
Advanced Quantum Mechanics course sequence · Interactive lesson
The Weisskopf-Wigner ansatz and its exact equations
This is where spontaneous emission stops being a rate you are handed and becomes dynamics you can solve. We find the one conservation law that makes an excited atom in the vacuum exactly solvable, write its equations without approximating anything, and fold 10²¹ coupled amplitudes into one equation with a memory.
UAQM-12.05 · Lesson 5 of 9 in this unit. Position is not a completion score.Build the model
Integrate out the photons and one exact equation remains, with a 10⁻¹⁶ s memory.
The key idea
The Weisskopf-Wigner ansatz writes the state of an initially excited two-level atom as cₑ(t)|e, 0⟩ plus a sum over modes of cₖ(t)|g, 1ₖ⟩; because the rotating-wave coupling conserves excitation number, this is the general state, and its amplitudes obey closed, exact equations.
Example. For hydrogen 2p, eliminating the photons leaves dcₑ/dt = −∫₀ᵗ K(t − t′)cₑ(t′) dt′, with a kernel that forgets in about 1/ω₀ = 6.5 × 10⁻¹⁷ s while cₑ changes over 1/A = 1.6 ns: a memory 4.0 × 10⁻⁸ of the lifetime.
The complete physical model and assumptions
Topic 12.02 took the A coefficient from the golden rule, which assumes the excited amplitude stays close to 1 and that the decay is already a rate. To watch a decay happen, solve the dynamics instead. Keep two atomic levels, the dipole approximation and the rotating-wave approximation, and one quantity is conserved: the number of excitations, atomic plus photonic. An excited atom in the vacuum therefore never leaves the states |e, 0⟩ and |g, 1ₖ⟩, so the Weisskopf-Wigner ansatz, cₑ|e, 0⟩ plus a sum of cₖ|g, 1ₖ⟩, is not a guess but the general state, and the Schrödinger equation becomes a closed and exact set of amplitude equations. Each cₖ is driven by cₑ alone, so the photons can be integrated out, leaving one equation in which the rate of change of cₑ depends on its whole past through a kernel K(τ), the sum over modes of |gₖ|²e(i(ω₀−ωₖ)τ). That kernel is the Fourier transform of the coupling-weighted mode density, which in free space varies only on the scale of ω₀, so it forgets in about 1/ω₀: 6 × 10⁻¹⁷ s against 1.6 ns for hydrogen 2p. Topic 12.06 spends that separation; this topic approximates nothing beyond the model. Where the model stops: the counter-rotating terms it drops break the closure, cost nothing in the rate and return in Unit 13's level shift, and a continuum narrower than the coupling, as a good cavity presents, gives the kernel a memory as long as the decay.
1The model, and the number it conserves
SI units throughout; no four-vector appears, and where the course writes one its metric is (+, −, −, −). From topic 11.08, quoted, build a model: two atomic levels |g⟩ and |e⟩ split by ħω₀, the dipole approximation, and the coupling −(q/m)A⋅p, with A² dropped, since in the dipole approximation it does not act on the atom. With real polarisation vectors the coupling is ħΣₖ(gₖσ₊ + gₖ*σ₋)(aₖ + aₖ†), where σ₊ = |e⟩⟨g|, σ₋ = σ₊† and k labels a wavevector and a polarisation. Two of its four products, σ₊aₖ and σ₋aₖ†, trade one photon for one atomic excitation. The other two, σ₊aₖ† and σ₋aₖ, raise or lower both at once. The rotating-wave approximation drops those, and then N̂ = σ₊σ₋ + Σₖaₖ†aₖ commutes with H, derived: each kept term moves one quantum across, each dropped one changes N̂ by ±2. In the free-field limit, every gₖ = 0, N̂ is conserved trivially; the content is that the coupling keeps it. Everything below is exact for this model until topic 12.06 takes the Markov step.
