University Quantum Mechanics II · Emission and Absorption of Radiation · 13.8
Beyond the dipole: magnetic-dipole and quadrupole lines
Parity closes the electric-dipole channel between any two levels of one configuration, yet those levels still radiate: the 21 cm line maps the Galaxy's hydrogen and green [O III] light outlines planetary nebulae. Here you find the couplings that do it, the rules they obey, and why their rates run from milliseconds to ten million years.
Build the model
Connect the measurement to the mechanism.
The electric-dipole approximation sets the wave's phase e(ik⋅r) to 1 across the atom. That is safe for visible light, where ka ≈ 10⁻³, but it keeps only the first term of a series; when parity forbids that term, as between any two levels of one configuration, the next is all that is left. It is i(k⋅r)(ε⋅p), joined at the same order by the spin's coupling to the wave's magnetic field, and it splits exactly in two: an antisymmetric part that becomes (μB/ħ)(L + 2S)⋅B, the magnetic dipole (M1), and a symmetric part that a commutator with H₀ turns into products of coordinates, the electric quadrupole (E2).
Both are even, connecting levels of equal parity. For equal-sized matrix elements an M1 rate is (αa₀/2a)², about 10⁻⁵, of an E1 rate at any frequency, since a Bohr magneton radiates like a charge e displaced by ħ/2mc = 0.19 pm; an E2 rate is down by a factor of order (ka)², 10⁻⁷ to 10⁻⁵ in the optical. Real M1 lines are slower still: L + 2S never touches r, so it links only levels of one multiplet, close in energy, where the ω³ law bites; hence 10⁷ years for the 21 cm flip.
The model stops where its numbers say so. A level living 41 s radiates only if nothing de-excites it first, so [O III] lines shine in nebulae, not in laboratory gas; and in heavy, highly charged ions ka grows with Zα, so the series converges slowly and equally large relativistic corrections must be quoted.
- Simple definition
- A magnetic-dipole (M1) or electric-quadrupole (E2) transition is driven by the first correction to the dipole approximation: the wave's magnetic field acting on (μB/ħ)(L + 2S), or its field gradient acting on products of the electron's coordinates. Both connect states of equal parity.
- Example
- Hydrogen's 2p₃/₂ → 2p₁/₂ keeps l and parity and changes J by 1, so it is an M1 line, at 10.97 GHz, with A = 4.4 × 10⁻¹³ s⁻¹. The same level's E1 decay to 1s runs at 6.3 × 10⁸ s⁻¹, about 10²¹ times faster.
Keeping only the 1 is 13.01's E1 approximation. Each further term costs one power of ka in the amplitude and (ka)² in the rate.
k = 2π/λ in m⁻¹, r the electron's position in m; ka = 6.6 × 10⁻⁴ for a = a₀ at 500 nm
An identity, derived: the antisymmetric half rotates the orbit and becomes M1; the symmetric half deforms it and becomes E2.
ε the unit polarisation, with ε⋅k = 0; p in kg m s⁻¹; L = r × p in J s
Derived: the rotation half gives the L term, the spin's own coupling to B gives 2S at the same order. It acts on angles and spin only, never on r.
μB = eħ/2m = 9.274 × 10⁻²⁴ J T⁻¹; B the wave's field in T; the 2 is g = 2, quoted from the Dirac equation
Derived from [H₀, (ε⋅r)(k⋅r)] = −(iħ/m)[(ε⋅p)(k⋅r) + (ε⋅r)(k⋅p)]. The operator becomes a product xᵢxⱼ: rank 2 and even under parity.
m the electron mass in kg; ωfi = (Ef − Eᵢ)/ħ in s⁻¹; (ε⋅r)(k⋅r) is traceless because ε⋅k = 0
Derived from 13.03's AE1 by d → μ/c, since |B| = |E|/c in the wave. For |⟨f|μ|i⟩| = μB it reads (α/3)(ħω/mc²)²ω.
μ = −(μB/ħ)(L + 2S) in J T⁻¹, ω in rad s⁻¹, μ₀ in T m A⁻¹; summed over final sublevels; A in s⁻¹
Quoted, not derived: the rank-2 average over directions and polarisations is not done here. Two more powers of ω than E1, so E2 fades fastest at long wavelength.
