Skip to main content

Physics IV course sequence · Interactive lesson

Selection rules, intensities and 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.

UPIV-11.08 · Lesson 8 of 9 in this unit. Position is not a completion score.
Starting model: Selection rules, intensities and lifetimesLog plot of spontaneous emission rate against photon energy. Solid: electric-dipole rate A = 1.06×10⁶ (ħω/eV)³(d/a₀)² per second. Dashed: next multipole, weaker by (2πa/λ)². At ħω = 10.2 eV and d = 0.75 a₀, the E1 rate is 10^8.80 s⁻¹ — a lifetime of 1.58 ns — while the forbidden route runs 10^5.13 times slower.1 s⁻¹10⁵10¹⁰spontaneous rate A / s⁻¹ (log)solid E1 · dashed forbiddennext/E1 = 10-5.13log₁₀A(E1) = 8.80photon energy ħω / eV 1 → 12
Start with a picture. Change the model in the lab below.
1

Build the model

A spectral line is a rate, and most rates are zero for a reason.

The key idea

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.

The complete physical model and assumptions

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 electric-dipole photon, survives 0.12 s.

1The 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.

2Why 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, whose valence electron spreads over about 4a₀, it is 2.3×10⁻³. 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. Keeping only the leading 1 is the electric-dipole approximation, and it works precisely because atoms are tiny compared with optical light. Push to the hard X-rays of heavy atoms, where the photon energy grows as Z² while the inner shell shrinks only as a₀/Z, so ka grows in proportion to Zα and reaches about 0.2 for lead, and higher multipoles stop being a footnote.

3Parity 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.

4From 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 steeply with nuclear charge (as Z⁴ in a hydrogen-like ion, more slowly for a screened valence electron), 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.

5Allowed 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/ea₀)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 on the sodium line at only 6×10⁻²² of the spontaneous rate: inversions must be built, never waited for.

6Lifetimes, 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 electric-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.

2

Key equations

Read the units, assumptions and limits before substituting.

Fermi's golden rule

R(i→f) = (2π/ħ) |⟨f|Ĥ′|i⟩|² ρ(Ef)

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

Electric-dipole matrix element

dfi = ⟨f|(−e r)|i⟩, nonzero only if parity changes

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

The rules: one electron, LS coupled

Δl = ±1, Δmₗ = 0, ±1 · ΔS = 0 · ΔL = 0, ±1 · ΔJ = 0, ±1

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

Spontaneous rate — the ω³ law

A₂₁ = ω³ |d₂₁|² / (3πε₀ħc³)

In practical units A ≈ 1.06×10⁶ (ħω/eV)³ (d/ea₀)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

Einstein relations

g₁B₁₂ = g₂B₂₁ · A₂₁/B₂₁ = 8πhν³/c³

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

Lifetime and natural linewidth

τ = 1 / Σf Aif · Δν = 1/(2πτ)

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
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

10.2 eV
1 to 12 eV; steps of 0.1.
0.75 a₀
0.2 to 3 a₀; steps of 0.05.
1.0 a₀
0.5 to 3 a₀; steps of 0.1.

Try this. 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.

Live results

LIFETIME τ = 1/A
1.58 ns
NATURAL LINEWIDTH
101.01 MHz
log₁₀ (AE1 / s⁻¹)
8.80
log₁₀ (Aₙₑₓₜ/AE1)
-5.13

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics IV modelLog plot of spontaneous emission rate against photon energy. Solid: electric-dipole rate A = 1.06×10⁶ (ħω/eV)³(d/a₀)² per second. Dashed: next multipole, weaker by (2πa/λ)². At ħω = 10.2 eV and d = 0.75 a₀, the E1 rate is 10^8.80 s⁻¹ — a lifetime of 1.58 ns — while the forbidden route runs 10^5.13 times slower.1 s⁻¹10⁵10¹⁰spontaneous rate A / s⁻¹ (log)solid E1 · dashed forbiddennext/E1 = 10-5.13log₁₀A(E1) = 8.80photon energy ħω / eV 1 → 12
Log plot of spontaneous emission rate against photon energy. Solid: electric-dipole rate A = 1.06×10⁶ (ħω/eV)³(d/a₀)² per second. Dashed: next multipole, weaker by (2πa/λ)². At ħω = 10.2 eV and d = 0.75 a₀, the E1 rate is 108.80 s⁻¹ — a lifetime of 1.58 ns — while the forbidden route runs 105.13 times slower.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Photon energy ħω: 10.2 eV; Dipole element |⟨f|r|i⟩|: 0.75 a₀; Radiator size a: 1.0 a₀

Baseline 10.2; choose 1 to 12, in steps of 0.1.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

"Forbidden" does not mean impossible; it means the electric-dipole term vanishes, so the transition takes whatever route is left. Hydrogen's 2s has no single-photon electric-dipole route to 1s, yet it gets there by two-photon emission at 8.2 s⁻¹ — a 0.12 s lifetime, not an infinite one — and the auroral red line at 630.0 nm is electric-dipole forbidden, radiates by a magnetic-dipole route, and still lights the sky. The mirror-image error is treating a selection rule as a law rather than a consequence: ΔS = 0 follows from the dipole operator ignoring spin, so it fails exactly where spin-orbit coupling mixes singlet and triplet, which is why mercury's 254 nm intercombination line is a hundred times slower than an allowed line rather than a hundred million. And there is no Δn rule anywhere: 5p → 1s is perfectly allowed.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

A 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

For 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

Sodium'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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

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?
7

Review & continue

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.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.