
Nuclear, particle & quantum-information sequence · Interactive lesson
Angular momentum, spin addition and selection rules
Because [Ĥ, Ĵ] = 0 for every isolated system, a reaction's total angular momentum — spins plus the orbital L of relative motion — is a label it cannot change. This lesson makes that ledger computable: couple the spins with Clebsch–Gordan coefficients, add L, and read off which totals and parities a decay can reach.
UPV-19.07 · Lesson 7 of 9 in this unit. Position is not a completion score.Nuclear, particle & quantum-information sequence. These units continue into nuclear and particle physics and quantum information. Follow your university’s prerequisites, mathematical depth and assessment; this independent pathway is not a universal programme specification.
Build the model
Rotations have a generator; that generator is what a reaction must balance.
The key idea
Angular momentum is conserved in a reaction because the total Ĵ — every spin plus every orbital L of relative motion — generates rotations and commutes with Ĥ, so the labels J and M of the initial state must be reachable by the final state.
Example. ρ⁰ (Jᴾ = 1⁻) → π⁺π⁻: two spin-0 pions reach J = 1 only through L = 1, and the parity (−1)(−1)(−1)¹ = −1 matches the ρ, so the decay is allowed — it is the ρ's dominant mode, with a 150 MeV width.
The complete physical model and assumptions
Rotate an isolated system and nothing measurable changes, so the rotation operators Û(n̂, φ) = exp(−iφ n̂⋅Ĵ/ħ) commute with Ĥ, and the Heisenberg equation d⟨Ĵ⟩/dt = (i/ħ)⟨[Ĥ, Ĵ]⟩ makes every component of the total Ĵ a constant of the motion. That is the entire source of the rule that angular momentum balances across a reaction. The price is set by the algebra [Ĵᵢ, Ĵⱼ] = iħ εᵢⱼₖ Ĵₖ: the components do not commute with one another, so a state carries only two labels, j and m, and Ĵ is not a number you add but an operator you couple. For two parts the product basis |j₁ m₁⟩|j₂ m₂⟩ and the coupled basis |J M⟩ span the same (2j₁+1)(2j₂+1)-dimensional space, related by Clebsch–Gordan coefficients, and J runs from |j₁ − j₂| to j₁ + j₂ in integer steps. In a reaction the spins are only part of the sum: two bodies flying apart also carry an orbital L of relative motion, integer-valued with parity (−1)L, so a decay is allowed only if some (S, L) pair reaches the parent's J, and, when the decay is strong or electromagnetic, with the right parity. Because L is integer, half-integer totals survive only with an odd count of fermions — which is why n → p + e⁻ is impossible and the neutrino had to exist. None of this fixes a rate; it says only which final states rotation permits.
1Rotations have a generator, and that is why J is conserved
A rotation by φ about n̂ acts on states as Û = exp(−iφ n̂⋅Ĵ/ħ); the three Hermitian generators Ĵₓ, Ĵy, Ĵz are defined by that role, and their algebra [Ĵᵢ, Ĵⱼ] = iħ εᵢⱼₖ Ĵₖ is the statement that rotations about different axes do not commute. If the physics is the same in every orientation, Û†ĤÛ = Ĥ for every φ, which for infinitesimal φ is [Ĥ, Ĵᵢ] = 0. The Heisenberg equation d⟨Ĵᵢ⟩/dt = (i/ħ)⟨[Ĥ, Ĵᵢ]⟩ then vanishes, and so does the one for Ĵ² = Ĵ⋅Ĵ, which commutes with every component. Ĥ, Ĵ² and Ĵz share an eigenbasis, and a state that starts as |J M⟩ stays inside that (J, M) sector under any interaction contained in Ĥ — strong, electromagnetic or weak. Two things must be said about which Ĵ. It is the total for the whole system, Ĵ = Ĵ₁ ⊗ 1 + 1 ⊗ Ĵ₂ + L̂ for two bodies, and only the total commutes with an interaction that couples the parts. And because Ĵₓ, Ĵy and Ĵz do not commute among themselves, only j and one m label a state, which is why a ledger balances J and M and never a vector.
