University Physics V · Multi-Electron Atoms · 12.9

Selection Rules from Wigner–Eckart

Every dipole matrix element in an atom splits into two pieces: a rotational geometry factor anyone can look up, and one number that only a radial integral can supply. Separate them, and the selection rules stop being a list to memorise and become something you derive in a line.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

An atomic Hamiltonian is rotationally invariant, and that single fact does most of the work of spectroscopy. The dipole operator d = −e r is not a scalar, but it is not arbitrary either: its three components can be rearranged into a rank-one spherical tensor T¹_q, an object that transforms among itself under rotation exactly the way a j = 1 multiplet of kets does. Schur's lemma then forces the answer.

Every one of the (2j+1)(2j′+1)(2k+1) matrix elements of T^k_q between two multiplets is one and the same number — the reduced matrix element ⟨α′j′‖T^k‖αj⟩ — multiplied by a purely geometric coefficient fixed by the m labels alone. A selection rule is a coefficient that is exactly zero, which is why a forbidden line is absent rather than faint. The cost is severe and worth stating out loud: the theorem supplies ratios, never rates.

The reduced element still needs a radial integral you have to do, and rotational symmetry is blind to reflection and blind to spin — so Δℓ = ±1 and ΔS = 0 have to be argued separately, one from an exact discrete symmetry and one from a coupling scheme that heavy atoms break.

Simple definition: The Wigner–Eckart theorem says that every matrix element of a rank-k spherical tensor between angular-momentum multiplets equals one m-independent reduced matrix element times a 3j symbol; a selection rule is a 3j symbol that vanishes.
Example: For sodium's D2 line the upper ²P₃/₂ has four M values and the lower ²S₁/₂ has two, so eight amplitudes share one reduced element ⟨²S₁/₂‖d‖²P₃/₂⟩; six survive with ΔM = 0, ±1 and the two needing ΔM = ±2 are exactly zero.

Wigner–Eckart theorem⟨α′j′m′|T^k_q|αjm⟩ = (−1)^{(j′−m′)} ( j′ k j−m′ q m ) ⟨α′j′‖T^k‖αj⟩

Change m, m′ or q and only the 3j symbol moves. The reduced element is the same number throughout, so every relative line strength inside a multiplet is table lookup.

α collects every non-angular label. The reduced element carries the units of T^k; the 3j symbol is a pure number.

Dipole as a rank-one tensorT¹₀ = z, T¹_(±1) = ∓(x ± iy)/√2, d = −e r

Cartesian components mix under rotation; the spherical ones transform as a j = 1 multiplet, which is what the theorem needs.

x, y, z in metres, d in C m. The label q is angular momentum about the quantisation axis, not a Cartesian direction.

Triangle and projection conditions|j − 1| ≤ j′ ≤ j + 1 and m′ = m + q

( 0 1 0 ; 0 0 0 ) is identically zero, so no rank-one operator of any kind links two J = 0 levels.

Both conditions are read off the 3j symbol. For k = 1: ΔJ = 0, ±1 with J = 0 ↛ J′ = 0, and ΔM = 0, ±1.

Parity, a separate exact ruleP† r P = −r ⟹ ⟨f|r|i⟩ = 0 unless π_f πᵢ = −1

Rotational symmetry is blind to reflection, so this gate is invisible to Wigner–Eckart and must be imposed separately.

π = (−1)^Σℓ for an atom. For a jump of one electron it sharpens to Δℓ = ±1 exactly, never Δℓ = 0.

Projection theorem and the Landé factorg_J = 1 + [J(J+1) + S(S+1) − L(L+1)] / [2J(J+1)]

Wigner–Eckart inside one multiplet: ²P₃/₂ gives 4/3, ²P₁/₂ gives 2/3, ²S₁/₂ gives 2 — the spacings of the D-line fans.

Dimensionless. Shift ΔE = g_J M_J μ_B B, with μ_B/h = 13.996 GHz T⁻¹ and μ_B = 5.788 × 10⁻⁵ eV T⁻¹.

Why higher multipoles are slowA(E2)/A(E1) ~ (a₀/λ)² ≈ 5 × 10⁻⁹ at λ = 729 nm

Ca⁺ 3d ²D₅/₂ can reach the ground state only by E2 and lives 1.17 s, against roughly 7 ns for an allowed E1 line.

a₀ = 5.29 × 10⁻¹¹ m. The expansion parameter is the atom's size divided by the wavelength of the light.

