University Quantum Mechanics II · Rotations and the Addition of Angular Momentum · 5.9

The Wigner-Eckart theorem and the projection theorem

This is the theorem that turns a table of matrix elements into one number. Compute a single component of a vector operator by the easiest route and rotations hand you all the others; inside one multiplet it goes further and trades the operator for its projection on J, which is how Unit 9 reads off Zeeman shifts.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A vector operator has three spherical components V_q, and the commutators that define it in 05.08 give them the same relations with J that the kets |1 q⟩ have. So V_q|j m⟩ turns under rotations exactly as the product |j m⟩|1 q⟩ does, and its piece in a multiplet j′ must carry the same Clebsch-Gordan coefficient ⟨j m; 1 q|j′ m′⟩ that the product carries. That is the Wigner-Eckart theorem: ⟨j′ m′|V_q|j m⟩ = ⟨j m; 1 q|j′ m′⟩⟨j′‖V‖j⟩, a geometric factor fixed by rotations times one reduced matrix element that holds every radial integral and every piece of dynamics. The lesson states it without proof, since the proof needs rotation matrices, and tests its predictions.

It gives Δj = 0 or ±1, with 0 to 0 barred, for any vector operator, spin included; it predicts all three components of ⟨1s|r|2p⟩ from one integral; and summed over q and m′ it makes the strength the same from every m, so all the sublevels of a level decay at one rate. Its diagonal case, the projection theorem, is derived from a double commutator: inside one multiplet V acts as (V⋅J)J/(j(j + 1)ħ²). The theorem stops where its inputs stop. It cannot size the reduced element, which a radial integral must supply.

It is blind to parity, so 2s₁/₂ to 1s₁/₂ passes its Δj = 0 test and still vanishes. And the projection theorem holds only while j is a good quantum number: a field that mixes 2p₃/₂ with 2p₁/₂ breaks it.

Simple definition: The Wigner-Eckart theorem says that every matrix element of a vector operator between two angular-momentum multiplets is a Clebsch-Gordan coefficient, fixed by rotations alone, times one reduced matrix element that holds all the dynamics and depends on none of m, m′ and q.
Example: In hydrogen ⟨100|z|210⟩ = 128√2/243 a = 3.94 × 10⁻¹¹ m. Dividing by ⟨1 0; 1 0|0 0⟩ = −1/√3 gives the reduced element −1.290a, and ⟨100|r₋₁|211⟩ = (1/√3)(−1.290a) = −0.745a follows with no second integral.

The Wigner-Eckart theorem for a vector operator⟨j′ m′|V_q|j m⟩ = ⟨j m1 q|j′ m′⟩ ⟨j′‖V‖j⟩

Stated, not proved: the proof needs rotation matrices. Geometry sits in the coefficient; radial integrals and dynamics sit in one reduced element shared by all 3(2j + 1) choices of m and q.

V₀ = V_z and V₊₁, V₋₁ = ∓(Vₓ ± iV_y)/√2; the coefficient is dimensionless and ⟨j′‖V‖j⟩ carries the units of V

The reduced element in other books' normalisation⟨j′‖V‖j⟩ = ⟨j′‖V‖j⟩_S/√(2j + 1) = ⟨j′‖V‖j⟩_E/√(2j′ + 1)

Ratios inside one pair of multiplets agree in every convention; absolute values differ by these roots, so check a table's convention before mixing its numbers with yours.

Quoted: S marks Sakurai's reduced element, E Edmonds' 3j-symbol form used by most atomic tables; all carry the units of V

Selection rules read off the coefficient⟨j′ m′|V_q|j m⟩ = 0 unless m′ = m + q and |j − 1| ≤ j′ ≤ j + 1, and always for j = j′ = 0

Derived from the coefficient: 0 ⊗ 1 contains only j′ = 1. It knows nothing of parity, so for r the rule Δl = ±1 of 05.08 still has to be applied on top.

