University Physics IV · Atomic Physics · 11.8
Selection Rules, Intensities & Lifetimes
Most of the lines an atom could emit, it does not. The electric-dipole matrix element decides which survive, Fermi's golden rule turns the survivors into rates, and those rates come back to you as brightness on a plate and nanoseconds on a scope.
Build the model
Connect the measurement to the mechanism.
A spectral line is not something an atom has; it is a rate at which an atom does something. Fermi's golden rule fixes that rate as (2π/ħ)|⟨f|Ĥ′|i⟩|²ρ(Ef), so everything hangs on one matrix element. Expand the field's e(ik⋅r) about the atom: an optical wavelength is thousands of atomic radii, so ka ≈ 2πa₀/λ ≈ 2.7×10⁻³ at Lyman-α, and only the leading term matters.
That term is the electric dipole −er, an odd operator, so ⟨f|r|i⟩ vanishes unless the two states have opposite parity; the angular integral then sharpens this to Δl = ±1 and Δmₗ = 0, ±1 for the one electron that jumps, and in LS coupling to ΔS = 0, ΔL = 0, ±1 (never 0 → 0) and ΔJ = 0, ±1 (never 0 → 0). What the rules buy is enormous economy: of the transitions a level diagram permits energetically, only a handful carry any strength. What they cost is the approximation underneath.
Drop the leading term and the next one, of relative strength (ka)² ≈ 10⁻⁵, is still there — so forbidden lines are slow, not absent. Hydrogen's 2p decays in 1.6 ns; its 2s, unreachable by any single dipole photon, survives 0.12 s.
- Simple definition
- An electric-dipole selection rule is the statement that the matrix element ⟨f|(−er)|i⟩ vanishes identically between two states, so the transition rate between them is zero at leading order and the line is missing from the spectrum.
- Example
- Hydrogen 3d → 1s has 12.1 eV to give away, but l drops by 2 and parity does not change, so ⟨1s|r|3d⟩ = 0: that photon never appears, and the atom cascades 3d → 2p → 1s at 656.3 nm then 121.6 nm instead.
Every rule below is this one matrix element being zero or not.
R in s⁻¹; Ĥ′ is the atom–field coupling in J; ρ(Ef) counts final states per joule
r is odd, so ψf* r ψᵢ integrates to zero unless the two states have opposite parity.
dfi in C m; r is the position operator, parity (−1)l for a one-electron state
No Δn rule at all: 5p → 1s is as allowed as 2p → 1s, merely weaker.
L = 0 ↛ 0 and J = 0 ↛ 0; configuration parity must change; exactly one electron jumps
In practical units A ≈ 1.06×10⁶ (ħω/eV)³ (d/a₀)² s⁻¹, so cubing the gap dominates.
A in s⁻¹, ω in rad s⁻¹, |d₂₁|² summed over the x, y, z components
One matrix element fixes absorption, stimulated and spontaneous emission together.
B in m³ J⁻¹ s⁻², paired with spectral energy density uν in J m⁻³ Hz⁻¹
1.6 ns gives 99 MHz; a 0.12 s metastable level gives 1.3 Hz.
sum over every open decay channel; Δν is the FWHM in Hz, and ΔE ≈ ħ/τ in joules
The rate comes first, the rule second
Time-dependent perturbation theory turns the atom–field coupling into a transition rate. With a continuum of final photon modes available, first order gives Fermi's golden rule, R = (2π/ħ)|⟨f|Ĥ′|i⟩|²ρ(Ef): a constant probability per unit time rather than an oscillation, because the density of final states smears the energy-conserving δ function. Everything else in this topic is bookkeeping on that one matrix element. A selection rule is therefore not an extra postulate — it is the observation that ⟨f|Ĥ′|i⟩ is identically zero for certain pairs of states, so the rate vanishes at this order and the line is missing. That framing tells you where to look when a forbidden line turns up anyway: either Ĥ′ was truncated too early, or the labels i and f were never exact quantum numbers in the first place.
Why only the dipole term survives
The coupling to a plane-wave field carries a factor e(ik⋅r), and r ranges over the atom, so the natural expansion is e(ik⋅r) = 1 + ik⋅r + …. The small parameter is ka ≈ 2πa/λ, the atom's size divided by the wavelength. For Lyman-α, 2π(0.0529 nm)/(121.6 nm) = 2.7×10⁻³; for the sodium D lines at 589 nm it is 5.6×10⁻⁴. Rates go as the square of the matrix element, so the next terms — magnetic dipole and electric quadrupole — arrive suppressed by (ka)², roughly 10⁻⁵ to 10⁻⁷. Keeping only the leading 1 is the electric-dipole approximation, and it works precisely because atoms are tiny compared with optical light. Push to hard X-rays or nuclear gamma rays, where ka is no longer small, and higher multipoles stop being a footnote.