Put the operator in a basis that rotations respect

Cartesian components are a fine basis for a vector and a bad basis for this argument: under a rotation about z, x and y mix into each other and neither is an eigenvector of anything useful. Combine them instead into T¹₊₁ = −(x + iy)/√2, T¹₀ = z and T¹₋₁ = +(x − iy)/√2. These obey [J_z, T¹_q] = qħ T¹_q and [J_±, T¹_q] = ħ√(2 − q(q ± 1)) T¹_(q±1) — the same two commutators that define how a j = 1 multiplet of kets behaves under the ladder operators. That pair of relations is the definition of a rank-one spherical tensor, and nothing has been approximated to get there; it is a change of basis inside the three-dimensional space of operators. The payoff is immediate: T¹_q|j m⟩ can now be handled exactly like a product of two angular momenta, |1 q⟩ ⊗ |j m⟩, so the whole Clebsch–Gordan apparatus you built for adding spins applies to an operator acting on a state.

One number for an entire block of matrix elements

Decompose that product. Coupling 1 to j gives j − 1, j and j + 1, and nothing else. Project onto ⟨α′j′m′| and only one term of the decomposition survives, weighted by a Clebsch–Gordan coefficient depending on m, m′ and q, times an overall scale that cannot depend on any of them — the reduced matrix element ⟨α′j′‖T^k‖αj⟩. Written with a 3j symbol for symmetry, ⟨α′j′m′|T^k_q|αjm⟩ = (−1)^{(j′−m′)} ( j′ k j ; −m′ q m ) ⟨α′j′‖T^k‖αj⟩. Count what that saves for sodium's D2 line: 4 upper M values × 2 lower M values × 3 values of q is 24 amplitudes, and all 24 follow from one radial integral of order e a₀. The theorem hands you every ratio inside the multiplet and not one absolute rate. That division of labour is the thing to remember — geometry from symmetry, magnitude from dynamics — and it is also why the theorem can never tell you whether a line is bright.

Read ΔJ and ΔM straight off the 3j symbol

A 3j symbol vanishes unless two conditions hold. Its bottom row must sum to zero, giving m′ = m + q; since q ∈ (−1, 0, +1) for a rank-one tensor, that is ΔM = 0, ±1, with q fixed by the light itself — q = 0 is π, linearly polarised along the quantisation axis, and q = ±1 is σ±, circular about it. Its top row must satisfy the triangle inequality |j − k| ≤ j′ ≤ j + k, giving ΔJ = 0, ±1. Both halves of that inequality earn their keep. The upper half is the familiar one; the lower half, 1 ≤ j + j′, is what forbids J = 0 → J′ = 0, because no triangle has sides 0, 1 and 0. Notice what is not forbidden: ΔJ = 0 is perfectly allowed away from 0 → 0. One further exact zero hides inside ΔJ = 0, though — ( j 1 j ; 0 0 0 ) = 0 — so M = 0 → M′ = 0 never appears on a J → J line.

Parity is a second, independent gate

Rotational symmetry is blind to reflection, so Wigner–Eckart cannot see parity at all; it has to be imposed by hand. Under P the position operator flips, P†rP = −r, so for parity eigenstates ⟨f|r|i⟩ = −π_f πᵢ ⟨f|r|i⟩. The element equals minus itself, hence zero, unless π_f πᵢ = −1. That is Laporte's rule, spotted in spectra before quantum mechanics could explain it. With π = (−1)^Σℓ for an atom, a one-electron jump sharpens it to Δℓ = ±1 exactly. The two gates are genuinely independent, and each catches what the other lets through. Sodium's 3s ²S₁/₂ → 4s ²S₁/₂ passes the triangle test (1/2, 1, 1/2 is a legal triangle) and passes ΔM, yet is dead, because s → s does not change parity. Mercury's 6s6p ³P₀ → 6s² ¹S₀ passes parity, odd to even, and is dead because 0, 1, 0 is not a triangle.

The spin rules are approximate, and heavy atoms show it

The dipole operator acts on spatial coordinates only: it is d ⊗ 1 in the spin factor. In an LS-coupled basis |L S J M⟩ the spin overlap factors straight out as δ_(S′S), giving ΔS = 0, and applying Wigner–Eckart inside the orbital space gives ΔL = 0, ±1 with L = 0 ↛ L′ = 0. Both rest on LS coupling being a good description, which in turn needs the spin–orbit term ξ(r)L⋅S to stay small next to the residual electrostatic splitting. ξ grows roughly as Z⁴, so in a heavy atom the labels leak: a nominal ³P₁ carries a small ¹P₁ admixture and borrows strength through it. Mercury, Z = 80, is the standard case — 6s6p ³P₁ → 6s² ¹S₀ at 253.7 nm has A ≈ 8 × 10⁶ s⁻¹, against A ≈ 8 × 10⁸ s⁻¹ for the fully allowed ¹P₁ line at 184.9 nm. Suppressed a hundredfold, not absent. Put the same pair of levels in helium, Z = 2: the 1s2p ³P₁ → 1s² ¹S₀ branch runs at A ≈ 2 × 10² s⁻¹ against A ≈ 1.8 × 10⁹ s⁻¹ for the allowed 1s2p ¹P₁ → 1s² ¹S₀ at 58.4 nm, a suppression of 10⁷ rather than 10². Take that comparison from ³P₁ and not from helium's famous 7900 s metastable 2 ³S₁, tempting though it is: 1s2s → 1s² is s → s, so parity blocks E1 there whatever the spin character of the state, and the lifetime measures both gates at once and neither one on its own.

