
Physics IV course sequence · Interactive lesson
Spin-orbit coupling and the Landé g factor
Alkali resonance lines come in close pairs, and in a weak magnet no alkali line splits into the textbook three. Both trace to the electron's spin moment sitting in the magnetic field of its own orbital motion. This lesson prices that internal coupling, then builds gJ — the one number fixing each level's response to a magnet.
UPIV-09.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Build the model
The atom magnetises itself: spin in the field of its own orbit.
The key idea
Spin-orbit coupling is the interaction between an electron's spin magnetic moment and the internal magnetic field of its own orbital motion; it splits each l > 0 level into fine-structure components with j = l + ½ and j = l − ½.
Example. Sodium's yellow line is really two: 589.0 nm and 589.6 nm, one 3p level split by 2.1 × 10⁻³ eV because the 3p electron's spin moment sits in an internal field of roughly 36 T.
The complete physical model and assumptions
Transform to the electron's rest frame and the nucleus becomes a circulating charge, generating a magnetic field of order a tesla in hydrogen and tens of teslas in heavier atoms — a field the atom supplies itself. The electron's spin moment sits in that field, so an interaction ξ(r)L⋅S appears with no apparatus involved; its coefficient carries a factor of one half that a naive rest-frame derivation misses, because the electron's frame is accelerating and a sequence of boosts amounts to a rotation — Thomas precession. The coupling costs the atom its old labels: Lz and Sz no longer commute with H, so mₗ and mₛ die as quantum numbers and the coupled pair j, mⱼ replaces them, with L⋅S = ½(J² − L² − S²) turning every shift into arithmetic. What it buys is fine structure — each l > 0 level split into j = l ± ½ by an interval scaling as Z⁴/n³ — and, hanging on it, the Landé factor gJ: the projection onto J of a magnetic moment that is not parallel to J, because spin counts double. gJ prices every weak-field Zeeman pattern, and it is only the weak-field answer: once μB B beats the fine-structure interval, the field decouples L from S and mₗ, mₛ come back to life.
1Ride with the electron and a magnet appears
In the lab frame the electron orbits a stationary nucleus. Transform to the electron's instantaneous rest frame and the charged nucleus circulates instead — a current loop threading the electron's position with a magnetic field Bᵢₙₜ ≈ −(v × E)/c², built from the same Coulomb field E that binds the atom and pointing along the orbital angular momentum L. The electron's spin moment μₛ = −gₛ μB S/ħ then has an orientation energy −μₛ⋅Bᵢₙₜ, and once the Thomas half of the next section is included, gₛ ≈ 2 makes reversing the spin in that field cost about μB Bᵢₙₜ — so a measured splitting reads off the field as Bᵢₙₜ ≈ ΔE/μB. Its size is not small: hydrogen's 2p splitting of 4.5 × 10⁻⁵ eV gives close to 0.8 T, a solid laboratory electromagnet's worth, and sodium's 3p doublet of 2.1 × 10⁻³ eV gives about 36 T, beyond all but the strongest magnets on Earth — supplied free by the atom's own motion. Since Bᵢₙₜ points along L, the orientation energy is proportional to L⋅S: the atom couples its own spin to its own orbit, with no external apparatus anywhere in the story.
2Thomas precession halves the naive answer
The rest-frame derivation hides an error: the electron's frame is not inertial. It accelerates continuously toward the nucleus, and in special relativity the composition of two non-parallel boosts is not a boost but a boost plus a rotation. The electron's frame therefore rotates — Thomas precession — at a rate exactly half the naive spin-precession rate and with the opposite sense, and the net effect is to halve the coupling: HSO = (1/(2mₑ²c²))(1/r)(dV/dr) L⋅S, the ½ being Thomas's. History pivoted on this factor. In 1926 the doublet intervals came out twice too large from the naive calculation; gₛ ≈ 2 and Thomas's ½ conspire so the observed splitting looks as if the spin g were 1, and the spin hypothesis of Uhlenbeck and Goudsmit was rescued within months. The Dirac equation later delivers the ½ automatically; at this level you insert it by hand and know why it is there.