2Couple two angular momenta: the triangle rule, built by hand
The states |j₁ m₁⟩|j₂ m₂⟩ span (2j₁+1)(2j₂+1) dimensions, and Ĵ² for the total is not diagonal in them. Build the coupled basis directly. The top state |j₁ j₁⟩|j₂ j₂⟩ has M = j₁ + j₂ and is the only state with that M, so it must be |J = j₁+j₂, M = j₁+j₂⟩. Apply Ĵ₋ = Ĵ₁₋ + Ĵ₂₋ to walk down its ladder; at each M the state orthogonal to the ladder starts the next multiplet, J = j₁ + j₂ − 1, and the process stops when the dimensions are used up, at J = |j₁ − j₂|. For 1 ⊗ ½ that is 6 = 4 + 2: J = 3/2 and J = 1/2, nothing else. For ½ ⊗ ½ it is 4 = 3 + 1, a triplet and a singlet. Two facts ride on this construction. J changes in integer steps, so whether the total is integer or half-integer is fixed by how many half-integer j's are coupled: an odd number gives a half-integer total, an even number an integer one, whatever orbital motion adds. And the maximum J = j₁ + j₂ is not the classical parallel sum: √(J(J+1)) is always less than √(j₁(j₁+1)) + √(j₂(j₂+1)), so even in the stretched state the two spins are not aligned — for ½ ⊗ ½ the triplet has ⟨Ŝ₁⋅Ŝ₂⟩ = +¼ħ², an angle of 70.5°, not 0°.
3Clebsch–Gordan coefficients are probability amplitudes
The coefficient ⟨j₁ m₁ j₂ m₂|J M⟩ is the overlap between a product state and a coupled one, so its square is the probability that a pair prepared with definite m₁ and m₂ is found with total J. Work 1 ⊗ ½ explicitly. Lowering |3/2, 3/2⟩ = |1,1⟩|½,½⟩ with Ĵ₋ gives √3 |3/2, ½⟩ = √2 |1,0⟩|½,½⟩ + |1,1⟩|½,−½⟩, so |3/2, ½⟩ = √(2/3)|1,0⟩|½,½⟩ + √(1/3)|1,1⟩|½,−½⟩, and the orthogonal combination is |½, ½⟩ = √(2/3)|1,1⟩|½,−½⟩ − √(1/3)|1,0⟩|½,½⟩ with the Condon–Shortley sign. Invert: a spin-1 particle in m₁ = +1 meeting a spin-½ in m₂ = −½ is in J = 3/2 with probability 1/3 and J = 1/2 with probability 2/3. That number sets a relative rate. If a reaction proceeds only through a J = 3/2 resonance, this spin combination feeds it at one third the strength of m₁ = +1, m₂ = +½, which is pure J = 3/2. The same coefficients run in every direction: a decaying J = 3/2, M = ½ state emits its two products with m₁ = 0, m₂ = +½ two thirds of the time and m₁ = 1, m₂ = −½ one third of the time, which is how a polarisation measurement reads a parent's spin.
4Two products also carry an orbital L, and L carries parity
Two products flying apart have a relative wavefunction whose angular part is a spherical harmonic YLm, with integer L and parity (−1)L. The ledger for a decay A → B + C therefore couples three things: sB ⊗ sC gives the possible channel spins S, then S ⊗ L must contain JA. For a strong or electromagnetic decay parity is conserved too, PA = PB PC (−1)L, which usually pins L to one value. ρ⁰ (Jᴾ = 1⁻) → π⁺π⁻: S = 0, so L = 1, and (−1)(−1)(−1)¹ = −1 matches. Δ⁺⁺ (3/2⁺) → pπ⁺: S = ½, and 3/2 ∈ ½ ⊗ L needs L = 1 or 2, while parity (+1)(−1)(−1)L = +1 needs L odd, so the decay is a pure p-wave. η (0⁻) → π⁺π⁻ would need S = 0, L = 0 and parity +1 against the η's −1: forbidden, and it is not seen, the η going to 3π and γγ instead. Identical bosons add an exchange rule: π⁰π⁰ must be symmetric, so L is even, and ρ⁰ → π⁰π⁰ is forbidden at L = 1 even though the ρ⁰ → π⁺π⁻ ledger balances. Weak decays drop the parity line but keep every angular-momentum line.
