University Physics IV · Atomic Physics · 11.3

Central Field & Screening

Hydrogen is solvable because one electron feels one centre; lithium is not, and the pair term is far too large to perturb away. This is the trade that saves the problem — average the repulsion into a single radial potential, solve one electron at a time, and pay in correlation.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Write down the Hamiltonian for N electrons and one term ruins it: the sum over pairs of e²/4πε₀rᵢⱼ, which ties every coordinate to every other so nothing separates. Nor is it small — switch it off in helium and you get −108.8 eV against a true −79.01 eV, a 38% error — so perturbing about bare Coulomb fails too. The central-field approximation is the trade that rescues the problem.

Most of that repulsion is structureless: each electron sees the others as a smeared, nearly spherical cloud, so replace the pair sum by a sum of one-electron potentials V(rᵢ), each the spherical average of what the rest of the atom does. The Hamiltonian separates again; the angular equation is untouched, because spherical symmetry is all Yₗₘ ever required, and only the radial equation changes. Since V is built from the orbitals it produces, you iterate to self-consistency — Hartree, or Hartree-Fock once antisymmetry's exchange term is included.

What you buy is a legitimate one-electron picture in which n, l, mₗ and mₛ are still good labels, and a V(r) running from −Ze²/4πε₀r at the nucleus to −e²/4πε₀r outside a neutral atom, whose r-dependence breaks the l-degeneracy the pure Coulomb problem had. What you pay is correlation: the instantaneous dodging of one electron by another, which no averaged potential can hold, and which is worth about 1.1 eV per pair.

Simple definition: The central-field approximation replaces the pairwise repulsion between electrons with a single spherically symmetric potential V(r) for each electron — the angular average of what all the others do — so the many-electron problem separates into one-electron problems again.
Example: For sodium's valence electron V(r) runs from −11e²/4πε₀r at the nucleus to −e²/4πε₀r outside the neon core; its measured binding of 5.14 eV needs Z_eff = 1.84, not 1.00, because the 3s orbital penetrates that core.

Many-electron HamiltonianH = Σᵢ [−ℏ²∇ᵢ²/2m − Ze²/4πε₀rᵢ] + Σᵢ<j e²/4πε₀rᵢⱼ

The pair term is the whole difficulty: not separable, and in helium worth 34 eV against a 79 eV total, so not perturbatively small either.

rᵢⱼ = |rᵢ − rⱼ| in metres; the pair sum has N(N−1)/2 terms and ties every coordinate to every other

Central-field replacementΣᵢ<j e²/4πε₀rᵢⱼ → Σᵢ V(rᵢ) + Hᵣₑₛ

Restores separability: ψ becomes a product of one-electron orbitals and the angular factors stay exactly the Yₗₘ.

V(rᵢ) in joules or eV, a function of the radius rᵢ alone — no direction, no other electron's coordinates

The two ends of V(r)V → −Ze²/4πε₀r as r → 0V → −(Z−N+1)e²/4πε₀r as r → ∞

Fixes both limits of the curve before any computation, and is the first check on any numerical V(r) you produce.

Z the nuclear charge and N the electron count, both pure numbers; for a neutral atom Z − N + 1 = 1

Hartree self-consistency loopρ(r) = −e Σⱼ |φⱼ(r)|² → ∇²V_H = −ρ/ε₀ → new φⱼ

The loop is the method, since V is built from the orbitals it produces. Hartree-Fock adds exchange and removes the self-interaction.

ρ in C m⁻³, V_H in volts; iterate until V shifts by less than tolerance, typically 10 to 30 passes

Slater's rules for the screening constantZ_eff = Z − s, Eₙₗ ≈ −13.6 eV × Z_eff²/n*²

Li 2s: s = 1.70, Z_eff = 1.30, E = −5.75 eV against 5.39 eV measured. Use n* = 3.7 for n = 4, 4.0 for n = 5.

s dimensionless: 0.35 per same-group electron (0.30 within 1s), 0.85 per electron in shell n−1, 1.00 deeper

Quantum defectEₙₗ = −13.6 eV / (n − δₗ)²

Reads screening straight off a spectrum: Na 3s −5.14 eV, 3p −3.03 eV, 3d −1.52 eV, hydrogen n = 3 −1.51 eV.

