University Quantum Mechanics II · Emission and Absorption of Radiation · 13.5
Selection rules applied: polarisation and metastable 2s
Selection rules earn their keep when they predict what a spectrometer records. Here they are read three ways: which lines exist at all, how each is polarised from each direction you look, and why hydrogen's 2s, with no dipole route down worth taking, waits a tenth of a second for two photons.
Build the model
Connect the measurement to the mechanism.
Unit 5 made one fact about r, that it is a polar vector, into rules: in the components r₀ = z, r₊₁ = −(x + iy)/√2 and r₋₁ = (x − iy)/√2, ⟨n′l′m′|rq|nlm⟩ vanishes unless l′ = l ± 1 and m′ = m + q, and r never touches spin. Applied to hydrogen they do three jobs. They choose the lines: of the six l-pairs joining n = 3 to n = 2 only 3s → 2p, 3p → 2s and 3d → 2p radiate, so Balmer-α is a blend, seven lines once fine structure is resolved.
They shape the light: ⟨f|r|i⟩ points along ẑ for Δm = 0 and along (x̂ ± iŷ)/√2 for Δm = ±1, a charge oscillating along z or circling it, and an observer receives only its projection across the line of sight, so π light is polarised along z and missing when viewed along z, while σ light is circular along z, linear across it, and elliptical between. They strand a state: 2s has no dipole route to 1s, and the one allowed level below it, 2P₁/₂, lies just 1057.8 MHz down, where ω³ leaves a rate of 8.0 × 10⁻¹⁰ s⁻¹. The level lives 0.122 s on a quoted two-photon rate of 8.23 s⁻¹.
The model ends where its premises do. A rule closes one operator's channel to leading order, so forbidden means slower by powers of α, not never; and it relies on parity, so a field of 0.054 V/cm, mixing 2p into 2s, empties the level as fast as the two photons do.
- Simple definition
- A selection rule says which matrix elements of the dipole operator r can be non-zero: in hydrogen Δl = ±1 and Δm = 0 or ±1 with spin unchanged, and Δm then fixes the light's polarisation, π for Δm = 0 and σ for Δm = ±1.
- Example
- 2p₀ → 1s is a π step, sending no Lyman-α along z and light polarised along z across it; 2p₊₁ → 1s is a σ step, circular along z and twice as bright there as across. 2s → 1s has Δl = 0 and emits no single dipole photon at all.
Quoted from Unit 5. It darkens three of the six l-pairs from n = 3 to n = 2 and leaves 7 of the 15 fine-structure pairs, so Balmer-α is a blend.
r₀ = z, r₊₁ = −(x + iy)/√2, r₋₁ = (x − iy)/√2; lengths in m; r acts on position only, so mₛ′ = mₛ
Derived: the dipole lies along ẑ, so the light is linear along z's projection, strongest across z and absent along it.
A the component's rate in s⁻¹, dA/dΩ in s⁻¹ sr⁻¹, θ between the emission direction and the quantisation axis z
Derived: circular along z, linear perpendicular to z across it, twice as bright along z. A π and two σ lines of equal strength sum to a constant.
Same symbols; seen at θ, the field traces an ellipse whose minor-to-major axis ratio is |cos θ|
Derived: 8.0 × 10⁻¹⁰ s⁻¹, one decay in 40 years. The channel is allowed; ω is 2.3 × 10⁶ times below Lyman-α's, and ω³ costs 1.3 × 10¹⁹.
ω = 2π × 1057.8 MHz (measured); |rfi|² = 9a², a = 5.29 × 10⁻¹¹ m; α ≈ 1/137; c in m s⁻¹; A in s⁻¹
Two photons share 10.2 eV as a continuum, not a line. The rate is 7.6 × 10⁷ times below 2p's: of order α⁸mc²/ħ against α⁵mc²/ħ for one photon.
