University Quantum Mechanics II · Time-Dependent Perturbation Theory · 12.8
The golden rule's time window, and non-exponential decay
Every lifetime in a table presumes that a decay rate exists. This lesson finds when it does: the stretch of times between the band's memory ħ/W and the depletion time 1/Γ, why every decay nonetheless starts quadratically, and how Weisskopf and Wigner carry the exponential past first order.
Build the model
Connect the measurement to the mechanism.
Fermi's golden rule is a statement about a window of times, not about all of them. Summed over a band, the first-order probability of having left a level is an integral of |V|²ρ against a kernel of area 2πt/ħ and width about 2πħ/t. Only when that width is small compared with the energy range W over which |V|²ρ is flat does the integral collapse to Γt with Γ = (2π/ħ)|V|²ρ, and only while Γt ≪ 1 is first order itself trustworthy. The rate therefore lives between two walls, ħ/W ≪ t ≪ 1/Γ, which leave room between them only if the level's own width ħΓ is far below W.
Outside the window the physics changes. Before it, any state of finite energy spread survives as 1 − (ΔH)²t²/ħ², a consequence of the Taylor series of e(−iHt/ħ) alone, so no decay begins exponentially, and checking the level often enough slows it: the quantum Zeno effect. After it, first order fails but decay does not. Weisskopf and Wigner feed the shrinking amplitude back into the coefficient equations and recover e(−Γt), with a principal-value shift of the level, for as long as ħΓ ≪ W.
Exponential decay is thus a middle-time law, bounded by the quadratic at the start and by a power-law tail at very late times, forced by the spectrum's lower bound. For hydrogen's 2p level ħΓ/W ≈ 4 × 10⁻⁸, about 0.1α³, so the linear stretch runs from 0.06 fs to 1.6 ns and the exponential long outlasts it.
- Simple definition
- The golden rule's time window is the range ħ/W ≪ t ≪ 1/Γ in which a level leaves at the constant rate Γ; before it every decay grows as t², and after it only a resummation beyond first order keeps the decay exponential.
- Example
- A level with Γ = 1.0 × 10⁸ s⁻¹ coupled to a band flat over W = 1.0 eV has ħ/W = 0.66 fs and 1/Γ = 10 ns: ħΓ = 66 neV lies 1.5 × 10⁷ times below W, so the window spans seven decades of time.
Fₜ peaks at t²/ħ², is about 2πħ/t wide and has area 2πt/ħ. Once it is narrower than the band only its area counts, and 1 − S = Γt.
Quoted from the golden-rule derivation. |V|² in J², ρ in J⁻¹, ε in J, Fₜ in J⁻², so 1 − S is dimensionless.
Room between the walls needs ħΓ ≪ W. For a flat band centred on Eᵢ, 10% accuracy needs t ≳ 11ħ/W below and Γt ≲ 0.19 above, so the window opens only for W above about 60ħΓ.
W in J is the energy range over which |V|²ρ is flat; Γ in s⁻¹. Both walls are derived here.
dS/dt = 0 at t = 0, while e(−Γt) starts with slope −Γ. No decay begins exponentially: the rate starts at zero.
(ΔH)² = ⟨H²⟩ − ⟨H⟩² in J², t in s. Derived here for any state of finite ΔH; the t⁴ term stays small only for t ≪ ħ/ΔH and, for a level in a band, t ≪ ħ/W, the stricter wall.
The parabola (ΔH)²t²/ħ² meets the line Γt at tc = 2πħ/W, whatever the coupling strength: the band alone sets when the rate switches on.
Vfi in J, ρ in J⁻¹, the sum over band states f ≠ i. Derived here, with H₀|i⟩ = Eᵢ|i⟩.
It resums the Dyson series. The 1/Γ wall bounds first order, not the exponential, which runs on for many lifetimes with corrections of order ħΓ/W.
Γ in s⁻¹ is the golden-rule rate, δE in J. Outlined here and quoted; needs ħΓ ≪ W and t ≫ ħ/W.
Band states above Eᵢ push the level down and those below push it up, so a band symmetric about Eᵢ gives zero. The width uses E = Eᵢ alone; the shift uses the whole band.
P.V. is the principal value; E and δE in J. Quoted: the second-order shift with the resonant pole removed.
