University Physics V · Multi-Electron Atoms · 12.8
Transition Rates, the Golden Rule & Lifetimes
First-order perturbation theory hands back a probability, not a rate: quadratic in time, reversible, and sharply peaked in detuning. This topic shows what has to be summed over before a constant rate exists at all, then marks the point where the continuum runs out and Rabi oscillation takes over.
Build the model
Connect the measurement to the mechanism.
Nothing in Ĥ₀ makes an excited atom decay: its eigenstates are stationary and |cₑ|² never moves. A transition is what a coupling does to that picture, so the whole subject is one perturbative move — expand |Ψ⟩ in the eigenbasis of Ĥ₀, integrate the coupled amplitude equations once, and read off |cf(t)|². The result is not a rate.
For a harmonic drive it is (|Vfi|/2ħ)²[sin(Δt/2)/(Δ/2)]², a probability quadratic in t at resonance and perfectly reversible. A rate appears only when that sharp function is summed over something: a continuum of final states, or a spread of drive frequencies. The sinc-squared has height ∝ t² and width ∝ 1/t, so its area grows as t alone, and that linear growth is the rate — Fermi's golden rule, W = (2π/ħ)|Vfi|²ρ(Ef), one factor naming the dynamics and one counting how many states sit on shell.
What it costs is generality: with a monochromatic drive on two sharp levels there is no continuum to sum, and the atom Rabi-oscillates instead of decaying. For spontaneous emission the continuum is supplied by the vacuum's photon modes, whose ω² mode density and √ω field amplitude combine into the ω³ of A = ω³|d|²/(3πε₀ħc³) — the single fact that makes optical lines nanoseconds long and the 21 cm line eleven million years.
- Simple definition
- Fermi's golden rule is the constant transition rate W = (2π/ħ)|⟨f|V̂|i⟩|²ρ(Ef) that first-order perturbation theory yields once the final states form a continuum dense enough to smear out the drive's energy uncertainty ħ/t.
- Example
- Hydrogen's 2p state coupled to the vacuum's photon modes decays at A = ω³|d|²/(3πε₀ħc³) = 6.26×10⁸ s⁻¹, with |d| = 0.745 e a₀ and λ = 121.6 nm — a lifetime of 1.60 ns and a natural linewidth of 99.6 MHz.
Quadratic in t and reversible — a probability, not yet a rate. The rotating-wave step has already dropped the ωfi + ω term.
Vfi = ⟨f|V̂|i⟩ in J for Ĥ′ = V̂ cos ωt; Δ in rad s⁻¹; peak height ∝ t², first zeros at Δ = ±2π/t
Splits the answer in two: a matrix element carrying the dynamics, and a state count carrying the bookkeeping.
ρ(Ef) in J⁻¹, evaluated at Ef = Eᵢ ± ħω; W in s⁻¹. For Ĥ′ = V̂ cos ωt insert |Vfi/2|² instead.
Makes the photon's spatial structure invisible to the atom. The next term gives M1 and E2 rates smaller by ~(k a₀)² ≈ 3×10⁻⁷.
k a₀ = 2π a₀/λ = 5.6×10⁻⁴ at 589 nm; dfi in C m, of order e a₀ = 8.48×10⁻³⁰ C m
One ω³ from mode density times vacuum field amplitude: optical lines live nanoseconds, radio lines megayears.
ω = (Eᵢ − Ef)/ħ in rad s⁻¹; A in s⁻¹; Δν the Lorentzian FWHM in Hz; the 1/3 is the polarisation average
Planck's law at equilibrium fixes all three from one: measure an absorption strength and the lifetime follows.
u(ν) is spectral energy density in J m⁻³ Hz⁻¹, so B is in m³ J⁻¹ s⁻²; n̄ = 1/(e(hν/kT) − 1)
Two sharp levels and one frequency: the population returns. No continuum, no ρ(Ef), no golden rule.
E₀ the field amplitude in V m⁻¹; Ω₀, Ω, Δ in rad s⁻¹. Small t gives P ≈ (Ω₀t/2)², the first-order result.
Integrate once: the amplitude for a harmonic drive
Nothing decays under Ĥ₀ alone, so add the coupling and expand in the unperturbed eigenbasis: |Ψ(t)⟩ = Σₙ cₙ(t) e(−iEₙ t/ħ)|n⟩. Substituting into iħ∂ₜ|Ψ⟩ = (Ĥ₀ + Ĥ′)|Ψ⟩ and projecting on ⟨f| gives the exact chain ċf = (1/iħ) Σₙ cₙ ⟨f|Ĥ′|n⟩ e(iω_fn t), with ωfn = (Ef − Eₙ)/ħ. First order means cutting the chain at one link: set cᵢ = 1 and every other amplitude to zero on the right-hand side. For a harmonic drive Ĥ′ = V̂ cos ωt = (V̂/2)(e(iωt) + e(−iωt)) the integral is elementary, and its two terms carry denominators ωfi − ω and ωfi + ω. Near resonance the first is small and the second is not, so drop the counter-rotating term — the rotating-wave approximation — and square what is left: Pᵢ→f(t) = (|Vfi|/2ħ)²[sin(Δt/2)/(Δ/2)]² with Δ = ωfi − ω. On resonance that is |Vfi|²t²/(4ħ²): quadratic in time, reversible, and not a rate.
