University Quantum Mechanics II · Time-Dependent Perturbation Theory · 12.6
Fermi's golden rule derived from a sum over a continuum
Nearly every transition rate you will meet, from atomic decay to electron scattering, is a golden-rule rate, and nearly every wrong one has lost a factor. Derive it from a sum over final states and you will know where the 2π comes from, what ρ counts, why a cosine drive costs a factor of 4, and why the box must drop out.
Build the model
Connect the measurement to the mechanism.
The golden rule is not a new postulate: it is the first-order probability of 12.03 summed over a continuum. For one final state that probability, (4|Vfi|²/ħ²) sin²(xt/2)/x² with ħx = Ef − Eᵢ, either oscillates or grows as t², so no single state ever has a rate. Sum it over final states with ρ(E) per unit energy, though, and everything that depends on time sits in the kernel sin²(xt/2)/x², whose peak rises as t² and narrows as 1/t while its area grows only as πt/2.
At long times it acts as (πt/2)δ(x): energy is conserved to within about ħ/t, the smooth factor |Vfi|²ρ is read at Ef = Eᵢ, and the probability grows linearly at Γ = (2π/ħ)|Vfi|²ρ. A drive V cos ωt splits into two rotating halves of amplitude V/2, so absorption and stimulated emission each run at a quarter of that, read at Eᵢ ± ħω. For free particles ρ comes from counting plane waves in a periodic box, and the box length cancels against their normalisation.
Three assumptions bound the result in time: |Vfi|²ρ must be smooth over the energy ħ/t, the lobe must still cover many levels, and the initial state must be barely depleted. Past Γt ≈ 1 first order has failed, and the exponential decay that replaces it needs the treatment of the next two topics.
- Simple definition
- Fermi's golden rule is the constant rate Γ = (2π/ħ)|Vfi|²ρ at which a weak, steady perturbation drains a state into a continuum of final states with ρ states per unit energy, read at Ef = Eᵢ.
- Example
- A level coupled by |Vfi| = 1.0 μeV to a band holding 1.0 × 10⁶ states per eV leaks at Γ = 2π × (1.0 × 10⁻⁶ eV)² × 1.0 × 10⁶ eV⁻¹ ÷ (6.58 × 10⁻¹⁶ eV s) = 9.5 × 10⁹ s⁻¹, a mean time of 105 ps.
Height |Vfi|²t²/ħ² at x = 0 and zeros at x = ±2π/t. One state oscillates or grows as t²; it never has a steady rate.
Vfi = ⟨f|V|i⟩ in J, x in rad s⁻¹, t in s, P dimensionless. Derived in 12.03 and quoted here.
The peak grows as t² and narrows as 1/t, so its area grows as t. That single power of t is what becomes a constant rate.
x in rad s⁻¹, δ(x) in s. The area πt/2 holds at every t, from the standard integral ∫ sin²u/u² du = π.
Dimensions: J² × J⁻¹ ÷ (J s) = s⁻¹. With |Vfi| in eV and ρ in states per eV, 2π/ħ = 9.55 × 10¹⁵ eV⁻¹ s⁻¹ turns the product straight into s⁻¹.
Γ in s⁻¹, |Vfi| in J, ρ in J⁻¹ (final states per joule). Derived here from the first-order amplitude, so Γ ∝ |Vfi|².
Each rotating half carries V/2, so each rate is a quarter of the static rule at the same V, provided ωt ≫ 2π keeps the two peaks apart.
+ for absorption, − for stimulated emission; V is the full amplitude of the cosine, ω in rad s⁻¹. Derived here.
ρ grows as the volume and as √E. An electron of 1.00 eV in a 1.00 μm box sees 3.41 × 10⁹ states per eV, 0.29 neV apart.
L in m, m in kg, k in m⁻¹, ρ in J⁻¹. One spin state; one state per (2π/L)³ of k-space. Derived here.