Parity halves the diagram; the angular integral does the rest
With the leading term kept, the operator is the electric dipole −er. Under r → −r it changes sign, while a one-electron state has parity (−1)l. The integrand ψf* r ψᵢ is then odd unless the two states have opposite parity, and an odd integrand over all space gives zero — Laporte's rule, which already deletes half of every level diagram. The angular integral sharpens it. The three Cartesian components of r are proportional to the l = 1 spherical harmonics, so ∫Y*(lf mf) Y(1q) Y(lᵢ mᵢ) dΩ survives only for lf = lᵢ ± 1 and mf = mᵢ + q with q = 0, ±1. Physically the photon carries one unit of angular momentum and the atom must absorb exactly that. The q = 0 case is light linearly polarised along z, q = ±1 the two circular polarisations — which is why a Zeeman pattern viewed along the field loses its π component.
From one electron to the whole atom
In a multi-electron atom the dipole operator is a sum over electrons, each term acting on one coordinate, so a single electron jumps and the configuration's parity must change. Written in term symbols under LS coupling the rules become ΔS = 0, because −er does not touch spin at all; ΔL = 0, ±1 with L = 0 → L = 0 barred; and ΔJ = 0, ±1 with J = 0 → J = 0 strictly barred, since one unit of photon angular momentum cannot be taken up by two states that both have none. Note what is absent: there is no rule on n, so hydrogen's 5p → 1s is as allowed as 2p → 1s. Note also which rule is softest. ΔS = 0 relies on S being a good quantum number, and spin-orbit coupling grows as Z⁴, so in mercury the 6s6p ³P₁ level reaches the ¹S₀ ground state at 254 nm with a lifetime near 120 ns — about 10² times slower than an allowed line, not 10⁸ times.
Allowed is not the same as bright: the ω³ law
Carrying the golden rule through to spontaneous emission gives A₂₁ = ω³|d₂₁|²/(3πε₀ħc³). Two things set the rate: the matrix element, and the cube of the photon frequency. In practical units A ≈ 1.06×10⁶ (ħω/eV)³ (d/a₀)² s⁻¹. For hydrogen 2p → 1s, |⟨1s|r|2p⟩| = 0.745 a₀ and ħω = 10.20 eV, giving A = 6.3×10⁸ s⁻¹. For the sodium D₂ line, ħω = 2.105 eV and the measured A is 6.2×10⁷ s⁻¹ — only ten times smaller, though the ω³ factor alone would predict 114 times. The difference is the matrix element: sodium's valence orbitals spread over several Bohr radii, so |d| is about 3.3 times hydrogen's. Absorption and stimulated emission share that same element through g₁B₁₂ = g₂B₂₁ and A₂₁/B₂₁ = 8πhν³/c³, which is why at 500 K a thermal field drives stimulated emission at 10⁻²² of the spontaneous rate in the visible: inversions must be built, never waited for.
Lifetimes, linewidths, and the price of forbidden
A level decays through every channel open to it, so τ = 1/Σf Aif, and the branching ratios are the individual A values divided by that sum. A finite lifetime is also a finite energy width: ΔE ≈ ħ/τ, or Δν = 1/(2πτ). Lyman-α's 1.6 ns gives 99 MHz on a 2.47×10¹⁵ Hz line, a fractional width of 4×10⁻⁸ — usually buried under Doppler broadening. Now close the dipole channel. Hydrogen's 2s cannot reach 1s by one dipole photon, so it decays by two-photon emission at 8.2 s⁻¹: a 0.12 s lifetime and a 1.3 Hz natural width. Atomic oxygen's ¹D₂ level radiates at 630.0 nm by a magnetic-dipole route with a lifetime near 110 s, and lights the red aurora only because the upper atmosphere is thin enough that collisions do not de-excite it first. The [O III] 500.7 nm nebular line is the same story. Forbidden lines are missing from a laboratory discharge for a density reason, not a rate-zero reason.
Change one variable at a time
Make the relationship visible.
Hold ħω at 10.2 eV and drag d from 0.75 a₀ down to 0.20 a₀: Lyman-α's 1.6 ns lifetime stretches to 22 ns, because A goes as d². Then move ħω instead and watch the far steeper cube law — and note that the dashed curve sinks but never reaches zero.
LIFETIME τ = 1/A1.58 ns
NATURAL LINEWIDTH101.01 MHz
log₁₀ (AE1 / s⁻¹)8.80
log₁₀ (Aₙₑₓₜ/AE1)-5.13
Live interpretationLIFETIME τ = 1/A: 1.58 ns. NATURAL LINEWIDTH: 101.01 MHz. log₁₀ (AE1 / s⁻¹): 8.80. log₁₀ (Aₙₑₓₜ/AE1): −5.13
Catch the common trap
Explain before calculating.
