Gamma decay, multipoles and isomers

The coupling is the nuclear current against the quantised field, H′ = −(1/c)∫ j⋅A d³r, and the golden rule needs a final photon state. Expanding A in plane waves hides angular momentum; expanding it in vector spherical harmonics diagonalises J² and parity instead, so every mode is stamped with a total angular momentum L, a projection μ, and a parity: electric EL with (−1)^L, magnetic ML with (−1)^(L+1). Two structural facts fall out. There is no L = 0 photon — the field is transverse and spin-1, and the L = 0 electric moment is the total charge, which cannot change. And in the long-wavelength regime kR ≪ 1, where kR = E_γR/ħc ≈ 0.02 for a 660 keV photon in a mid-heavy nucleus, only the leading term of each expansion survives, leaving M(EL, μ) = ∫ρ(r) r^L Y_L^μ* d³r and its convection-plus-magnetisation partner for ML. Everything downstream is a matrix element of one of those two operators.

M(σL, μ) is a spherical tensor of rank L, so ⟨J_f m_f|M(σL, μ)|Jᵢ mᵢ⟩ = ⟨Jᵢ mᵢ L μ|J_f m_f⟩⟨J_f‖M(σL)‖Jᵢ⟩/√(2J_f+1). Three consequences arrive at once. The Clebsch–Gordan coefficient vanishes unless μ = m_f − mᵢ and |Jᵢ − J_f| ≤ L ≤ Jᵢ + J_f, so the selection rule is geometry, not dynamics — no nuclear model is involved in it. The (2Jᵢ+1)(2L+1)(2J_f+1) matrix elements collapse onto one number, so a rate summed over m_f and averaged over mᵢ depends only on B(σL) = |⟨J_f‖M(σL)‖Jᵢ⟩|²/(2Jᵢ+1); this is why a rate never depends on how the source is oriented. And the m-dependence that was factored out is not lost — it is precisely what a γ–γ angular correlation, or the angular distribution from an aligned source, measures.

Subtract the spins, multiply the parities, then take the smallest surviving L. Ba-137: the 661.66 keV level is 11/2⁻ and the ground state 3/2⁺, so 4 ≤ L ≤ 7; the parity product is −1, which lets EL live only at odd L and ML only at even L, leaving M4, E5, M6, E7 — the answer is M4. Fe-57: 3/2⁻ → 1/2⁻ at 14.4 keV gives 1 ≤ L ≤ 2 with no parity change, so M1 and E2 both survive; the transition is M1 with a measured mixing ratio δ = 0.0022, an E2 intensity fraction δ² ≈ 5×10⁻⁶. Mixing is the normal case whenever L and L+1 are both allowed, and the mixing ratio δ is measured, not predicted. The clean case is an even-even ground-state band: 2⁺ → 0⁺ forces L = 2 exactly and no parity change makes it pure E2, which is why B(E2) values are the standard currency of collective structure.

Both factors in the rate punish large L. The operator carries r^L integrated over a nucleus of radius R = 1.2A¹⁄³ fm, and the photon, through its (kr)^L multipole field and its phase space, carries k^(2L+1). Collecting them in the Weisskopf estimate, neighbouring orders of the same family stand in the ratio λ(L+1)/λ(L) ≈ (kR)²⋅L(L+2)(L+3)²/[(L+1)²(L+4)²(2L+3)²] — in practice (kR)² divided by something close to (2L+3)². For the 661.66 keV transition in A = 137, kR = 0.0207, so (kR)² = 4.3×10⁻⁴ and one extra unit of L within a family costs about 10⁻⁵. Run the ladder at that energy, where the magnetic factor 0.307A^(−2/3) ≈ 10⁻² makes the steps alternate: E1 near 10⁻¹⁵ s, M2 near 10⁻⁸ s, E3 near 2×10⁻⁵ s, M4 near 450 s. That is isomerism in one line. A level whose lowest allowed multipole is M4 — and an isomer is usually low in energy too, which shrinks kR further — simply has no fast way to shed its angular momentum.

λ_W is what you get by assuming a single proton hops between two shell-model orbitals with a uniform-density radial integral and every geometric factor set to unity: B_W(EL) = (1/4π)[3/(L+3)]²R^(2L)e², with the magnetic estimate at the same L fixed at 0.307A^(−2/3) times the electric one, since μ_N/(e⋅fm) = ħc/(2m_pc²⋅fm) = 0.105. Nothing there describes a real nucleus, which is exactly why the ratio λ_γ/λ_W — the strength in Weisskopf units, W.u. — is useful as a diagnostic. Collective E2 transitions in deformed nuclei run 100–300 W.u. because many nucleons move coherently; M1 transitions typically run 10⁻² to 10⁻³ W.u.; E1 transitions are hindered to 10⁻⁴ or below by isospin and by the centre-of-mass constraint; the M4 in Ba-137m comes out near 2.6 W.u. Enhancement and hindrance are the structure the single-particle picture threw away.

Two things stop a photon. First, no allowed L: for 0 → 0 both |Jᵢ − J_f| and Jᵢ + J_f are zero, so the triangle admits only L = 0, which no photon has. Such a level empties by E0 internal conversion, whose matrix element ⟨f|Σ rₖ²|i⟩ is weighted by the electron density inside the nuclear volume, or by internal pair creation once the energy exceeds 2m_ec² = 1.022 MeV. Second, competition: internal conversion runs alongside every γ, so a measured half-life gives λₜₒₜₐₗ = λ_γ(1 + α), and only λ_γ may be set against a Weisskopf estimate. The coefficient α grows with Z³, with L, and with falling transition energy. For the 661.66 keV M4 in Ba-137 it is 0.1124, which is exactly why 94.7% of Cs-137 decays feed the isomer while only 94.7/1.1124 = 85.1% of them yield the 661.66 keV photon a detector sees.