Qᵢⱼ = e(3xᵢxⱼ − r²δᵢⱼ) in C m², summed over all nine i, j and over final sublevels; ω in rad s⁻¹
The term the dipole approximation drops
In 13.01 the coupling (e/m)A⋅p became E1 once e(ik⋅r) was set to 1 across the atom. With a = a₀ and λ = 500 nm, ka = 6.6 × 10⁻⁴, and the correction is negligible whenever E1 is allowed. But ε⋅r is odd under parity, so between two levels of one configuration, such as the ³P and ¹D terms of 2p², every E1 matrix element vanishes exactly and the next term is all there is. Keep one more power: the absorption operator gains i(k⋅r)(ε⋅p), a product of two vectors and therefore even. A second coupling enters at the same order. The spin term (e/m)S⋅B, with g = 2 quoted from the Dirac equation, uses B = ik × A, already one power of k beyond A, so its size relative to (e/m)A⋅p ≈ (e/m)A(ħ/a) is again ka. Both couplings are first order in the field, so Unit 12's golden rule and 13.03's detailed balance apply unchanged: only the matrix element is new.
Rotation and deformation: splitting (k⋅r)(ε⋅p)
Write (k⋅r)(ε⋅p) as half its antisymmetric part plus half its symmetric part. The antisymmetric part, (k⋅r)(ε⋅p) − (ε⋅r)(k⋅p), is (k × ε)⋅(r × p) by the identity for (a × b)⋅(c × d) with r kept left of p; ε⋅k = 0 makes the ordering harmless. For A = A₀ε e(ik⋅r) the field at the atom is B = iA₀(k × ε), so (e/m)A₀(i/2)(k × ε)⋅L is (e/2m)L⋅B, and the spin adds (e/m)S⋅B: together (μB/ħ)(L + 2S)⋅B, the energy −μ⋅B of the atom's moment. For the symmetric part use [p², xᵢ] = −2iħpᵢ on each factor: [H₀, (ε⋅r)(k⋅r)] = −(iħ/m)[(ε⋅p)(k⋅r) + (ε⋅r)(k⋅p)], so between eigenstates of H₀ that symmetric sum equals imωfi⟨f|(ε⋅r)(k⋅r)|i⟩. That operator is Σεᵢkⱼxᵢxⱼ, traceless because ε⋅k = 0: a product of coordinates coupled to the gradient of E, the electric quadrupole. All of it is derived; only g = 2 is quoted.
What each coupling can connect
Both operators are even, so parity must not change: exactly the case E1 leaves open. They differ in rank. M1 is a vector: ΔJ = 0, ±1, never 0 → 0. It is also blind to r, so in a one-electron atom ⟨n′l′|L + 2S|nl⟩ needs l′ = l and carries the radial overlap ∫Rₙ′ₗRₙₗr² dr = δₙ′ₙ: M1 only reorients spins and orbits within one n and l, which confines it to fine- and hyperfine-structure intervals. In a many-electron atom in LS coupling, L + 2S commutes with L² and S², so M1 stays inside one term. E2 is rank 2: ΔJ = 0, ±1, ±2 with J + J′ ≥ 2, and Δl = 0, ±2 for one electron, so hydrogen 3d → 1s is E2. The rules explain a hydrogen oddity: 2s → 1s has no E1, no E2 (l = 0 to 0) and a non-relativistic M1 of exactly zero. Its tabulated M1 rate, 2.5 × 10⁻⁶ s⁻¹, is a relativistic correction, quoted, and the two-photon decay of 13.05, 8.2 s⁻¹ and also quoted, wins.