5Integer L means fermion number changes by an even amount
Orbital angular momentum is integer, so it can never turn a half-integer total into an integer one. Whether each side of a reaction carries a half-integer J is therefore set by its count of half-integer spins alone — odd gives half-integer, even gives integer — and rotational invariance demands the two sides agree. This is the argument that made the neutrino necessary. A neutron (½) decaying to a proton (½) and an electron (½) would take a half-integer J to an integer one, since ½ ⊗ ½ ⊗ L is integer for every L: the reaction is closed by angular momentum, not only by the continuous energy spectrum. Add a third spin-½ product and ½ ⊗ ½ ⊗ ½ ⊗ L reaches ½ and 3/2 already at L = 0, and the ledger closes: n → p + e⁻ + ν̄ₑ. The same count runs through π⁻ → μ⁻ ν̄μ (0 fermions to 2), K⁺ → π⁺ π⁰ (0 to 0) and Λ → p π⁻ (1 to 1), where the Λ's spin ½ reaches ½ ⊗ 0 ⊗ L with L = 0 or 1, and the weak interaction, ignoring parity, uses both — the interference of those two waves is how the Λ came to reveal parity violation.
6Photon selection rules come from the same coefficients
An electric-dipole photon carries one unit of angular momentum, so the transition operator is a rank-1 tensor and the Wigner–Eckart theorem writes ⟨J′ M′|T̂¹q|J M⟩ = ⟨J M 1 q|J′ M′⟩ × ⟨J′‖T̂¹‖J⟩. The Clebsch–Gordan factor vanishes unless J′ ∈ J ⊗ 1, so ΔJ = 0, ±1 with J = 0 → J′ = 0 excluded because 0 ⊗ 1 contains no 0, and M′ = M + q with q = 0, ±1. The dipole operator also has odd parity, so parity must flip: Δℓ = ±1 for a single electron. Everything about the rate that depends on orientation sits in the coefficient; everything about the atom sits in the reduced matrix element, one number per pair of levels. Hydrogen 2p → 1s satisfies every line and lives 1.6 ns; 2s → 1s fails Δℓ = ±1 and the parity flip, and the 2s state survives 0.12 s until two-photon emission, a second-order process, takes it. That factor of 7.5 × 10⁷ in lifetime is what the dipole selection rule looks like in the laboratory: not an absolute prohibition, only the loss of the leading term.
Key equations
Read the units, assumptions and limits before substituting.
Rotation generator and its algebra
Û(n̂, φ) = exp(−i φ n̂⋅Ĵ/ħ), [Ĵᵢ, Ĵⱼ] = iħ εᵢⱼₖ Ĵₖ
The commutator alone fixes the spectrum j(j+1)ħ², the ladder of m, and every coupling rule that follows.
φ in radians about the unit axis n̂; Ĵ in J s; εᵢⱼₖ the Levi-Civita symbolConservation from rotational invariance
[Ĥ, Ĵᵢ] = 0 ⇒ d⟨Ĵᵢ⟩/dt = (i/ħ)⟨[Ĥ, Ĵᵢ]⟩ = 0
J and M are good labels before and after the reaction, however violent the interaction in between.
Ĥ the full Hamiltonian in J, every interaction included; Ĵ the total for the whole systemEigenvalues of Ĵ² and Ĵz
Ĵ²|j m⟩ = j(j+1)ħ²|j m⟩, Ĵz|j m⟩ = mħ|j m⟩, m = −j, …, j
The length is √(j(j+1))ħ, never jħ — the leftover Ĵₓ and Ĵy are what non-commutation costs a state.
j = 0, ½, 1, … for spin; orbital ℓ is integer only; 2j+1 values of mTriangle rule for coupling
j₁ ⊗ j₂ = |j₁ − j₂| ⊕ (|j₁ − j₂| + 1) ⊕ … ⊕ (j₁ + j₂)
Every reachable total of a pair before any dynamics — the first line of a reaction ledger.