2The ansatz is the whole sector, and its equations are exact
The excited atom in the vacuum, |e, 0⟩, has N̂ = 1, so it can reach only itself and the one-photon states |g, 1ₖ⟩. Hence |ψ(t)⟩ = cₑ(t)e(−iω₀t)|e, 0⟩ + Σₖ cₖ(t)e(−iωₖt)|g, 1ₖ⟩ is not a trial function but the general state; the phases remove the free evolution, with energies counted from |g, 0⟩. Put it into iħ∂|ψ⟩/∂t = H|ψ⟩ and project on ⟨e, 0| and ⟨g, 1ₖ|: dcₑ/dt = −iΣₖ gₖcₖe(i(ω₀−ωₖ)t) and dcₖ/dt = −igₖ*cₑe(−i(ω₀−ωₖ)t), derived, with cₑ(0) = 1 and every cₖ(0) = 0. Checks: gₖ is in s⁻¹, as both sides need. In d/dt(|cₑ|² + Σₖ|cₖ|²) the photon terms are minus the complex conjugates of the atomic ones, so the real part vanishes and probability is conserved. With every gₖ = 0, cₑ stays 1. The price is the number of equations, one per mode: a cubic metre holds about 5 × 10²¹ modes below the Lyman-α frequency alone.
3Integrating out the photons leaves an equation with memory
Each cₖ is driven by cₑ alone, so integrate it: cₖ(t) = −igₖ*∫₀ᵗ cₑ(t′)e(−i(ω₀−ωₖ)t′) dt′, still exact. Substituting folds the whole set into one equation, dcₑ/dt = −∫₀ᵗ K(t − t′)cₑ(t′) dt′, with K(τ) = Σₖ|gₖ|²e(i(ω₀−ωₖ)τ), derived. Reabsorption is included: the right side is the photon amplitude emitted at t′ acting back on the atom. K is in s⁻², so the right side is in s⁻¹. Two limits check it. One resonant mode alone gives K = g², so d²cₑ/dt² = −g²cₑ and cₑ = cos(gt): topic 11.09's vacuum Rabi oscillation at 2g, observed for a Rydberg atom in a microwave cavity near 2g/2π = 50 kHz (Brune et al., 1996, quoted). For any set of modes, differentiating at t = 0 gives d²cₑ/dt² = −K(0), so Pₑ = |cₑ|² = 1 − K(0)t² + O(t⁴), the t³ term vanishing because dK/dτ at τ = 0 is imaginary: Quantum Mechanics II's quadratic start, now exact, with K(0) = Σₖ|gₖ|².
4From a sum over modes to a spectral density
Write K(τ) = ∫₀^∞ J(ω)e(i(ω₀−ω)τ) dω, with J(ω) = Σₖ|gₖ|²δ(ω − ωₖ), the coupling-weighted density of modes. The coupling −(q/m)A⋅p with peg = imω₀reg gives |gₖ|² = ω₀²|êₖ⋅d|²/(2ε₀ħωₖV), d = q⟨e|r|g⟩. Topic 12.02's sum, Vω²/(8π³c³) modes per unit solid angle and angular frequency for each polarisation, over both polarisations and all directions, gives J(ω) = ω₀²ω|d|²/(6π²ε₀ħc³) = (A/2π)(ω/ω₀), derived. Units: ε₀ħc³ is C² m² s⁻², so J is in s⁻¹. V has cancelled, and J(ω₀) = A/2π = 9.98 × 10⁷ s⁻¹ for 2p. Off resonance the two-level model is gauge-sensitive: the length form −d⋅E gives |gₖ|² ∝ ωₖ and J ∝ ω³, equal at ω₀, where the rate is set, and different elsewhere, where the shift is. The integral replaces the box sum only while no photon has come back from a wall: light covers c/A = 0.48 m in one 2p lifetime, so in a smaller mirrored box the kernel would show echoes.