3Evaluate L⋅S where it is diagonal
HSO does not commute with Lz or Sz separately — the coupling makes L and S precess about their sum — so mₗ and mₛ stop being good quantum numbers. It does commute with J² = (L+S)², L², S² and Jz, so the coupled basis |l, s; j, mⱼ⟩ diagonalises it. Square J = L + S to get L⋅S = ½(J² − L² − S²), and every matrix element becomes arithmetic: ⟨L⋅S⟩ = (ħ²/2)[j(j+1) − l(l+1) − s(s+1)]. For a p electron (l = 1, s = ½): j = 3/2 gives +ħ²/2 and j = 1/2 gives −ħ². Two checks come free. Degeneracy-weighted, 4 × (+½) + 2 × (−1) = 0: the coupling splits the level about an unmoved centre of gravity, so an average over the multiplet feels nothing. And the gap between adjacent j levels is proportional to the larger j — the Landé interval rule, your first diagnostic for whether a measured multiplet really is spin-orbit split.
4Term symbols: the labels that survive
With HSO switched on, the labels that still commute with the Hamiltonian for one electron are n, l, s, j and mⱼ, and spectroscopy compresses the level part into the term symbol n²ˢ⁺¹LJ: the superscript is the spin multiplicity 2s + 1, the letter S, P, D, F codes l = 0, 1, 2, 3, and the subscript is j. Sodium's 3p level becomes 3²P₃/₂ and 3²P₁/₂, both decaying to 3²S₁/₂, and every fine-structure diagram you will meet is drawn in this notation. Two readings of it trip students. The superscript counts spin orientations, not levels: ²S₁/₂ is a doublet term with a single level, because with l = 0 there is only j = ½ and no orbit for the spin to couple to. And the subscript is the only label that knows about the coupling — mₗ and mₛ are no longer good, so a term symbol never carries them. How far apart the levels sit is a separate, radial question: in a hydrogenic atom the doublet gap scales as Z⁴/n³, and the larger effective charge a penetrating valence electron feels is why sodium's 3p gap of 2.1 × 10⁻³ eV dwarfs hydrogen's 2p gap of 4.5 × 10⁻⁵ eV. UPIV-11.07 derives the gap along with the relativistic and Darwin terms that complete hydrogen's fine structure.
5Project the moment: where gJ comes from
A level's magnetic moment is μ = −μB(L + 2S)/ħ — spin counts double because gₛ ≈ 2 — and that 2 means μ is not antiparallel to J = L + S. In a weak field, L and S precess rapidly about J at the fine-structure frequency while J precesses slowly about B. The component of μ perpendicular to J therefore averages to zero, and only the projection along J couples to the field: ⟨μ⟩ = −gJ μB J/ħ with gJ = ⟨(L + 2S)⋅J⟩/⟨J²⟩. Write L + 2S = J + S, so gJ = 1 + ⟨S⋅J⟩/[j(j+1)ħ²], and square L = J − S to get S⋅J = ½(J² + S² − L²). The result is gJ = 1 + [j(j+1) + s(s+1) − l(l+1)]/[2j(j+1)], and each level moves by gJ μB B mⱼ. Test it on limits you already trust: l = 0 leaves pure spin and gJ = 2; s = 0 leaves pure orbit and gJ = 1. For one electron it collapses to gJ = 1 ± 1/(2l + 1) for j = l ± ½, tending to the orbital value as l grows. For sodium: g(²S₁/₂) = 2, g(²P₁/₂) = 2/3, g(²P₃/₂) = 4/3 — three levels, three different magnetic prices per unit mⱼ. That is why alkali Zeeman patterns are anomalous: a normal three-line pattern needs upper and lower levels to shift at the same rate, which s = 0 guarantees by forcing gJ = 1 everywhere; the line patterns themselves are counted in UPIV-11.07.