The rate is an area, not a height
Watch that function as t grows. Its peak at Δ = 0 rises as t²; its first zeros sit at Δ = ±2π/t, so its width falls as 1/t; and its area is fixed by ∫[sin(Δt/2)/(Δ/2)]² dΔ = 2πt, growing only linearly. Height times width, not height alone, is what survives — which is why a rate exists only when something integrates over Δ. Give the final states a smooth density ρ(Ef) spanning the peak; with Ef = Eᵢ + ħω + ħΔ the energy integral supplies one factor of ħ, and the sum becomes Pₜₒₜ(t) = (|Vfi|²/4ħ²)⋅ħ⋅2πt⋅ρ(Ef) = (2π/ħ)|Vfi/2|²ρ(Ef) t. Keep that ½: it is the amplitude of each rotating half of V̂ cos ωt, and it is why the golden-rule card carries the same warning. The result is linear in t, so W = dP/dt is constant: Fermi's golden rule. Energy is conserved only to within ħ/t, which is the sinc's width read as an energy. The price is a time window with two walls: the sinc must be narrow compared with the band, t ≫ ħ/ΔEband, and first order must still hold, P ≪ 1, so t ≪ 1/W. For a band 1 THz wide, ħ/ΔEband = 1/(2π × 10¹² Hz) = 0.16 ps, so feeding a transition of rate 10⁸ s⁻¹ the window reads 0.16 ps ≪ t ≪ 10 ns — almost five decades wide, and exponential decay is what the rule predicts inside it, not outside.
Two sharp levels and one frequency: Rabi, not a rate
Remove the continuum and the rate goes with it. Truncate to span(|g⟩, |e⟩), make the rotating-wave approximation, and in the frame rotating at the drive frequency the Hamiltonian is time-independent: Ĥᵣₒₜ = (ħ/2)(Ω₀ σ̂ₓ − Δ σ̂z), with Ω₀ = |dfi|E₀/ħ. Its eigenvalues are ±ħΩ/2 with Ω = √(Ω₀² + Δ²), so the state precesses about a tilted axis on the Bloch sphere and Pₑ(t) = (Ω₀²/Ω²) sin²(Ωt/2). Expand for small t and (Ω₀t/2)² comes back — first-order theory is the tangent parabola at the origin, nothing more. At long times the population returns to zero rather than settling at any rate. Sodium's D₂ transition has |dfi| = 2.99×10⁻²⁹ C m, so a field of 5.0 kV m⁻¹ gives Ω₀ = 1.42×10⁹ rad s⁻¹ and a Rabi period 2π/Ω₀ = 4.4 ns — about three and a half cycles inside the 16.2 ns lifetime, which is why the oscillation is visible at all. The golden rule returns only when the drive carries bandwidth, or the final states form a continuum.
The dipole approximation, and why almost every line is E1
Which V̂ goes into the matrix element? In the Coulomb gauge the atom–field coupling is Ĥ′ = (e/mₑ) Â⋅p̂, and a mode of the field carries the factor e(i k⋅r). Expand it as 1 + i k⋅r + … and compare kr with 1 across an atom. For the sodium D line k a₀ = 2π a₀/λ = 2π(0.0529 nm)/(589 nm) = 5.6×10⁻⁴, so truncating at the leading 1 costs almost nothing. That truncation is the electric-dipole approximation. Using [r̂, Ĥ₀] = iħ p̂/mₑ converts ⟨f|p̂|i⟩ into i mₑ ωfi ⟨f|r̂|i⟩, so the surviving operator is the dipole d̂ = −e r̂ — odd under parity, a rank-one spherical tensor, and the source of every E1 selection rule. Keep the next term instead and you get M1 and E2 amplitudes down by k a₀, hence rates down by (k a₀)² ≈ 3×10⁻⁷. Hydrogen's 2s state is where that matters: E1-forbidden by parity, since ⟨1s|r̂|2s⟩ vanishes between two even states, it decays by two-photon emission with τ = 0.12 s, some 7.5×10⁷ times slower than 2p.