Any vector operator, spin included; j and j′ integer or half-integer; a statement about zeros, so it carries no units

Hydrogen check between 1s and 2p⟨100|r_q|21m⟩ = ⟨1 m1 q|0 0⟩ ⟨0‖r‖1⟩, with ⟨0‖r‖1⟩ = −(128√6/243)a = −1.290a

Derived from Unit 2's R₁₀ and R₂₁: the q = 0 integral fixes ⟨0‖r‖1⟩, and the m = ±1 components come out equal to it in size, as ⟨1 m; 1 −m|0 0⟩ = (−1)^{(1 − m)}/√3 demands.

a = 5.29 × 10⁻¹¹ m, the Bohr radius for infinite nuclear mass; nonzero only for q = −m, each of size 128√2/243 a = 0.745a

Strength summed over q and m′Σ_q Σₘ′ |⟨j′ m′|V_q|j m⟩|² = [(2j′ + 1)/(2j + 1)] |⟨j′‖V‖j⟩|²

Derived from rotational invariance and the normalisation of each coupled state. Every sublevel of a level has the same total strength, so the level has one lifetime: 1.60 ns for hydrogen 2p.

The same for every m, in the units of V squared; from each 2p sublevel to 1s it is (1/3)(1.290a)² = 0.555a²

The projection theorem, inside one multiplet⟨j m′|V|j m⟩ = ⟨V⋅J⟩ ⟨j m′|J|j m⟩ / [j(j + 1)ħ²]

Derived from [J², [J², V]] = 2ħ²(J²V + VJ²) − 4ħ²(V⋅J)J, whose same-j elements vanish. Inside the multiplet V acts as its component along J; the rest connects to other multiplets.

⟨V⋅J⟩ is one value for the whole multiplet, in units of V times J s; j(j + 1)ħ² is in J² s²; bra and ket must share the multiplet

What the theorem says, and what it leaves to you

05.08's definition gives, in a line each, [J_z, V_q] = qħV_q and [J₊, V_q] = ħ√((1 − q)(2 + q)) V_(q+1), with its mirror for J₋: the rules J_z and J₊ obey on the kets |1 q⟩. So J₊ and J₋ acting on V_q|j m⟩ generate the recursion that the total J₊ and J₋ generate on the product |j m⟩|1 q⟩, and 05.06 showed that recursion fixes Clebsch-Gordan coefficients up to one constant per multiplet. That is the reason, not a proof, and the lesson states the result: ⟨j′ m′|V_q|j m⟩ = ⟨j m; 1 q|j′ m′⟩⟨j′‖V‖j⟩. The coefficient is pure geometry, the same for r, p, L, S or a magnetic moment. The reduced element carries the radial integrals and the dynamics, one number for all 3(2j + 1) choices of m and q. Books disagree on what to pull out of it. This lesson pulls out nothing, Sakurai divides by √(2j + 1), and Edmonds, followed by most atomic tables, uses a 3j symbol, equivalent to dividing by √(2j′ + 1). Ratios inside one pair of multiplets agree in all three; absolute values do not.

Selection rules read off one coefficient

The coefficient ⟨j m; 1 q|j′ m′⟩ vanishes unless m′ = m + q and j′ is one of the values j ⊗ 1 contains, |j − 1| up to j + 1. So every vector operator obeys Δm = q and Δj = 0 or ±1, and j = 0 to j′ = 0 is barred because 0 ⊗ 1 contains only j′ = 1. These rules govern j, not only l, and any vector operator, not only r: 3d₅/₂ to 2p₁/₂ has Δj = 2 and no dipole element at all, while 3d₃/₂ to 2p₁/₂ is allowed. What the theorem cannot see is parity. The rule Δl = ±1 of 05.08 came from r being odd, and it is a separate condition: 2s₁/₂ to 1s₁/₂ passes the Δj = 0 test and still has a zero dipole element, while 2p₁/₂ to 1s₁/₂, also Δj = 0, is a genuine component of Lyman α. For L or S inside one multiplet the theorem allows Δj = 0 elements, and that is the case the projection theorem pins down.

