University Physics V · Atomic Spectroscopy · 13.6
Spin–Orbit Coupling: LS against jj
Spin–orbit coupling in a many-electron atom starts as a sum over every electron and collapses, inside one term, to a single number. Learn the projection that does it, the interval rule that makes it falsifiable against a printed level table, and the ratio that says when to stop writing ³P₂ and start writing (3/2, ½)₂.
Build the model
Connect the measurement to the mechanism.
Each electron sees, in its own rest frame, a magnetic field proportional to l from the self-consistent central field, and coupling it to the spin gives the one-body sum HSO = Σᵢ ξ(rᵢ) lᵢ⋅sᵢ with ξ(r) = (1/2m²c²)(1/r)(dV/dr). It is diagonal in no basis a term analysis uses, since it touches every electron separately. The move that rescues it is the projection theorem applied twice: inside one term, where L and S are both fixed, any vector operator built from the lᵢ must be proportional to L and any built from the sᵢ proportional to S, so the whole sum collapses to a single scalar A L⋅S.
N one-electron operators become one number per term, and it is falsifiable at once: EJ − E_(J−1) = A J, the Landé interval rule, which a printed level table either obeys or does not. Carbon's 2p² ³P intervals stand at 1.64 : 1 against the predicted 2 : 1, tin's at 1.03, lead's at 0.36. The cost is in the derivation.
Projection is legal only within one term, so once the spin–orbit matrix elements between terms — which the same operator also has — reach the electrostatic separations between them, A stops existing and every level is a mixture. That single ratio, ζₙₗ against the Slater-integral scale, is the whole content of LS against jj: ζ grows as Zeff⁴/n³ while the repulsion scale barely moves down a group, so light atoms are LS, heavy atoms jj, and everything between must be diagonalised.
- Simple definition
- Inside a single LS term, the many-electron spin–orbit sum Σᵢ ξ(rᵢ) lᵢ⋅sᵢ reduces to one constant times L⋅S, so the entire fine structure of that term is the single number A — provided the term has not yet mixed with its neighbours.
- Example
- Carbon's 2p² ³P has A = +16.42 cm⁻¹, read straight off E(³P₁) − E(³P₀); the same A then predicts E(³P₂) − E(³P₁) = 2A = 32.83 cm⁻¹ against a measured 27.00 cm⁻¹.
Names the operator, and names it as a sum of one-electron pieces — which is exactly what lets the projection theorem act on it.
V(r) the self-consistent central field in J; ξ in J per ħ², so ξ l⋅s is an energy; the ½ is the Thomas factor
Hydrogen 2p returns 0.365 cm⁻¹, the measured splitting; sodium's 17.20 cm⁻¹ D doublet inverts to Zeff = 3.55.
ζₙₗ in cm⁻¹; Zeff is a fitted parameter absorbing screening and the gap between n and the effective n*
Collapses N one-electron operators to one constant per term. It is the only reason a term has an interval rule at all.
P projects onto one term of fixed L and S > 0; + for a shell less than half full, − for more than half; A in cm⁻¹
A parameter-free ratio test. A ³P triplet must give 2 : 1; carbon returns 1.64, tin 1.03, lead 0.36.
J, L, S dimensionless, A carrying the energy; each gap scales with the upper J of the pair, not the lower
X ≪ 1 leaves LS labels good, X ≫ 1 leaves jj labels good, X ≈ 1 leaves only eigenvalues — so diagonalise.
F₂ (or G₁ for unlike shells) is the Slater integral setting the residual electrostatic term splitting, in cm⁻¹
The smallest complete crossover model: it runs ³P₁/¹P₁ into (½,½)₁/(3/2,½)₁ and hands you the singlet borrowing.
Δ = E(¹P) − E(³P centroid) in cm⁻¹; the off-diagonal element is fixed by the ζ → ∞ limit, with no tables needed
Where ξ(r) comes from, and why the sum is one-body
Boost into the electron's rest frame and the nucleus and core orbit it, so the electron sits in a magnetic field B = −(v × E)/c², where eE = ∇V is the force from the self-consistent central field. Coupling that to the spin moment and applying the Thomas factor of ½ for the frame's precession leaves ξ(r) = (1/2m²c²)(1/r)(dV/dr), positive for any attractive V that rises with r. For a hydrogenic field ξ ∝ Z/r³, and ⟨1/r³⟩ = Z³/[a₀³n³l(l+½)(l+1)], which is where the Z⁴/n³ of the next card is born. Two features matter downstream. First, the operator is a sum of terms each acting on one electron, so it can be projected shell by shell. Second, 1/r³ weights the region inside the core, which is why sodium's valence 3p electron reports Zeff = 3.55 rather than the 1 it feels at large r. The genuinely two-body magnetic interactions — spin-other-orbit and spin–spin, both of order α² from the Breit reduction — are dropped here, and they are the first thing to blame when a light atom misses the interval rule by ten per cent.
