University Physics V · Atomic Spectroscopy · 13.4
Selection Rules: Parity & Wigner-Eckart
Most pairs of levels in an atom never make a line, and this is where you learn to say which before measuring anything. Two symmetry arguments do it: inversion kills every pair that keeps its parity, and Wigner-Eckart reduces whatever survives to one number times a table of 3-j symbols that also gives the relative strengths.
Build the model
Connect the measurement to the mechanism.
The rate of an electric-dipole line is set by one number, |⟨f|d⋅ε̂|i⟩|², and two symmetries decide when that number must vanish identically. Inversion is the blunt one: the dipole d = −e Σᵢ rᵢ is odd, so sandwiching it between two states of the same parity forces the element to equal minus itself, hence zero — the Laporte rule, which for one jumping electron reads Δl = ±1. Rotation is the sharp one.
The operator d is a rank-1 spherical tensor, and the Wigner-Eckart theorem says every one of its 3(2J+1)(2J′+1) matrix elements is the same reduced element ⟨γ′J′‖d‖γJ⟩ times a Clebsch-Gordan coefficient that depends on nothing but angular-momentum bookkeeping. That single statement does two jobs: the coefficient vanishes unless |J−1| ≤ J′ ≤ J+1 and m′ = m + q, which is the selection rule, and where it does not vanish its square is the relative strength of every Zeeman and fine-structure component — obtained without a single radial integral. The cost is honesty about scope.
These are theorems about the symmetry of the operator, not about the atom: ΔS = 0 and ΔL = 0, ±1 hold only while the eigenstates are genuinely LS states, and forbidden never means impossible, only that the leading term of e(ik⋅r) = 1 + ik⋅r + … gave zero and the next term takes over, suppressed by (ka₀)² ≈ 10⁻⁵ rather than switched off.
- Simple definition
- A selection rule is the statement that a transition's matrix element vanishes identically by symmetry — parity under inversion, the Wigner-Eckart 3-j factor under rotation — so the line is absent whatever the radial wavefunctions happen to be.
- Example
- Hydrogen 2p → 1s has Δl = −1 and opposite parity, so it runs at A = 6.27 × 10⁸ s⁻¹, a lifetime of 1.60 ns; 2s → 1s has Δl = 0 with both states even, its dipole element is exactly zero, and the level survives 0.12 s until two photons carry it down.
Collapses 3(2J+1)(2J′+1) matrix elements of a vector operator onto one number times a table of angles.
C is a Clebsch-Gordan coefficient, pure geometry; the reduced element carries T's units (C m for d) and all the radial physics.
q = 0 is π light polarised along B, q = ±1 is σ± circular in the plane normal to it — so ΔmJ = q is a statement about your polariser.
q labels the photon's angular-momentum projection on the quantisation axis; x, y, z in metres and e = 1.602 × 10⁻¹⁹ C.
One photon carries one unit of angular momentum: it can reorient J but cannot create it from nothing, which is why 0 → 0 has no coupling at all.
The 3-j (J′ 1 J; −m′ q m) needs |J−1| ≤ J′ ≤ J+1 and −m′ + q + m = 0, or it is identically zero.
A test Wigner-Eckart cannot see. It kills Δl = 0, so s → s, p → p and 2s → 1s stay dark however favourable the ΔJ arithmetic looks.
Non-zero only if πᵢ πf = −1. Atomic parity is π = (−1)(Σ lᵢ), so one jumping electron must have Δl = ±1.
The dipole touches no spin coordinate, so spin factors out — until spin-orbit, growing as Z⁴, spoils the labels and leaks intercombination lines.
Exact only for pure LS eigenstates. A mixed level |³P₁⟩ + ε|¹P₁⟩ radiates to ¹S₀ at ε² times the allowed rate.
Forbidden means the leading term vanished, not that the line cannot occur — and a rate 10⁻⁵ down is exactly what makes nebular lines diagnostic.
k = 2π/λ and a₀ = 5.29 × 10⁻¹¹ m; for Lyman-α, k a₀ = 2.7 × 10⁻³ and the suppression is 7.5 × 10⁻⁶.
