University Physics IV · Angular Momentum and Spin · 9.8
Adding Angular Momenta & Clebsch-Gordan Coefficients
Fine structure, the helium spectrum, every term symbol you will ever assign: each begins by asking what totals two coupled angular momenta can form. This topic hands you the allowed j values, the audit proving none are missing, and the coefficients that translate between the two honest ways of labelling the same states.
Build the model
Connect the measurement to the mechanism.
Add two angular momentum operators, Ĵ = Ĵ₁ + Ĵ₂, and the sum obeys the same commutation relations as any angular momentum — so its eigenvalues must be j(j+1)ħ² and mħ with no new derivation needed. The only open question is which j survive, and ladder bookkeeping answers it: j runs from |j₁ − j₂| to j₁ + j₂ in integer steps, and the count Σ(2j+1) = (2j₁+1)(2j₂+1) confirms the coupled multiplets exactly exhaust the product states. The subtlety is that this is a change of basis, not a relabelling. Ĵz = Ĵ₁z + Ĵ₂z, so m = m₁ + m₂ holds exactly; but Ĵ² contains the cross term 2Ĵ₁⋅Ĵ₂, which fails to commute with Ĵ₁z and Ĵ₂z separately — so a state of definite total j is in general a superposition of several products, and a product like ↑↓ has no definite total j at all.
The Clebsch-Gordan coefficients are the entries of the unitary matrix connecting the two bases, and they cost nothing beyond ladder algebra: write the unique stretched state |j₁+j₂, j₁+j₂⟩, lower it with Ĵ₋ = Ĵ₁₋ + Ĵ₂₋, and take an orthogonal combination whenever a new multiplet opens. The price of coupling is definiteness of the parts — in a coupled state m₁ and m₂ are individually uncertain — which is exactly the trade an internal Ĵ₁⋅Ĵ₂ interaction forces you to make.
- Simple definition
- Coupling angular momenta j₁ and j₂ yields total quantum numbers j from |j₁ − j₂| up to j₁ + j₂ in integer steps, and the Clebsch-Gordan coefficients are the overlaps that expand each coupled state |j, m⟩ in the product states |j₁ m₁⟩|j₂ m₂⟩.
- Example
- Two spin-½ electrons couple to s = 1 or s = 0 only — 3 + 1 = 4 = 2 × 2 states — and the m = 0 triplet is (↑↓ + ↓↑)/√2: two Clebsch-Gordan coefficients of 1/√2 ≈ 0.707.
Names every multiplet before any coefficient is computed — the steps are integers even when the j values themselves are half-integers.
j₁, j₂, j are dimensionless quantum numbers; the momentum magnitudes are √(j(j+1)) ħ, with ħ in J s
The audit that the coupled multiplets exactly exhaust the products: j₁ = 2, j₂ = 3/2 gives 2+4+6+8 = 20 = 5 × 4.
sum over j from |j₁−j₂| to j₁+j₂; both sides count basis states, so both are pure numbers
Kills most coefficients on sight and organises every Clebsch-Gordan table into blocks of constant total m.
follows from Ĵz = Ĵ₁z + Ĵ₂z; each m is a z-projection in units of ħ
The unitary dictionary between the bases — each squared coefficient is the probability of finding that product.
sum over m₁ + m₂ = m; coefficients real and dimensionless in the Condon–Shortley convention
Applied repeatedly from the stretched state |j₁+j₂, j₁+j₂⟩, this single relation generates every table entry.