How much slower: α² for M1, (ka)² for E2
In 13.03's AE1 = ω³|dfi|²/(3πε₀ħc³), d⋅E becomes μ⋅B, and in the wave |B| = |E|/c with the same one-third angular average, so d → μ/c gives AM1 = μ₀ω³|μfi|²/(3πħc³). Against E1 with |d| = ea the ratio is (αa₀/2a)², 1.3 × 10⁻⁵ at a = a₀, at every frequency. The quoted AE2 carries ω⁵, so E2 sits a factor of order (ka)² below E1, 10⁻⁷ to 10⁻⁵ for optical lines and far less in the radio. In hydrogen ka₀ ≈ (3/8)α at Lyman α, so both penalties are order α²: E1 rates scale as α⁵mc²/ħ, optical E2 as α⁷mc²/ħ, and an M1 across a fine-structure gap, ħω of order α⁴mc², as α¹³mc²/ħ, crushed by its small ω. Dimensions: μ₀μB²/(ħc³) is s², so times ω³ it is s⁻¹. Limits: as ω → 0 every rate vanishes, E2 fastest; as a → 0 E1 and E2 vanish but M1 survives, since a spin's moment needs no orbit.
Checking the M1 rate against a measured lifetime
Boron-like Ar¹³⁺ has one 2p electron outside closed shells, and its ²P₃/₂ level can reach only ²P₁/₂ of the same configuration: an M1 line at 441.38 nm in vacuum (441.26 nm in air), and the vacuum value is the one that fixes ω. Write L + 2S = J + S. J cannot connect different J, so only S acts. From |3/2, 3/2⟩ = |mₗ = 1, ↑⟩ the only route is S₋ to |1, ↓⟩, whose overlap with |1/2, 1/2⟩ is √(2/3), so Σ|⟨f|S|i⟩|² = ½ × ħ² × 2/3 = ħ²/3 and |μfi|² = μB²/3. Then AM1 = 105 s⁻¹, a lifetime of 9.56 ms. Lapierre and colleagues measured 9.573 ms in an electron-beam ion trap in 2005. The 0.1% agreement is partly luck: the electron's anomalous moment, quoted from quantum electrodynamics, raises the rate by 0.46% and the prediction falls to 9.52 ms, relativistic corrections of order (Zα)² move it again, and full calculations give 9.53 ms, 0.45% short of experiment. The formula is right to a fraction of a percent; the last digits belong to QED.
[O III]: magnetic-dipole lines that also flip the spin
The green lines at 495.9 and 500.7 nm puzzled astronomers from Huggins's first nebular spectrum in 1864 until Bowen showed in 1927 that they are ¹D₂ → ³P₁ and ³P₂ in O²⁺, inside Unit 7's 2p² terms. They are doubly hindered: equal parity forbids E1, and in pure LS coupling L + 2S cannot change S, so M1 works only through a spin-orbit admixture of ³P₂ into ¹D₂, of order 10⁻² in amplitude. The tabulated rates, quoted from intermediate-coupling calculations, are 1.81 × 10⁻² s⁻¹ at 500.7 nm and 6.21 × 10⁻³ s⁻¹ at 495.9 nm, against 215 s⁻¹ for a full-strength M1: |μfi| ≈ 0.009 μB. The ¹D₂ level lives 41 s. Electron collisions, at nₑ q with q ≈ 4 × 10⁻¹⁴ m³ s⁻¹ at 10⁴ K (quoted), match that at ncrit ≈ 6 × 10¹¹ m⁻³; nebulae hold 10⁹ to 10¹⁰ m⁻³, so there the ion radiates. Measured check: sharing an upper level, the lines keep an energy ratio set by atomic physics alone, 2.89 from these rates; relativistic M1 corrections (quoted) raise it to 2.98, and AGN spectra give 2.993 ± 0.014.
Change one variable at a time
Make the relationship visible.
Drag the cursor from 100 nm to 21 cm at a = 1 and watch the E1–M1 gap hold at 4.9 decades while E2 falls away as λ⁻⁵; then raise a to 3, where E1 climbs 1.0 decade, E2 1.9 and M1 not at all, and note that [O III] is M1, held 10⁴ below its line by the spin flip, not an E2 line.
E1 RATE, log₁₀(A / s⁻¹)7.21
M1 RATE, log₁₀(A / s⁻¹)2.33
E2 RATE, log₁₀(A / s⁻¹)-0.45
log₁₀ (ka)² AT THE CURSOR-6.36
Live interpretationE1 RATE, log₁₀(A / s⁻¹): 7.21. M1 RATE, log₁₀(A / s⁻¹): 2.33. E2 RATE, log₁₀(A / s⁻¹): −0.45. log₁₀ (ka)² AT THE CURSOR: −6.36
Catch the common trap
Explain before calculating.
