University Quantum Mechanics II · Rotations and the Addition of Angular Momentum · 5.6
Clebsch-Gordan coefficients by lowering and orthogonality
Nobody should memorise a Clebsch-Gordan table. You need one state you can write down without thinking, one operator that walks down a multiplet, and one orthogonality condition for each new multiplet. Build the 1 × ½ table once by hand and every published table becomes something you can read and check rather than trust.
Build the model
Connect the measurement to the mechanism.
The product states |j₁ m₁⟩|j₂ m₂⟩ and the coupled states |j m⟩ are two orthonormal bases of one (2j₁ + 1)(2j₂ + 1)-dimensional space, and the Clebsch-Gordan coefficients ⟨j₁ m₁; j₂ m₂|j m⟩ are the matrix between them. Because Jz = J₁z + J₂z, that matrix only joins states with m = m₁ + m₂, so it breaks into one small block per M. The top block has size one: |j₁ j₁⟩|j₂ j₂⟩ is the only product state with M = j₁ + j₂, so it is the stretched state |j₁ + j₂, j₁ + j₂⟩.
Lowering it with J₋ = J₁₋ + J₂₋, using only Quantum Mechanics I's √((j + m)(j − m + 1))ħ, generates its whole multiplet. One step down, the block has one more product state than that multiplet uses, so the top of the next multiplet is the orthogonal direction, fixed up to an overall phase that physics cannot fix, since |ψ⟩ and e(iχ)|ψ⟩ are one state; with real coefficients that phase is a sign. The Condon-Shortley convention, the coefficient with m₁ = j₁ real and positive, makes that choice once per multiplet, and lowering carries it down.
Every step is real, so the change of basis is a real orthogonal matrix, and swapping j₁ with j₂ costs only the sign (−1)(j₁ + j₂ − j). The coefficients carry no ħ and no α: they are exact geometry. The method runs out of patience, not validity, at large j, where Racah's closed formula (quoted) or a computer takes over, and it never says which basis a Hamiltonian needs; that is physics, settled in the next topics.
- Simple definition
- A Clebsch-Gordan coefficient ⟨j₁ m₁; j₂ m₂|j m⟩ is the overlap of a product state with a coupled state: the amplitude for finding z-components m₁ and m₂ when the pair is known to have total angular momentum j and z-component m.
- Example
- For one p electron, ⟨1, 0; ½, ½|3/2, ½⟩ = √(2/3) = 0.816: in the state j = 3/2, mⱼ = ½ the spin is up with mₗ = 0 with probability 2/3, and down with mₗ = +1 with probability 1/3.
Both cross terms of J² = J₁² + J₂² + 2J₁zJ₂z + J₁₊J₂₋ + J₁₋J₂₊ annihilate it, leaving (j₁ + j₂)(j₁ + j₂ + 1)ħ²: the top of the table costs nothing.
The only product state with M = j₁ + j₂; its coefficient is 1 by choice of phase and, like every coefficient here, dimensionless.
Apply J₋ = J₁₋ + J₂₋ to |j m⟩ and project onto a product state: each new coefficient needs at most two old ones, so a multiplet's top generates all of it.
N(j, m) = √((j + m)(j − m + 1)) from QM I's J₋; ⟨m₁, m₂|j, m⟩ is short for ⟨j₁ m₁; j₂ m₂|j m⟩; one ħ per term cancels.
The change of basis splits into one block per M. For 1 × ½ the blocks have sizes 1, 2, 2, 1, so only j = 3/2 and ½ occur, and 4 + 2 = 6 = 3 × 2 states in all.
From Jz = J₁z + J₂z between a product bra and a coupled ket, (m − m₁ − m₂)ħ times the coefficient must vanish; the triangle is the previous topic's count of states at each M.
Two free checks on any published table, whose columns are coupled states: the squares down a column sum to 1, and so do the squares along a row, which says the coupled basis is complete.
Real entries in a unitary matrix make it orthogonal: in each M-block, rows and columns are unit vectors at right angles.
Impose it at the top of each multiplet and let lowering carry it down; that is why every published table agrees with yours entry by entry.
Chosen, not derived: physics fixes each multiplet only up to an overall phase, since |j m⟩ and e(iχ)|j m⟩ are one state; real and positive removes it.