Ĵ₋ = Ĵ₁₋ + Ĵ₂₋ acts on both parts at once; the factor carries the ħ
Singlet and triplet for two spins; j = l ± ½ for one electron's orbit and spin — the input to every term symbol.
dimension checks: 4 = 3 + 1 and 2(2l+1) = (2l+2) + 2l; j = ½ alone when l = 0
Two complete label sets, one space
Four commuting operators label the product states: Ĵ₁², Ĵ₂², Ĵ₁z, Ĵ₂z give the uncoupled basis |j₁ m₁⟩|j₂ m₂⟩, and for fixed j₁ and j₂ there are (2j₁+1)(2j₂+1) of them. The total Ĵ = Ĵ₁ + Ĵ₂ obeys the standard angular-momentum commutators, so its own labels j and m are also available — but Ĵ² = Ĵ₁² + Ĵ₂² + 2Ĵ₁⋅Ĵ₂, and the cross term contains Ĵ₁₊Ĵ₂₋ + Ĵ₁₋Ĵ₂₊, which shuffles m₁ against m₂. So Ĵ² commutes with Ĵ₁² and Ĵ₂² but not with Ĵ₁z or Ĵ₂z separately, and you must choose a set: (Ĵ₁², Ĵ₂², Ĵ₁z, Ĵ₂z) or (Ĵ₁², Ĵ₂², Ĵ², Ĵz). Which one is good is decided by the Hamiltonian. An internal coupling proportional to Ĵ₁⋅Ĵ₂ — spin-orbit, spin-spin, hyperfine — is diagonal in the coupled basis; a strong external field that grips each moment separately prefers the uncoupled one. That competition is exactly the weak-field versus Paschen-Back divide of the next topic.
The triangle rule comes from counting projections
The largest possible m is j₁ + j₂, and only one product reaches it: m₁ = j₁ with m₂ = j₂. One state means one multiplet passes through it, so jₘₐₓ = j₁ + j₂ and j never exceeds the sum. Step down to m = j₁ + j₂ − 1 and two products qualify; the jₘₐₓ ladder claims one combination, and the single leftover must start a new multiplet with j = j₁ + j₂ − 1. Each unit step inward adds one more product and opens one more ladder, until the smaller momentum runs out of projections at m = |j₁ − j₂|; below that the count is flat and nothing new opens. Hence j runs from |j₁ − j₂| to j₁ + j₂ in integer steps — 2 min(j₁, j₂) + 1 multiplets, never a half step between neighbours even when the j values are half-integers. The audit confirms it: for j₁ = 2, j₂ = 3/2 the totals are j = 7/2, 5/2, 3/2, 1/2 with 8 + 6 + 4 + 2 = 20 states, exactly the 5 × 4 products you started with.
m adds exactly; j does not add at all
Ĵz = Ĵ₁z + Ĵ₂z is an operator identity, so m = m₁ + m₂ holds for every state with no exception — any Clebsch-Gordan coefficient violating it vanishes, which is why the tables are organised by total m. Nothing similar holds for j, because the magnitudes are not arrows of length jħ. Two electron spins each have |S| = (√3/2)ħ ≈ 0.87ħ; laid parallel as arrows they would reach 1.73ħ, yet the allowed totals are √2 ħ ≈ 1.41ħ for s = 1 and exactly 0 for s = 0 — neither is the arrow sum. The honest statement is weaker and stranger: a product state of definite m₁ and m₂ generally has no definite j at all. ↑↓ has m = 0 but is an equal superposition of s = 1 and s = 0. The only products that are simultaneously coupled states are the stretched ones at the ends of each m ladder, where a single product has nothing to mix with.
Generate the table: top rung, lower, orthogonalise
Every Clebsch-Gordan table is built with one tool. Start at the unique top: for two spin-½ particles, |1, 1⟩ = ↑↑, forced because no other product has m = 1. Apply Ĵ₋ = Ŝ₁₋ + Ŝ₂₋. The left side gives ħ√(1⋅2 − 1⋅0) = √2 ħ times |1, 0⟩; on the right each one-particle lowering carries the factor ħ√(¾ + ¼) = ħ, producing ħ(↓↑ + ↑↓). Dividing through, |1, 0⟩ = (↑↓ + ↓↑)/√2 — both coefficients 1/√2, and no integral was ever computed. The m = 0 subspace is two-dimensional, so one combination is left over; orthogonality forces |0, 0⟩ = (↑↓ − ↓↑)/√2, with the overall sign fixed by the Condon–Shortley convention that the coefficient of the highest m₁ is positive. One more lowering gives |1, −1⟩ = ↓↓, completing the 4 × 4 unitary matrix. Larger tables need nothing new: lower until each ladder ends, orthogonalise whenever a new multiplet opens, and the phase convention keeps every entry real.