Two walls around linear growth
The golden-rule derivation, quoted here, gives the first-order probability of having left a level coupled by V to a band: 1 − S(t) = ∫ dE |V|²ρ Fₜ(E − Eᵢ), with Fₜ(ε) = 4 sin²(εt/2ħ)/ε². The kernel peaks at t²/ħ², has its first zeros at ε = ±2πħ/t, and has area 2πt/ħ. When it is much narrower than the range W over which |V|²ρ is flat, |V|²ρ leaves the integral and 1 − S = (2π/ħ)|V|²ρ t = Γt. That is the lower wall, t ≫ ħ/W, and it is derived rather than assumed: for a flat band centred on Eᵢ, (1 − S)/Γt is 0.16 at t = ħ/W and 0.77 at 2πħ/W, and stays within 10% of 1 only beyond about 11ħ/W. The upper wall belongs to first order itself, which set cᵢ = 1 on the right-hand side and so needs Γt ≪ 1. At Γt = 0.1 it overshoots the true loss by 5%; at Γt = 1 it predicts certain departure where 0.63 is correct. Between the walls there is room only if ħΓ ≪ W; asking for 10% accuracy at both ends, only if W exceeds about 60ħΓ. A harmonic perturbation V cos ωt obeys the same walls, with V/2 in place of V and Eᵢ + ħω in place of Eᵢ.
Every decay starts with zero slope
Take any normalised |ψ⟩ with finite ⟨H²⟩ and expand its survival amplitude: A(t) = ⟨ψ|e(−iHt/ħ)|ψ⟩ = 1 − i⟨H⟩t/ħ − ⟨H²⟩t²/2ħ² + O(t³). Re A holds only even powers of t and Im A only odd ones, so |A|² has no odd powers, and Im A supplies the ⟨H⟩²t²/ħ² term: S(t) = 1 − (ΔH)²t²/ħ² + O(t⁴). That is the whole derivation, and it uses no perturbation theory. Hence dS/dt = 0 at t = 0, while e(−Γt) starts with slope −Γ: no decay begins exponentially. First order agrees. For t ≪ ħ/W the kernel is t²/ħ² across the whole band, so 1 − S = (t²/ħ²)Σf |Vfi|², and for |i⟩ an eigenstate of H₀ the variance of H₀ + V is exactly Σf |Vfi|² over f ≠ i. Exponential decay from t = 0 would need infinite ΔH, and a Lorentzian energy distribution has exactly that: no finite variance. The flat start has been measured. In 1997 Wilkinson and co-workers watched sodium atoms tunnel out of an accelerated optical lattice and found survival that was non-exponential at first, before settling into exponential decay.
The crossover time, and the quantum Zeno effect
For a flat band (ΔH)² = |V|²ρW = ħΓW/2π, so the parabola (ΔH)²t²/ħ² meets the line Γt at tc = 2πħ/W, a time in seconds since ħ/W is J s over J. The coupling sets Γ; the band's width alone sets when the rate switches on, and as W → ∞ the quadratic start shrinks to nothing while ΔH diverges. Suppose the level is checked by a projective measurement every τ ≪ tc. Each check finds it with probability 1 − (ΔH)²τ²/ħ² and restarts the clock, so after N = T/τ checks S = [1 − (ΔH)²τ²/ħ²]N ≈ exp(−(ΔH)²τT/ħ²). The effective rate (ΔH)²τ/ħ² = Γτ/tc falls as the checks come faster: this is the quantum Zeno effect, named here rather than developed. For hydrogen 2p, W ≈ 10 eV gives tc ≈ 0.4 fs, so slowing its decay tenfold would need a check every 40 as, which is why spontaneous emission is not frozen this way in practice. It has been seen where the energy scale is small: in 1990 Itano and co-workers suppressed a 256 ms radio-frequency transition in trapped ⁹Be⁺ ions with up to 64 laser pulses, and in 2001 Raizen's group slowed the tunnelling of cold sodium atoms.
Past first order: the Weisskopf–Wigner outline
Let V couple the level only to the band. Integrate the band equations, iħċf = Vfi e(iωfi t)cᵢ with ωfi = (Ef − Eᵢ)/ħ, and substitute into the level's own equation. That is derived in two lines and is exact: ċᵢ(t) = −(1/ħ²)∫₀ᵗ K(t − t′)cᵢ(t′)dt′, with the memory kernel K(τ) = ∫ dE |V|²ρ e(i(Eᵢ − E)τ/ħ). K(0) = (ΔH)², and K dies away after about ħ/W as the band's phases spread. If cᵢ barely changes over that memory time, which is the condition ħΓ ≪ W again, take cᵢ(t) outside the integral and extend it to infinity, using ∫₀^∞ e(ixτ)dτ = πδ(x) + i P.V.(1/x). The result, quoted here rather than derived with error bounds, is ċᵢ = −(Γ/2 + iδE/ħ)cᵢ, so S(t) = e(−Γt) at the golden-rule rate. It holds for t ≫ ħ/W and for many lifetimes: the 1/Γ wall belonged to first order, not to the decay. For a Lorentzian band of full width W centred on Eᵢ the kernel is exponential, and the equation can be solved exactly, as the figure does. The late-time rate is Γ(1 + ħΓ/W + …): 1.127Γ at W = 10ħΓ and 1.001Γ at W = 1000ħΓ, errors of first order in ħΓ/W.