Final plane waves carry L(−3/2), so |Vfi|² ∝ L⁻³ cancels this L³. A rate that still depends on L has lost a normalisation.
Ω in sr, dρ/dΩ in J⁻¹ sr⁻¹. Derived here; the form Unit 15 uses to turn a rate into a differential cross-section.
A single final state never gives a rate
Start from what 12.03 derived and this lesson quotes: a perturbation V switched on at t = 0 and then held constant takes |i⟩ to |f⟩ with probability P(t) = (4|Vfi|²/ħ²) sin²(xt/2)/x², where ħx = Ef − Eᵢ. Follow one state. Detuned, it oscillates and never exceeds 4|Vfi|²/(ħx)²: with |Vfi| = 1.0 μeV and ħx = 10 μeV that ceiling is 0.040. On resonance it grows as (|Vfi|t/ħ)², reaching 0.023 after 100 ps and still accelerating, until the exact Rabi solution of 12.05 bends it over. Neither behaviour is a rate. Yet an excited atom in the radiation field, or an electron leaking from a quantum dot into a metal, shows exactly that: dP/dt constant, independent of t. The only way to get it is from many final states at once.
Sum over a continuum and the kernel becomes a delta function
Let the final states lie so close together that the sum over them becomes ∫ρ(Ef) dEf, with ρ the number of states per unit energy. Then P(t) = ∫ρ(Ef)|Vfi|²(4/ħ²) sin²(xt/2)/x² dEf, and with dEf = ħ dx everything that depends on time sits in the kernel sin²(xt/2)/x². Its height at x = 0 is t²/4 and its first zeros are at x = ±2π/t. Its area comes from the standard integral ∫sin²u/u² du = π over the whole line (Parseval's theorem for a square pulse; quoted): putting u = xt/2 gives ∫sin²(xt/2)/x² dx = πt/2. So as t grows the peak rises as t², narrows as 1/t, and holds an area that grows only as t, 90.3% of it inside the central lobe. At long times it acts as (πt/2)δ(x). After 1.0 ps the lobe spans ±4.1 meV of final energy; after 1.0 ns, ±4.1 μeV.
Pull out the smooth factor and read off the rate
If ρ(Ef)|Vfi|² hardly changes across the lobe, it can be taken outside the integral at x = 0, that is at Ef = Eᵢ: P(t) = (4/ħ²)|Vfi|²ρ(Eᵢ) × ħ × πt/2 = (2π/ħ)|Vfi|²ρ(Eᵢ) t. The probability grows linearly, and Γ = dP/dt = (2π/ħ)|Vfi|²ρ(Eᵢ) is Fermi's golden rule, derived. Check it four ways. Dimensions: J² × J⁻¹ ÷ (J s) = s⁻¹. Limits: Γ vanishes as |Vfi|² and as ρ, and Ef = Eᵢ is not imposed but emerges from the long-time limit, sharp only to about 2πħ/t. Order: Γ is built from the first-order amplitude, so it goes as |Vfi|²; when V couples an electron to light, |Vfi|² ∝ e², and a one-photon rate carries one explicit factor of α. Size: |Vfi| = 1.0 μeV into 1.0 × 10⁶ states per eV gives 9.5 × 10⁹ s⁻¹, so the first-order statement P = Γt is trustworthy only while Γt ≪ 1, here for t well below 100 ps.