The projection theorem spaces a Zeeman fan

Specialise to k = 1 with j′ = j: a vector operator acting inside one multiplet. Both V and J are rank one on the same space, so their reduced elements differ only by a number, and ⟨jm′|V|jm⟩ = (⟨J⋅V⟩/[ħ²j(j+1)]) ⟨jm′|J|jm⟩. Apply that to μ = −μ_B(L + 2S)/ħ and the number that falls out is the Landé factor g_J = 1 + [J(J+1) + S(S+1) − L(L+1)]/[2J(J+1)]. For ²P₃/₂, 1 + (3.75 + 0.75 − 2)/7.50 = 4/3; for ²P₁/₂, 1 + (0.75 + 0.75 − 2)/1.50 = 2/3; for ²S₁/₂, 2. A weak-field component then shifts by (gᵤ Mᵤ − gₗ Mₗ) μ_B B, so D2 opens into six lines at ±1/3, ±1 and ±5/3 in units of μ_B B, while D1 opens into four at ±2/3 and ±4/3. At 0.50 T one unit is μ_B B/h = 6.998 GHz, so D2 spans 23.3 GHz — safely inside the weak-field regime, since sodium's 3p fine-structure interval is 515 GHz. How many lines there are is the ΔM rule; where they sit is the projection theorem.

Change one variable at a time

Make the relationship visible.

Interactive model

Magnetic field B0.50 T

View angle θ from B90 °

Set theta to 90 degrees and the two pi sticks stand tallest, 4 to 3 over the inner sigma pair; drag theta to 0, looking straight down the field, and they vanish entirely, because a photon travelling along B cannot be polarised along B. Raising B stretches the pattern without ever reordering it.

π PAIR SEPARATION4.67 GHz

OUTER σ SHIFT11.66 GHz

π / INNER σ HEIGHT1.33

OUTER σ IN WAVELENGTH13.5 pm

Live interpretationπ PAIR SEPARATION: 4.67 GHz. OUTER σ SHIFT: 11.66 GHz. π / INNER σ HEIGHT: 1.33. OUTER σ IN WAVELENGTH: 13.5 pm

Catch the common trap

Explain before calculating.

Misconception check

The error is to credit the Wigner–Eckart theorem with the whole list. It is a statement about rotations and nothing else, so it yields ΔJ = 0, ±1 with 0 ↛ 0, and ΔM = 0, ±1, and stops there. Δℓ = ±1 comes from parity, a separate exact symmetry; ΔS = 0 and ΔL = 0, ±1 come from the dipole being blind to spin in an LS-coupled basis, which is only as good as LS coupling itself. That difference has consequences you can measure. A vanishing 3j symbol is an exact zero that no perturbation of the atom can leak, so a J = 0 → J = 0 line is absent at any sensitivity. An approximate rule leaks in proportion to how badly its assumption fails, which is why mercury's intercombination line at 253.7 nm is merely a hundred times weaker than the allowed one rather than missing. "Forbidden" names two different situations, and only one of them survives a better spectrometer.

Mercury's 6s6p ³P₀ level lies 4.67 eV above the 6s² ¹S₀ ground state. No electric-dipole line connects them at any intensity, however sensitive the detector. Which statement gives the reason?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySodium's D2 line is 3p ²P₃/₂ → 3s ²S₁/₂. Taking both levels as full M multiplets, count how many upper-to-lower M pairs carry a non-zero electric-dipole amplitude, and name the rule that kills the rest.

Upper ²P₃/₂ has Mᵤ = +3/2, +1/2, −1/2, −3/2; lower ²S₁/₂ has Mₗ = +1/2, −1/2. That is 4 × 2 = 8 pairs to test.
The 3j symbol ( 1/2 1 3/2 ; −Mₗ q Mᵤ ) is non-zero only when its bottom row sums to zero, so q = Mₗ − Mᵤ with q restricted to −1, 0, +1. The amplitude therefore survives only for ΔM = Mᵤ − Mₗ = 0, ±1.
Test the pairs. Mᵤ = +3/2 → Mₗ = −1/2 would need ΔM = +2, and Mᵤ = −3/2 → Mₗ = +1/2 would need ΔM = −2. Both lie outside the allowed set, so both 3j symbols are exactly zero.
Every other pair works: 3/2→1/2, 1/2→1/2, 1/2→−1/2, −1/2→1/2, −1/2→−1/2 and −3/2→−1/2.
Check the other gate separately: the triangle condition |3/2 − 1/2| = 1 ≤ 1 ≤ 2 is satisfied for every pair, so ΔJ removes nothing here, and parity is fine because p → s changes ℓ by one.