A selection rule is a matrix element that must vanish
Fermi's golden rule gives the electric-dipole rate as Γ ∝ |⟨f|d⋅ε̂|i⟩|² times a density of final states, so a forbidden transition is one whose integral is not small but identically zero, for every radial function you could put in it. Write the operator down and read the rules off it: d = −e Σᵢ rᵢ is a one-body, spin-free, parity-odd, rank-1 object. Each of those four words is a symmetry statement, and each closes a different set of lines. One-body forbids two electrons jumping at once. Spin-free gives ΔS = 0. Parity-odd gives the Laporte rule. Rank-1 gives the ΔJ and ΔmJ rules through Wigner-Eckart. Nothing about Z, screening or the radial overlap appears anywhere in that list; those quantities set how bright an allowed line is, never whether it exists. Inversion and rotation are logically independent symmetries, so a line must clear both gates, and most level pairs in an atom fail at least one.
Parity: a one-line argument that kills half the pairs
Let Π be the inversion operator, Π² = 1, and let the atomic states be parity eigenstates, Π|i⟩ = πᵢ|i⟩ with π = ±1 — they always are, because the Coulomb Hamiltonian commutes with Π. Insert Π†Π = 1 on both sides of the dipole: ⟨f|d|i⟩ = ⟨f|Π†(Π d Π†)Π|i⟩ = πf πᵢ(−1)⟨f|d|i⟩, since d is odd. Either πf πᵢ = −1 or the element is zero. There is no third case and no small residue. For an atom π = (−1)(Σ lᵢ), so a single active electron must change l by an odd amount, and the rank-1 triangle |l−1| ≤ l′ ≤ l+1 then narrows that to Δl = ±1 exactly. The consequence is visible in the hydrogen lifetimes: 2p → 1s has Δl = −1 and decays in 1.60 ns, while 2s → 1s has Δl = 0 with both states even, its dipole element is zero to all orders in the radial integral, and the level survives 0.12 s — a factor of 7.6 × 10⁷ — before two-photon emission takes it down. Parity is the reason the metastable level exists at all.
Wigner-Eckart: geometry out, dynamics into one number
Under rotation the three spherical components dq mix among themselves exactly as the l = 1 spherical harmonics do; that is what rank 1 means. The theorem then reads ⟨γ′J′m′|dq|γJm⟩ = ⟨J m; 1 q | J′ m′⟩ ⟨γ′J′‖d‖γJ⟩/√(2J′+1). Read the two factors separately. The reduced element holds every dynamical detail — radial integrals, screening, configuration mixing — and carries no magnetic quantum number at all. The Clebsch-Gordan coefficient (equivalently a 3-j symbol) is pure geometry, and it vanishes unless the triangle |J−1| ≤ J′ ≤ J+1 holds and m′ = m + q. That is precisely where ΔJ = 0, ±1 and ΔmJ = 0, ±1 come from, and it is where 0 → 0 dies: coupling J = 0 to k = 1 can only produce J′ = 1, the triangle has no solution, and there is nothing left to reduce. Note what is not implied. ΔJ = 0 is perfectly allowed once J ≥ ½ — sodium's D1 line is a J = ½ → ½ transition — while Δl = 0 is never allowed, because that is parity's rule and not this one.
The m-dependence is geometry you can read straight off
Because the reduced element is m-independent, the entire Zeeman pattern of a line is fixed by 3-j symbols alone, before any atomic structure calculation. Two readings are worth memorising. For a ΔJ = 0 line the π branch goes as m², so the m = 0 → m′ = 0 component is exactly missing: a Zeeman fan with a hole in the middle is the fingerprint of J → J. For a ΔJ = +1 line at the stretched state m = J, the σ⁺ branch reaches the largest value any single branch can have, exactly 1, against 1/(J+1) for π and 1/[(2J+1)(J+1)] for σ⁻ — with J = 2 that is 1, ⅓ and 1/15. Drive σ⁺ light on that transition and the excited |J+1, J+1⟩ state has no lower sublevel to reach but |J, J⟩, so the cycle closes and you have the transition every magneto-optical trap runs on. Finally Σq |⟨J m; 1 q|J′ m+q⟩|² = (2J′+1)/(2J+1), independent of m: the sum rule that makes every sublevel of an upper level decay at the same rate, which is why a lifetime is a property of a level and not of a sublevel.