5How long the kernel remembers
Model the dipole approximation's failure near ωc = c/a₀ by a factor e(−ω/ωc) on J, which lowers J(ω₀) by only 0.3%. Then K(τ) = (A/(2πω₀))e(iω₀τ)/(ωc⁻¹ + iτ)², derived, with |K| = (A/(2πω₀))/(ωc⁻² + τ²). It has two parts. A spike of width 1/ωc = 1.8 × 10⁻¹⁹ s and height K(0) = Aωc²/(2πω₀) = 2.1 × 10²⁹ s⁻² sets the quadratic start; to leading order its integral is −iAωc/(2πω₀), a phase rate linear in the cutoff and no decay at all, the divergent shift that topic 12.06 meets and Unit 13 treats. Then a tail falling as τ⁻² and rotating at ω₀, which cancels itself against a slowly varying cₑ: integrating by parts, lags beyond T add a fraction (ω₀T)⁻²/π of the A/2 the decay needs, 0.3% at T = 10/ω₀. So the past acts on cₑ over about 1/ω₀ = 6.45 × 10⁻¹⁷ s, while cₑ changes over 1/A = 1.595 ns: a ratio A/ω₀ = 4.0 × 10⁻⁸. Topic 12.06 uses that small number.
6What the closure cost, and where it returns
Restore σ₊aₖ† and σ₋aₖ and the sector leaks: |g, 1ₖ⟩ is driven to |e, 1ₖ1ₖ′⟩ with N̂ = 3, and |g, 0⟩ itself mixes with every |e, 1ₖ⟩, so the true ground state carries virtual photons. Since ħω₀ + ħωₖ is never zero, the dropped terms cannot conserve energy and add nothing to the rate at order e². They do move the energies. By second-order perturbation theory, used as Quantum Mechanics II derived it, the shift of |e, 0⟩ in the two-level model comes only from kept terms, Σₖ|gₖ|²/(ω₀ − ωₖ) in angular frequency, and that of |g, 0⟩ only from dropped ones, −Σₖ|gₖ|²/(ω₀ + ωₖ). Their difference carries 1/(ω₀ − ω) + 1/(ω₀ + ω) = 2ω₀/(ω₀² − ω²), falling as ω⁻² where each term falls as ω⁻¹, so the counter-rotating terms turn a transition shift linear in the cutoff into one logarithmic in it. Dropping them keeps the decay right and gets the shift wrong. They return with A², which changes photon number by zero or two, with the atom's other levels and with the cutoff, in Unit 13, where Bethe's calculation sums over every level.
Key equations
Read the units, assumptions and limits before substituting.
Excitation number is conserved
[HRWA, N̂] = 0, with N̂ = σ₊σ₋ + Σₖ aₖ†aₖ
Derived: gₖσ₊aₖ and gₖ*aₖ†σ₋ each move one quantum between atom and field. The dropped σ₊aₖ† and σ₋aₖ change N̂ by ±2, so dropping them is what closes the sector.
σ₊ = |e⟩⟨g| and σ₋ = σ₊† flip the atom; aₖ† adds a photon to mode k, a wavevector and a polarisation; N̂ is dimensionless, with integer eigenvalues.The exact amplitude equations
dcₑ/dt = −i Σₖ gₖcₖ e(i(ω₀−ωₖ)t), dcₖ/dt = −i gₖ* cₑ e(−i(ω₀−ωₖ)t)
Derived by projecting the Schrödinger equation on each state. Nothing beyond the model is approximated, and |cₑ|² + Σₖ|cₖ|² = 1 at every time.
cₑ, cₖ: amplitudes of |e, 0⟩ and |g, 1ₖ⟩ with the free phases removed; ω₀, ωₖ, gₖ in s⁻¹; cₑ(0) = 1, cₖ(0) = 0.One equation with memory
dcₑ/dt = −∫₀ᵗ K(t − t′) cₑ(t′) dt′, K(τ) = Σₖ |gₖ|² e(i(ω₀−ωₖ)τ)
Derived by integrating each cₖ, which only cₑ drives, and substituting. Reabsorption is kept: the right side is the photon amplitude emitted at t′ acting back on the atom.