δₗ dimensionless and shrinking fast with l; sodium has δₛ = 1.37, δₚ = 0.88, δ_d = 0.01

The pair term is neither separable nor small

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.

Average it into one spherically symmetric potential

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 pair sum by a sum of one-electron potentials, 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.

Iterate until the potential stops moving

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.

Slater's rules put a number on it in one line

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.

Penetration is what lifts the l-degeneracy

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 never 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 entire content 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.

What the average throws away

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 why orbital energies are not ionisation energies, and why quantum chemistry spends all its effort past this point.

Change one variable at a time

Make the relationship visible.

Interactive model

Nuclear charge Z11

Core screening radius d0.7 a₀

Marker radius r2.0 a₀

Set Z = 11 and d = 0.7 a₀, then drag the marker in from 6 a₀ to 0.5 a₀: Z_eff climbs from 1.00 to 5.90 and the pull deepens from −4.5 eV to −321 eV. That climb is penetration, and it is why 3s lies below 3p lies below 3d.

Z_eff AT THE MARKER1.57

SCREENING s = Z − Z_eff9.43

SCREENED V(r)-21.4 eV

UNSCREENED −Ze²/4πε₀r-149.7 eV

Live interpretationZ_eff AT THE MARKER: 1.57. SCREENING s = Z − Z_eff: 9.43. SCREENED V(r): −21.4 eV. UNSCREENED −Ze²/4πε₀r: −149.7 eV

Catch the common trap

Explain before calculating.

Misconception check

The error is to treat screening as a headcount: sodium has ten core electrons, so they cancel ten protons and the valence electron sees Z_eff = 1. That would be right only if the core were a point charge sitting at the nucleus and the valence orbital lay entirely outside it, and neither holds. Screening is a function of position, not a number — Z_eff(r) runs from 11 at the nucleus down to 1 far outside — and every valence orbital keeps some amplitude inside the core. Sodium's 3s is bound at 5.14 eV, which requires Z_eff = 1.84, not 1.00. If perfect screening were right, 3s, 3p and 3d would all sit at hydrogen's −1.51 eV and sodium would have no D lines at all. The same headcount fails inside a shell too: a same-shell electron screens only about 0.35 of a charge, which is why Z_eff keeps climbing across a period even though no new shell has opened.

In sodium (Z = 11) the 3s electron is bound by 5.14 eV, while the 3d electron is bound by 1.52 eV — essentially hydrogen's 1.51 eV for n = 3. What is responsible for the 3.6 eV difference?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyLithium is 1s²2s¹ with Z = 3. Use Slater's rules to find the effective nuclear charge seen by the 2s electron, estimate its binding energy from −13.6 eV × Z_eff²/n², and compare with the measured first ionisation energy of 5.39 eV.

Group the orbitals as (1s)(2s2p). The 2s electron has no partner in its own group, and two electrons sit in the shell below it, n − 1 = 1.
Slater's coefficients for an s electron: 0.35 per same-group electron and 0.85 per electron in shell n − 1. So s = 0 × 0.35 + 2 × 0.85 = 1.70.
Z_eff = Z − s = 3 − 1.70 = 1.30. With n = 2: E = −13.6 eV × 1.30²/2² = −13.6 × 1.69/4 = −5.75 eV.
Bracket it with the two extremes. No screening at all (Z_eff = 3) gives −30.6 eV; perfect screening by both core electrons (Z_eff = 1) gives −3.40 eV. The measured −5.39 eV lies between them, and Slater's partial screening lands 0.36 eV, or 6.6%, too deep.

AnswerZ_eff = 1.30 and E ≈ −5.75 eV against the measured −5.39 eV — 6.6% too deep, and comfortably inside the −30.6 eV to −3.40 eV bracket set by no screening and perfect screening.