6Which labels survive a field
Switch on a field B along z and H gains (μB B/ħ)(Lz + 2Sz), a term with its own preferences. HSO keeps J² and Jz but not Lz or Sz, as the third section showed; the Zeeman term commutes with Lz and Sz but not with J², because the 2 on Sz makes it Jz + Sz rather than a multiple of Jz. Apart from L² and S², only Jz commutes with both, so mⱼ = mₗ + mₛ is the one label that stays exact at every field. Whether j or the pair mₗ, mₛ is approximately good is settled by the ratio μB B/ΔEFS. Well below 1, j survives and the projection argument holds; well above 1, the field decouples L from S, mₗ and mₛ become good again, and each state shifts by (mₗ + 2mₛ)μB B — the Paschen-Back regime. The handover sits where the applied field matches the internal one from the first section, B ≈ ΔEFS/μB: about 0.8 T for hydrogen 2p and 36 T for sodium 3p, so the same magnet is strong for one atom and weak for another. Between the limits only mⱼ labels anything; the Hard example diagonalises one mⱼ block exactly and finds gJ in its low-field slopes.
Key equations
Read the units, assumptions and limits before substituting.
Spin-orbit Hamiltonian (Thomas ½ included)
HSO = (1/(2mₑ²c²)) (1/r)(dV/dr) L⋅S
The ½ is Thomas precession: the electron's rest frame rotates, and the naive derivation gives exactly twice the observed coupling.
V(r) = potential energy in J; mₑ in kg; L⋅S carries ħ², so HSO lands in joulesL⋅S in the coupled basis
⟨L⋅S⟩ = (ħ²/2)[j(j+1) − l(l+1) − s(s+1)]
Turns the operator product into arithmetic — the shift of every fine-structure level without a single integral.
from squaring J = L + S; j, l, s are pure numbers and ħ² carries the units. p electron: +ħ²/2 (j = 3/2), −ħ² (j = 1/2)Why mₗ and mₛ stop being good
[L⋅S, Lz] = iħ(L × S)z = −[L⋅S, Sz]
L⋅S trades a unit of mₗ for a unit of mₛ and back, so neither survives while their sum mⱼ does: the operator reason j and mⱼ replace mₗ and mₛ.
L and S in J s, so both sides carry units of ħ³; the two commutators cancel in the sum, giving [L⋅S, Jz] = 0, and L⋅S = ½(J² − L² − S²) commutes with J²Projecting μ onto J
gJ = ⟨(L + 2S)⋅J⟩ / ⟨J²⟩ = 1 + ⟨S⋅J⟩ / [j(j+1)ħ²]
Writing L + 2S = J + S leaves only the spin's extra copy to project, and the dot product it needs is one more piece of coupled-basis arithmetic.
μ = −μB(L + 2S)/ħ; ⟨S⋅J⟩ = (ħ²/2)[j(j+1) + s(s+1) − l(l+1)] from squaring L = J − S; every ratio is dimensionlessLandé g factor
gJ = 1 + [j(j+1) + s(s+1) − l(l+1)] / [2j(j+1)]
The projection of the tilted moment μ = −μB(L + 2S)/ħ onto J — the only part that survives the fast precession of L and S about J.
dimensionless; assumes gₗ = 1, gₛ = 2. ²S₁/₂: 2; ²P₁/₂: 2/3; ²P₃/₂: 4/3Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. At l = 1, flip d between +1 and −1: for j = 3/2 the spin leans along J and the projection of L + 2S overshoots |J| (gJ = 4/3); for j = 1/2 it leans against J and falls short (gJ = 2/3). Then raise l to 3 — the orbit dominates, the tilt shrinks, and gJ = 1 ± 1/(2l + 1) closes in on the orbital value 1.
Live results
- TOTAL j
- 1.5
- ⟨L⋅S⟩ / ħ²
- 0.50
- ⟨S⋅J⟩ / ħ²
- 1.25
- LANDÉ gJ
- 1.333
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Orbital quantum number l: 1; Coupling branch d, j = l + d/2: 1
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The reflex is to treat μB B as the universal Zeeman quantum — g = 1, three lines, done. It fails because the atom's moment is μ = −μB(L + 2S)/ħ: the spin's double weight tilts μ away from −J, so a level does not shift by μB B mⱼ. The perpendicular part of μ precesses away at the fine-structure frequency and only the projection survives, worth gJ μB B per unit mⱼ — 2 for ²S₁/₂, 2/3 for ²P₁/₂, 4/3 for ²P₃/₂. Upper and lower levels of a line then split at different rates, and the pattern is anomalous. The normal triplet is the special case, not the rule: it needs gJ identical in both levels, which s = 0 (singlet terms) guarantees — or a field so strong the atom is in the Paschen-Back regime, where the question itself changes.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A single p electron has l = 1 and s = ½. Find the allowed j values, evaluate ⟨L⋅S⟩ in each, and show that spin-orbit coupling leaves the degeneracy-weighted centre of gravity of the level unmoved.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Allowed j run from |l − s| to l + s in integer steps: j = 1/2 and j = 3/2, holding 2j+1 = 2 and 4 states — six in all, matching the (2l+1)(2s+1) = 6 uncoupled products.
