L for box orbitals 1 and 2. Parallel spins pay nothing for the interaction; antiparallel spins pay g/L. The same antisymmetry is why a spin-polarised gas of fermionic atoms hardly collides at low temperature, and why the first degenerate ⁴⁰K gas, in 1999, had to be evaporatively cooled as a mixture of two spin states.

Set i = j in the double sum: ⟨ii|v|ii⟩ − ⟨ii|v|ii⟩ = 0 term by term, because an electron is always parallel to itself. The restriction to i ≠ j that Hartree imposed by hand in Quantum Mechanics II is therefore automatic, and the sum may run freely over all occupied pairs, which is what will let topic 10.06 give every orbital the same operator. Two checks. In the one-particle limit N = 1 the only term is i = j, which cancels, so hydrogen's determinant energy is just ⟨1s|h|1s⟩ = −13.606 eV, and the reduced-mass factor turns that into the measured −13.598 eV. For helium's 1s↑1s↓ the four ordered pairs give ½(0 + 0 + J + J) = J = (5/8)Z Eₕ = 34.0 eV at Z = 2, so the determinant energy is −108.8 + 34.0 = −74.8 eV, with no exchange from the antiparallel pair. Drop the exchange term but keep i = j, as the Hartree energy ½∫∫ ρ ρ v of the total density does, and the repulsion doubles to 68.0 eV: a 34 eV self-interaction error. In position space the same cancellation says the exchange hole around each electron holds exactly one electron's worth of charge.

With v = 0, H = Σ h is a one-body operator; if the occupied orbitals are eigenfunctions of h with eigenvalues εᵢ, then H|Φ⟩ = (Σᵢ εᵢ)|Φ⟩ and the determinant is an exact eigenstate, the non-interacting limit of this calculation. With v present, |Φ⟩ is only a normalised trial state, so Quantum Mechanics II's variational theorem gives E[Φ] ≥ E₀ for every choice of orbitals. Helium shows the size of the gap. Hydrogenic Z = 2 orbitals give −74.8 eV; varying the orbitals to self-consistency, as topic 10.08 will, gives −77.87 eV (quoted); minus the sum of the two measured ionisation energies is −79.00 eV. The last 1.1 eV is correlation, out of reach of any single determinant because it holds no term that makes antiparallel electrons avoid each other: its opposite-spin pair density is the plain product ρ↑(r₁) ρ↓(r₂). The bound also has a label: a determinant with M_S = 1 has no overlap with any singlet, so it bounds the lowest triplet rather than the ground state.