K in s⁻², so both sides are in s⁻¹; τ = t − t′ is how far back the atom's own amplitude still acts on it.The kernel is a Fourier transform
K(τ) = ∫₀^∞ J(ω) e(i(ω₀−ω)τ) dω, J(ω) = Σₖ |gₖ|² δ(ω − ωₖ)
Derived. A J that varies only over a frequency range B gives a kernel that forgets in about 1/B; in free space B is of order ω₀ itself.
J(ω), in s⁻¹, is the coupling-weighted density of modes: |gₖ|² summed over the modes per unit angular frequency.Free-space spectral density, A⋅p form
J(ω) = ω₀²ω|d|²/(6π²ε₀ħc³) = (A/2π)(ω/ω₀)
Derived. J(ω₀) = A/2π, 9.98 × 10⁷ s⁻¹ for 2p, holds the rate. The length form −d⋅E gives (A/2π)(ω/ω₀)³: equal at ω₀, different where the shift is decided.
From |gₖ|² = ω₀²|êₖ⋅d|²/(2ε₀ħωₖV) and topic 12.02's mode sum; d = q⟨e|r|g⟩ in C m; the box volume V cancels.The exact short-time start
Pₑ(t) = 1 − K(0) t² + O(t⁴), K(0) = Σₖ |gₖ|²
Derived from d²cₑ/dt² = −K(0) at t = 0: Quantum Mechanics II's quadratic start, now exact. With ωc = c/a₀, K(0) = 2.1 × 10²⁹ s⁻² for 2p, and it grows with the cutoff.
Pₑ = |cₑ|²; valid for t ≪ 1/ωc, with ωc the frequency at which the coupling is cut off; K(0) in s⁻².Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. At λ = 0.25 γ₀, the Hard example, the kernel outlives the would-be lifetime: Pₑ empties at γ₀t = 5.8 and revives to 0.09 at 9.5. Raise λ past 2 γ₀ and the oscillation stops; by 10 γ₀ the solid curve runs close to the dashed one. Free-space 2p, about 2.5 × 10⁷ γ₀ wide, lies far beyond.
Live results
- MEMORY TIME 1/λ
- 4.00 γ₀⁻¹
- COUPLING g = √(γ₀λ/2)
- 0.354 γ₀
- LATE-TIME DECAY RATE OF Pₑ
- 0.250 γ₀
- MEMORYLESS e(−γ₀t) AT MARKER
- 0.135
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Continuum half-width λ/γ₀: 0.25; Marker time γ₀t: 2.0
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting reading is that the Weisskopf-Wigner ansatz is an approximation because it keeps only one photon: surely an excited atom could emit two, and those states have been thrown away. They have not. For the model as stated, two levels, dipole coupling and the rotating-wave terms, the excitation number N̂ is conserved, and |e, 0⟩ has N̂ = 1; any state with the atom in |g⟩ and two photons has N̂ = 2 and cannot be reached. The ansatz holds every state the evolution can visit, and its equations are exact. The approximations sit elsewhere: the two-level truncation, the dipole approximation, dropping A², and the rotating-wave approximation; the last two are what make the sector close, since A² also holds terms that create or destroy photons in pairs. Real two-photon decay, such as hydrogen 2s to 1s at about 8.2 s⁻¹, needs a third level, which the truncation removed, not the ansatz. Nor is reabsorption neglected; it is the gₖcₖ term in dcₑ/dt.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Compare the kernel's memory, about 1/ω₀, with the lifetime for three transitions: hydrogen 2p to 1s (ħω₀ = 10.20 eV, A = 6.27 × 10⁸ s⁻¹), the rubidium D2 line (780.24 nm, measured lifetime 26.24 ns) and hydrogen's 21 cm line (1420.4 MHz, A = 2.9 × 10⁻¹⁵ s⁻¹). How many optical cycles fit in each lifetime?
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.