Z the nuclear charge number; quoted, not derived: Unit 12's second-order amplitude summed over all p states; τ = 0.122 s at Z = 1
Equals the two-photon rate at 0.054 V/cm and cuts the lifetime to 0.35 ms at 1 V/cm: the field breaks the parity the rule relied on.
F the static electric field; Γq in s⁻¹; from the first-order 2P admixture into 2S₁/₂, with Γ₂ₚ = 6.27 × 10⁸ s⁻¹
Three rules, and the lines they leave
Unit 5 derived the rules; here they are only applied. Nothing restricts n, and spin is a spectator, because r acts on position alone. Run Δl = ±1 over n = 3 → n = 2. Of six l-pairs, three radiate: 3s → 2p, 3p → 2s and 3d → 2p. The other three are dark: 3s → 2s and 3p → 2p have Δl = 0, and 3d → 2s has Δl = 2. So Balmer-α at 656.3 nm is three transitions sharing one Bohr energy. With fine structure the rule reads Δj = 0, ±1 with a change of parity, and of the fifteen pairs of n = 3 and n = 2 levels exactly seven radiate; 3D₅/₂ → 2P₁/₂ passes on l and dies on j. The rules also route the cascade. 3s and 3d cannot reach 1s, so every atom in them emits Balmer-α on the way down. 3p can, and tabulated rates, quoted, send 88% of it down Lyman-β and 11.8% into 2s: the only one-photon route into 2s from n = 3.
The matrix element is a vector with a direction
A rule says whether ⟨f|r|i⟩ vanishes; its direction says what the light looks like. From Unit 2's functions, ⟨1s|z|2p₀⟩ = (2⁷√2/3⁵)a = 0.745a, and for m = ±1 the azimuthal integrals ∫cos φ e(±iφ) dφ = π and ∫sin φ e(±iφ) dφ = ±iπ give ⟨1s|r|2p, m = ±1⟩ = ∓0.745a(x̂ ± iŷ)/√2. Each has squared length 0.555a², the value behind the 2p lifetime, so all three sublevels decay at 6.27 × 10⁸ s⁻¹, as they must, being rotations of one another. The vector is the atom's dipole. In a superposition c₁|lower⟩ + c₂|upper⟩, ⟨r⟩(t) = 2Re[c₁*c₂⟨lower|r|upper⟩e(−iωt)]. For Δm = 0 that is a charge oscillating along z. For Δm = +1, upper minus lower, it is x̂ cos ωt + ŷ sin ωt up to phase and size: a charge circling anticlockwise about +z, and clockwise for Δm = −1. The photon leaves with Jz = Δm ħ, and the circling is that angular momentum made visible.
What an observer at angle θ receives
Light leaving along n̂ with polarisation ε̂ is emitted at a rate proportional to |ε̂*⋅⟨f|r|i⟩|². Summed over the two polarisations across n̂, that is |⟨f|r|i⟩|² − |n̂⋅⟨f|r|i⟩|², the part of the dipole seen side-on. For π, along ẑ, this is sin²θ: nothing along z, most across it, linear along z's projection. For σ, project (x̂ ± iŷ)/√2 on the observer's two axes: 1/√2 perpendicular to both z and n̂, cos θ/√2 along z's projection, a quarter-period apart. The field traces an ellipse of axis ratio |cos θ|, a circle along z and a line perpendicular to z across it, with intensity (1 + cos²θ)/2. Each pattern integrates over 4π to 8π/3, so direction redistributes a rate without changing it. Seen from −z the rotation looks reversed, since the light now travels the other way. Cadmium's red 643.8 nm line, the textbook normal triplet, shows all of this, and in 1897 the sense of rotation of σ light told Zeeman and Lorentz that the radiating charge was negative.