Dimension check: Σ(2J+1) over the range equals (2j₁+1)(2j₂+1)Clebsch–Gordan expansion
|J M⟩ = Σ ⟨j₁ m₁ j₂ m₂ | J M⟩ |j₁ m₁⟩|j₂ m₂⟩, m₁ + m₂ = M
|⟨j₁ m₁ j₂ m₂|J M⟩|² is the probability that a product state is found with total J.
Real in the Condon–Shortley convention; the matrix is orthogonal, so the inverse uses the same numbersTwo-body selection rule
JA ∈ S ⊗ L with S ∈ sB ⊗ sCPA = PB PC (−1)L
Some (S, L) must reach the parent's J with its parity, or the decay is forbidden whatever its Q value.
L the orbital angular momentum of relative motion, in units of ħ; P intrinsic parities ±1; parity line for strong and EM onlyInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. With j₁ = 1 and j₂ = ½, slide J from 3/2 down to 1: the triangle still closes but the ladder mismatch reads 0.5, so J = 1 falls between rungs; at J = ½ the dot lights again with θ₁₂ ≈ 145°. Push J to 2 and the arc runs past every tip — cos θ₁₂ would have to be 1.33.
Live results
- COS θ₁₂ (RAW)
- 0.41
- ⟨Ĵ₁⋅Ĵ₂⟩
- 0.50 ħ²
- TRIANGLE GAP
- 0.00 ħ
- LADDER MISMATCH
- 0.0
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. First angular momentum j₁: 1.0; Second angular momentum j₂: 0.5; Trial total J: 1.5
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The error is to add angular momenta as numbers — to write ρ⁰ (J = 1) → π⁺π⁻ as 1 → 0 + 0 and call it forbidden, or to say that two spin-½ particles 'have spin 1'. Ĵ is a vector operator whose components do not commute, so a product state |j₁ m₁⟩|j₂ m₂⟩ is in general not an eigenstate of Ĵ² at all: it is a superposition over every J from |j₁ − j₂| to j₁ + j₂, weighted by squared Clebsch–Gordan coefficients. And the spins are never the whole sum — two products separating carry an orbital L of relative motion that can supply any integer, which is exactly how the ρ's one unit reaches two spinless pions. The one thing the arithmetic gets right, it gets right for the right reason: L is integer, so it cannot repair a mismatch between half-integer and integer totals, and that mismatch, not the spin sum, is what forbids n → p + e⁻.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Two spin-½ particles are prepared as |↑⟩|↓⟩, the first up and the second down along z. List the allowed total S, check the dimension count, and find the probability that a measurement of Ŝ² returns S = 1.
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.
- Triangle rule: S runs from |½ − ½| = 0 to ½ + ½ = 1 in integer steps, so S = 0 or S = 1. Dimension check: (2⋅½+1)² = 4 = 3 + 1, a triplet and a singlet.
- Top state: |1, 1⟩ = |↑⟩|↑⟩, the only state with M = 1. Lower it with Ŝ₋ = Ŝ₁₋ + Ŝ₂₋: Ŝ₋|1,1⟩ = √2 ħ|1,0⟩, while (Ŝ₁₋ + Ŝ₂₋)|↑↑⟩ = ħ(|↓↑⟩ + |↑↓⟩), so |1,0⟩ = (|↑↓⟩ + |↓↑⟩)/√2.
- The M = 0 state orthogonal to it starts the next multiplet: |0,0⟩ = (|↑↓⟩ − |↓↑⟩)/√2. Invert: |↑↓⟩ = (|1,0⟩ + |0,0⟩)/√2, so ⟨1 0|↑↓⟩ = 1/√2.
- Probability of S = 1 is |1/√2|² = ½, and of S = 0 also ½. The product state is not an eigenstate of Ŝ²: ⟨Ŝ²⟩ = ½ × 2ħ² + ½ × 0 = ħ², which equals S(S+1)ħ² for neither allowed S.
Answer. S = 0 or 1 (4 = 3 + 1); P(S = 1) = ½ and P(S = 0) = ½ — |↑↓⟩ is an equal superposition of triplet and singlet.
MediumExample 2
Audit two strong decays with the full ledger — channel spin S, orbital L, and parity PA = PB PC (−1)L. (a) Δ⁺⁺ (Jᴾ = 3/2⁺) → p (½⁺) + π⁺ (0⁻). (b) ρ⁰ (1⁻) → π⁰ + π⁰. Which L does each need, and is each allowed?