A cosine drive: two resonances, a quarter each
For H′ = V cos ωt write cos ωt = (e(iωt) + e(−iωt))/2. The first-order amplitude splits into two copies of the constant case, each carrying Vfi/2: one with x replaced by x − ω, peaked at Ef = Eᵢ + ħω (absorption), and one with x + ω, peaked at Ef = Eᵢ − ħω (stimulated emission). Once ωt ≫ 2π the peaks lie many lobe widths apart; their cross term stays bounded while each direct term grows as t, and each gives the static rule with Vfi → Vfi/2: Γ± = (2π/ħ)|Vfi/2|²ρ(Eᵢ ± ħω) = (π/(2ħ))|Vfi|²ρ(Eᵢ ± ħω), derived. The factor 4 is bookkeeping, not an approximation: texts that write H′ = V e(−iωt) + V† e(iωt) keep 2π/ħ because their V is already the half-amplitude. The limit ω → 0 checks it: the peaks merge, the amplitudes V/2 + V/2 add before squaring, and the static rule returns, four times Γ₊, not the sum Γ₊ + Γ₋.
Counting plane waves in a periodic box, and why L cancels
A free particle needs a countable continuum, so put it in a cube of side L with periodic boundaries: ⟨r|k⟩ = e(ik⋅r)/L³⁄², with k = (2π/L)(nₓ, ny, nz) for integers n. Each state owns a cell (2π/L)³ of k-space, so N(k) = (4πk³/3)/(2π/L)³ = L³k³/(6π²) states lie inside radius k. With E = ħ²k²/(2m), dE = ħ²k dk/m and ρ = dN/dE = L³mk/(2π²ħ²), derived; one direction's share is ρ/(4π) per steradian. Check: m³ × kg m⁻¹ ÷ (J² s²) = J⁻¹, and ρ falls as √E to zero at the bottom of the continuum. An electron of 1.00 eV has k = 5.12 × 10⁹ m⁻¹, and in a 1.00 μm box ρ = 3.41 × 10⁹ states per eV. Spin is not counted, since a V that leaves spin alone reaches one spin state. ρ grows as L³, but each final plane wave carries L(−3/2), so for a localised |i⟩ the product |Vfi|²ρ, and Γ, is independent of L. If L survives in a rate, a normalisation has been lost.
The window of times the rule lives in
Each approximation in the derivation becomes a bound on t. Pulling |Vfi|²ρ out needs it smooth across the lobe: if it changes over an energy W, then t ≫ ħ/W, and before that every state in reach is still in its t² phase, so P starts quadratically. Turning the sum into an integral needs many levels inside the lobe, 2πħ/t ≫ 1/ρ, so t ≪ 2πħρ; later a finite box shows its discrete spectrum, a bound that retreats as L³. Keeping the initial amplitude at 1 needs Γt ≪ 1; the decay beyond that belongs to 12.07 and 12.08. A wide window needs ħΓ ≪ W, and atoms supply it. Unit 13 turns the cosine rule, with Einstein's argument, into a hydrogen 2p lifetime of 1.60 ns, quoted here and measured at 1.60 ns. There Γ ∝ α⁵mc²/ħ, so ħΓ/(E₂p − E₁s) = 0.104α³ = 4.0 × 10⁻⁸.
Change one variable at a time
Make the relationship visible.
With d = 0, drag t from 1 to 5 ps: the kernel narrows from ±4.1 to ±0.83 meV and dP/dt climbs from 0.78 to 1.00 of the golden rule. Narrow the band and the climb slows; move it off Eᵢ and the ratio rings. Either way it settles only once t is several ħ/w.
KERNEL ZEROS ±2πħ/t4.14 meV
t IN UNITS OF ħ/w1.5
BAND AT Eᵢ ÷ PEAK1.000
dP/dt ÷ GOLDEN RULE0.781
Live interpretationKERNEL ZEROS ±2πħ/t: 4.14 meV. t IN UNITS OF ħ/w: 1.5. BAND AT Eᵢ ÷ PEAK: 1.000. dP/dt ÷ GOLDEN RULE: 0.781
Catch the common trap
Explain before calculating.