Spontaneous emission: the vacuum supplies the continuum
Spontaneous emission has no external drive, so where is its continuum? It is the electromagnetic vacuum. Quantise the field in a box of volume V: the accessible final states are one-photon states, and their density is ρ(ω) dω = V ω² dω/(π²c³) with both polarisations counted. The coupling to an empty mode carries the vacuum field amplitude √(ħω/2ε₀V), whose square supplies one more power of ω, and V cancels between the two. Feed both into the golden rule and average |ε̂⋅dfi|² over directions and polarisations: Afi = ω³|dfi|²/(3πε₀ħc³). That ω³ is the most useful fact here. Lyman-α at ω = 1.55×10¹⁶ rad s⁻¹ with |d| = 0.745 e a₀ gives A = 6.26×10⁸ s⁻¹ and τ = 1.60 ns. Hydrogen's 21 cm line sits at ω = 8.93×10⁹ rad s⁻¹, lower by 1.74×10⁶, so ω³ alone costs 5.2×10¹⁸; being magnetic dipole costs a further (μB/c e a₀)² = 1.3×10⁻⁵. Together they predict A ≈ 1.6×10⁻¹⁵ s⁻¹ against a measured 2.85×10⁻¹⁵ s⁻¹ — a lifetime of 11 million years.
Einstein's three coefficients under one thermal constraint
Einstein assembled the same accounting in 1917 with no quantum field theory to hand. Put two levels of degeneracy g₁ and g₂ in a bath of spectral energy density u(ν): absorption removes population at B₁₂N₁u, stimulated emission returns it at B₂₁N₂u, and spontaneous emission at A₂₁N₂. Impose steady state, insert N₂/N₁ = (g₂/g₁) e(−hν/kT), and demand the result reproduce Planck's u(ν) at every temperature. Two identities are forced: g₁B₁₂ = g₂B₂₁ and A₂₁/B₂₁ = 8πhν³/c³. The first says absorption and stimulated emission are one matrix element seen twice. The second says neither emission channel can be left out, and the two failures are different. Set A₂₁ = 0 and the balance collapses to B₁₂N₁u = B₂₁N₂u, which the ratio identity satisfies only at u = 0: with nothing to seed emission there is no thermal field at all. Keep A₂₁ but drop the stimulated term instead and the balance gives u = (8πhν³/c³) e(−hν/kT), Wien's law — right in the high-frequency tail and wrong at low frequency, because stimulated emission is precisely what turns the Wien exponential into Planck's 1/(e(hν/kT) − 1), whose hν ≪ kT limit is Rayleigh–Jeans. The practical reading is the ratio B₂₁u/A₂₁ = 1/(e(hν/kT) − 1) = n̄, the photon occupation of a single mode. At 589 nm a 3000 K filament has n̄ = 2.9×10⁻⁴, so a lamp emits almost entirely spontaneously; a laser beats that by loading one mode, not by being hot.
Change one variable at a time
Make the relationship visible.
Push t to 3 ns: the peak climbs as t² while the lobe collapses inside the dashed band, so only the area — linear in t — survives the sum, and that linear growth is the golden-rule rate. Pull t back to 0.5 ns, or squeeze w to 0.2 GHz, and the lobe spills past the band with no rate to define; note too that by t = 3 ns the peak has reached P = 0.36, so first order is already straining against the window's second wall.
PEAK P at resonance0.090
MAIN LOBE 2/t1.33 GHz
AREA ∫P dΔν0.060 GHz
LOBE ÷ BAND WIDTH0.44
Live interpretationPEAK P at resonance: 0.090. MAIN LOBE 2/t: 1.33 GHz. AREA ∫P dΔν: 0.060 GHz. LOBE ÷ BAND WIDTH: 0.44
Catch the common trap
Explain before calculating.
A monochromatic laser is tuned exactly onto resonance between two sharp atomic levels, and the interval is short enough that spontaneous emission can be ignored. First-order perturbation theory gives an excitation probability growing as t². What does the exact two-level solution give at long times?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasySodium's 3p level decays only to the 3s ground state, with Einstein coefficient A = 6.17×10⁷ s⁻¹ at λ = 589.0 nm. Find the natural lifetime, the natural linewidth in hertz, and the fractional width Δν/ν₀. Then compare it with the 1.70 GHz Doppler width of a 500 K vapour cell.
- One decay channel, so the sum 1/τ = Σf Afi has a single term: τ = 1/(6.17×10⁷ s⁻¹) = 1.62×10⁻⁸ s = 16.2 ns.
- The population decays as e(−t/τ), so the emitted field decays as e(−t/2τ) and its Fourier transform is a Lorentzian of FWHM Δν = 1/(2πτ) = A/2π = 6.17×10⁷/6.283 = 9.82×10⁶ Hz = 9.82 MHz.
- The line centre is ν₀ = c/λ = 2.998×10⁸/5.890×10⁻⁷ = 5.090×10¹⁴ Hz, so Δν/ν₀ = 9.82×10⁶/5.090×10¹⁴ = 1.93×10⁻⁸ — a quality factor of 5.2×10⁷.