- Square J = L + S: J² = L² + S² + 2L⋅S, so ⟨L⋅S⟩ = (ħ²/2)[j(j+1) − l(l+1) − s(s+1)], with l(l+1) = 2 and s(s+1) = 3/4.
- j = 3/2: ⟨L⋅S⟩ = (ħ²/2)(15/4 − 2 − 3/4) = +ħ²/2. j = 1/2: (ħ²/2)(3/4 − 2 − 3/4) = −ħ². With HSO = ξL⋅S and ξ > 0, j = 3/2 rises and j = 1/2 falls, split by (3/2)ξħ².
- Weight each shift by its degeneracy: 4(+ħ²/2) + 2(−ħ²) = 2ħ² − 2ħ² = 0. The coupling splits the level symmetrically about its old position — a check worth running on any multiplet.
Answer. j = 3/2 with ⟨L⋅S⟩ = +ħ²/2 (4 states) and j = 1/2 with ⟨L⋅S⟩ = −ħ² (2 states); 4(½) + 2(−1) = 0, so the centre of gravity is unshifted.
MediumExample 2
A single d electron (l = 2, s = ½) forms a ²D term with levels ²D₅/₂ and ²D₃/₂. For each level evaluate ⟨L⋅J⟩ and ⟨S⋅J⟩ in units of ħ², project μ = −μB(L + 2S)/ħ onto J to get gJ, and test the pair against the sum rule for mⱼ = +½: the weak-field slopes gJ mⱼ of the two levels must add up to the strong-field slopes mₗ + 2mₛ of the two uncoupled states with mₗ + mₛ = +½.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Square S = J − L and L = J − S: L⋅J = ½(J² + L² − S²) and S⋅J = ½(J² + S² − L²), with l(l+1) = 6 and s(s+1) = 3/4 here.
- ²D₅/₂, j(j+1) = 35/4: ⟨S⋅J⟩ = ½(35/4 + 3/4 − 6)ħ² = (7/4)ħ² and ⟨L⋅J⟩ = ½(35/4 + 6 − 3/4)ħ² = 7ħ², which add to (35/4)ħ² = ⟨J²⟩ as they must. Projection: gJ = (⟨L⋅J⟩ + 2⟨S⋅J⟩)/⟨J²⟩ = (7 + 7/2)/(35/4) = 6/5.
- ²D₃/₂, j(j+1) = 15/4: ⟨S⋅J⟩ = ½(15/4 + 3/4 − 6)ħ² = −(3/4)ħ² and ⟨L⋅J⟩ = ½(15/4 + 6 − 3/4)ħ² = (9/2)ħ², summing to (15/4)ħ². gJ = (9/2 − 3/2)/(15/4) = 4/5: the spin leans against J, so its double weight drags gJ below the orbital value 1.
- Both agree with the Landé formula and with its one-electron form gJ = 1 ± 1/(2l + 1) = 1 ± 1/5.
- Sum rule for mⱼ = +½: weak field gives (6/5)(½) + (4/5)(½) = 1. Strong field: (mₗ, mₛ) = (1, −½) gives 1 − 1 = 0 and (0, +½) gives 0 + 1 = 1, total 1. They must agree, because the trace of Lz + 2Sz over a fixed-mⱼ block is the same in either basis.
Answer. ²D₅/₂: ⟨L⋅J⟩ = 7ħ², ⟨S⋅J⟩ = (7/4)ħ², gJ = 6/5. ²D₃/₂: ⟨L⋅J⟩ = (9/2)ħ², ⟨S⋅J⟩ = −(3/4)ħ², gJ = 4/5. For mⱼ = +½ both bases give a total slope of 1.
