Central field and screening

The N-electron Hamiltonian splits into two kinds of term. The one-electron parts — kinetic energy and the nuclear attraction −Ze²/4πε₀rᵢ — are already a sum over electrons, so they separate and a product wavefunction works. The pair sum Σᵢ<j e²/4πε₀rᵢⱼ does not: rᵢⱼ couples every coordinate to every other, and with it in place no product is an eigenfunction. Nor can you demote it to a perturbation. Switch it off in helium and two hydrogenic 1s electrons at Z = 2 give E = 2 × (−54.4 eV) = −108.8 eV, against a measured −79.01 eV. That is an error of 29.8 eV, or 38% — the neglected term is more than a third of the answer. So the repulsion can be neither kept exactly nor thrown away, and the central-field approximation is what you do instead.

Look at what one electron actually experiences. The others are spread over the whole atom, and averaged over their positions the force they exert is very nearly radial and very nearly independent of direction. So replace the nuclear attraction and the pair sum together by a sum of one-electron potentials, −Σᵢ Ze²/4πε₀rᵢ + Σᵢ<j e²/4πε₀rᵢⱼ → Σᵢ V(rᵢ), where V(r) is the nuclear attraction plus the spherical average of the repulsion from the smeared cloud of everyone else. The Hamiltonian becomes a sum of identical one-electron operators, so it separates and each electron gets its own orbital. Because V depends on r alone, the angular equation is exactly the one hydrogen had: the spherical harmonics Yₗₘ survive untouched and l and mₗ stay good quantum numbers. Only the radial equation changes. Two limits pin the new V(r) down with no calculation at all: at r → 0 the electron is inside everything else and sees the bare nucleus, V → −Ze²/4πε₀r; at r → ∞ it sees the nucleus screened by the other N − 1 electrons, V → −(Z−N+1)e²/4πε₀r, which for a neutral atom is simply −e²/4πε₀r.

V(r) is built from the orbitals, and the orbitals are the solutions in V(r) — a circular definition you break by iteration. Guess a V₀(r), solve the radial equation for the N lowest orbitals, build the charge density ρ(r) = −eΣⱼ|φⱼ|², solve Poisson's equation for the potential that density makes, spherically average it, add the nuclear term, and feed the result back in. Stop when the potential shifts by less than your tolerance; a light atom converges in ten to thirty passes. That is the Hartree method. Hartree-Fock adds the term antisymmetry demands: exchange, which is non-local — it acts on φᵢ(r) through an integral over φᵢ(r') — and which incidentally cancels the spurious self-interaction Hartree leaves behind, the piece in which electron i repels its own charge cloud. The output is not a formula but a table: orbital energies εₙₗ and radial functions, computed one atom at a time.

You do not always need the computation. Slater fitted screening constants to it: write Z_eff = Z − s and build s by grouping the orbitals as (1s)(2s2p)(3s3p)(3d)(4s4p)(4d)(4f). Each other electron in the same group contributes 0.35, except within 1s where it contributes 0.30; for an s or p electron each electron in shell n−1 contributes 0.85 and each one deeper contributes 1.00; for a d or f electron everything to its left contributes the full 1.00. Lithium's 2s electron: s = 2 × 0.85 = 1.70, so Z_eff = 1.30 and −13.6 eV × 1.30²/2² = −5.75 eV, against a measured 5.39 eV. Two lessons hide in those coefficients. Same-shell electrons screen only about a third of a charge, so Z_eff climbs by roughly 0.65 per proton across a period — 1.30 at lithium to 5.85 at neon — and since ⟨r⟩ ≈ n²a₀/Z_eff, the atom contracts by a factor of 4.5 across period 2. And the 0.30 for 1s is no accident: the variational helium calculation gives 5/16 = 0.3125.

In a pure Coulomb potential every l at a given n shares one energy. A screened V(r) is not pure Coulomb, and that accident dies. The mechanism is penetration: the centrifugal barrier l(l+1)ℏ²/2mr² holds high-l orbitals out at large r, where the core has screened the nucleus down to Z_eff ≈ 1, while a low-l orbital keeps amplitude close in, where Z_eff is still large. More time at small r means a deeper well, so E(ns) < E(np) < E(nd). Sodium reads it straight off: 3s at −5.14 eV, 3p at −3.03 eV, 3d at −1.52 eV, and hydrogen's n = 3 level is −1.51 eV. The 3d is hydrogenic to within 0.01 eV because it barely reaches the core; the 3s is bound 3.4 times deeper, and the 2.10 eV gap between 3s and 3p is the 589 nm D line. As quantum defects, E = −13.6 eV/(n − δₗ)², those energies are δₛ = 1.37, δₚ = 0.88, δ_d = 0.01. This is also the physical root of the n + l filling order: potassium's penetrating 4s is bound at 4.34 eV while its 3d sits at 1.67 eV, so 4s fills first.

The central field is exact only for the spherically symmetric part of the repulsion, and two things are left outside it. The non-spherical residue Hᵣₑₛ is now a genuine perturbation — small enough to treat as one — and it is what splits a single configuration into its terms, the ¹S, ³P and the rest of an open shell. Correlation is the worse loss, because it is not a correction on top of a good state but a defect in the state itself: electrons dodge one another instant by instant, and a single determinant of one-electron orbitals cannot describe that however well V(r) is chosen. Helium sizes it. The best screened hydrogenic guess gives −77.49 eV, the Hartree-Fock limit — the best any single central field can reach — is −77.87 eV, and the truth is −79.01 eV. The missing 1.14 eV is the correlation energy: only 1.4% of the total, but a quarter of the 4.75 eV that holds H₂ together. That is one reason orbital energies are not exact ionisation energies, and why quantum chemistry spends all its effort past this point.