- Against the Doppler width the natural width is smaller by 1.70×10⁹/9.82×10⁶ = 173, so an ordinary absorption scan measures the oven temperature and not the lifetime; recovering A needs a Doppler-free method.
Answerτ = 16.2 ns, natural FWHM Δν = 9.82 MHz, Δν/ν₀ = 1.93×10⁻⁸. The 500 K Doppler width is 173 times larger, so the natural line is buried unless Doppler broadening is suppressed.
MediumFor hydrogen the radial and angular integrals give ⟨1s|ẑ|2p₀⟩ = 0.7449 a₀, and Lyman-α lies at 121.57 nm. Use A = ω³|dfi|²/(3πε₀ħc³) to find the spontaneous emission rate of the 2p state, its lifetime and its natural linewidth. Take a₀ = 5.292×10⁻¹¹ m and e = 1.602×10⁻¹⁹ C.
- Angular frequency of the emitted photon: ω = 2πc/λ = 2π(2.998×10⁸)/(1.2157×10⁻⁷) = 1.5495×10¹⁶ rad s⁻¹.
- Dipole matrix element: |dfi| = e × 0.7449 a₀ = (1.602×10⁻¹⁹)(0.7449)(5.292×10⁻¹¹) = 6.315×10⁻³⁰ C m, so |dfi|² = 3.988×10⁻⁵⁹ C² m².
- Numerator: ω³|dfi|² = (1.5495×10¹⁶)³ × 3.988×10⁻⁵⁹ = 3.720×10⁴⁸ × 3.988×10⁻⁵⁹ = 1.484×10⁻¹⁰.
- Denominator: 3πε₀ħc³ = 9.425 × 8.854×10⁻¹² × 1.0546×10⁻³⁴ × 2.6946×10²⁵ = 2.371×10⁻¹⁹ in SI units.
- A = 1.484×10⁻¹⁰/2.371×10⁻¹⁹ = 6.26×10⁸ s⁻¹, so τ = 1/A = 1.60 ns and Δν = A/2π = 9.96×10⁷ Hz = 99.6 MHz.
- The same A holds for all three 2p states: for m = 0 only ⟨1s|ẑ|2p₀⟩ survives, while for m = ±1 the x and y elements are each 0.7449a₀/√2, so Σ|⟨1s|r̂|2pₘ⟩|² = 0.7449²a₀² in every case.
AnswerA(2p→1s) = 6.26×10⁸ s⁻¹, τ = 1.60 ns, natural linewidth 99.6 MHz — identical for m = −1, 0 and +1, as rotational invariance requires.
HardFor the sodium D line at 589.0 nm the upper level has A₂₁ = 6.17×10⁷ s⁻¹. (a) Use A₂₁/B₂₁ = 8πhν³/c³ to find B₂₁. (b) Show that stimulated emission matches spontaneous emission when the mean photon number per mode reaches one, and find the blackbody temperature that achieves it. (c) Evaluate the same ratio inside a 3000 K filament lamp.
- ν = c/λ = 2.998×10⁸/5.890×10⁻⁷ = 5.090×10¹⁴ Hz, so hν = 6.626×10⁻³⁴ × 5.090×10¹⁴ = 3.373×10⁻¹⁹ J = 2.105 eV.
- A₂₁/B₂₁ = 8πhν³/c³ = (1.665×10⁻³²)(1.319×10⁴⁴)/(2.695×10²⁵) = 8.15×10⁻¹⁴ J s m⁻³, hence B₂₁ = 6.17×10⁷/8.15×10⁻¹⁴ = 7.57×10²⁰ m³ J⁻¹ s⁻².
- In a thermal field u(ν) = (8πhν³/c³)/(e(hν/kT) − 1), so the ratio of stimulated to spontaneous emission is B₂₁u/A₂₁ = 1/(e(hν/kT) − 1) = n̄, the mean photon number in one mode. It equals 1 when e(hν/kT) = 2.
- hν/kT = ln 2 gives T = hν/(k ln 2) = 3.373×10⁻¹⁹/(1.381×10⁻²³ × 0.6931) = 3.52×10⁴ K, far hotter than any lamp filament.
- At T = 3000 K, kT = 4.142×10⁻²⁰ J, so hν/kT = 3.373×10⁻¹⁹/4.142×10⁻²⁰ = 8.143. Then e8.143 = 3439, and n̄ = 1/(3439 − 1) = 1/3438 = 2.91×10⁻⁴.
AnswerB₂₁ = 7.57×10²⁰ m³ J⁻¹ s⁻²; stimulated emission matches spontaneous only at T = 3.5×10⁴ K, and a 3000 K lamp emits 3.4×10³ times more spontaneously than by stimulation. A laser wins by loading one mode, not by being hot.