HardExample 3
A p electron (l = 1, s = ½) in a field B along z has H = ξL⋅S + (μB B/ħ)(Lz + 2Sz). Work in the mⱼ = +½ block spanned by |mₗ = 1, mₛ = −½⟩ and |mₗ = 0, mₛ = +½⟩, writing a = ξħ² and b = μB B. (a) Build the 2 × 2 matrix and find its eigenvalues. (b) Show that their slopes are the Landé values gJ mⱼ at small b and the uncoupled values mₗ + 2mₛ at large b. (c) Find where the two straight-line limits cross, and evaluate the exact levels there for hydrogen 2p, whose fine-structure interval (3/2)a is 4.5 × 10⁻⁵ eV. (μB = 5.79 × 10⁻⁵ eV/T.)
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Diagonal: on a product state only LzSz in L⋅S = LzSz + ½(L₊S₋ + L₋S₊) survives, giving −a/2 for (1, −½) and 0 for (0, +½); the Zeeman term adds (mₗ + 2mₛ)b = 0 and b. The diagonal entries are −a/2 and b.
- Off-diagonal: ½L₊S₋ carries (0, +½) into (1, −½), with L₊ contributing ħ√(l(l+1) − 0) = √2 ħ and S₋ contributing ħ√(s(s+1) + ¼) = ħ, so the element is (√2/2)a = a/√2. The Zeeman term is diagonal in this basis and adds nothing off the diagonal.
- The matrix has rows (−a/2, a/√2) and (a/√2, b), so E± = (b − a/2)/2 ± √[((b + a/2)/2)² + a²/2]. At b = 0 the root is √(a²/16 + a²/2) = 3a/4, giving +a/2 and −a — the j = 3/2 and j = 1/2 values of ⟨L⋅S⟩, recovered without ever assuming j.
- Slopes: the root's derivative is [(b + a/2)/2]/(2√…), which at b = 0 is (a/4)/(3a/2) = 1/6, so dE±/db = ½ ± 1/6 = 2/3 and 1/3 — exactly gJ mⱼ = (4/3)(½) and (2/3)(½). At large b the root tends to (b + a/2)/2, so E₊ → b and E₋ → −a/2: slopes 1 and 0, the values of mₗ + 2mₛ for (0, +½) and (1, −½).
- Crossing: the Landé line a/2 + (2/3)b meets the Paschen-Back line b at b = 3a/2, and the lower pair, −a + b/3 and −a/2, meet at the same b — the regimes hand over exactly where μB B equals the fine-structure interval. There E± = a/2 ± √(a² + a²/2) = a/2 ± 1.22a, so each exact level sits 0.22a outside both straight lines, pushed apart by the off-diagonal a/√2.
- Hydrogen 2p: a = (2/3)(4.5 × 10⁻⁵ eV) = 3.0 × 10⁻⁵ eV, and the crossing field is B = 4.5 × 10⁻⁵/5.79 × 10⁻⁵ = 0.78 T. There E₊ = 1.72a = 5.2 × 10⁻⁵ eV and E₋ = −0.72a = −2.2 × 10⁻⁵ eV, each 0.22a = 6.7 × 10⁻⁶ eV from where either straight-line picture would put it.
Answer. Rows (−a/2, a/√2) and (a/√2, b); E± = (b − a/2)/2 ± √[((b + a/2)/2)² + a²/2]. Low-field slopes 2/3 and 1/3 are gJ mⱼ; high-field slopes 1 and 0 are mₗ + 2mₛ. The limits cross at b = 3a/2, the fine-structure interval, where the exact levels lie 0.22a outside them — for hydrogen 2p, at B = 0.78 T and by 6.7 × 10⁻⁶ eV.
Quick check
Select an answer and explain it before checking.
Review & continue
Spin-orbit coupling is the interaction between an electron's spin magnetic moment and the internal magnetic field of its own orbital motion; it splits each l > 0 level into fine-structure components with j = l + ½ and j = l − ½.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.