University Physics IV · Atomic Physics · 11.6
LS Coupling & Term Symbols
A configuration names which orbitals are filled and then stops. This is the machinery that turns 2p² into real levels — which terms antisymmetry allows, how spin-orbit splits each one, in what order, and when the labels stop meaning anything.
Build the model
Connect the measurement to the mechanism.
The central field leaves two things out, and their contest decides everything. What remains is the non-spherical part of the electron-electron repulsion and the spin-orbit interaction, both acting inside a configuration whose states are still degenerate. Degenerate perturbation theory does not let you pick an order: the larger term must be diagonalised first, because only then is the smaller one a genuine small correction.
In a light atom repulsion wins, so the orbital momenta couple into a total L and the spins into a total S, the configuration splits into terms separated by electronvolts, and antisymmetry has already deleted the combinations two equivalent electrons cannot form. Spin-orbit then acts inside a term as a multiple of L⋅S, which is diagonal in J = L + S, splitting the term into levels running from |L − S| to L + S with gaps that grow linearly in J. The price is that L and S are approximations, good only while mixing between terms stays small next to the term separation.
The spin-orbit parameter climbs as roughly Z⁴ while the repulsion barely moves, so past about Z = 50 the ordering reverses: each electron's own l and s couple first into j, the j add to J, and the term symbol becomes a name with nothing behind it. J and parity survive both schemes. Nothing else does.
- Simple definition
- LS coupling adds all the electrons' orbital momenta into one L and all their spins into one S, then couples those two into J; the level is named by a term symbol whose three labels are exactly S, L and J.
- Example
- Sodium's excited 3p electron has L = 1 and S = 1/2, so J = 1/2 or 3/2 and the term ²P splits into ²P₁/₂ and ²P₃/₂ — 17.2 cm⁻¹ apart, which is the D-line doublet at 589.0 and 589.6 nm.
Counts the fine-structure levels a term carries, and shows why the multiplicity is not always that count.
L, S and J are pure numbers; the vector length is √(J(J+1)) ℏ, and there are min(2S+1, 2L+1) values
One label holding every quantum number a level still has once its configuration is fixed.
2S+1 is the multiplicity; a superscript o is added for odd parity, meaning the sum of the lᵢ is odd
One constant fixes a whole multiplet, so measuring one gap predicts every other gap in it.
A is the term's spin-orbit constant in cm⁻¹ or J; it follows from L⋅S = ½(J² − L² − S²)
Ratios of gaps read straight off a spectrum give the J values, and from them the L and S you never measured.
A > 0 for a subshell less than half filled (normal), A < 0 past half filling (inverted)
Splitting moves levels but creates none: a check that no state was lost or invented on the way.
each level holds 2J+1 states labelled by mJ, and the degeneracy-weighted mean shift is zero
The same J values and the same parity, regrouped into different levels with quite different spacings.
used once ζ exceeds the residual repulsion; ζ climbs as roughly Z⁴ while the repulsion barely moves
Two leftovers, and the bigger one is diagonalised first
The self-consistent central field has already swallowed the spherical average of the electron-electron repulsion. Two things are left over: the non-spherical remainder of that repulsion, and the spin-orbit interaction, a sum of terms in l⋅s for each electron. Both act on a configuration whose states are still degenerate — 2p² carries fifteen of them before either switches on — and degenerate perturbation theory does not let you choose an order. The dominant term must be diagonalised first, because only the basis it picks out makes the other one a genuine small correction. In carbon the ordering is not close: repulsion spreads 2p² over 21648 cm⁻¹ while spin-orbit moves things by tens of cm⁻¹, a ratio near 500. In lead the same two are within a couple of per cent of each other, and the question has no clean answer at all.
Which L and S survive: antisymmetry does half the work
For non-equivalent electrons every combination occurs. The excited configuration 2p3p holds 6 × 6 = 36 states and gives six terms — ¹S, ³S, ¹P, ³P, ¹D, ³D — carrying 1 + 3 + 3 + 9 + 5 + 15 = 36 states between them. Equivalent electrons are different. Two 2p electrons share six spin-orbitals, so there are only C(6,2) = 15 antisymmetric states, and the Pauli principle keeps just the terms with L + S even: ¹S, ³P and ¹D, carrying 1 + 9 + 5 = 15. That is not a convention to memorise; it is the same Slater determinant that gives exclusion, doing more work than the slogan about two electrons in one state. Always count states before and after: if the term list does not reproduce the configuration's state count, a term has been invented or lost.
