University Physics V · Multi-Electron Atoms · 12.7
LS Coupling, Term Symbols & Hund's Rules
A configuration is not a level. Antisymmetry decides which L and S survive, the leftover electron repulsion decides which term lies lowest, and ξ(r) L⋅S splits that term into a J ladder whose spacings you can check against a measured spectrum.
Build the model
Connect the measurement to the mechanism.
LS coupling is not a new physical law; it is a claim about which of three energies is biggest. The central field the self-consistent calculation supplies is the largest and fixes the configuration. What that central field failed to absorb of Σ e²/4πε₀rᵢⱼ comes next.
The spin-orbit term Σ ξ(rᵢ) lᵢ⋅sᵢ is smallest. Because the residual repulsion is a scalar under rotating every orbital coordinate at once, and contains no spin operator at all, it commutes with L = Σ lᵢ and with S = Σ sᵢ separately, so both stay good quantum numbers and the configuration's states break into terms ²ˢ⁺¹L, each (2L+1)(2S+1)-fold degenerate. Antisymmetry then decides which (L, S) pairs exist at all: two equivalent nl electrons keep only L + S even, which is how p² keeps ³P, ¹D and ¹S out of the six terms two inequivalent p electrons would give.
Hund's three rules rank the terms of that one configuration, and the smallest energy in the hierarchy resolves each term into levels labelled by J, spaced by the Landé rule EJ − E_(J−1) = AJ. The cost is the hierarchy itself. ζₙₗ climbs roughly as Z⁴/n³ while the residual electrostatic splitting hardly moves down a column, so in heavy atoms the ordering that justified L and S collapses, term symbols become labels on states that are really mixtures, and jj coupling takes over at the far end.
- Simple definition
- In LS (Russell–Saunders) coupling the electrons' orbital angular momenta are summed into a total L and their spins into a total S, and only then are L and S coupled into J = L + S — an order that is justified only while the residual electrostatic repulsion outweighs the spin-orbit interaction.
- Example
- Carbon's 2p² subshell holds 15 antisymmetric microstates, which LS coupling sorts into ³P (9 states), ¹D (5) and ¹S (1); spin-orbit then splits ³P alone, putting ³P₁ 16.40 cm⁻¹ and ³P₂ 43.40 cm⁻¹ above the ground level ³P₀.
Fixes how many states hide behind one term label and how many levels it can split into.
Only L is printed as a letter: L = 0,1,2,3,4 become S, P, D, F, G, while the multiplicity 2S+1 and the total J are printed as numbers.
Gives the full term list before antisymmetry: 2p3p yields ¹S ¹P ¹D ³S ³P ³D, 36 states.
Integer steps in each series, and every L pairs with every S while the electrons are inequivalent.
Cuts p² down to ³P, ¹D and ¹S — 9 + 5 + 1 = 15, not the 36 of 2p3p.
C(6,2) = 15 for p², C(10,2) = 45 for d²; holes work too, so p⁴ carries the terms of p².
Turns the operator into arithmetic: J² = L² + S² + 2 L⋅S, so ⟨L⋅S⟩ is fixed once J is.
For the maximum-multiplicity term of one open subshell A = ±ζₙₗ/(2S), plus below half filling and minus above, so carbon's A = 16.40 cm⁻¹ means ζ₂p = 2A = 32.8 cm⁻¹. A is not ζ.
One measured interval predicts the rest, and the size of the failure measures the departure from pure LS.
A in cm⁻¹ or eV; the second identity is the centre-of-gravity theorem for a single term.
Hydrogen 2p: ζ = 3.02×10⁻⁵ eV = 0.244 cm⁻¹, and since E(j=3/2) − E(j=1/2) = (3/2)ζ the doublet is 0.365 cm⁻¹. The Z⁴ is why LS holds in carbon and breaks in lead.
α² = 5.325×10⁻⁵, R_∞hc = 13.606 eV; ζ = 0 for l = 0, and screening replaces Z by Zeff.