One integral fixes all three 1s–2p components

Take ψ₁₀₀ = R₁₀Y₀₀ and ψ₂₁ₘ = R₂₁Y₁ₘ from Unit 2. For q = 0 the integral splits: ∫R₁₀R₂₁r³dr = (2/√24)(4!)(2/3)⁵a = 128√6/243 a, and ∫Y₀₀ cos θ Y₁₀ dΩ = 1/√3, so ⟨100|z|210⟩ = 128√2/243 a = 0.745a. Dividing by ⟨1 0; 1 0|0 0⟩ = −1/√3 gives ⟨0‖r‖1⟩ = −1.290a. The theorem then predicts ⟨100|r₋₁|211⟩ and ⟨100|r₊₁|21, −1⟩ as (1/√3)(−1.290a) = −0.745a each, and the direct integrals, with r_q = r√(4π/3)Y₁q, agree in size and sign. Each is a length, and Σ_q|⟨100|r_q|21m⟩|² = 0.555a² for every m. With Unit 13's rate A = ω³e²|⟨1s|r|2p⟩|²/(3πε₀ħc³), quoted, and ħω = 10.20 eV, this becomes A = (2/3)⁸α⁵mₑc²/ħ = 6.27 × 10⁸ s⁻¹, fifth order in α: a lifetime of 1.60 ns, against 1.60 ± 0.01 ns measured by beam-foil timing.

The strength summed over q and m′ does not depend on m

Fix |j m⟩ and add the squares of every element into the multiplet j′. The sum is ⟨j m|V†⋅P V|j m⟩, with P the projector onto that multiplet, and V†⋅P V is a scalar: it commutes with J₊ and J₋, so its expectation value cannot change as they step m up or down. Now sum over m as well. Each coupled state |j′ m′⟩ is normalised, so Σ over m and q of |⟨j m; 1 q|j′ m′⟩|² = 1, and the grand total is (2j′ + 1)|⟨j′‖V‖j⟩|²; each of the 2j + 1 values of m carries the share (2j′ + 1)/(2j + 1). This is derived, and it is why a level has one lifetime rather than one per sublevel. The 1s–2p numbers pass: (1/3)(1.290a)² = 0.555a², as found component by component. Because r acts on the orbit alone, the same total holds for every state of a p level, coupled or not, so sodium's D₂ line, with four upper sublevels, is twice as strong as D₁, with two: the oscillator strengths from the measured lifetimes are 0.6405 and 0.3199.

The projection theorem from a double commutator

Inside one multiplet V and J are both vector operators with the same coefficients, so the theorem makes them proportional. The constant can be derived without the unproved theorem. Two applications of [J², V] = iħ(V × J − J × V), with J × J = iħJ and J × V = −V × J + 2iħV, give, after a page of algebra only sketched here, [J², [J², V]] = 2ħ²(J²V + VJ²) − 4ħ²(V⋅J)J: 05.08's identity for r, with the term that r⋅L = 0 had removed now restored. Take its element between ⟨j m′| and |j m⟩. Each J² gives j(j + 1)ħ² on either side, so the left side is zero, and 4j(j + 1)ħ⁴⟨j m′|V|j m⟩ = 4ħ²⟨j m′|(V⋅J)J|j m⟩. J keeps the state in the multiplet and V⋅J is a scalar, constant there, so ⟨j m′|V|j m⟩ = ⟨V⋅J⟩⟨j m′|J|j m⟩/(j(j + 1)ħ²). Inside the multiplet V acts as its component along J; the perpendicular part has no elements there at all. Checks: both sides carry the units of V; V = J returns an identity; V = r, with r⋅L = 0, gives zero, as parity requires.

Checked on the 05.07 states, and where it stops

For |l, ½, j, mⱼ⟩, S⋅J = (J² + S² − L²)/2 is a number: (2l + 3)ħ²/4 for j = l + ½ and −(2l − 1)ħ²/4 for j = l − ½. The projection theorem gives ⟨S_z⟩ = ±mⱼħ/(2l + 1), and 05.07's coefficients give the same directly, (ħ/2)(c↑² − c↓²): +ħ/6 in 2p₃/₂ and −ħ/6 in 2p₁/₂ at mⱼ = ½, and ħ/2 in the stretched state |1, ↑⟩, as it must be. Unit 9 turns this into the Landé factor: ⟨L + gₛS⟩ = ⟨J + (gₛ − 1)S⟩ is proportional to ⟨J⟩, and with gₛ = 2.0023 quoted it gives 1.3341 for P₃/₂ and 0.6659 for P₁/₂, beside sodium's measured 1.3342 and 0.6658. The theorem is exact, but only inside one multiplet, because S also connects the two: ⟨2p₁/₂|S_z|2p₃/₂⟩ = −√2ħ/3 at mⱼ = ½, larger than either diagonal value. Once μ_B B nears hydrogen's 45 μeV n = 2 fine-structure splitting that element mixes the multiplets, and the projection fails.