Inside a term: L⋅S, and the interval rule that tests it
Within a term L and S are fixed, so squaring J = L + S gives L⋅S = ½[J(J+1) − L(L+1) − S(S+1)]ℏ² and the spin-orbit energy collapses to one constant: E = (A/2)[J(J+1) − L(L+1) − S(S+1)]. Take ³P, with L = S = 1. Then E(J=0) = −2A, E(J=1) = −A and E(J=2) = +A, so the gaps are A and 2A — the Landé rule, E(J) − E(J−1) = A J, with each gap proportional to the upper J. Carbon's first gap is 16.40 cm⁻¹, so the rule places ³P₂ at 16.40 + 32.80 = 49.20 cm⁻¹ above ³P₀. Measurement says 43.40, low by 13%. That failure is informative: only J = 2 has a partner of the same J in this configuration, the ¹D₂ level 10192.63 cm⁻¹ above, and the two repel, pushing ³P₂ down while ³P₀ and ³P₁ have nothing to mix with.
Normal, inverted, and the centre that never moves
A is positive for a subshell less than half filled and negative past half filling, which is Hund's third rule stated in one symbol. Carbon's 2p² puts two electrons in six places, so its ³P is normal and ³P₀ is the ground level. Oxygen's 2p⁴ puts in four, so the same ³P is inverted: ³P₂ sits at 0, ³P₁ at 158.27 cm⁻¹ and ³P₀ at 226.98 cm⁻¹. The interval rule still applies, now read downward — the gaps 158.27 and 68.71 cm⁻¹ stand in the ratio 2.30 against the predicted 2. What never moves is the degeneracy-weighted mean. For ³P, 1(−2A) + 3(−A) + 5(+A) = 0, so the split levels straddle the unsplit term exactly. That sum is the fastest check on any level scheme you write down.
When LS coupling fails, and what survives it
The spin-orbit parameter ζ grows roughly as Zeff⁴/n³ while the electrostatic parameters barely change down a group, so the ratio of spin-orbit to repulsion climbs steeply. Compare the group-14 np² sequence. Carbon's whole ³P spread is 43.40 cm⁻¹ against a ³P-to-¹D separation of 10149.23, a ratio of 0.0043. Lead's lowest triplet spreads over 10650.33 cm⁻¹ against a 10807.47 cm⁻¹ gap to the next J = 2 level: a ratio of 0.985. Lead's first two gaps, 7819.26 and 2831.07 cm⁻¹, stand in the ratio 0.36 where the interval rule demands 2, so the LS labels have stopped describing anything. In the jj limit each electron's own l and s couple first into j, and the j add to J. What survives the change of scheme is J and parity; L and S were only ever names for a choice of basis.
Reading a multiplet backwards
A measured group of close levels can be decoded without calculating anything about the atom. Count the levels: that number is min(2S+1, 2L+1). Then take successive gaps and divide them by the smallest, because the interval rule makes those ratios the J values of the upper members. A five-level group whose gaps stand as 2 : 3 : 4 : 5 therefore has J = 1, 2, 3, 4, 5, so |L − S| = 1 and L + S = 5, giving L = 3 and S = 2 — a ⁵F term. Check it on the state count: 3 + 5 + 7 + 9 + 11 = 35 = (2L+1)(2S+1) = 7 × 5. And if the ratios refuse to come out as consecutive integers, that is not a bad measurement; it is the spectrum telling you this atom is not LS coupled.
Change one variable at a time
Make the relationship visible.
Set L = 1 and A = 16 for carbon's ³P: gaps of 16 and 32 cm⁻¹, in the ratio 1 : 2. Now drag A negative — past half filling the ladder inverts, as it does in oxygen's 2p⁴ — while the rung lengths never move, because degeneracy does not care about the sign of A.
LOWER GAP A L16.0 cm⁻¹
UPPER GAP A (L+1)32.0 cm⁻¹
GAP RATIO2.00
STATES IN THE TERM9 states
Live interpretationLOWER GAP A L: 16.0 cm⁻¹. UPPER GAP A (L+1): 32.0 cm⁻¹. GAP RATIO: 2.00. STATES IN THE TERM: 9 states
Catch the common trap
Explain before calculating.
A term symbol reads ⁴P. How many fine-structure levels does it contain, and what are their J values?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasySodium's valence electron is promoted from 3s to 3p. Write the term symbols for both configurations, say how many levels each has, and use the observed D lines at 588.995 nm and 589.592 nm to find the splitting of the upper term and its constant A.
- Closed shells contribute L = 0 and S = 0, so the one valence electron sets the term. For 3s: l = 0 and s = 1/2 give L = 0, S = 1/2 and J = 1/2 only, so the term is ²S₁/₂ — a doublet with a single level, because min(2S+1, 2L+1) = min(2, 1) = 1.
- For 3p: l = 1 gives L = 1, S = 1/2, and J runs from |1 − 1/2| = 1/2 to 1 + 1/2 = 3/2. Two levels: ²P₁/₂ and ²P₃/₂.
- Convert both lines to wavenumbers: 10⁷/588.995 = 16978.07 cm⁻¹ and 10⁷/589.592 = 16960.88 cm⁻¹.