J₊|j j⟩ = 0 makes the top state's coefficients alternate in sign along m₁, with j₁ + j₂ − j changes end to end; the singlet's minus sign is the j = 0 case.
j₁ + j₂ − j is a whole number: 0 for the stretched multiplet, 1 for the next, and so on down to the smallest j.
The top of the table is free
Start with what Jz already knows. Sandwich Jz = J₁z + J₂z between a product bra ⟨j₁ m₁; j₂ m₂| and a coupled ket |j m⟩: acting left gives (m₁ + m₂)ħ, acting right gives mħ, so (m − m₁ − m₂)ħ times the coefficient is zero. Only m = m₁ + m₂ survives, and the change of basis splits into one block per M, which is why nobody ever solves the full problem at once. Count along each M: for an l = 1 orbit and a spin-½, the blocks have 1, 2, 2 and 1 states at M = 3/2, ½, −½ and −3/2. The top block has size one, so |1, 1⟩|↑⟩ must be |3/2, 3/2⟩, and in general |j₁ j₁⟩|j₂ j₂⟩ is |j₁ + j₂, j₁ + j₂⟩. Nothing needs solving: J₁₊J₂₋ and J₁₋J₂₊ both annihilate that state, so J² gives j₁(j₁ + 1) + j₂(j₂ + 1) + 2j₁j₂ = (j₁ + j₂)(j₁ + j₂ + 1) in units of ħ², which is 15/4 for 1 × ½. Its coefficient is 1 by choice of the overall phase.
Lowering walks down one multiplet
Quantum Mechanics I supplies the only input: J₋|j m⟩ = ħ√((j + m)(j − m + 1))|j, m − 1⟩, true for J₁, J₂ and J alike. Apply J₋ = J₁₋ + J₂₋ to both sides of |3/2, 3/2⟩ = |1, 1⟩|↑⟩. The left gives √3 ħ|3/2, ½⟩. On the right, J₁₋ turns |1, 1⟩ into √2 ħ|1, 0⟩ and J₂₋ turns |↑⟩ into ħ|↓⟩. Divide by √3 ħ: |3/2, ½⟩ = √(2/3)|1, 0⟩|↑⟩ + √(1/3)|1, 1⟩|↓⟩. Lower again. The left gives 2ħ|3/2, −½⟩ and the right gives ħ[(2/√3)|1, −1⟩|↑⟩ + 2√(2/3)|1, 0⟩|↓⟩], so |3/2, −½⟩ = √(1/3)|1, −1⟩|↑⟩ + √(2/3)|1, 0⟩|↓⟩. A third step lands on |1, −1⟩|↓⟩ with coefficient exactly 1, as it must. Every term carried one ħ, which cancelled, so the coefficients are dimensionless. Nothing was normalised by hand; each state came out with squares summing to 1, and that is the check that no factor was dropped.
Orthogonality finds the next top, up to a sign
At M = ½ there are two product states and the j = 3/2 multiplet used one combination of them. The other must be |½, ½⟩, because eigenstates of the Hermitian J² with different eigenvalues are orthogonal. Write it a|1, 0⟩|↑⟩ + b|1, 1⟩|↓⟩ with a and b taken real. Orthogonality gives √(2/3)a + √(1/3)b = 0 and normalisation gives a² + b² = 1, so (a, b) = ±(−√(1/3), √(2/3)). Physics cannot choose between the two: they differ by an overall −1, and |ψ⟩ and −|ψ⟩ are the same ray. The Condon-Shortley convention decides: the coefficient with m₁ = j₁, here that of |1, 1⟩|↓⟩, is positive. So |½, ½⟩ = √(2/3)|1, 1⟩|↓⟩ − √(1/3)|1, 0⟩|↑⟩. Check it independently, because J₊ must annihilate the top of any multiplet: J₊ gives ħ(√(2/3) − √2 × √(1/3))|1, 1⟩|↑⟩ = 0. One lowering step then gives |½, −½⟩ = √(1/3)|1, 0⟩|↓⟩ − √(2/3)|1, −1⟩|↑⟩. For larger j₁ and j₂ the rule repeats: at each new M, the new top is orthogonal to every multiplet already built. The block gains one product state per step down only as far as M = |j₁ − j₂|, so the last new multiplet starts there and no j outside |j₁ − j₂| to j₁ + j₂ ever appears: every coefficient that breaks this triangle condition is zero.