Change one variable at a time

Make the relationship visible.

Interactive model

Initial multiplet j1

Change j′ − j−1

Initial sublevel m0

Start at j = 1 and j′ − j = −1, the 2p to 1s check: one bar of 1/3 at every m. Then set j′ − j = 0 and step m from −1 to 1: the q = 0 bar, m²/(j(j + 1)), vanishes at m = 0, as the projection theorem's ⟨V_z⟩ ∝ m says, while the sum bar never moves.

COEFFICIENT, q = +10.000

COEFFICIENT, q = 0-0.577

COEFFICIENT, q = −10.000

SUM (2j′ + 1)/(2j + 1)0.333

Live interpretationCOEFFICIENT, q = +1: 0.000. COEFFICIENT, q = 0: −0.577. COEFFICIENT, q = −1: 0.000. SUM (2j′ + 1)/(2j + 1): 0.333

Catch the common trap

Explain before calculating.

An electron is in the 2p₁/₂ state with mⱼ = +½, and the projection theorem is used to find the expectation value of its spin along z. What is ⟨S_z⟩?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA vector operator V with the units of angular momentum acts inside an l = 1 multiplet, and one element is known: ⟨1, 1|V_z|1, 1⟩ = 0.60ħ. Using ⟨1 m; 1 0|1 m⟩ = m/√2 and ⟨1 0; 1 1|1 1⟩ = −1/√2 from the 1 ⊗ 1 table, find the reduced element, then ⟨1, 0|V_z|1, 0⟩, ⟨1, −1|V_z|1, −1⟩ and ⟨1, 1|V₊₁|1, 0⟩.

Wigner-Eckart with j = j′ = 1 and q = 0: ⟨1, 1|V₀|1, 1⟩ = ⟨1 1; 1 0|1 1⟩⟨1‖V‖1⟩ = (1/√2)⟨1‖V‖1⟩, so ⟨1‖V‖1⟩ = √2 × 0.60ħ = 0.849ħ.
Same reduced element, new coefficients: ⟨1, 0|V_z|1, 0⟩ = (0/√2)(0.849ħ) = 0, and ⟨1, −1|V_z|1, −1⟩ = (−1/√2)(0.849ħ) = −0.60ħ. The diagonal elements go as m.
For q = +1, m = 0 and m′ = 1: ⟨1, 1|V₊₁|1, 0⟩ = ⟨1 0; 1 1|1 1⟩⟨1‖V‖1⟩ = (−1/√2)(0.849ħ) = −0.60ħ.
Check with an operator you can compute: V = 0.60L. ⟨1, 1|L_z|1, 1⟩ = ħ, and ⟨1, 1|L₊₁|1, 0⟩ = −⟨1, 1|L₊|1, 0⟩/√2 = −√2ħ/√2 = −ħ, so 0.60L reproduces every number. Inside one multiplet a vector operator acts as a multiple of the angular momentum, which is the projection theorem.

Answer⟨1‖V‖1⟩ = 0.85ħ; ⟨1, 0|V_z|1, 0⟩ = 0, ⟨1, −1|V_z|1, −1⟩ = −0.60ħ and ⟨1, 1|V₊₁|1, 0⟩ = −0.60ħ.

MediumHydrogen's R₃₁ = (8/(27√6)) a^(−3/2) (r/a)(1 − r/(6a)) e^(−r/(3a)) is quoted from the table of radial functions. Compute ⟨100|z|310⟩, use the Wigner-Eckart theorem to predict ⟨100|r_q|31m⟩ for m = ±1, then scale the 2p to 1s rate, A = 6.27 × 10⁸ s⁻¹ at ħω = 10.20 eV, to the 3p to 1s rate, taking A ∝ ω³|⟨1s|r|np⟩|² from Unit 13.