The case atoms use daily: l ⊗ ½
An electron in an orbital of quantum number l couples its orbit to its spin, giving j = l + ½ or l − ½ — two fine-structure levels — with j = ½ alone when l = 0, which is why s terms show no doublet. The general coefficients are worth knowing: |l+½, m⟩ = √((l+m+½)/(2l+1)) |mₗ = m−½⟩↑ + √((l−m+½)/(2l+1)) |mₗ = m+½⟩↓, and the orthogonal combination builds j = l−½. For a p electron, l = 1: |3/2, ½⟩ = √(2/3)|mₗ = 0⟩↑ + √(1/3)|mₗ = 1⟩↓. Read the structure off the square roots: at the stretched rungs m = ±(l+½) one of them vanishes and the state is a single product, while mid-ladder the blend is strongest. The dimension audit holds as always: 2(2l+1) = (2l+2) + 2l, so for l = 1 the six products regroup into four j = 3/2 states and two j = 1/2 states.
What the coefficients buy, and what coupling costs
Each squared coefficient is a measurement probability. In the triplet |1, 0⟩ the amplitude on ↑↓ is 1/√2, so measuring S₁z returns +ħ/2 exactly half the time: coupling to definite total spin has made the individual projections maximally uncertain. That is the cost. The payoff is that in the coupled basis Ĵ₁⋅Ĵ₂ = (Ĵ² − Ĵ₁² − Ĵ₂²)/2 collapses to a number. For l ⊗ ½ it equals +lħ²/2 when j = l+½ and −(l+1)ħ²/2 when j = l−½; a p electron gives +ħ²/2 and −ħ², so a spin-orbit Hamiltonian ξ L̂⋅Ŝ splits the level into a doublet of separation ξ(2l+1)ħ²/2 — for l = 1, precisely 3ξħ²/2. No wavefunction integral was needed: the angular content of fine structure is pure coupling algebra, and the radial physics enters only through ξ. That is the hand-off to the next topic, where term symbols, the interval rule, and the Lande g factor all consume j, l and s through exactly this machinery.
Change one variable at a time
Make the relationship visible.
Set j₁ = j₂ = 2: the kinks meet at m = 0, so jₘᵢₙ = 0 and five multiplets open. Slide j₂ to ½ and the plateau drops to 2 — only j = j₁ ± ½ survive, the l ⊗ ½ split every atomic term uses. The corners always sit at count 1: a stretched top rung is a single product, never a mixture.
j MAX = j₁ + j₂3.0
j MIN = |j₁ − j₂|1.0
MULTIPLETS3
TOTAL STATES15
Live interpretationj MAX = j₁ + j₂: 3.0. j MIN = |j₁ − j₂|: 1.0. MULTIPLETS: 3. TOTAL STATES: 15
Catch the common trap
Explain before calculating.
Two electrons are prepared in the product state ↑↓ — the first definitely spin up along z, the second definitely down. What values can a measurement of the total S² return, and with what probabilities?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA particle with j₁ = 3/2 couples to a second with j₂ = 1. List every allowed total j, and verify that the coupled states exactly account for the product states.
- Triangle-rule bounds: j runs from |j₁ − j₂| = |3/2 − 1| = 1/2 up to j₁ + j₂ = 5/2.
- Integer steps between the bounds: j = 1/2, 3/2, 5/2 — three multiplets, matching 2⋅min(j₁, j₂) + 1 = 2(1) + 1 = 3. Values such as j = 1 or 2 never appear: they sit half a step off the ladder.
- Count the coupled states: Σ(2j+1) = 2 + 4 + 6 = 12.
- Count the product states: (2j₁+1)(2j₂+1) = 4 × 3 = 12. The audit balances, so no state has been lost or invented.
Answerj = 1/2, 3/2, 5/2 — and 2 + 4 + 6 = 12 coupled states, exactly the 4 × 3 = 12 product states regrouped.