The finished table is a rotation in each block
The M = ±3/2 blocks are the number 1. The M = ½ block, rows j = 3/2 and ½, columns |1, 0⟩|↑⟩ and |1, 1⟩|↓⟩, is (√(2/3), √(1/3); −√(1/3), √(2/3)): a rotation matrix with cos θ₀ = √(2/3), θ₀ = 35.26°. Reversing every m gives the M = −½ block, leaving the j = 3/2 entries alone and flipping the sign of the j = ½ ones. Every step used real square roots and a real sign choice, so every coefficient is real, and a unitary matrix with real entries is orthogonal: rows and columns are unit vectors at right angles. That gives two free checks. In this layout the row of |3/2, ½⟩ has 2/3 + 1/3 = 1, its normalisation, and the column of |1, 1⟩|↓⟩ has 1/3 + 2/3 = 1, which is the completeness of the coupled basis; a published table prints the transpose. Limiting cases test the pattern: for j₁ × ½ at M = j₁ − ½, one lowering step of |j₁ j₁⟩|↑⟩ gives [√(2j₁)|j₁, j₁ − 1⟩|↑⟩ + |j₁ j₁⟩|↓⟩]/√(2j₁ + 1), so the angle is arctan(1/√(2j₁)), which is 45° at j₁ = ½, reproducing the triplet and singlet, and falls towards 0 as j₁ grows, because one spin flip barely disturbs a large orbit. No α appears anywhere, so these numbers hold exactly at every order in α.
Swap the order and pay (−1)(j₁ + j₂ − j)
Tables list j₁ first, so coupling spin before orbit changes some signs, and the reason sits in each multiplet's top state. Write |j j⟩ = Σ c(m₁)|j₁ m₁⟩|j₂, j − m₁⟩ and demand J₊|j j⟩ = 0. The coefficient of each product state in the result gives c(m₁ + 1)N₊(j₂, j − m₁ − 1) = −c(m₁)N₊(j₁, m₁), with N₊(j, m) = √((j − m)(j + m + 1)). Both roots are positive, so neighbouring coefficients alternate in sign. From m₁ = j₁ down to m₁ = j − j₂ there are j₁ + j₂ − j steps, so the two end coefficients differ in sign by (−1)(j₁ + j₂ − j). Condon-Shortley makes the m₁ = j₁ end positive; in the swapped order it makes the other end positive. The whole top state therefore changes by that factor, and J₋ = J₁₋ + J₂₋, symmetric in the two, carries the factor down the multiplet. Check it on 1 × ½: in ½ × 1 order, |½, ½⟩ = √(1/3)|↑⟩|1, 0⟩ − √(2/3)|↓⟩|1, 1⟩, both signs flipped for exponent 1, while the j = 3/2 states, exponent 0, are unchanged. The singlet of two spin-halves is the smallest case: j = 0, exponent 1, antisymmetric.
Reading a published table, and checking it
Published tables, the Particle Data Group's being the standard, print each coefficient squared, with its sign outside: read an entry −1/3 as −√(1/3) = −0.577, never −0.333. Each block is one j₁ × j₂ and one M, with rows labelled (m₁, m₂) and columns labelled j. For 1 × ½ at M = +½, row (+1, −½) reads 1/3 under j = 3/2 and 2/3 under j = ½, and row (0, +½) reads 2/3 and −1/3: exactly the numbers built above. Check any entry by one lowering step instead of trusting it. The j = 3/2, M = −½ column reads 2/3 on (0, −½) and 1/3 on (−1, +½). Lowering gives √(2/3) × √2 + √(1/3) = √3 on |1, −1⟩|↓⟩, which must equal √((j + m)(j − m + 1)) = √3 at j = 3/2, m = −½, and it does. The table also meets experiment. In |3/2, ½⟩ the spin is up with probability 2/3, so ⟨Lz + 2Sz⟩ = (1/3 + 1/3)ħ and the Landé factor is 4/3, against 1.3342 measured for sodium's 3p ²P₃/₂ level; the small gap is the electron's order-α anomalous moment.
Change one variable at a time
Make the relationship visible.
Rotate φ until the left bar empties. It happens twice, 180° apart, at φ = θ₀ ± 90°, and the bars cannot tell the two apart; only the dot, where the |j₁, j₁⟩|↓⟩ coefficient is positive, picks one. Then step j₁ from ½ up to 4 and watch θ₀ fall from 45° to 19.5°.