Why L and S survive as separate labels
Split the Hamiltonian as H = Hcf + Hᵣₑₛ + HSO. Hcf 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. HSO = Σᵢ ξ(rᵢ) lᵢ⋅sᵢ, with ξ(r) = (1/2mₑ²c²r) dV/dr. LS coupling is the assumption Hᵣₑₛ ≫ HSO, 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, Jz, Π) and the coupled basis |L S J MJ⟩, assembled from |L ML⟩|S MS⟩ by Clebsch-Gordan coefficients. HSO commutes with J² and Jz but not with Lz or Sz, which is exactly why the uncoupled basis stops being useful the moment fine structure matters.
Clebsch-Gordan first, antisymmetry second
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 ML and MS, 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.
Hund's rules are electrostatics, not magnetism
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.
Turning L⋅S into a number, and the interval rule
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 ESO(J) = (A/2)[J(J+1) − L(L+1) − S(S+1)]. Subtract neighbours: J(J+1) − (J−1)J = 2J, so EJ − 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 ΣJ (2J+1) ESO(J) = 0. Spin-orbit coupling redistributes a term's energy among its levels; it never moves the term.
The interval rule as a measurement: carbon to lead
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.
Where LS coupling hands over to jj
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.
Change one variable at a time
Make the relationship visible.
Drag A negative: past half filling the multiplet inverts, ³P₂ drops below ³P₀ and LOWEST LEVEL J flips from 0 to 2. Then push |A| to 50 and D down to 40 — ³P₂ rises past ¹D₂ once |A| exceeds D, that is once 3|A|/D passes 3, and the LS labels stop meaning much. Real carbon is off this scale entirely: A = 16.4 cm⁻¹ against D ≈ 10160 cm⁻¹, a ratio of 0.005.
SPREAD E(³P₂)−E(³P₀)60 cm⁻¹
GAP ³P₂ → ¹D₂180 cm⁻¹
RATIO 3|A| / D0.30
LOWEST LEVEL J0
Live interpretationSPREAD E(³P₂)−E(³P₀): 60 cm⁻¹. GAP ³P₂ → ¹D₂: 180 cm⁻¹. RATIO 3|A| / D: 0.30. LOWEST LEVEL J: 0
Catch the common trap
Explain before calculating.
Oxygen's ground configuration is 1s² 2s² 2p⁴. Which term symbol is its ground level, and on what grounds?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyCarbon's ground configuration is 1s² 2s² 2p². Count the antisymmetric microstates of the 2p² subshell, list the terms LS coupling allows, verify the count, and give the ground-state term symbol.
- The 2p subshell holds 2(2l+1) = 6 spin-orbitals and takes 2 electrons, so the number of antisymmetric microstates is C(6,2) = 15 — not the 36 that two inequivalent p electrons, 2p3p, would give.
- Clebsch-Gordan alone would allow L = 0, 1, 2 with S = 0, 1. Antisymmetry keeps only L + S even for two equivalent nl electrons, leaving ³P (L=1, S=1), ¹D (L=2, S=0) and ¹S (L=0, S=0).
- Count check with (2L+1)(2S+1): 3×3 = 9 for ³P, 5×1 = 5 for ¹D, 1×1 = 1 for ¹S. 9 + 5 + 1 = 15, so nothing is missing and nothing is double-counted.
- Hund 1 takes the largest S, which is S = 1 and selects ³P outright; Hund 2 then has nothing left to choose. Hund 3: 2 electrons in a capacity of 6 is less than half full, so A > 0 and the smallest J = |L − S| = 0 lies lowest.
Answer15 microstates → ³P, ¹D, ¹S (9 + 5 + 1); ground level ³P₀, with ³P₁ and ³P₂ measured 16.40 and 43.40 cm⁻¹ above it.
MediumCarbon's ³P term is measured with ³P₁ 16.40 cm⁻¹ and ³P₂ 43.40 cm⁻¹ above ³P₀. Fit the spin-orbit constant A from the first interval, use the Landé rule to predict the second, compare with the measurement, and express A in meV. Take 1 cm⁻¹ = 1.2398 × 10⁻⁴ eV.
- Inside a term, ESO(J) = (A/2)[J(J+1) − L(L+1) − S(S+1)]. For ³P, L = S = 1 so L(L+1) + S(S+1) = 4, giving ESO(J) = (A/2)[J(J+1) − 4]: E₀ = −2A, E₁ = −A, E₂ = +A.