Two uses of the projection theorem collapse the sum
Fix a term: L and S given, the subspace of dimension (2L+1)(2S+1). The projection theorem, itself a corollary of Wigner–Eckart for rank-1 tensors, says that within a manifold of fixed total angular momentum any vector operator V has the same matrix elements as (⟨V⋅J⟩/ħ²J(J+1)) J. Apply it in the orbital space first: Σᵢ ξᵢ lᵢ is a vector operator with respect to L, so inside the term it may be replaced by a L. Apply it again in the spin space, where each sᵢ is a vector operator with respect to S, replaced by b S. The scalar Σᵢ ξᵢ lᵢ⋅sᵢ therefore acts inside the term exactly as A L⋅S with the single constant A = ab, one reduced matrix element for the whole term. For one open shell A = ±ζₙₗ/2S, the sign positive when the shell is less than half filled. Note what the theorem never claimed: nothing at all about matrix elements between different terms. Those exist, they are of the same size ζ, and the model's whole validity is the statement that they are small compared with the electrostatic gaps they connect.
The interval rule, its sign, and how a level table tests it
With A fixed, L⋅S = (J² − L² − S²)/2 gives E(J) = (A/2)[J(J+1) − L(L+1) − S(S+1)], so adjacent levels differ by EJ − E_(J−1) = A J. All the multiplicity structure is geometry and the gaps carry no new parameter, which makes the ratio of successive intervals a clean test. Carbon's 2p² ³P sits at 0, 16.417 and 43.413 cm⁻¹: the first gap gives A = 16.42 cm⁻¹, the second should be 2A = 32.83 cm⁻¹ and measures 27.00, a ratio of 1.64 against 2. Sign is the other half. A shell more than half full is cheaper to describe with holes, whose contribution enters the projection with reversed sign, so A < 0 and the multiplet inverts. Oxygen's 2p⁴ ³P has ³P₂ lowest, with ³P₁ at 158.265 and ³P₀ at 226.977 cm⁻¹; |A| = 79.13 cm⁻¹ from the first gap predicts 79.13 for the second against a measured 68.71, again about 15 per cent high. Both atoms are deep in LS coupling, so those misses are the neglected two-body magnetic terms and configuration interaction, not a failure of the scheme.
ζ grows as Zeff⁴/n³, and that is why heavy atoms leave LS
Alkali doublets read ζ off directly, because a lone p electron splits by (l+½)ζ = 3ζ/2. Sodium's D lines at 16956.170 and 16973.366 cm⁻¹ differ by 17.196 cm⁻¹, so ζ₃p = 11.46 cm⁻¹; inverting the central-field formula gives Zeff⁴ = ζ n³ l(l+½)(l+1)/α²R∞ = 158.9, so Zeff = 3.55. Caesium's 6p doublet spans 554.08 cm⁻¹, so ζ₆p = 369.4 cm⁻¹ and Zeff = 14.23. Between them ζ rises 32-fold, of which the 1/n³ supplies a factor of 8 and the fitted charge the rest. In the sixth row it is larger still: thallium's 6p doublet at 7792.7 cm⁻¹ gives ζ₆p = 5.2 × 10³ cm⁻¹, more than a hundred and fifty times carbon's 2p value of 32.8 cm⁻¹. The competing quantity, the residual electrostatic splitting between terms, is set by Slater integrals that change by well under a factor of two down a group. One quantity moves by two orders of magnitude, the other barely moves, and the coupling scheme follows the ratio.
