Magnetic-dipole emission and the 21 cm rate computed

Units are SI throughout, and the atom moves slowly, so no four-vectors and no metric signature enter. Unit 11 ordered the atom-field Hamiltonian in powers of e; at first order it holds (e/m)A⋅p and the spin term −μ⋅B. Take the curl of the quantised A: each mode contributes a magnetic field of amplitude √(ħ/2ε₀ωV) × ω/c = √(μ₀ħω/2V), directed along k̂ × ε. Emission into an empty mode uses the a† term, whose √(n + 1) is 1, so ⟨f, 1|−μ⋅B|i, 0⟩ has magnitude √(μ₀ħω/2V)|⟨f|μ|i⟩⋅(k̂ × ε)|. The orbital moment comes out of A⋅p at first order in k⋅r: (k⋅r)(ε⋅p) = ½[(k⋅r)(ε⋅p) + (ε⋅r)(k⋅p)] + ½(k × ε)⋅L, and the last piece is exactly −μ_L⋅B with μ_L = −μ_B L/ħ. Spin and orbit join in μ = −μ_B(L + 2S)/ħ, the 2 being the Dirac equation's; the symmetric piece is the electric quadrupole. The spin term itself needs only e^(ik⋅r) ≈ 1, and for 21 cm ka₀ = 1.6 × 10⁻⁹.

Topic 12.02's machinery now applies unchanged. For a right-handed triad (ε₁, ε₂, k̂), k̂ × ε₁ = ε₂ and k̂ × ε₂ = −ε₁: the magnetic polarisation vectors are the electric pair turned through 90°, still orthonormal and perpendicular to k. So Σ_λ|μ_fi⋅(k̂ × ε_λ)|² = |μ_fi|² − |k̂⋅μ_fi|², whose integral over directions is (8π/3)|μ_fi|². With Vω²/(8π³c³) modes per unit ω and solid angle, the golden rule gives A = (2π/ħ²)(Vω²/8π³c³)(μ₀ħω/2V)(8π/3)|μ_fi|² = μ₀ω³|μ_fi|²/(3πħc³), derived, with V cancelling: topic 12.02's result with d replaced by μ/c. Classical limit: a moment m₀ cos ωt radiates μ₀ω⁴m₀²/(12πc³), quoted from electromagnetism, and with m₀ = 2|μ_fi| that power divided by ħω is the same rate. A Δm = 0 line again has the sin²θ pattern, but its electric vector lies along μ × k̂, turned 90° from the electric-dipole case: polarisation, not angular shape, tells M1 from E1.

Divide the two rates at one frequency: A_M1/A_E1 = |μ_fi|²/(c²|d_fi|²). The atomic sizes are μ_B = eħ/2mₑ and e a₀, and since a₀ = ħ/(mₑcα), μ_B/(e a₀ c) = α/2 exactly, so a one-magneton moment radiates (α/2)² = 1.33 × 10⁻⁵ as fast as an e a₀ dipole. The reason is kinematic: a charge e circling at radius r and speed v has μ = evr/2 and d = er, so μ/(cd) = v/2c, and Bohr's orbit has v = αc. Selection rules then put M1 lines where ω³ hurts most. μ is an axial vector, even under parity, so it links states of equal parity, exactly where electric-dipole decay is forbidden; and −μ_B(L + 2S)/ħ never touches the radial coordinate, so without relativity it connects only states sharing a radial function: members of one fine- or hyperfine-structure multiplet, with ΔJ or ΔF = 0, ±1 and 0 → 0 excluded. Those splittings are small. Hydrogen's 1s and 2s radial functions are orthogonal, so ⟨1s|μ|2s⟩ = 0; relativistic corrections give the 2s a magnetic rate of 2.5 × 10⁻⁶ s⁻¹, quoted, far below its two-photon rate of 8.2 s⁻¹.