In O²⁺ the ¹D₂ level decays to ³P₂ at 500.7 nm, to ³P₁ at 495.9 nm and to ³P₀ at 493.1 nm, with tabulated rates of 1.8 × 10⁻² s⁻¹, 6.2 × 10⁻³ s⁻¹ and 2.4 × 10⁻⁶ s⁻¹. Why is the third line thousands of times slower than the other two?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyFor each hydrogen decay, name the lowest multipole, E1, M1 or E2, that can drive it directly, or say that none can: 3d → 2p; 3d → 1s; 1s F = 1 → F = 0; 3s → 2s. Then evaluate (ka₀)² for the E2 case, whose wavelength is 102.6 nm.
- Parity is (−1)l: 3d, 1s, 3s and 2s are even, 2p is odd. 3d → 2p changes parity with Δl = −1, and 3d₃/₂ or 3d₅/₂ can reach a 2p level with |ΔJ| ≤ 1, so it is E1.
- 3d → 1s keeps parity, so E1 is out. M1 cannot change l, because L + 2S acts only on angles and spin and L² commutes with it. E2 allows Δl = 2 and ΔJ = 2, so 3d → 1s is E2.
- 1s F = 1 → F = 0 stays in one n and l and flips the electron spin relative to the proton's: ΔF = 1, not 0 → 0, so it is M1. E2 is closed because F + F′ = 1 is less than 2.
- 3s → 2s: E1 is out by parity; M1 carries ∫R₃₀R₂₀r²dr = 0, since radial functions of equal l are orthogonal; E2 needs ⟨Y₀₀|Y₂q|Y₀₀⟩, which is zero. None of the three works, and 3s empties instead by E1 to 2p.
- For 3d → 1s: k = 2π/(102.6 × 10⁻⁹ m) = 6.124 × 10⁷ m⁻¹, so ka₀ = 6.124 × 10⁷ × 5.292 × 10⁻¹¹ = 3.24 × 10⁻³ and (ka₀)² = 1.05 × 10⁻⁵.
Answer3d → 2p is E1, 3d → 1s is E2, the 1s hyperfine flip is M1, and 3s → 2s has no single-photon channel of these three; (ka₀)² = 1.05 × 10⁻⁵ at 102.6 nm.
MediumCompute the spontaneous rate and lifetime of hydrogen's 1s F = 1 level for the 1420.406 MHz decay to F = 0, taking the moment operator as μ = −(2μB/ħ)Sₑ and ignoring the proton's moment. Then say why the line is nevertheless bright in radio maps of the Galaxy.
- With the electron's arrow first, |1, 0⟩ = (|↑⇓⟩ + |↓⇑⟩)/√2 and |0, 0⟩ = (|↑⇓⟩ − |↓⇑⟩)/√2. Sez turns |1, 0⟩ into (ħ/2)|0, 0⟩, so ⟨0, 0|Sez|1, 0⟩ = ħ/2; its x and y components flip the electron and reach only triplet states.
- So Σ|⟨f|μ|i⟩|² = (2μB/ħ)² × ħ²/4 = μB², and rotational symmetry gives the same from mF = ±1. The proton's moment is 658 times smaller and is dropped.
- ω = 2π × 1.420406 × 10⁹ Hz = 8.925 × 10⁹ rad s⁻¹, so ω³ = 7.108 × 10²⁹ s⁻³, and 3πħc³ = 2.678 × 10⁻⁸ J m³ s⁻². Then A = μ₀ω³μB²/(3πħc³) = (1.2566 × 10⁻⁶ × 7.108 × 10²⁹ × 8.601 × 10⁻⁴⁷)/(2.678 × 10⁻⁸) = 2.87 × 10⁻¹⁵ s⁻¹.