A localised level is driven by H′ = V cos ωt into a band of states that extends both above and below it. For the upward transition |Vfi| = 0.30 μeV, and the band holds 4.0 × 10⁷ states per eV at Eᵢ + ħω but only 1.0 × 10⁷ states per eV at Eᵢ. What is the golden-rule absorption rate?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAt t = 0 a constant perturbation couples a localised electron state to a band of extended states running 60 meV either side of it. Near Eᵢ every band state has |Vfi| = 1.5 μeV, and the band holds 8.0 × 10⁵ states per eV. Find the golden-rule rate and mean time, the first-order probability that the electron has left after 5.0 ps, and check the rule's assumptions at that time.
- The perturbation is constant, so the static rule applies with ρ read at Ef = Eᵢ: Γ = (2π/ħ)|Vfi|²ρ = 2π × (1.5 × 10⁻⁶ eV)² × 8.0 × 10⁵ eV⁻¹ ÷ 6.582 × 10⁻¹⁶ eV s = 1.131 × 10⁻⁵ eV ÷ 6.582 × 10⁻¹⁶ eV s = 1.72 × 10¹⁰ s⁻¹, so 1/Γ = 58 ps.
- In the linear regime P = Γt = 1.72 × 10¹⁰ s⁻¹ × 5.0 × 10⁻¹² s = 0.086.
- Smoothness: the kernel's first zeros sit at Ef − Eᵢ = ±2πħ/t = ±4.136 × 10⁻¹⁵ eV s ÷ 5.0 × 10⁻¹² s = ±0.83 meV, far inside the ±60 meV band; equivalently t ≫ ħ/W = 6.582 × 10⁻¹⁶ eV s ÷ 0.060 eV = 11 fs.
- Continuum: the central lobe, 1.65 meV wide, covers 1.65 × 10⁻³ eV × 8.0 × 10⁵ eV⁻¹ ≈ 1.3 × 10³ levels, and the recurrence time 2πħρ = 4.136 × 10⁻¹⁵ eV s × 8.0 × 10⁵ eV⁻¹ = 3.3 ns is far off.
- Depletion: Γt = 0.086 is small but not negligible. If the decay is exponential, as 12.07 finds, the true loss is 1 − e(−0.086) = 0.082, so the first-order answer is 4% high; at 50 ps it would read 0.86 and mean nothing.
AnswerΓ = 1.7 × 10¹⁰ s⁻¹ and 1/Γ = 58 ps; P(5.0 ps) = 0.086, with the kernel (±0.83 meV) far narrower than the band and about 1.3 × 10³ levels inside it.
MediumA localised state sits 0.50 eV above the bottom of a free-electron continuum and is coupled to it by a static potential V(r). With the continuum taken as plane waves in a periodic box of side L, ⟨k|V|i⟩ = F/L³⁄², where F = ∫e(−ik⋅r)V(r)φᵢ(r) d³r has |F|² = 4.0 × 10⁻³² eV² m³ at this energy. Find the decay rate for L = 0.50 μm, show that L = 1.0 μm gives the same, and find ħΓ.
- Wavenumber at 0.50 eV = 8.011 × 10⁻²⁰ J: k = √(2mE)/ħ = √(2 × 9.109 × 10⁻³¹ kg × 8.011 × 10⁻²⁰ J) ÷ 1.0546 × 10⁻³⁴ J s = 3.62 × 10⁹ m⁻¹.
- Density of states: ρ = L³mk/(2π²ħ²) = (1.25 × 10⁻¹⁹ m³)(9.109 × 10⁻³¹ kg)(3.62 × 10⁹ m⁻¹) ÷ (2π² × 1.112 × 10⁻⁶⁸ J² s²) = 1.88 × 10²⁷ J⁻¹ = 3.01 × 10⁸ states per eV.
- Matrix element: |Vfi|² = |F|²/L³ = 4.0 × 10⁻³² eV² m³ ÷ 1.25 × 10⁻¹⁹ m³ = 3.2 × 10⁻¹³ eV², so |Vfi| = 0.57 μeV.