Radial integral: ∫R₁₀R₃₁r³dr = (16/(27√6))a⁻⁴ ∫r⁴(1 − r/(6a))e^(−4r/(3a))dr. With ∫rⁿe^(−r/b)dr = n! bⁿ⁺¹ and b = 3a/4, the integral is 24b⁵ − 20b⁶/a = 9b⁵, so the radial factor is 81/(64√6) a = 0.517a.
The angular factor is 1/√3, as for 2p, so ⟨100|z|310⟩ = 27√2/128 a = 0.298a. Dividing by ⟨1 0; 1 0|0 0⟩ = −1/√3 gives ⟨0‖r‖1⟩ = −0.517a for this pair of levels.
Prediction: ⟨100|r₋₁|311⟩ = ⟨100|r₊₁|31, −1⟩ = (1/√3)(−0.517a) = −0.298a, every other component zero. The sum over q is (0.298a)² = 0.0890a² from each of the three 3p sublevels.
Scale from 2p: ħω = 13.606 eV × 8/9 = 12.09 eV against 10.20 eV, so (ω₃/ω₂)³ = (32/27)³ = 1.665, and the strength ratio is 0.0890/0.555 = 0.160. A(3p to 1s) = 6.27 × 10⁸ s⁻¹ × 1.665 × 0.160 = 1.67 × 10⁸ s⁻¹.
Checks: each element is a length; the rate has the same α⁵mₑc²/ħ scaling as 2p's; the tabulated Lyman-β rate is 1.6725 × 10⁸ s⁻¹. The smaller overlap of 1s with 3p, not the frequency, makes Lyman β the weaker line.

Answer⟨100|z|310⟩ = 0.298a; ⟨100|r₋₁|311⟩ = ⟨100|r₊₁|31, −1⟩ = −0.298a; A(3p to 1s) = 1.67 × 10⁸ s⁻¹.

HardAn electron in a 3d level of a one-electron atom sits in the j = 3/2 multiplet built in 05.07. Use the projection theorem to find ⟨S_z⟩ and ⟨L_z⟩ for each mⱼ and the element ⟨3/2, 3/2|S₊|3/2, 1/2⟩, check both against the 05.07 coefficients, then find ⟨μ_z⟩ = −(μ_B/ħ)⟨L_z + 2S_z⟩ in the mⱼ = 3/2 state, taking gₛ = 2.

S⋅J = (J² + S² − L²)/2 = (15/4 + 3/4 − 6)ħ²/2 = −3ħ²/4, one value for all four states, and j(j + 1)ħ² = 15ħ²/4. Inside the multiplet S acts as (−3/4)/(15/4) J = −J/5.
Diagonal: ⟨S_z⟩ = −mⱼħ/5, that is −0.30ħ, −0.10ħ, +0.10ħ and +0.30ħ for mⱼ = 3/2, 1/2, −1/2 and −3/2, and ⟨L_z⟩ = mⱼħ − ⟨S_z⟩ = (6/5)mⱼħ. The spin leans against J, as j = l − ½ says it must.
Off-diagonal: ⟨3/2, 3/2|S₊|3/2, 1/2⟩ = −(1/5)⟨3/2, 3/2|J₊|3/2, 1/2⟩ = −(1/5)ħ√(15/4 − (1/2)(3/2)) = −√3ħ/5 = −0.346ħ.
Check with 05.07: |3/2, 3/2⟩ = −√(1/5)|1, ↑⟩ + √(4/5)|2, ↓⟩ and |3/2, 1/2⟩ = −√(2/5)|0, ↑⟩ + √(3/5)|1, ↓⟩. Then ⟨S_z⟩ = (ħ/2)(2/5 − 3/5) = −ħ/10 at mⱼ = 1/2, and S₊ turns √(3/5)|1, ↓⟩ into √(3/5)ħ|1, ↑⟩, whose overlap with the top state is −√3ħ/5. Both agree.
Moment: L_z + 2S_z = J_z + S_z acts as (1 − 1/5)J_z = (4/5)J_z, so ⟨μ_z⟩ = −(4/5)(3/2)μ_B = −1.2μ_B = −1.11 × 10⁻²³ J T⁻¹, with μ_B = 9.274 × 10⁻²⁴ J T⁻¹. The factor 4/5 is what Unit 9 calls g_J for a d₃/₂ level.

Answer⟨S_z⟩ = −mⱼħ/5 and ⟨L_z⟩ = (6/5)mⱼħ; ⟨3/2, 3/2|S₊|3/2, 1/2⟩ = −√3ħ/5 = −0.346ħ; ⟨μ_z⟩ = −1.2μ_B = −1.11 × 10⁻²³ J T⁻¹ at mⱼ = 3/2.