- Both transitions end on the same ²S₁/₂ level, so the difference is the splitting of the upper term: 16978.07 − 16960.88 = 17.19 cm⁻¹, which is 17.19 × 1.2398 × 10⁻⁴ = 2.13 × 10⁻³ eV.
- The interval rule gives E(J) − E(J−1) = A J with the upper J = 3/2, so A = 17.19 ÷ 1.5 = 11.46 cm⁻¹. For a single electron outside closed shells this A is the spin-orbit parameter ζ₃ₚ itself.
Answer3s gives ²S₁/₂, one level; 3p gives ²P₁/₂ and ²P₃/₂. The doublet splitting is 17.19 cm⁻¹, or 2.13 meV, and A = ζ₃ₚ = 11.46 cm⁻¹.
MediumFor the ground configuration 1s²2s²2p² of carbon, list the terms the Pauli principle allows, check the state count, order them with Hund's rules, and test the Landé interval rule against the observed ³P levels at 0, 16.40 and 43.40 cm⁻¹.
- Two equivalent p electrons occupy 2 of 6 spin-orbitals, so the configuration holds C(6,2) = 15 antisymmetric states, not the 36 that two non-equivalent p electrons would give.
- For two equivalent p electrons only the terms with L + S even survive: ¹S (L=0, S=0), ³P (L=1, S=1) and ¹D (L=2, S=0). Count them: 1 + 9 + 5 = 15, so nothing is missing.
- Hund's first rule takes the largest S, putting ³P lowest. Hund's third rule: 2p² is less than half filled, so A > 0 and the smallest J lies lowest — the ground level is ³P₀.
- Interval rule with L = S = 1: the gaps are A × 1 and A × 2. The first observed gap fixes A = 16.40 cm⁻¹, so ³P₂ is predicted at 16.40 + 32.80 = 49.20 cm⁻¹ above ³P₀.
- Observed is 43.40 cm⁻¹, so the prediction is high by 5.80 cm⁻¹, or 13%. Only ³P₂ has a same-J partner in this configuration — ¹D₂ at 10192.63 cm⁻¹ — and the two repel, pushing ³P₂ down while ³P₀ and ³P₁ are untouched.
- Check the centre of gravity with E = −2A, −A, +A: 1(−32.80) + 3(−16.40) + 5(+16.40) = −32.80 − 49.20 + 82.00 = 0.
Answer¹S, ³P and ¹D — fifteen states in five levels, ground level ³P₀. The interval rule predicts ³P₂ at 49.20 cm⁻¹ against 43.40 observed, 13% high, because ¹D₂ mixes with ³P₂ and pushes it down.
HardThe five lowest levels of carbon (2p²) lie at 0, 16.40, 43.40, 10192.63 and 21648.01 cm⁻¹; those of lead (6p²) lie at 0, 7819.26, 10650.33, 21457.80 and 29466.83 cm⁻¹. Decide which atom is LS coupled, and show that the set of J values does not depend on the answer.
- Both are np², so both must produce the same five J values in either scheme. In LS they are ³P₀, ³P₁, ³P₂, ¹D₂, ¹S₀, that is J = 0, 1, 2, 2, 0.
- Interval-rule test on the lowest three. Carbon: gaps 16.40 and 43.40 − 16.40 = 27.00, ratio 27.00/16.40 = 1.65 against the predicted 2. Lead: gaps 7819.26 and 10650.33 − 7819.26 = 2831.07, ratio 0.36. Carbon nearly obeys; lead does not.
- Size test. Carbon's triplet spans 43.40 cm⁻¹ while the gap up to ¹D₂ is 10192.63 − 43.40 = 10149.23, a ratio of 0.0043. Lead's triplet spans 10650.33 while the gap to its next J = 2 level is 21457.80 − 10650.33 = 10807.47, a ratio of 0.985.
- So carbon has spin-orbit at 0.4% of the electrostatic splitting, which is Russell-Saunders coupling; in lead the two are equal to within 2%, which is intermediate coupling, far closer to jj than to LS.
- In the jj limit each 6p electron has j = 1/2 or 3/2. Two equivalent electrons of the same j allow only even J, so (1/2, 1/2) gives J = 0 and (3/2, 3/2) gives J = 0 and 2, while the mixed pair (1/2, 3/2) gives J = 1 and 2. The set is again 0, 1, 2, 2, 0.
- The consequence is spectroscopic. The rule ΔS = 0 is only as good as S is, so where the singlet-triplet distinction is fictional the intercombination lines turn on — which is why mercury's 253.7 nm line, 6s6p ³P₁ down to 6s² ¹S₀, is strong enough to run a fluorescent lamp.
AnswerCarbon is LS coupled: interval ratio 1.65 against 2, spin-orbit 0.4% of the term splitting. Lead is not: ratio 0.36, spin-orbit 98% of it. Both carry J = 0, 1, 2, 2, 0; only the labels and the spacings change.