LOWERED-STATE ANGLE θ₀35.3 °
OVERLAP WITH |j₁+½, j₁−½⟩0.816
⟨J²⟩/ħ² IN THE TRIAL STATE2.750
COEFFICIENT ON |j₁, j₁⟩|↓⟩0.000
Live interpretationLOWERED-STATE ANGLE θ₀: 35.3 °. OVERLAP WITH |j₁+½, j₁−½⟩: 0.816. ⟨J²⟩/ħ² IN THE TRIAL STATE: 2.750. COEFFICIENT ON |j₁, j₁⟩|↓⟩: 0.000
Catch the common trap
Explain before calculating.
For an l = 1 orbit and a spin-½, lowering |1, 1⟩|↑⟩ gives |3/2, ½⟩ = √(2/3)|1, 0⟩|↑⟩ + √(1/3)|1, 1⟩|↓⟩. Which state is |½, ½⟩ in the Condon-Shortley convention?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasySodium-23 has nuclear spin I = 3/2 and a 3s electron with J = ½, so its ground level splits into F = 2 and F = 1. Taking I as j₁ and J as j₂, build |F = 2, M = 1⟩ and |F = 1, M = 1⟩ from the product states |mI⟩|mJ⟩ by one lowering step and orthogonality, and find the probability that the electron spin is up in each.
- The stretched state is |2, 2⟩ = |3/2⟩|↑⟩, the only product state with M = 2.
- Apply F₋ = I₋ + J₋. The left gives ħ√((2 + 2)(2 − 2 + 1)) = 2ħ times |2, 1⟩. On the right, I₋|3/2⟩ = ħ√(3 × 1)|½⟩ = √3 ħ|½⟩ and J₋|↑⟩ = ħ|↓⟩. So |2, 1⟩ = (√3/2)|½⟩|↑⟩ + (1/2)|3/2⟩|↓⟩, and 3/4 + 1/4 = 1 confirms the norm.
- M = 1 holds exactly two product states, so |1, 1⟩ = ±[(√3/2)|3/2⟩|↓⟩ − (1/2)|½⟩|↑⟩], the orthogonal combination. Condon-Shortley makes the mI = I = 3/2 coefficient positive, which selects the upper sign.
- Check with F₊, which must annihilate the top of the F = 1 multiplet: J₊ on the first term gives (√3/2)ħ|3/2⟩|↑⟩ and I₊ on the second gives −(1/2)√3 ħ|3/2⟩|↑⟩, which cancel exactly. A published 3/2 × ½ table prints 3/4 and −1/4 in the F = 1 column, the squares with the sign outside.
- The spin-up probability is the squared coefficient on |½⟩|↑⟩: 3/4 in |2, 1⟩ and 1/4 in |1, 1⟩. They sum to 1, as the row of an orthogonal matrix must.
Answer|2, 1⟩ = (√3/2)|½⟩|↑⟩ + ½|3/2⟩|↓⟩ and |1, 1⟩ = (√3/2)|3/2⟩|↓⟩ − ½|½⟩|↑⟩; the electron spin is up with probability 0.75 in F = 2 and 0.25 in F = 1.
MediumTwo p electrons, as in carbon's 2p² configuration, each have l = 1. Build the M = 0 member of the L = 2, 1 and 0 multiplets from the product states |m₁⟩|m₂⟩, find the probability that both electrons have m = 0 in each, and say which L are symmetric when the two orbital labels are exchanged.
- Write coupled states |L, M⟩ and product states |m₁⟩|m₂⟩. Top: |2, 2⟩ = |1⟩|1⟩. One lowering step, with √((2 + 2)(2 − 2 + 1)) = 2 on the left and √2 from each l₋ on the right, gives |2, 1⟩ = (|1⟩|0⟩ + |0⟩|1⟩)/√2.
- At M = 1 the orthogonal combination, with the m₁ = +1 coefficient positive by Condon-Shortley, is |1, 1⟩ = (|1⟩|0⟩ − |0⟩|1⟩)/√2.
- Lower |2, 1⟩: the left gives √(3 × 2) = √6 and the right gives |1⟩|−1⟩ + 2|0⟩|0⟩ + |−1⟩|1⟩, so |2, 0⟩ = (|1⟩|−1⟩ + 2|0⟩|0⟩ + |−1⟩|1⟩)/√6. Lower |1, 1⟩: the left gives √2, the two |0⟩|0⟩ terms cancel, and |1, 0⟩ = (|1⟩|−1⟩ − |−1⟩|1⟩)/√2.