The level shift, checked against hydrogen
The exponent's imaginary part gives δE = P.V.∫ dE |V|²ρ/(Eᵢ − E), in J since J² × J⁻¹ × J ÷ J = J. It is the second-order shift Σf |Vfi|²/(Eᵢ − Ef) made finite at resonance: 1/(x + i0) = P.V.(1/x) − iπδ(x) sends the principal value to the shift and the delta function to the decay. Band states above Eᵢ push the level down and those below push it up, so a symmetric band gives no shift. The width is local, set by |V|²ρ at Eᵢ alone; the shift draws on the whole band. Both are of order |V|²ρ, so δE and ħΓ are usually comparable. Hydrogen tests this. The golden rule for Lyman α, applied in Unit 13 and quoted here, gives Γ = 6.265 × 10⁸ s⁻¹, a 2p lifetime of 1.596 ns to set beside the measured 1.60 ns. Because |V|²ρ ∝ ω³|d|² changes on the scale of ħω = 10.2 eV, ħΓ/W ≈ 4.0 × 10⁻⁸ ≈ 0.10α³, and the window runs from 0.06 fs to 1.6 ns. The radiative shift is the non-relativistic seed of the Lamb shift; renormalised by quantum electrodynamics, quoted here, it grows into the measured 1057.8 MHz, or 4.37 μeV. The 2p width and the Lamb shift are both of order α⁵mc² = 10.6 μeV: 0.039 and 0.41 of it.
Where the exponential ends
An S(t) that is exactly exponential at all times needs an energy distribution that is Lorentzian over every energy, arbitrarily negative ones included. A real Hamiltonian is bounded below, and Khalfin showed in 1957, from the Paley–Wiener theorem, that the survival amplitude must then eventually fall more slowly than any exponential. Near a threshold Eₜₕ the spectral weight vanishes as (E − Eₜₕ)ν, and the end point of the Fourier integral leaves a tail A(t) ∝ t(−(ν+1)), so S ∝ t(−2(ν+1)), which is t⁻³ for ν = ½. That is quoted, not derived. The tail wins only after many lifetimes, once the exponential has sunk beneath it, which is why it went unobserved until 2006, in the late luminescence of dissolved organic molecules. The whole history of a decaying level is therefore a quadratic start lasting about ħ/W, exponential decay at the golden-rule rate for many lifetimes, and a power law at the end. What gathers in the band states meanwhile, a Lorentzian of full width ħΓ centred on Eᵢ + δE, is the line shape, left to Unit 13.
Change one variable at a time
Make the relationship visible.
At W = 1000ħΓ the solid curve rises near ħ/W, rides the golden-rule line inside the band, then follows the dashed exponential past 1/Γ. Below about 57ħΓ it never reaches the band. Park the dot left of ħ/W and the Zeno rate collapses. At the narrowest band, W ≈ 5.6ħΓ, even the late decay runs 30% fast.
ZENO RATE AT THE DOT ÷ Γ0.938
SURVIVAL ÷ PURE EXPONENTIAL1.002
LATE DECAY RATE ÷ Γ1.001
WINDOW, DECADES, ×10 RULE1.00
Live interpretationZENO RATE AT THE DOT ÷ Γ: 0.938. SURVIVAL ÷ PURE EXPONENTIAL: 1.002. LATE DECAY RATE ÷ Γ: 1.001. WINDOW, DECADES, ×10 RULE: 1.00
Catch the common trap
Explain before calculating.
A level coupled to a band that is flat over W = 2.0 meV decays with golden-rule rate Γ = 5.0 × 10⁹ s⁻¹. Over which times does its survival S fall at the rate Γ, that is, with −(dS/dt)/S = Γ?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA discrete level sits in the middle of a band over which |V|²ρ is flat for W = 4.0 meV, and its golden-rule rate is Γ = 1.0 × 10¹⁰ s⁻¹. Find the level's energy spread ΔH, the probability that it has left after t = 50 fs, and compare that with the 1 − e(−Γt) of pure exponential decay.
- The coupling enters through |V|²ρ = ħΓ/2π, and the band holds (ΔH)² = Σf |Vfi|² = |V|²ρW = ħΓW/2π. Here ħΓ = 6.582 × 10⁻¹⁶ eV s × 1.0 × 10¹⁰ s⁻¹ = 6.58 μeV.
- (ΔH)² = 6.58 × 10⁻⁶ eV × 4.0 × 10⁻³ eV ÷ 2π = 4.19 × 10⁻⁹ eV², so ΔH = 64.7 μeV.
- Check the regime: ħ/W = 6.582 × 10⁻¹⁶ eV s ÷ 4.0 × 10⁻³ eV = 165 fs, so t = 50 fs is 0.30 ħ/W, and ħ/ΔH = 10.2 ps. The short-time law applies.
