LS coupling, term symbols and Hund's rules

Split the Hamiltonian as H = H_cf + Hᵣₑₛ + H_SO. H_cf is the central field the self-consistent calculation supplies, and it fixes the configuration. Hᵣₑₛ is whatever part of Σ_(i<j) e²/4πε₀rᵢⱼ the central field failed to absorb. H_SO = Σᵢ ξ(rᵢ) lᵢ⋅sᵢ, with ξ(r) = (1/2mₑ²c²r) dV/dr. LS coupling is the assumption Hᵣₑₛ ≫ H_SO, and nothing else. Now ask what Hᵣₑₛ commutes with. It is a scalar under rotating every orbital coordinate simultaneously, so [Hᵣₑₛ, L] = 0; it contains no spin operator at all, so [Hᵣₑₛ, S] = 0. It does not commute with any individual lᵢ, so the one-electron labels are already gone — only the totals survive. That leaves the commuting set (H, L², S², J², J_z, Π) and the coupled basis |L S J M_J⟩, assembled from |L M_L⟩|S M_S⟩ by Clebsch-Gordan coefficients. H_SO commutes with J² and J_z but not with L_z or S_z, which is exactly why the uncoupled basis stops being useful the moment fine structure matters.

Two inequivalent p electrons — 2p3p — are told apart by n, so nothing is forbidden. L runs 0, 1, 2 and S runs 0, 1, giving ¹S, ¹P, ¹D, ³S, ³P and ³D; the states count 1 + 3 + 5 + 3 + 9 + 15 = 36 = (3×2)². Make them equivalent — 2p² — and both electrons must share the same six spin-orbitals, so only C(6,2) = 15 antisymmetric states exist and more than half the terms have to go. The survivors are those whose spatial and spin functions carry opposite exchange symmetry: for two electrons of the same l the spatial function has exchange parity (−1)^L and the spin function (−1)^(S+1), so an antisymmetric product needs L + S even. That leaves ³P, ¹D and ¹S, with 9 + 5 + 1 = 15. If you do not trust the shortcut, run the microstate table instead — list every allowed (mₗ, mₛ) pair, tabulate M_L and M_S, then strip off the largest available term and repeat until the table is empty. Holes work as well as electrons, so p⁴ carries the terms of p² and d⁸ those of d², which is the fastest route to oxygen and to nickel.

The first rule takes the largest S, and the mechanism is not magnetic. A high-spin state forces an antisymmetric spatial function, which vanishes at r₁₂ = 0 and digs a Fermi hole, so the electrons see less of 1/r₁₂ and the high-spin term sits 2K below the low-spin term built on the same configuration, K being the exchange integral. Helium shows the size of it: 1s2s ³S lies 0.80 eV below 1s2s ¹S, four orders of magnitude more than a magnetic dipole-dipole coupling between the two spins could manage. The second rule takes the largest L available at that S, for the related reason that electrons circulating the same way meet each other less often. The third rule is different in kind — it is nothing but the sign of A, positive below half filling and negative above, so the multiplet is regular in carbon and inverted in oxygen. Note what the rules do not do. They order the terms of a single configuration, and reliably only the lowest of them; they do not order the excited terms, and they say nothing about which configuration wins, which is why chromium is 3d⁵4s¹ rather than 3d⁴4s².

J = L + S gives J² = L² + S² + 2 L⋅S, so inside a single term, where L² and S² are already fixed numbers, ⟨L⋅S⟩ = (ħ²/2)[J(J+1) − L(L+1) − S(S+1)]. Project the one-electron sum Σ ξ(rᵢ) lᵢ⋅sᵢ onto that one term and it collapses to a single constant multiplying L⋅S: for the maximum-multiplicity term of one open subshell A = ±ζₙₗ/(2S), the sign following the filling. The level shift is then E_SO(J) = (A/2)[J(J+1) − L(L+1) − S(S+1)]. Subtract neighbours: J(J+1) − (J−1)J = 2J, so E_J − E_(J−1) = AJ. That is the Landé interval rule, and it says the successive gaps of a multiplet stand in the ratio 1 : 2 : 3 …, counting up from the smallest J. For ³P, with L = S = 1, the bracket is J(J+1) − 4 and the three levels sit at −2A, −A and +A. Check the weighted sum: 1(−2A) + 3(−A) + 5(+A) = 0, the general centre-of-gravity theorem Σ_J (2J+1) E_SO(J) = 0. Spin-orbit coupling redistributes a term's energy among its levels; it never moves the term.

Carbon's 2p² ³P has ³P₁ at 16.40 cm⁻¹ and ³P₂ at 43.40 cm⁻¹ above ³P₀. Read A off the first interval: A = +16.40 cm⁻¹, so ζ₂p = 2A = 32.8 cm⁻¹. It is positive, so the multiplet is regular and J = 0 lies lowest, exactly as Hund's third rule requires for 2 electrons in a capacity of 6. The rule then predicts E₂ − E₀ = 3A = 49.20 cm⁻¹ against the measured 43.40 cm⁻¹, high by 5.80 cm⁻¹ or 13.4%; equivalently the observed interval ratio is 27.00/16.40 = 1.65 rather than 2. That deficit is physics, not sloppiness, but name the right physics. It is not configuration interaction: second-order spin-orbit mixing of ³P₂ with the ¹D₂ term 10192.6 cm⁻¹ above it shifts the level by at most ζ²/ΔE = 32.8²/10192.6 = 0.11 cm⁻¹, some fifty times too small, and mixing with the far more distant 2p3p configuration is smaller still. What a one-body ξ(r) l⋅s form cannot contain is the two-body magnetic interaction between the electrons — the spin-spin and spin-other-orbit terms — and in a light atom those are exactly the size of the discrepancy. Silicon's 3p² does better, with intervals of 77.1 and 146.0 cm⁻¹ for a ratio of 1.89. Lead's 6p² gives 7819 and 2831 cm⁻¹, a ratio of 0.36 — not a strained rule but a broken one, and the A fitted from lead's first interval means nothing.

The hierarchy that justified LS coupling is a contest between two numbers. The residual electrostatic splitting hardly changes down a column, because the valence orbitals stay similar in shape; ζₙₗ climbs steeply, roughly as Z⁴/n³. In group 14 the ³P spread measured against the ³P-to-¹D gap runs 43.4/10193 = 0.0043 for carbon and 10650/21458 = 0.50 for lead — a factor near 120, while the electrostatic gap merely doubles. When ζ wins, couple each electron first, lᵢ + sᵢ = jᵢ, then J = Σ jᵢ, and label the levels (j₁, j₂)_J. Nothing about the state space changes: p² gives (½,½)₀, (3/2,3/2) with J = 0, 2 and (½,3/2) with J = 1, 2, which is 1 + 6 + 8 = 15 states carrying J = {0, 0, 1, 2, 2} — exactly the multiset LS coupling produced. Only the labels and the energies differ. Real heavy atoms sit in between, in intermediate coupling, which is why mercury's nominally spin-forbidden 6³P₁ → 6¹S₀ line at 254 nm is strong enough to run a lamp on.