Choosing the basis, and why the eigenvalues do not care
Both |(L S) J M⟩ and |(j₁ j₂) J M⟩ are complete orthonormal bases for the same configuration, related by a unitary transformation whose coefficients are 9-j symbols. The eigenvalues of the full Hamiltonian are therefore identical in either, and the choice is purely about which zeroth-order labels survive as approximations. J and parity survive in both, exactly, because H commutes with J², Jz and Π for any strength of ζ. A useful check: the level count by J must agree. An sp configuration gives ³P₀,₁,₂ and ¹P₁, that is J = {0, 1, 1, 2}; the jj scheme gives (½,½) → J = 0, 1 and (3/2,½) → J = 1, 2, the same multiset. Group 14 shows the migration in one column, the ³P interval ratio against the predicted 2: carbon 1.64, silicon 1.89, germanium 1.53, tin 1.03, lead 0.36. Silicon sitting closer to 2 than carbon is not noise — carbon's ζ is so small that the dropped two-body terms are a large fraction of it. Numerically the recipe is fixed: build Hₑₛ diagonal in LS and ζ Σ l⋅s diagonal in jj, transform one into the other's basis, and hand the matrix to scipy.linalg.eigh.
Intermediate coupling worked out on sp, and what it buys
Take an sp configuration and the J = 1 sector, spanned by ³P₁ and ¹P₁. Put the ³P centroid at zero, so ¹P sits at Δ. The diagonal spin–orbit elements are −ζ/2 for ³P₁, from (A/2)[J(J+1) − 4] with A = ζ/2, and zero for the singlet. The off-diagonal element needs no table: at Δ = 0 the eigenvalues must be the one-electron values ζ⟨l⋅s⟩/ħ², namely −ζ for j = ½ and +ζ/2 for j = 3/2, and a 2 × 2 matrix with that trace and determinant forces the off-diagonal to be ζ/√2. Its eigenvector there is √(2/3)|³P₁⟩ − √(1/3)|¹P₁⟩, so a jj level is exactly one third singlet. At the other extreme the admixture is first order in ζ/√2Δ, and it is what lets an intercombination line exist: magnesium's 3s3p gives ζ = 40.12 cm⁻¹ and Δ = 13160 cm⁻¹, hence a ¹P₁ weight of 4.6 × 10⁻⁶ in ³P₁ and a 457 nm line running roughly a million times slower than the 285 nm resonance line it borrows from. J = 0 and J = 2 have one state each and cannot mix, so the entire crossover lives in this 2 × 2.
Change one variable at a time
Make the relationship visible.
Slide ζ down to 0.05 and the three triplet markers merge on the electrostatic scale while the interval readout settles near 2, the Landé rule. Now push ζ past Δ: J = 0 closes on the lower J = 1 and J = 2 on the upper — the (½,½) and (3/2,½) pairs — as the ¹P₁ weight climbs towards 33%.
ζ / Δ0.50
(E₂−E₁)/(E₁−E₀), LS: 23.78
¹P₁ WEIGHT, LOWER J = 16.481 %
LOWER J = 1 LEVEL-1.37 10³ cm⁻¹
Live interpretationζ / Δ: 0.50. (E₂−E₁)/(E₁−E₀), LS: 2: 3.78. ¹P₁ WEIGHT, LOWER J = 1: 6.481 %. LOWER J = 1 LEVEL: −1.37 10³ cm⁻¹
Catch the common trap
Explain before calculating.
Tin's 5p² ground term is measured at ³P₀ = 0, ³P₁ = 1691.8 and ³P₂ = 3427.7 cm⁻¹, so its two intervals stand in the ratio 1.03 rather than the 2 that the Landé rule predicts for a ³P triplet. What does that tell you?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasySodium's D lines sit at 16956.170 cm⁻¹ (D₁) and 16973.366 cm⁻¹ (D₂). Extract the 3p spin–orbit constant ζ, then invert the central-field expression ζₙₗ = α²R∞hc Zeff⁴/[n³ l(l+½)(l+1)] with α²R∞ = 5.844 cm⁻¹ to find the effective charge the valence electron reports.
- The doublet is one p electron split by j. With E(j) = (ζ/2)[j(j+1) − l(l+1) − ¾], j = 3/2 gives +ζ/2 and j = 1/2 gives −ζ, so the splitting is (l + ½)ζ = 3ζ/2.
- Measured splitting: 16973.366 − 16956.170 = 17.196 cm⁻¹, so ζ₃p = (2/3)(17.196) = 11.464 cm⁻¹.
- Invert with n = 3 and l = 1, where l(l+½)(l+1) = 3: Zeff⁴ = ζ n³ l(l+½)(l+1)/α²R∞ = 11.464 × 27 × 3 / 5.844 = 158.9.
- Fourth root: Zeff = 158.9¹⁄⁴ = 3.55.
Answerζ₃p = 11.46 cm⁻¹ and Zeff = 3.55 — far above the charge of 1 the electron feels far outside the core, because ⟨1/r³⟩ weights the part of the 3p orbital that penetrates the neon shell, and far below the bare Z = 11.