ΔS = 0 is only as good as your LS basis
The dipole acts on spatial coordinates only, so in the |L S J MJ⟩ basis the spin overlap factors out as δ_(S′S), giving ΔS = 0; running the same rank-1 argument on L gives ΔL = 0, ±1 with 0 → 0 barred. But those are statements about labels, and labels are good only while the Hamiltonian is close to LS-diagonal. Spin-orbit, whose one-body coefficient scales roughly as Z⁴/n³, mixes states of the same J and parity across different S, so the physical level is |6³P₁⟩ + ε|6¹P₁⟩ and only the ε piece reaches the ¹S₀ ground state. Wigner-Eckart makes the comparison exact: both lines share the same reduced element ⟨¹S₀‖d‖¹P₁⟩, so it cancels and the rate ratio is ε² times the ω³ factor. In mercury the 253.7 nm intercombination line runs at 8.0 × 10⁶ s⁻¹ against 7.6 × 10⁸ s⁻¹ for the allowed 184.9 nm line, which unwinds to ε ≈ 0.16. In helium the same admixture is of order 10⁻⁵ and 2³S₁ is metastable for about 7900 s. ΔS = 0 is not a law; it is a measurement of how good your basis is.
Forbidden means slow: the next term in e(ik⋅r)
The dipole approximation came from truncating e(ik⋅r) ≈ 1 inside the (e/m)A⋅p coupling. When that leading term's matrix element vanishes, the next order does the work: the antisymmetric part of ik⋅r gives the magnetic dipole M1, the symmetric part the electric quadrupole E2. Their rates are down by roughly (ka)², with a the size of the state. For hydrogenic transitions ħω ≈ α²mc² and a ≈ a₀ = ħ/(αmc), so ka₀ ≈ α/2 = 3.6 × 10⁻³; for Lyman-α exactly, k = 2π/121.6 nm gives ka₀ = 2.74 × 10⁻³ and (ka₀)² = 7.5 × 10⁻⁶. This is why forbidden lines are the astronomer's instrument. At nebular densities near 10² cm⁻³ collisional de-excitation is slower than a forbidden rate of order 10⁻² s⁻¹, so the [O III] and [N II] lines dominate the optical spectrum of an H II region although no laboratory discharge shows them; and hydrogen's 21 cm hyperfine line, an M1 transition with A = 2.9 × 10⁻¹⁵ s⁻¹ and a lifetime near 11 million years, maps the whole Galaxy.
Change one variable at a time
Make the relationship visible.
Set ΔJ = 0 with J = 0 and all three bars vanish: 0 → 0 has no coupling at all. Keep ΔJ = 0, raise J to 2 and slide m to 0 — now only the π bar dies, because that branch goes as m². Push ΔJ to 2 and everything drops to zero: the rank-1 triangle has no solution.
σ⁺ BRANCH Δm = +10.400
π BRANCH Δm = 00.600
σ⁻ BRANCH Δm = −10.400
SUM OVER q1.400
Live interpretationσ⁺ BRANCH Δm = +1: 0.400. π BRANCH Δm = 0: 0.600. σ⁻ BRANCH Δm = −1: 0.400. SUM OVER q: 1.400
Catch the common trap
Explain before calculating.
Hydrogen's 2s level lies 10.2 eV above 1s. Both have l = 0 and J = ½, and the electric-dipole rate between them is exactly zero. Which statement gives the reason?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyDecide which of these hydrogen transitions are electric-dipole allowed, and name the rule that closes each of the others: 3d → 2p, 3s → 2s, 3p → 2p, 3d → 1s. Then apply the ΔJ test to the surviving line's fine-structure components.
- Parity first. A one-electron state has π = (−1)l and d is odd, so the two levels must differ in parity: only an odd Δl can survive, and the rank-1 triangle |l−1| ≤ l′ ≤ l+1 narrows that to Δl = ±1.
- 3d → 2p: l goes 2 → 1, so Δl = −1 and the parities are even → odd. Allowed.
- 3s → 2s and 3p → 2p both have Δl = 0, so πᵢ πf = +1 and ⟨f|d|i⟩ = −⟨f|d|i⟩ = 0. Dark at E1, whatever the radial overlap.
- 3d → 1s has Δl = −2: parity is unchanged, and the triangle fails too since |2−1| ≤ 0 is false. Doubly barred, and it appears only as a weak E2 line.