AnswerSix of the eight pairs survive. The two needing ΔM = ±2 vanish exactly, because a rank-one tensor shifts M by at most one unit; the triangle condition removes nothing here.

MediumA cadmium ¹P₁ → ¹S₀ line is observed in a field of 0.80 T. Both terms are singlets, so the upper level has g = 1 and the lower has J = 0 and does not split. Find how many Zeeman components appear, where they sit in frequency, and what their squared 3j symbols are — then say what changes when the light is collected along the field instead of across it.

Upper Mᵤ = +1, 0, −1; lower Mₗ = 0 only. All three pairs have ΔM = 0, ±1, so all three survive — the normal Zeeman triplet.
Shifts: ΔE = (gᵤ Mᵤ − gₗ Mₗ) μ_B B = Mᵤ μ_B B. With μ_B/h = 13.996 GHz T⁻¹ and B = 0.80 T, μ_B B/h = 11.20 GHz, so the components sit at −11.20, 0 and +11.20 GHz.
Strengths: ( 0 1 1 ; 0 q m ) has magnitude 1/√3 for every allowed q, so all three squared 3j symbols equal 1/3, and they sum to 1. The three components are intrinsically equal in strength — geometry alone, no dynamics.
Equal strength is not what a detector records, because the emission pattern depends on q. The ΔM = 0 component is π and radiates as sin²θ from the field axis; the ΔM = ±1 components are σ and radiate as (1 + cos²θ)/2.
Across the field, θ = 90°, the ratio is π : σ : σ = 1 : 0.5 : 0.5 — the familiar triplet with a strong central line. Along the field, θ = 0°, the π factor is sin²0 = 0, so the central line disappears and only two oppositely circularly polarised σ lines remain.

AnswerThree components at 0 and ±11.20 GHz, with equal squared 3j symbols of 1/3. Across the field they appear as 1 : 0.5 : 0.5; along it the central π line vanishes and two oppositely circular σ lines remain.

HardThe sodium D2 line, 3p ²P₃/₂ → 3s ²S₁/₂ at 589.0 nm, is observed in a field of 0.50 T. Use the projection theorem to get the Landé factors, then find every Zeeman component's shift, the total span of the pattern in frequency, and the resolving power a spectrograph needs to separate the extreme components.

Landé factors from g_J = 1 + [J(J+1) + S(S+1) − L(L+1)]/[2J(J+1)]. Upper ²P₃/₂ (L = 1, S = 1/2, J = 3/2): 1 + (3.75 + 0.75 − 2)/7.50 = 1 + 1/3 = 4/3. Lower ²S₁/₂ (L = 0, S = 1/2, J = 1/2): 1 + (0.75 + 0.75 − 0)/1.50 = 2.
ΔM = 0, ±1 leaves the same six M-pairs as before, and each shifts by (gᵤ Mᵤ − gₗ Mₗ) μ_B B.
In units of μ_B B: (3/2→1/2) gives 2 − 1 = +1; (1/2→1/2) gives 2/3 − 1 = −1/3; (1/2→−1/2) gives 2/3 + 1 = +5/3; and the mirror set −1, +1/3, −5/3. The two ΔM = 0 lines are the inner pair at ∓1/3 and are π; the four at ±1 and ±5/3 have ΔM = ±1 and are σ.
Scale it: μ_B/h = 13.996 GHz T⁻¹, so at 0.50 T one unit is 6.998 GHz. The six components sit at ±2.33, ±7.00 and ±11.66 GHz.
Span between the extreme components: 2 × 11.66 = 23.33 GHz. The zero-field line is at ν₀ = c/λ = 2.998 × 10⁸ / (589.0 × 10⁻⁹) = 5.090 × 10¹⁴ Hz.
Δλ/λ = Δν/ν, so the span in wavelength is 589.0 nm × (23.33 × 10⁹ / 5.090 × 10¹⁴) = 589.0 × 4.58 × 10⁻⁵ = 0.0270 nm, and the required resolving power is λ/Δλ = 589.0/0.0270 ≈ 2.2 × 10⁴.

AnswerSix components at ±2.33, ±7.00 and ±11.66 GHz; a span of 23.3 GHz, or 0.027 nm at 589.0 nm, needing a resolving power near 2.2 × 10⁴ — beyond a small grating, easy for a Fabry–Pérot etalon.