MediumFor two spin-½ particles, construct |1, 0⟩ by lowering from |1, 1⟩ = ↑↑, find the singlet |0, 0⟩, and give the probability that a measurement of S₁z in the state |1, 0⟩ returns +ħ/2.
- Apply Ĵ₋ = Ŝ₁₋ + Ŝ₂₋ to |1, 1⟩. On the left, the ladder factor is ħ√(s(s+1) − m(m−1)) = ħ√(1⋅2 − 1⋅0) = √2 ħ, so Ĵ₋|1, 1⟩ = √2 ħ |1, 0⟩.
- On the right, Ŝ₁₋ lowers only the first spin with factor ħ√(¾ − (½)(−½)) = ħ, giving ħ ↓↑; likewise Ŝ₂₋ gives ħ ↑↓.
- Equate and divide by √2 ħ: |1, 0⟩ = (↑↓ + ↓↑)/√2 — both Clebsch-Gordan coefficients equal to 1/√2.
- The m = 0 subspace is two-dimensional, so |0, 0⟩ is the orthogonal combination: |0, 0⟩ = (↑↓ − ↓↑)/√2, the positive sign on ↑↓ fixed by the Condon–Shortley convention.
- In |1, 0⟩ the amplitude for the first spin up (the ↑↓ term) is 1/√2, so P(S₁z = +ħ/2) = (1/√2)² = 1/2: the coupled state leaves the individual projection completely undetermined.
Answer|1, 0⟩ = (↑↓ + ↓↑)/√2 and |0, 0⟩ = (↑↓ − ↓↑)/√2; P(S₁z = +ħ/2) = (1/√2)² = 1/2.
HardA 2p electron has l = 1 and s = ½. Construct |j = 3/2, mⱼ = ½⟩ by lowering from the stretched state, write its orthogonal partner |j = ½, mⱼ = ½⟩, give the mₗ probabilities in the j = 3/2 state, and evaluate L⋅S for both j values.
- The stretched state is the unique product with mⱼ = 3/2: |3/2, 3/2⟩ = |mₗ = 1⟩↑. Apply Ĵ₋ = L̂₋ + Ŝ₋. Left-side factor: ħ√(j(j+1) − m(m−1)) = ħ√(15/4 − 3/4) = √3 ħ.
- Right side: L̂₋|mₗ = 1⟩ = ħ√(l(l+1) − mₗ(mₗ−1))|0⟩ = √2 ħ|0⟩, so the orbital part gives √2 ħ|0⟩↑; the spin part gives ħ|1⟩↓.
- Divide by √3 ħ: |3/2, ½⟩ = √(2/3)|mₗ = 0⟩↑ + √(1/3)|mₗ = 1⟩↓. Measuring mₗ in this state: P(mₗ = 0) = 2/3, P(mₗ = 1) = 1/3.
- The orthogonal mⱼ = ½ combination is the j = ½ state: |1/2, ½⟩ = √(2/3)|mₗ = 1⟩↓ − √(1/3)|mₗ = 0⟩↑, the positive coefficient on the highest mₗ fixed by Condon–Shortley. Check: overlap √(2/3)⋅√(1/3) − √(1/3)⋅√(2/3) = 0.
- L⋅S = (Ĵ² − L̂² − Ŝ²)/2 is a number in the coupled basis. For j = 3/2: ½(15/4 − 2 − 3/4)ħ² = +ħ²/2. For j = 1/2: ½(3/4 − 2 − 3/4)ħ² = −ħ². A spin-orbit term ξ L̂⋅Ŝ therefore splits the p level into a doublet separated by 3ξħ²/2.
Answer|3/2, ½⟩ = √(2/3)|mₗ = 0⟩↑ + √(1/3)|mₗ = 1⟩↓ with P(mₗ = 0) = 2/3, P(mₗ = 1) = 1/3; |1/2, ½⟩ = √(2/3)|mₗ = 1⟩↓ − √(1/3)|mₗ = 0⟩↑; L⋅S = +ħ²/2 for j = 3/2 and −ħ² for j = 1/2.