- Now the ΔJ test inside 3d → 2p. The 3d term splits into ²D₅⁄₂ and ²D₃⁄₂, the 2p into ²P₃⁄₂ and ²P₁⁄₂. ΔJ = 0, ±1 kills ²D₅⁄₂ → ²P₁⁄₂, which needs ΔJ = −2; the other three pairings survive.
AnswerOnly 3d → 2p is E1-allowed. 3s → 2s and 3p → 2p keep their parity; 3d → 1s changes l by 2. Inside 3d → 2p, three of the four fine-structure pairings survive — ²D₅⁄₂ → ²P₁⁄₂ is barred by ΔJ = −2.
MediumSodium's D lines both end on 3s ²S₁⁄₂: D2 from 3p ²P₃⁄₂ at 589.0 nm, D1 from 3p ²P₁⁄₂ at 589.6 nm. The recoupling 6-j symbol (j 1 j′; 0 ½ 1) has the same magnitude 1/√6 for j = 3/2 and j = 1/2, and both lines share one radial integral. Find the ratio of line strengths and then of the two lifetimes.
- Line strength is the squared reduced element, S = (2j+1)(2j′+1)|6-j|²|⟨l‖C¹‖l′⟩|²R². With the 6-j magnitude and the radial part common to both, only the degeneracy factor differs.
- D2: (2⋅3/2+1)(2⋅1/2+1) = 4 × 2 = 8. D1: (2⋅1/2+1)(2⋅1/2+1) = 2 × 2 = 4. So S(D2) : S(D1) = 2 : 1 — the strength ratio is pure angular-momentum counting.
- The decay rate divides that strength among the upper sublevels: A = ω³S/(3πε₀ħc³ gᵤ) with gᵤ = 2Jᵤ + 1 = 4 for D2 and 2 for D1.
- A(D2)/A(D1) = (2/4) ÷ (1/2) × (ω₂/ω₁)³ = 1 × (589.6/589.0)³ = (1.001019)³ = 1.00306.
- So the predicted lifetime ratio is τ(D2)/τ(D1) = 1/1.00306 = 0.99695. Tabulated values are 16.25 ns and 16.30 ns, a ratio of 0.99693.
AnswerS(D2) : S(D1) = 2 : 1, yet A(D2)/A(D1) = 1.0031, so the lifetimes match to 0.3%: 16.25 ns against 16.30 ns. The doubled strength is shared among twice as many upper sublevels, so it never reaches the lifetime.
HardMercury's 253.7 nm line, 6 ³P₁ → 6 ¹S₀, is ΔS = 0 forbidden yet runs at A = 8.0 × 10⁶ s⁻¹, while the fully allowed 184.9 nm line 6 ¹P₁ → 6 ¹S₀ runs at A = 7.6 × 10⁸ s⁻¹. Model the upper level as |6³P₁⟩ + ε|6¹P₁⟩ and extract ε.
- In pure LS the dipole cannot connect S = 1 to S = 0: d carries no spin operator, so the spin overlap ⟨S′ = 0|S = 1⟩ = 0 factors the element to zero.
- Spin-orbit at Z = 80 mixes the two J = 1, odd-parity levels, so the physical state is |6³P₁⟩ + ε|6¹P₁⟩ (normalised to first order). Only the admixed singlet reaches ¹S₀: ⟨¹S₀|d|6³P₁⟩ = ε⟨¹S₀|d|6¹P₁⟩.
- Both lines therefore share the reduced element ⟨¹S₀‖d‖¹P₁⟩, which cancels in the ratio. With A ∝ ω³|⟨d⟩|², Aᵢₙₜ/Aₐₗₗ = ε²(λₐₗₗ/λᵢₙₜ)³.
- The wavelength factor: (184.9/253.7)³ = (0.72881)³ = 0.38712.
- The measured ratio: Aᵢₙₜ/Aₐₗₗ = 8.0 × 10⁶ / 7.6 × 10⁸ = 1.0526 × 10⁻².
- So ε² = 1.0526 × 10⁻² / 0.38712 = 2.719 × 10⁻², giving ε = 0.165. The corresponding lifetimes are 1/(8.0 × 10⁶) = 125 ns and 1/(7.6 × 10⁸) = 1.32 ns.
Answerε = 0.165, so the nominal triplet carries about 2.7% singlet character by probability — enough to turn a strictly forbidden line into the 125 ns, 253.7 nm transition that every mercury lamp radiates.