- Static rule: Γ = (2π/ħ)|Vfi|²ρ = 2π × 3.2 × 10⁻¹³ eV² × 3.01 × 10⁸ eV⁻¹ ÷ 6.582 × 10⁻¹⁶ eV s = 6.05 × 10⁻⁴ eV ÷ 6.582 × 10⁻¹⁶ eV s = 9.2 × 10¹¹ s⁻¹.
- At L = 1.0 μm, ρ grows eightfold to 2.41 × 10⁹ per eV and |Vfi|² falls eightfold to 4.0 × 10⁻¹⁴ eV²; their product stays 9.63 × 10⁻⁵ eV, so Γ is unchanged. In symbols Γ = |F|²mk/(πħ³), with no L left.
- ħΓ = 6.582 × 10⁻¹⁶ eV s × 9.2 × 10¹¹ s⁻¹ = 0.61 meV, about a thousandth of the 0.50 eV over which the √E density of states changes, so the smoothness condition holds with room to spare.
AnswerΓ = 9.2 × 10¹¹ s⁻¹ (1/Γ = 1.1 ps) for either box, and ħΓ = 0.61 meV.
HardAn electron bound at Eᵢ = −2.0 eV is driven by H′ = V cos ωt with ħω = 3.5 eV. In a periodic box of side L, its matrix element to a plane wave at angle θ to the drive's axis has |Vfi|² = (|F|²/L³) cos²θ, with |F| = 4.0 × 10⁻¹⁷ eV m³⁄² at the final energy. Find the rate per unit solid angle along the axis, the total rate, the fraction ejected within 30° of the axis either way, and the window of times in which the rate holds.
- Branches: absorption puts the electron at Eᵢ + ħω = +1.5 eV, inside the free continuum; stimulated emission would need a final state at −5.5 eV, where there is none, so only Γ₊ acts. At 1.5 eV, k = √(2mE)/ħ = 6.27 × 10⁹ m⁻¹.
- Per unit solid angle, with the cosine's quarter: dΓ/dΩ = (π/(2ħ))|Vfi|² dρ/dΩ = (π/(2ħ))(|F|²cos²θ/L³)(L³mk/(8π³ħ²)) = |F|²mk cos²θ/(16π²ħ³). The L³ cancels before any number goes in.
- Along the axis: |F|² = 1.6 × 10⁻³³ eV² m³ = 4.11 × 10⁻⁷¹ J² m³ and mk = 5.72 × 10⁻²¹ kg m⁻¹, so their product is 2.35 × 10⁻⁹¹ J² kg m²; with 16π²ħ³ = 1.85 × 10⁻¹⁰⁰ J³ s³, dΓ/dΩ(0) = 1.27 × 10⁹ s⁻¹ sr⁻¹.
- Total: ∫cos²θ dΩ = 4π/3, so Γ = (4π/3) × 1.27 × 10⁹ s⁻¹ = 5.31 × 10⁹ s⁻¹ and 1/Γ = 0.19 ns. The static rule with the full cosine amplitude would give 2.1 × 10¹⁰ s⁻¹, four times too much.
- Cones: the fraction within θ₀ of the axis, both ways, is 2 × 2π∫cos²θ sinθ dθ (from 0 to θ₀) ÷ (4π/3) = 1 − cos³θ₀ = 1 − 0.650 = 0.35 for θ₀ = 30°.
- Window: the two resonances sit 2ħω = 7.0 eV apart and separate once t ≫ 2π/ω = 1.2 fs; depletion caps t ≪ 1/Γ = 0.19 ns. Since ħΓ = 3.5 μeV is 1.0 × 10⁻⁶ of ħω, the window spans five decades.
AnswerdΓ/dΩ = 1.3 × 10⁹ s⁻¹ sr⁻¹ along the axis; Γ = 5.3 × 10⁹ s⁻¹ (1/Γ = 0.19 ns), independent of L; 35% ejected within 30° of the axis; valid for 1.2 fs ≪ t ≪ 0.19 ns.