Selection rules: parity and Wigner-Eckart

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 closes a different set of lines. One-body forbids two electrons jumping at once, as far as a single configuration describes each level. Spin-free gives ΔS = 0. Parity-odd gives the Laporte rule. Rank-1 gives the ΔJ and Δm_J 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.

Let Π be the inversion operator, Π² = 1, and let the atomic states be parity eigenstates, Π|i⟩ = πᵢ|i⟩ with π = ±1 — they can always be chosen so, 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⟩ = π_f πᵢ(−1)⟨f|d|i⟩, since d is odd. Either π_f πᵢ = −1 or the element is zero. Within electromagnetism there is no third case and no small residue; the weak interaction's parity violation leaves an E1 amplitude near 10⁻¹¹ e a₀ between caesium's 6s and 7s, far below anything an emission spectrum shows. 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.

The factorisation is quoted here, not rederived: by UPV-11.09 and UPV-12.09, ⟨γ′J′m′|d_q|γJm⟩ is one reduced element times ⟨J m; 1 q | J′ m′⟩/√(2J′+1), and that coefficient vanishes unless ΔJ = 0, ±1 with 0 → 0 barred and m′ = m + q. What a spectroscopist adds is that q is not bookkeeping. It is the photon's angular-momentum projection on the field axis, so every Zeeman component leaves the atom in a definite polarisation. A π component, q = 0, radiates like a dipole oscillating along B: intensity ∝ sin²θ at angle θ from the field, polarised linearly along B. A σ± component, q = ±1, radiates like a dipole rotating in the plane normal to B: intensity ∝ (1 + cos²θ)/2, circular when seen along the field and linear, perpendicular to B, when seen across it. Two viewing directions therefore sort any Zeeman pattern. Across the field every component is present, and a linear polariser separates them, passing π when set along B and σ when set across it. Along the field the π components vanish, because an oscillating dipole sends nothing down its own axis, and σ⁺ and σ⁻ arrive with opposite circular handedness — which is how Hale read magnetic fields in sunspots in 1908, and how a solar magnetograph still reads the sign of the line-of-sight field.

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 Σ_q |⟨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.

Quote the LS rules in one line from UPV-12.09: the dipole acts on no spin coordinate, so between pure LS states ΔS = 0 and ΔL = 0, ±1 with 0 → 0 barred. On a term diagram this splits the singlet and triplet ladders apart, and the spectroscopist's question is how bright the lines that cross between them are. Spin-orbit mixes levels of equal J and parity, so to first order the nominal ³P₁ is |³P₁⟩ + ε|¹P₁⟩ with ε = ⟨¹P₁|H_SO|³P₁⟩/(E(³P₁) − E(¹P₁)), the spin-orbit matrix element over the singlet-triplet gap. The crossing line borrows the allowed line's reduced element, so Aᵢₙₜ/Aₐₗₗ = ε²(λₐₗₗ/λᵢₙₜ)³, and only J = 1 can borrow: ³P₀ → ¹S₀ is 0 → 0 and ³P₂ → ¹S₀ needs ΔJ = 2, so neither has an E1 route while J is a good quantum number. Down a column the numerator grows far faster than the gap — spin-orbit rises as Z⁴ in a hydrogenic ion and more slowly, but still steeply, for a screened valence electron — so the crossing line brightens by orders of magnitude. Strontium's 689 nm line runs at 4.7 × 10⁴ s⁻¹, about 4000 times slower than its 461 nm singlet line; mercury's 253.7 nm line runs at 8.0 × 10⁶ s⁻¹, only 95 times slower than its 184.9 nm line, and the Hard example turns that ratio into ε. A kilohertz-wide intercombination line is a tool rather than a nuisance: strontium's is what laser-cools the atoms below a microkelvin.

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. Compared at the same frequency, M1 is down by (μ_B/(e a₀ c))² = (α/2)², since μ_B/(e a₀) = αc/2 is half the Bohr-orbit speed, and E2 by roughly (ka)², with a the size of the state. For hydrogenic transitions ħω ≈ ½α²mc² and a ≈ a₀ = ħ/(αmc), so ka₀ ≈ α/2 = 3.6 × 10⁻³ and both multipoles pay a factor of order α²; 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.