Polarisation needs something to pick out z
Give every upper sublevel the same population. Rotational symmetry then hands the π, σ⁺ and σ⁻ components equal total strength, and their patterns sum to sin²θ + 2 × (1 + cos²θ)/2 = 2 at every θ. The polarisations cancel as well: along z's projection sin²θ + cos²θ = 1, across it ½ + ½ = 1, and the opposite circular senses annul. An unresolved line from an isotropic source is isotropic and unpolarised, which is the limiting-case check on both patterns. Polarised light therefore needs an axis chosen from outside: a magnetic field that separates the components in frequency so a spectrometer can take one, or an excitation that fills the sublevels unequally, such as a directed electron beam or a polarised laser. A j = ½ upper level resists even that. 2P₁/₂ with mⱼ = ½ is ⅓ of |mₗ = 0, ↑⟩ and ⅔ of |mₗ = 1, ↓⟩, radiating π and σ to 1s, and (⅓)sin²θ + (⅔)(1 + cos²θ)/2 = ⅔ with no linear polarisation, however the level was filled.
2s: one exit shut by symmetry, the other by ω³
Two levels lie below 2S₁/₂. 1S₁/₂ has l = 0 too, so ⟨1s|r|2s⟩ = 0. 2P₁/₂ is fully allowed, but QED puts it only 1057.8 MHz lower, the measured Lamb shift, quoted. Its rate is derived here. Unit 9's ⟨2s|z|2p₀⟩ = −3a gives 9a² to each 2p sublevel. From 2S₁/₂(mⱼ = ½) = |0, ↑⟩, r reaches |0, ↑⟩, |−1, ↑⟩ and |+1, ↑⟩, whose shares in 2P₁/₂ are ⅓, ⅔ and 0, so |rfi|² = 9a², a third of the total. With ω = 6.65 × 10⁹ s⁻¹, A = 4αω³(9a²)/(3c²) = 8.0 × 10⁻¹⁰ s⁻¹, one decay in 40 years. Dimensions: ω³a²/c² is s⁻¹. Limit: A → 0 with ω, since low frequencies offer fewer modes, each coupling more weakly, which is also why radio lines are slow. Order: with ħω near α⁵mc², A is of order α¹⁴mc²/ħ = 9 × 10⁻¹⁰ s⁻¹. Against Lyman-α, ω³ costs (2.33 × 10⁶)³ = 1.3 × 10¹⁹, and the same formula with Lyman-α's ω gives the measured 1.60 ns 2p lifetime. Symmetry shuts one exit; ω³ shuts the other.
The slow ways out, and the fast one a field opens
What does empty 2s is quoted, not derived. Unit 12's second-order amplitude, summed over every p state including the continuum, gives two photons sharing 10.2 eV in any split, a continuum peaked at 5.1 eV each, at 8.23 s⁻¹: τ = 0.122 s, 7.6 × 10⁷ times the 2p lifetime. Counting α, a one-photon rate is of order α⁵mc²/ħ = 1.6 × 10¹⁰ s⁻¹ and a two-photon rate of order α⁸mc²/ħ = 6 × 10³ s⁻¹, with true factors 0.039 and 0.0013. A relativistic magnetic-dipole rate, 2.5 × 10⁻⁶ s⁻¹, is quoted and irrelevant; electric quadrupole cannot join two s states. The long life pays: the 1S–2S two-photon line, 1.3 Hz wide, is measured to a few parts in 10¹⁵. It is also fragile. Unit 9's field mixes in 2P and the state inherits its decay, 2.81 × 10³ s⁻¹ at 1 V/cm, matching two photons at 0.054 V/cm. Lamb and Retherford used both in 1947: 2s atoms crossed the apparatus intact, and microwaves, stimulated and so free of the ω³ penalty, drove them into 2P and quenched them.
Change one variable at a time
Make the relationship visible.
Start at θ = 0: π vanishes and both σ lines are circles, twice their side-on brightness. Swing to 90°: π peaks and each σ ellipse flattens to a line perpendicular to B, giving 1 : 2 : 1. Past 90° the σ⁺ circular fraction turns negative: the same circling charge, seen from the other end, turns clockwise.