- Check with (α/3)(ħω/mc²)²ω: ħω/mc² = 5.874 μeV/511.0 keV = 1.150 × 10⁻¹¹, and (7.297 × 10⁻³/3) × 1.322 × 10⁻²² × 8.925 × 10⁹ = 2.87 × 10⁻¹⁵ s⁻¹. The lifetime is 1/A = 3.49 × 10¹⁴ s = 1.1 × 10⁷ years.
- gₑ = 2.00232, quoted from QED, and the proton's moment add in the same matrix element, a factor 1.0054, giving the tabulated 2.88 × 10⁻¹⁵ s⁻¹.
- Why it is bright: hν/k = 0.068 K, so three quarters of the atoms sit in F = 1, and a Galactic column of 10²⁵ atoms m⁻² emits 0.75 × 10²⁵ × 2.9 × 10⁻¹⁵ ≈ 2 × 10¹⁰ photons m⁻² s⁻¹. Numbers, not speed, make the line.
AnswerA = 2.87 × 10⁻¹⁵ s⁻¹ (2.88 × 10⁻¹⁵ s⁻¹ with g − 2 and the proton's moment), a lifetime of 1.1 × 10⁷ years; the line is seen because columns hold about 10²⁵ atoms per square metre.
HardCompute the E2 rate for hydrogen 3d → 1s from R₁₀, R₃₂ and the Yₗₘ, using the quoted AE2 = ω⁵Σ|⟨f|Qᵢⱼ|i⟩|²/(360πε₀ħc⁵). Find what fraction of 3d decays take this route, given the quoted 3d → 2p E1 rate of 6.47 × 10⁷ s⁻¹.
- Take m = 0. Qzz = e(3z² − r²) = 2er²√(4π/5) Y₂₀, and ∫Y₀₀Y₂₀Y₂₀ dΩ = 1/√(4π), so ⟨1s|Qzz|3d, 0⟩ = (2/√5) e I with I = ∫R₁₀R₃₂r⁴dr. Tracelessness and axial symmetry give Qₓₓ = Qyy = −Qzz/2, and xy, xz, yz carry m = ±1, ±2 and vanish: Σ|Qᵢⱼ|² = (1 + ¼ + ¼)(4/5)e²I² = 1.2e²I².
- R₁₀ = 2a₀(−3/2)e(−r/a₀) and R₃₂ = (4/(81√30))a₀(−3/2)(r/a₀)²e(−r/3a₀), so I = (8/(81√30))a₀⁻⁵∫r⁶e(−4r/3a₀)dr = (8/(81√30)) × 720 × (3/4)⁷ a₀² = 1.733a₀². Hence Σ|Qᵢⱼ|² = 3.604e²a₀⁴ = 3.604 × 2.013 × 10⁻⁷⁹ = 7.255 × 10⁻⁷⁹ C² m⁴.
- ħω = 13.606 eV × (1 − 1/9) = 12.094 eV, so ω = 1.8374 × 10¹⁶ rad s⁻¹ and ω⁵ = 2.094 × 10⁸¹ s⁻⁵.
- AE2 = (2.094 × 10⁸¹ × 7.255 × 10⁻⁷⁹)/(360π × 8.854 × 10⁻¹² × 1.0546 × 10⁻³⁴ × 2.4216 × 10⁴²) = 1519/2.557 = 594 s⁻¹. Dimensions: s⁻⁵ C² m⁴ over F m⁻¹ J s m⁵ s⁻⁵ is s⁻¹. The reduced mass lowers it 0.05% to 593.8 s⁻¹; the table gives 593.7 and 593.8 s⁻¹ for the two J levels.
- Branch: 594/(6.47 × 10⁷ + 594) = 9.2 × 10⁻⁶ of 3d decays go straight to 1s. Scaling check: Lyman β, the E1 line 3p → 1s at the same wavelength, runs at 1.67 × 10⁸ s⁻¹ (quoted), so E2/E1 = 3.6 × 10⁻⁶, about a third of the Easy example's (ka₀)² = 1.05 × 10⁻⁵.
AnswerAE2(3d → 1s) = 594 s⁻¹ (593.8 s⁻¹ with the reduced mass, against a tabulated 593.7 s⁻¹), a branch of 9.2 × 10⁻⁶ of all 3d decays.