Quantum Mechanics II built hydrogen's ground hyperfine states from the two spins, quoted here with ↑ for the electron and ⇑ for the proton: |1, 1⟩ = |↑⇑⟩, |1, 0⟩ = (|↑⇓⟩ + |↓⇑⟩)/√2 and |0, 0⟩ = (|↑⇓⟩ − |↓⇑⟩)/√2. Both are 1s, so L = 0 and the moment is the spins': μ = −μ_B σₑ + (gₚ/2)μ_N σₚ, with gₑ = 2. The z component of σₑ turns |1, 0⟩ into (|↑⇓⟩ − |↓⇑⟩)/√2 = |0, 0⟩, so the electron gives ⟨0, 0|μ_z|1, 0⟩ = −μ_B. From |1, 1⟩ the x and y components of σₑ give |↓⇑⟩ and i|↓⇑⟩, each of magnitude 1/√2 on |0, 0⟩, so the squared moment is ½μ_B² + ½μ_B² = μ_B² again: every substate decays at one rate, as rotational symmetry demands. The proton spin's element is −1 instead of +1, and its charge is opposite, so the two moments add: the amplitude grows by gₚmₑ/2mₚ = 0.15% and the rate by 0.30%. For a nucleus of spin I the same steps from the stretched state |↑, m_I = I⟩ give Σ|⟨f|σₑ|i⟩|² = 4I/(2I + 1), which is 1 for the proton.

The measured 1420.405752 MHz gives ω = 8.925 × 10⁹ rad s⁻¹ and ω³ = 7.108 × 10²⁹ s⁻³. With μ₀μ_B²/(3πħc³) = 4.036 × 10⁻⁴⁵ s², A₁₀ = 2.869 × 10⁻¹⁵ s⁻¹ and τ = 3.49 × 10¹⁴ s = 11.0 million years: the 2.9 × 10⁻¹⁵ s⁻¹ Quantum Mechanics II quoted, now derived. The proton's moment and gₑ = 2.00232 raise it to 2.884 × 10⁻¹⁵ s⁻¹, the value radio astronomy uses. The natural width A₁₀/2π = 4.6 × 10⁻¹⁶ Hz is invisible beside a 10 kHz Doppler width at 100 K. No laboratory can wait eleven million years, but hydrogen-like ²⁰⁹Bi⁸²⁺, with I = 9/2, has the same transition at 243.87 nm, and its lifetime was measured in a storage ring as 0.397 ± 0.016 ms. Scale from hydrogen: the frequency ratio cubed is 6.48 × 10¹⁷ and 4I/(2I + 1) = 1.8, giving 0.299 ms. At Z = 83 the bound electron's g is 1.728, Breit's Dirac-equation result, quoted, and (g/2)² = 0.746 lengthens the estimate to 0.400 ms, inside the error bar.

Interstellar hydrogen does not sit in empty space. At 2.725 K, hν/kT = 0.0682 K/2.725 K = 0.0250, so each 1420 MHz mode holds n̄ = 1/(e^0.0250 − 1) = 39.5 photons, close to kT/hν − ½. Topic 12.03's rates then apply: an F = 1 atom emits at A₁₀(n̄ + 1) = 1.16 × 10⁻¹³ s⁻¹, stimulated 39.5 times as often as spontaneously, and an F = 0 atom absorbs at 3A₁₀n̄ = 3.40 × 10⁻¹³ s⁻¹, the 3 counting the upper level's substates. With radiation alone the populations relax at A₁₀(4n̄ + 1), in 2.2 × 10¹² s or about 7 × 10⁴ years, to N₁/N₀ = 3e^(−hν/kT) = 2.93: the background's own temperature, with emission and absorption cancelling. A 21 cm line seen against the background therefore needs something else to set the populations. In a cloud of one atom per cubic centimetre, spin-exchange collisions at roughly 10⁻¹⁰ s⁻¹ per atom, quoted, outpace the radiative rates and tie the spin temperature to the gas.