MediumPotassium has Z = 19 and an argon core, 1s²2s²2p⁶3s²3p⁶. Use Slater's rules to compare one extra electron placed in 4s with one placed in 3d, taking the effective quantum numbers n* = 3.7 for n = 4 and n* = 3 for n = 3, and decide which is the ground configuration.

For the 4s electron the groups to its left are (1s)(2s2p)(3s3p)(3d), and its own group (4s4p) holds no other electron.
Apply the s and p rule: the eight electrons of shell n − 1 = 3 give 8 × 0.85 = 6.80, and the ten in shells 1 and 2 give 10 × 1.00 = 10.00. So s = 16.80 and Z_eff(4s) = 19 − 16.80 = 2.20.
For the 3d electron the d rule is harsher — every electron to its left counts a full 1.00 — so s = 18 × 1.00 = 18.00 and Z_eff(3d) = 1.00. The 3d is completely screened, that is, hydrogenic.
Energies with the effective quantum numbers: E(4s) = −13.6 × 2.20²/3.7² = −13.6 × 4.84/13.69 = −4.81 eV, and E(3d) = −13.6 × 1.00²/3² = −1.51 eV.
The 4s lies 3.3 eV below the 3d despite its larger n. Measured: potassium's 4s is bound at 4.34 eV and its 3d at 1.67 eV. The penetrating 4s beats a 3d that the centrifugal barrier keeps outside the core — which is the whole content of the n + l filling order.

AnswerZ_eff = 2.20 for the 4s against 1.00 for the 3d, giving −4.81 eV and −1.51 eV. The 4s wins by 3.3 eV, so [Ar]4s¹ is the ground state; the measured values are −4.34 eV and −1.67 eV.

HardHelium, Z = 2. (i) Switch the repulsion off and use two hydrogenic 1s orbitals. (ii) Add it to first order, using ⟨Vₑₑ⟩ = (5/8)Z × 27.211 eV. (iii) Let the orbital relax to the best screened 1s, Z' = Z − 5/16, for which E(Z') = (Z'² − 2ZZ' + (5/8)Z') × 27.211 eV. Compare all three with the Hartree-Fock limit of −77.87 eV and the exact −79.01 eV.

(i) Two independent hydrogenic 1s electrons at Z = 2: E = 2 × (−13.6 eV × 2²) = −108.8 eV. The exact value is −79.01 eV, the sum of the two measured ionisation energies 24.59 + 54.42 eV, so this is 29.8 eV or 38% too deep. The repulsion is not a small correction.
(ii) First order: ⟨Vₑₑ⟩ = (5/8) × 2 × 27.211 eV = 34.01 eV, so E ≈ −108.8 + 34.0 = −74.8 eV. The error has fallen from 29.8 eV to 4.2 eV — one averaging step removes 86% of it — but it now overshoots, because the orbitals were held rigid while the repulsion was switched on.
(iii) Let them relax: Z' = 2 − 5/16 = 1.6875. Then E = (1.6875² − 2 × 2 × 1.6875 + 0.625 × 1.6875) × 27.211 eV = (2.8477 − 6.7500 + 1.0547) × 27.211 = −2.8477 × 27.211 = −77.49 eV, an error of 1.5 eV.
Note that the fitted screening 5/16 = 0.3125 is exactly what Slater's 1s rule rounds to 0.30 — the rule is this calculation, tabulated.
The true self-consistent solution does a little better: the Hartree-Fock limit is −77.87 eV. That is the best any single central field can reach, whatever shape V(r) is allowed to take.
The gap that survives, −79.01 − (−77.87) = −1.14 eV, is the correlation energy — the instantaneous dodging that no averaged potential contains.

Answer(i) −108.8 eV, (ii) −74.8 eV, (iii) −77.49 eV, against the Hartree-Fock limit −77.87 eV and the exact −79.01 eV. Correlation is the residual 1.14 eV: 1.4% of the total energy, but a quarter of the 4.75 eV that binds H2.