- ΔH t/ħ = 6.47 × 10⁻⁵ eV × 5.0 × 10⁻¹⁴ s ÷ 6.582 × 10⁻¹⁶ eV s = 4.92 × 10⁻³, dimensionless as it must be, so 1 − S = (4.92 × 10⁻³)² = 2.42 × 10⁻⁵.
- Pure exponential decay would give 1 − e(−Γt) ≈ Γt = 5.0 × 10⁻⁴. The ratio is Wt/2πħ = 0.048: the true loss is about a twentieth, because the rate has not yet switched on.
AnswerΔH ≈ 65 μeV. After 50 fs the level has lost 2.4 × 10⁻⁵ of its probability, about 1/21 of the 5.0 × 10⁻⁴ that exponential decay at Γ would predict.
MediumThe level of the previous problem is checked every τ = 20 fs by a measurement that projects it back onto the level whenever it is found there. Find its survival after T = 1.0 ns and its effective decay rate, and compare with the level left unchecked.
- Each check restarts the evolution from the level, so each interval contributes the short-time factor 1 − (ΔH)²τ²/ħ². That needs τ ≪ ħ/W, and τ = 20 fs is 0.12 ħ/W.
- ΔH τ/ħ = 6.47 × 10⁻⁵ eV × 2.0 × 10⁻¹⁴ s ÷ 6.582 × 10⁻¹⁶ eV s = 1.97 × 10⁻³, whose square is 3.87 × 10⁻⁶.
- N = T/τ = 1.0 × 10⁻⁹ s ÷ 2.0 × 10⁻¹⁴ s = 5.0 × 10⁴ checks, so S = (1 − 3.87 × 10⁻⁶)(5.0 × 10⁴) = e(−0.193) = 0.824.
- The effective rate is (ΔH)²τ/ħ² = 3.87 × 10⁻⁶ ÷ 2.0 × 10⁻¹⁴ s = 1.93 × 10⁸ s⁻¹, in s⁻¹ as a rate must be. As a check, its ratio to Γ is Wτ/2πħ = τ/tc = 0.0193.
- Unchecked, W/ħΓ = 608 puts the level well inside the window, and the Weisskopf–Wigner exponential gives S = e(−ΓT) = e(−10) = 4.5 × 10⁻⁵.
AnswerChecked every 20 fs, the level survives 1.0 ns with probability 0.82, decaying at 1.9 × 10⁸ s⁻¹; unchecked, it survives with probability 4.5 × 10⁻⁵. The checks slow the decay 52-fold.
HardA level lies in a band whose |V|²ρ is flat from its bottom E₋ to its top E₊ = E₋ + 4.0 meV, at Eᵢ = E₋ + 0.50 meV, with golden-rule rate Γ = 2.0 × 10⁹ s⁻¹. Find the Weisskopf–Wigner level shift δE with its sign, compare it with ħΓ, decide whether the decay is exponential, and say what happens as the level approaches the band edge.
- With |V|²ρ = ħΓ/2π constant, δE = (ħΓ/2π) P.V.∫ dE/(Eᵢ − E) from E₋ to E₊. The principal value cuts a symmetric gap around Eᵢ whose two sides cancel, leaving δE = (ħΓ/2π) ln[(Eᵢ − E₋)/(E₊ − Eᵢ)].
- ħΓ = 6.582 × 10⁻¹⁶ eV s × 2.0 × 10⁹ s⁻¹ = 1.316 μeV, so ħΓ/2π = 0.2095 μeV, and ln(0.50/3.50) = −1.946. Then δE = −0.408 μeV, or δE/h = −98.6 MHz.
- Sign and size: 3.50 meV of band lies above the level and 0.50 meV below, and states above push down, so the level moves down; a centred level would not move at all. |δE| = 0.31 ħΓ, comparable with the width, since both are of order |V|²ρ.
- The flat stretch around Eᵢ ends at the nearer edge, 0.50 meV away, so take W ≈ 0.50 meV: ħ/W = 1.32 ps against 1/Γ = 500 ps, and ħΓ/W = 2.6 × 10⁻³. Linear growth holds from about 13 ps to 50 ps, and the exponential continues beyond, with corrections near 0.3%.
- Limit: put the level 1.0 μeV above the edge and the logarithm gives δE = −1.74 μeV, larger than ħΓ, while ħΓ/W ≈ 1.3. The window has closed, the Markov step fails, and the decay near threshold is no longer exponential.
AnswerδE = −0.41 μeV (−99 MHz), downward and 0.31 ħΓ in size. With ħΓ/W = 2.6 × 10⁻³ the decay is exponential from about 13 ps on, but the shift formula and the exponential both fail as the level reaches the band edge.