- Write |0, 0⟩ = a|1⟩|−1⟩ + b|0⟩|0⟩ + c|−1⟩|1⟩. Orthogonality to |2, 0⟩ and |1, 0⟩ gives a + 2b + c = 0 and a − c = 0, so a = c = −b. Normalise and make a positive: |0, 0⟩ = (|1⟩|−1⟩ − |0⟩|0⟩ + |−1⟩|1⟩)/√3.
- P(m₁ = m₂ = 0) is the squared coefficient on |0⟩|0⟩: 4/6 = 2/3 for L = 2, 0 for L = 1 and 1/3 for L = 0. The three sum to 1, the row check.
- Exchanging the labels swaps m₁ and m₂. L = 2 and L = 0 are unchanged and L = 1 changes sign, exactly the factor (−1)(1 + 1 − L).
Answer|2, 0⟩ = (|1⟩|−1⟩ + 2|0⟩|0⟩ + |−1⟩|1⟩)/√6, |1, 0⟩ = (|1⟩|−1⟩ − |−1⟩|1⟩)/√2, |0, 0⟩ = (|1⟩|−1⟩ − |0⟩|0⟩ + |−1⟩|1⟩)/√3; P(both m = 0) = 2/3, 0 and 1/3; L = 2 and 0 are symmetric, L = 1 antisymmetric.
HardSodium's 3p electron has l = 1 and s = ½, and its magnetic moment is μz = −μB(Lz + gₛ Sz)/ħ, from Unit 3. A level's Landé factor gJ is defined by ⟨μz⟩ = −gJ mⱼ μB. Using only the 1 × ½ table, find gJ for j = 3/2 and j = ½, first with Dirac's gₛ = 2 and then with the QED value gₛ = 2.00232 (both quoted), and compare with the measured 1.3342(2) and 0.66581(12).
- Lz and Sz are diagonal in the product basis, so their expectation values in a coupled state are averages over its product components weighted by the squared coefficients; no cross terms survive. The signs in the table never enter.
- |3/2, ½⟩ = √(2/3)|1, 0⟩|↑⟩ + √(1/3)|1, 1⟩|↓⟩ gives ⟨Lz⟩ = ħ/3 and ⟨Sz⟩ = (2/3 − 1/3)(ħ/2) = ħ/6. Then gJ × ½ = 1/3 + gₛ/6, so gJ = 2/3 + gₛ/3 = 4/3 for gₛ = 2.
- Cross-check on the stretched state |3/2, 3/2⟩ = |1, 1⟩|↑⟩: ⟨Lz + gₛ Sz⟩/ħ = 1 + gₛ/2 = 2, and 2 ÷ 3/2 = 4/3 again. Two values of mⱼ give one gJ, which the projection theorem at the end of this unit explains.
- |½, ½⟩ = √(2/3)|1, 1⟩|↓⟩ − √(1/3)|1, 0⟩|↑⟩ gives ⟨Lz⟩ = 2ħ/3 and ⟨Sz⟩ = (1/3 − 2/3)(ħ/2) = −ħ/6. Then gJ = 2(2/3 − gₛ/6) = 4/3 − gₛ/3 = 2/3 for gₛ = 2.
- Against measurement, 1.33333 and 0.66667 miss 1.3342 and 0.66581 by 0.0009, four and seven standard errors. With gₛ = 2.00232, gJ = 1.33411 and 0.66589, within 0.0001 of both. The residual was gₛ − 2 ≈ α/π = 0.00232, first order in α; the coefficients themselves carry no α.
- Size and units: gJ is dimensionless, and for |3/2, ½⟩ ⟨μz⟩ = −(1.3341)(½)(9.274 × 10⁻²⁴ J T⁻¹) = −6.19 × 10⁻²⁴ J T⁻¹. In B = 1.00 T its energy −⟨μz⟩B is 6.19 × 10⁻²⁴ J, or 38.6 μeV.
AnswergJ = 4/3 for j = 3/2 and 2/3 for j = ½ with gₛ = 2, or 1.3341 and 0.6659 with gₛ = 2.00232, matching the measured 1.3342(2) and 0.66581(12); ⟨μz⟩ = −6.19 × 10⁻²⁴ J T⁻¹ in |3/2, ½⟩.