- The first interval fixes A: E₁ − E₀ = (−A) − (−2A) = A = 16.40 cm⁻¹. It is positive, a regular multiplet — as Hund's third rule requires for a subshell 2 electrons into a capacity of 6. The one-electron constant behind it is ζ₂p = 2SA = 2 × 16.40 = 32.8 cm⁻¹; A and ζ differ by 2S and must not be interchanged.
- The Landé rule EJ − E_(J−1) = AJ then predicts E₂ − E₁ = 2A = 32.80 cm⁻¹, so E₂ − E₀ = 3A = 49.20 cm⁻¹.
- Measured, E₂ − E₀ = 43.40 cm⁻¹. The prediction is high by 5.80 cm⁻¹, or 5.80/43.40 = 13.4%, and the observed interval ratio is 27.00/16.40 = 1.65 instead of 2. Blame the two-body magnetic terms — spin-spin and spin-other-orbit — that a one-body ξ(r) l⋅s cannot contain; second-order spin-orbit mixing with ¹D₂ contributes only about ζ²/ΔE = 32.8²/10192.6 = 0.11 cm⁻¹, fifty times too little to explain the gap.
- Convert: A = 16.40 × 1.2398 × 10⁻⁴ eV = 2.033 × 10⁻³ eV = 2.03 meV. That is 1/620 of the ³P-to-¹D electrostatic gap of 10192.6 cm⁻¹ (1.264 eV), which is why LS coupling is an excellent approximation in carbon.
AnswerA = +16.40 cm⁻¹ = 2.03 meV (so ζ₂p = 32.8 cm⁻¹), a regular multiplet. Landé predicts ³P₂ at 49.20 cm⁻¹ against 43.40 measured — 13.4% high, an interval ratio of 1.65 rather than 2.
HardShow that jj coupling gives a p² configuration exactly the same set of J values as LS coupling, then use lead's measured 6p² levels — ³P₁ at 7819 cm⁻¹ and ³P₂ at 10650 cm⁻¹ above ³P₀, with ¹D₂ at 21458 cm⁻¹ — to decide which scheme its labels deserve. Compare with carbon, whose ³P spread is 43.4 cm⁻¹ against a ¹D₂ at 10193 cm⁻¹.
- LS side: p² gives ³P, ¹D and ¹S, so the levels are J = 0, 1, 2 from ³P, J = 2 from ¹D and J = 0 from ¹S — the multiset {0, 0, 1, 2, 2}, holding 1 + 3 + 5 + 5 + 1 = 15 states.
- jj side: each p electron has j = l ± ½ = ½ or 3/2. For two electrons in the same j orbital antisymmetry allows even J only, so (½,½) gives J = 0 and (3/2,3/2) gives J = 0 and 2; the mixed pair (½,3/2) is unrestricted and gives J = 1 and 2.
- Count the jj states: 1 + (1 + 5) + (3 + 5) = 15, and the J multiset is {0} ∪ {0,2} ∪ {1,2} = {0, 0, 1, 2, 2} — identical to LS. Changing the coupling scheme changes labels and energies, never the number of levels or their J values.
- Lead's diagnostic: the Landé ratio should be (E₂ − E₁)/(E₁ − E₀) = 2. Measured, it is (10650 − 7819)/7819 = 2831/7819 = 0.36, low by a factor of 5.5. The fine-structure spread against the electrostatic gap is 10650/21458 = 0.50.
- Carbon's same two numbers are 27.00/16.40 = 1.65 and 43.4/10193 = 0.0043. Spin-orbit therefore weighs about 120 times more heavily in lead than in carbon while the electrostatic gap only doubles — the steep climb ζ ∝ Z⁴/n³ predicts, softened by screening and by lead's larger n.
- Verdict: carbon's ³P₀, ³P₁, ³P₂ are genuine LS levels. Lead's are labels of convenience on states that are strong mixtures, closer to (½,½)₀ and (½,3/2)₁,₂ — which is why lead's intercombination transitions are not weak.
AnswerBoth schemes give J = {0, 0, 1, 2, 2} across 15 states. Carbon's Landé ratio 1.65 and spin-orbit fraction 0.0043 mark a clean LS case; lead's 0.36 and 0.50 do not, so its ³P labels are nominal.