π INTENSITY sin²θ0.75
EACH σ (1 + cos²θ)/20.63
σ AXIS RATIO |cos θ|0.50
σ⁺ CIRCULAR FRACTION V/I0.80
Live interpretationπ INTENSITY sin²θ: 0.75. EACH σ (1 + cos²θ)/2: 0.63. σ AXIS RATIO |cos θ|: 0.50. σ⁺ CIRCULAR FRACTION V/I: 0.80
Catch the common trap
Explain before calculating.
Hydrogen in a 5.0 T field along z emits the Balmer-α Zeeman triplet. A spectrometer looking along the x axis, perpendicular to the field, isolates the σ⁺ component, Δmₗ = +1. Compared with the same component viewed along z, what does it record?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA hydrogen discharge sits in a 5.0 T field along z, strong enough for Unit 9's Paschen–Back pattern, in which every Balmer-α component is shifted by μB B Δmₗ. With the upper sublevels evenly populated, find the wavelength offset of each component, and describe what a spectrometer records looking along z and looking along x.
- r ignores spin, so Δmₛ = 0 and the spin energy 2μB B mₛ is the same above and below; it drops out. The photon energy moves by μB B Δmₗ, with Δmₗ, upper minus lower, equal to 0 or ±1: three components, however many sublevels take part.
- μB B = 5.788 × 10⁻⁵ eV T⁻¹ × 5.0 T = 2.89 × 10⁻⁴ eV, a frequency shift Δν = 70.0 GHz. At λ = 656.3 nm, Δλ = λ²Δν/c = (656.3 × 10⁻⁹ m)² × (7.00 × 10¹⁰ s⁻¹)/(3.00 × 10⁸ m s⁻¹) = 1.00 × 10⁻¹⁰ m. σ⁺, with Δmₗ = +1, carries more energy and sits at 656.2 nm; σ⁻ at 656.4 nm.
- Along z, θ = 0: π has sin²θ = 0 and is missing; each σ has (1 + cos²θ)/2 = 1. Two equal lines, circularly polarised in opposite senses, σ⁺ turning anticlockwise for an observer at the +z end facing the source.
- Along x, θ = 90°: π has intensity 1 and is linear along z; each σ has ½ and is linear along y. Three lines in the ratio 1 : 2 : 1.
- Check: the totals are 0 + 1 + 1 = 2 along z and ½ + 1 + ½ = 2 along x, as an isotropic source requires. The fine-structure spread inside each component, about a sixth of the 70 GHz spacing at this field, is neglected.
AnswerOffsets of ±0.100 nm, σ⁺ on the short-wavelength side. Along the field: two equal σ lines, circular in opposite senses, and no π. Across it: three lines in the ratio 1 : 2 : 1, π polarised parallel to the field and σ perpendicular to it.
MediumHydrogen's 2S₁/₂ lies 1057.8 MHz above 2P₁/₂ (measured). Using ⟨2s|z|2p₀⟩ = −3a from Unit 9 and Unit 5's coupled states, find |rfi|², the squared dipole element summed over the 2P₁/₂ sublevels, then the spontaneous rate of this allowed decay, and compare it with the 2p → 1s rate of 6.27 × 10⁸ s⁻¹ and the quoted two-photon rate of 8.23 s⁻¹.
- Spinless first: 2s is spherically symmetric, so every 2p sublevel receives the same squared vector element as 2p₀, (−3a)² = 9a², or 27a² over all three.
- With spin, start from 2S₁/₂(mⱼ = ½) = |mₗ = 0, ↑⟩. The components of r reach |0, ↑⟩, |−1, ↑⟩ and |+1, ↑⟩, each with 9a². Their shares in 2P₁/₂ are ⅓, ⅔ and 0, so |rfi|² = (⅓ + ⅔) × 9a² = 9a² = 2.52 × 10⁻²⁰ m². The other 18a² belongs to 2P₃/₂, which lies above.
- ω = 2π × 1.0578 × 10⁹ Hz = 6.646 × 10⁹ s⁻¹. A = 4αω³|rfi|²/(3c²) = 4 × (7.297 × 10⁻³) × (2.936 × 10²⁹ s⁻³) × (2.520 × 10⁻²⁰ m²)/(3 × 8.988 × 10¹⁶ m² s⁻²) = 8.0 × 10⁻¹⁰ s⁻¹, a mean wait of 1.25 × 10⁹ s, about 40 years.