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.
- (a) Channel spin: ½ ⊗ 0 = ½ only. Then 3/2 ∈ ½ ⊗ L requires |½ − L| ≤ 3/2 ≤ ½ + L, so L = 1 (giving ½ and 3/2) or L = 2 (giving 3/2 and 5/2).
- (a) Parity: Pₚ Pπ (−1)L = (+1)(−1)(−1)L must equal +1, so (−1)L = −1 and L is odd. Only L = 1 survives: the Δ decays in a pure p-wave, which is what its measured angular distribution shows.
- (b) Channel spin S = 0, so J = 1 forces L = 1. Parity: (−1)(−1)(−1)¹ = −1 matches the ρ. Angular momentum and parity both pass — the same audit that allows ρ⁰ → π⁺π⁻.
- (b) But the two π⁰ are identical bosons, so the state must be symmetric under exchange, which for spinless particles means (−1)L = +1 and L even. L = 1 is odd: the only L that reaches J = 1 is the one Bose symmetry forbids.
- Verdict: Δ⁺⁺ → pπ⁺ is allowed with L = 1; ρ⁰ → π⁰π⁰ is forbidden. Neither audit says anything about the Δ's 117 MeV width or the ρ's 150 MeV — that needs the matrix element.
Answer. (a) Allowed, L = 1 only (p-wave). (b) Forbidden: J = 1 needs L = 1, but identical π⁰ demand even L — and ρ⁰ → π⁰π⁰ is not observed.
HardExample 3
A spin-1 beam meets a spin-½ target, and the reaction proceeds only in an s-wave (L = 0) through a J = 3/2 resonance whose reduced amplitude is the same for every M. Using the 1 ⊗ ½ coefficients, find the fraction of beam–target pairs in J = 3/2 when (a) both are unpolarised, (b) the beam is pure m₁ = +1 and the target unpolarised, (c) both are polarised, m₁ = +1 and m₂ = +½. What does each buy in rate?
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.
- Coupled multiplets: 1 ⊗ ½ = 3/2 ⊕ ½, dimensions 6 = 4 + 2. From the lesson, |3/2, ½⟩ = √(2/3)|1,0⟩|½,½⟩ + √(1/3)|1,1⟩|½,−½⟩ and |½, ½⟩ = √(2/3)|1,1⟩|½,−½⟩ − √(1/3)|1,0⟩|½,½⟩.
- (a) Unpolarised: all six product states are equally likely, and the J = 3/2 subspace is 4 of the 6 dimensions, so the fraction is 4/6 = 2/3.
- (b) Beam m₁ = +1, target half +½ and half −½. With m₂ = +½, M = 3/2 belongs to J = 3/2 alone: probability 1. With m₂ = −½, |1,1⟩|½,−½⟩ = √(1/3)|3/2,½⟩ + √(2/3)|½,½⟩: probability 1/3. Average ½(1 + 1/3) = 2/3.
- Polarising the beam alone changed nothing. That is the sum rule Σ over m₂ of |⟨1 m₁ ½ m₂|3/2 M⟩|² = (2J+1)/(2j₁+1) = 4/3 for every m₁; averaged over the two target states it is 2/3 whatever the beam does.
- (c) m₁ = +1 and m₂ = +½ give M = 3/2, which only J = 3/2 can carry: fraction 1. With an M-independent reduced amplitude the resonant rate rises by 1 ÷ 2/3 = 1.5 over the unpolarised case; flipping the target to m₂ = −½ would instead cut it to 1/3 ÷ 2/3 = ½.
Answer. (a) 2/3, (b) 2/3, (c) 1. Polarising one partner buys nothing; polarising both gives a 50% gain in the J = 3/2 rate, or a halving with the target reversed.
Quick check
Select an answer and explain it before checking.
Review & continue
Angular momentum is conserved in a reaction because the total Ĵ — every spin plus every orbital L of relative motion — generates rotations and commutes with Ĥ, so the labels J and M of the initial state must be reachable by the final state.
- 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.