Each pair below connects two configurations of opposite parity, so the parity requirement is already met. Which one of these transitions is electric-dipole allowed in LS coupling?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA hydrogen atom sits in the 3d level. Decide which of 3d → 2p, 3d → 2s, 3d → 1s and 3p → 2p are electric-dipole allowed, and state how a 3d atom actually reaches the ground state.
- Parity of a one-electron state is (−1)l, and the dipole operator is odd, so l must change by an odd number. The angular integral tightens this to exactly Δl = ±1.
- 3d → 2p: l goes 2 → 1, so Δl = −1 and parity flips. Allowed.
- 3d → 2s: l goes 2 → 0, Δl = −2 with no parity change. Forbidden. 3d → 1s: same Δl = −2, also forbidden, even though 12.1 eV is available.
- 3p → 2p: Δl = 0, so parity is unchanged and the matrix element vanishes — the energy drop of 1.89 eV is irrelevant. Note there is no rule on n: 3p → 1s (Δn = −2) is allowed.
- With only 2p reachable, the 3d atom must cascade: 3d → 2p at 1.889 eV, then 2p → 1s at 10.20 eV.
AnswerOnly 3d → 2p is allowed; the other three all fail on Δl. A 3d atom reaches the ground state in two steps, emitting 656.3 nm (Hα) and then 121.6 nm (Lyman-α).
MediumFor hydrogen the 2p–1s dipole element is |⟨1s|r|2p⟩|² = 0.555 a₀², summed over the three Cartesian components, and the gap is 10.20 eV. Using A = ω³|d|²/(3πε₀ħc³) with |d| = e|⟨1s|r|2p⟩|, find the spontaneous emission rate, the lifetime of the 2p level, and the natural linewidth of Lyman-α.
- Angular frequency: ω = ΔE/ħ = (10.20 × 1.602×10⁻¹⁹ J)/(1.055×10⁻³⁴ J s) = 1.550×10¹⁶ rad s⁻¹, so ω³ = 3.72×10⁴⁸ s⁻³.
- Dipole squared: |d|² = e² × 0.555 a₀² = (1.602×10⁻¹⁹)² × 0.555 × (5.292×10⁻¹¹)² = 3.99×10⁻⁵⁹ C² m².
- Denominator: 3πε₀ħc³ = 9.425 × 8.854×10⁻¹² × 1.055×10⁻³⁴ × 2.694×10²⁵ = 2.371×10⁻¹⁹ in SI.
- A = (3.72×10⁴⁸ × 3.99×10⁻⁵⁹)/(2.371×10⁻¹⁹) = 6.3×10⁸ s⁻¹.
- The 2p level has only one open channel, so τ = 1/A = 1.6×10⁻⁹ s, and Δν = 1/(2πτ) = 1/(2π × 1.60×10⁻⁹ s) = 1.0×10⁸ Hz.
AnswerA = 6.3×10⁸ s⁻¹, τ = 1.6 ns, Δν ≈ 99 MHz. That is 4×10⁻⁸ of the 2.47×10¹⁵ Hz line frequency, which is why Doppler and collision broadening normally hide it.
HardSodium's 3p ²P₃/₂ level decays only to the 3s ²S₁/₂ ground level at 589.0 nm, with a measured lifetime of 16.2 ns. Find A and the natural linewidth; compare the rate with Lyman-α's 6.27×10⁸ s⁻¹ and say what the comparison reveals about the matrix element; then find how stimulated emission compares with spontaneous emission in a 500 K thermal field.
- One channel only, so A = 1/τ = 1/(16.2×10⁻⁹ s) = 6.17×10⁷ s⁻¹, and Δν = 1/(2πτ) = 9.8×10⁶ Hz, that is 9.8 MHz.
- Photon energies: 1239.8/589.0 = 2.105 eV against 10.20 eV for Lyman-α. Since A ∝ ω³|d|², the frequency factor alone predicts a ratio of (10.20/2.105)³ = 4.846³ = 114.
- Measured ratio: 6.27×10⁸/6.17×10⁷ = 10.2. Dividing, |d(Na)|²/|d(H)|² = 114/10.2 = 11.2, so |d| is about 3.3 times larger — roughly 2.5 a₀ against hydrogen's 0.745 a₀, because sodium's 3s and 3p orbitals overlap over several Bohr radii.
- Stimulated versus spontaneous emission in a thermal field is the photon occupation number: B uν/A = 1/(e(hν/kT) − 1). Here hν/kT = 2.105/(8.617×10⁻⁵ × 500) = 2.105/0.0431 = 48.9.
- e48.9 = 1.7×10²¹, so the ratio is 6.1×10⁻²². Spontaneous emission wins by twenty-two orders of magnitude, which is why a hot sodium lamp glows rather than lases: an inversion has to be built, never thermally waited for.
AnswerA = 6.17×10⁷ s⁻¹ and Δν = 9.8 MHz; sodium's dipole matrix element is about 3.3× hydrogen's, and stimulated emission is only 6×10⁻²² of spontaneous emission at 500 K.