- Against 2p → 1s: the frequency ratio is 10.20 eV/4.375 μeV = 2.33 × 10⁶, whose cube is 1.27 × 10¹⁹, partly offset by the larger matrix element, 9a²/0.555a² = 16.2. The rates differ by 1.27 × 10¹⁹/16.2 = 7.8 × 10¹⁷.
- Against the two-photon decay the channel is 8.23/(8.0 × 10⁻¹⁰) = 1.0 × 10¹⁰ times slower, so it shortens the 0.122 s lifetime by one part in 10¹⁰.
Answer|rfi|² = 9a² = 2.52 × 10⁻²⁰ m² and A = 8.0 × 10⁻¹⁰ s⁻¹, about 40 years: allowed, but 7.8 × 10¹⁷ times slower than 2p → 1s and 1.0 × 10¹⁰ times slower than the two-photon decay.
HardHydrogen-like He⁺ (Z = 2) has a measured 2S₁/₂–2P₁/₂ Lamb shift of 14.04 GHz and a 2P fine-structure interval of 175.5 GHz, Z⁴ times hydrogen's 10.97 GHz. Scale this lesson's hydrogen results to find the 2p and two-photon rates, the one-photon rate across the Lamb shift, and the static field at which Stark quenching equals the two-photon rate. Compare that field with hydrogen's 0.054 V/cm.
- Lengths scale as a/Z and energies as Z², so a one-photon rate, ∝ ω³|r|², goes as Z⁶ × Z⁻² = Z⁴: A₂ₚ = 16 × 6.27 × 10⁸ s⁻¹ = 1.003 × 10¹⁰ s⁻¹. The two-photon rate is quoted as 8.23 Z⁶ s⁻¹ = 527 s⁻¹, so 2s lives 1.90 ms.
- Lamb channel: |rfi|² falls to 9a²/Z² while ω grows by 14.04/1.0578 = 13.27, so A = 8.0 × 10⁻¹⁰ s⁻¹ × 13.27³/4 = 4.7 × 10⁻⁷ s⁻¹, still 1.1 × 10⁹ times below the two-photon rate: ω³ shuts it for He⁺ too.
- Quenching: repeat Unit 9's admixture with the element 3a/Z. eaF/(Zh) = 0.640 MHz per V/cm, so with F in V/cm, η₁² = 3(0.640/14 040)²F² = 6.23 × 10⁻⁹F² for 2P₁/₂, and η₂² = 6(0.640/161 460)²F² = 9.4 × 10⁻¹¹F² for 2P₃/₂, which lies 175.5 − 14.04 = 161.46 GHz above 2S₁/₂.
- Γq = (η₁² + η₂²)A₂ₚ = 6.32 × 10⁻⁹ × 1.003 × 10¹⁰ s⁻¹ × F² = 63.4 s⁻¹ × F². Setting it equal to 527 s⁻¹ gives F = √(527/63.4) V/cm = 2.88 V/cm.
- Ratio to hydrogen: 2.88/0.054 = 53. Scaling check: the matching field obeys F² ∝ (two-photon rate) × L²/(|r|²A₂ₚ) ∝ Z⁶L²/(Z⁻² × Z⁴) = Z⁴L², so F ∝ Z²L, and 4 × 13.27 = 53, the small 2P₃/₂ term aside. The Lamb shift grows almost as Z⁴, which is what makes He⁺ metastables so much harder to quench.
AnswerA₂ₚ = 1.00 × 10¹⁰ s⁻¹, two-photon 527 s⁻¹ (τ = 1.90 ms), Lamb channel 4.7 × 10⁻⁷ s⁻¹; quenching matches the two-photon rate at 2.9 V/cm, about 53 times hydrogen's 0.054 V/cm.