MediumCarbon's 2p² ground term is measured at ³P₀ = 0, ³P₁ = 16.417 and ³P₂ = 43.413 cm⁻¹, with ¹D₂ at 10192.63 cm⁻¹. Fit A from the first interval, test the prediction for the second, extract ζ₂p, and decide whether any discrepancy is a failure of LS coupling.
- Landé rule: EJ − E_(J−1) = A J. The J = 1 ← 0 gap is A × 1, so A = 16.417 cm⁻¹.
- Predict the next gap as A × 2 = 32.834 cm⁻¹, putting ³P₂ at 16.417 + 32.834 = 49.251 cm⁻¹. Measured 43.413 cm⁻¹, so the prediction is 5.84 cm⁻¹ or 13.4% too high, and the interval ratio is 26.996/16.417 = 1.64 rather than 2.
- Convert to the shell constant: 2p² is less than half full with S = 1, so A = +ζ/2S = ζ/2, giving ζ₂p = 2A = 32.83 cm⁻¹. (As a check, that inverts to Zeff = 3.41 in the central-field formula.)
- Test the scheme: the residual electrostatic scale is the ³P–¹D separation, 10192.63 cm⁻¹, so X = ζ/ΔE = 32.83/10192.63 = 3.2 × 10⁻³.
- With X of order 10⁻³ the terms are nowhere near mixing, so LS labels are excellent and the 13% miss cannot be a coupling-scheme failure. It belongs to the two-body magnetic terms dropped from HSO and to configuration interaction with higher even-parity configurations, neither of which respects the A L⋅S form.
AnswerA = 16.42 cm⁻¹ and ζ₂p = 32.83 cm⁻¹; the rule predicts ³P₂ at 49.25 cm⁻¹ against 43.41 measured, 13% high, while X = 3 × 10⁻³ shows LS is intact — so the residual is corrections, not the scheme.
HardMagnesium's 3s3p levels are ³P₀ = 21850.405, ³P₁ = 21870.464, ³P₂ = 21911.178 and ¹P₁ = 35051.264 cm⁻¹. Confirm the interval rule, build the J = 1 matrix, find the ¹P₁ weight in the ³P₁ state, and use it to predict the ratio of the 457.1 nm intercombination rate to the 285.2 nm resonance rate.
- Intervals: 21870.464 − 21850.405 = 20.059 and 21911.178 − 21870.464 = 40.714 cm⁻¹, a ratio of 2.030 against the predicted 2 — LS coupling is clean. The first gap is A × 1, so A = 20.06 cm⁻¹ and, with S = 1 and only the p electron contributing, ζ = 2A = 40.12 cm⁻¹.
- Locate ¹P above the triplet centroid: (1 × 21850.405 + 3 × 21870.464 + 5 × 21911.178)/9 = 21890.85 cm⁻¹, so Δ = 35051.264 − 21890.85 = 13160.4 cm⁻¹ and X = ζ/Δ = 3.05 × 10⁻³.
- The J = 1 matrix on (³P₁, ¹P₁) is [[−ζ/2, ζ/√2], [ζ/√2, Δ]]. With X this small, first-order perturbation theory suffices: the singlet amplitude in the lower root is (ζ/√2)/(0 − Δ) = −28.369/13160.4 = −2.156 × 10⁻³.
- Weight = amplitude squared = 4.65 × 10⁻⁶. The ³P₁ level is that much ¹P₁, and that admixture carries the entire electric-dipole strength of the ³P₁ → ¹S₀ line, since the pure triplet has none.
- Rates go as |d|²ω³, so A(457.1)/A(285.2) = 4.65 × 10⁻⁶ × (285.21/457.11)³ = 4.65 × 10⁻⁶ × 0.2429 = 1.13 × 10⁻⁶.
- Compare measurement: A(285.2) = 4.95 × 10⁸ s⁻¹ and A(457.1) ≈ 2.5 × 10² s⁻¹, a ratio of 5.1 × 10⁻⁷. The two-level estimate lands within a factor of two, the remainder coming from the other ¹P₁ states left out of the 2 × 2.
Answerζ = 40.12 cm⁻¹, Δ = 13160 cm⁻¹, X = 3.0 × 10⁻³; the ¹P₁ weight in ³P₁ is 4.6 × 10⁻⁶, predicting a rate ratio of 1.1 × 10⁻⁶ against a measured 5 × 10⁻⁷.