- Angular frequencies: ω₀ = 10.20 eV ÷ ħ = 1.550 × 10¹⁶ s⁻¹ for 2p; 2πc/λ = 2.414 × 10¹⁵ s⁻¹ for rubidium; 2π × 1420.4 MHz = 8.925 × 10⁹ s⁻¹ for 21 cm. The memory times 1/ω₀ are 6.45 × 10⁻¹⁷ s, 4.14 × 10⁻¹⁶ s and 1.12 × 10⁻¹⁰ s.
- Lifetimes 1/Γ: 1.595 ns for 2p; 26.24 ns for rubidium, so Γ = 3.811 × 10⁷ s⁻¹; 3.4 × 10¹⁴ s for 21 cm, about 11 million years.
- Memory over lifetime is Γ/ω₀: 4.05 × 10⁻⁸, 1.58 × 10⁻⁸ and 3.2 × 10⁻²⁵. Across the kernel's reach |cₑ|² changes by about that fraction, so cₑ(t′) under the integral is constant to that accuracy.
- Cycles per lifetime, ω₀/(2πΓ): 3.93 × 10⁶ for 2p, 1.01 × 10⁷ for rubidium and 4.9 × 10²³ for 21 cm.
- Checks: Γ/ω₀ is a ratio of two rates, so dimensionless. The 2p value uses topic 12.02's derived A, whose 1.6 ns agrees with measured 2p lifetimes; the rubidium value uses a measured lifetime directly.
Answer. Memory over lifetime: 4.0 × 10⁻⁸ for hydrogen 2p, 1.6 × 10⁻⁸ for the rubidium D2 line and 3.2 × 10⁻²⁵ for 21 cm; that is 3.9 × 10⁶, 1.0 × 10⁷ and 4.9 × 10²³ cycles per lifetime.
MediumExample 2
For hydrogen 2p take J(ω) = (A/2π)(ω/ω₀) times e(−ω/ωc), with ωc = c/a₀ modelling where the dipole approximation fails. From the exact equation for cₑ, find the survival probability at very short times, the total coupling K(0), how long the result holds and how much population is at stake. Compare with the golden-rule loss At.
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.
- Differentiate dcₑ/dt = −∫₀ᵗ K(t − t′)cₑ(t′) dt′. At t = 0 the integral is empty, so dcₑ/dt = 0, and d²cₑ/dt² = −K(0)cₑ(0) = −K(0). So cₑ ≈ 1 − K(0)t²/2 and Pₑ ≈ 1 − K(0)t². No t³ term survives in Pₑ, because dK/dτ at 0, iΣₖ|gₖ|²(ω₀ − ωₖ), is imaginary.
- K(0) = Σₖ|gₖ|² = ∫₀^∞ J dω = (A/(2πω₀))∫₀^∞ ω e(−ω/ωc) dω = Aωc²/(2πω₀). Units: s⁻¹ × s⁻² ÷ s⁻¹ = s⁻², as a kernel needs.
- Numbers: ωc = c/a₀ = 5.665 × 10¹⁸ s⁻¹ (ħωc = 3.73 keV), ω₀ = 1.550 × 10¹⁶ s⁻¹ and A/(2πω₀) = 6.44 × 10⁻⁹, so K(0) = 2.07 × 10²⁹ s⁻².
- The quadratic form needs K(τ) ≈ K(0) across the interval, so t ≪ 1/ωc = 1.77 × 10⁻¹⁹ s. At t = 0.1/ωc the loss is K(0)t² = 6.4 × 10⁻¹¹ against At = 1.1 × 10⁻¹¹. The two cross at t = A/K(0) = 3.0 × 10⁻²¹ s; before that the exact decay is slower than the exponential.
- Whatever the cutoff, the population at stake in the quadratic window is of order K(0)/ωc² = A/(2πω₀) = 6 × 10⁻⁹. Limits: one mode gives K(0) = g² and cos²(gt) ≈ 1 − g²t², topic 11.09's start; as ωc → ∞ the window shrinks to nothing while K(0) diverges.
Answer. Pₑ ≈ 1 − K(0)t² with K(0) = Aωc²/(2πω₀) = 2.1 × 10²⁹ s⁻², valid for t ≪ 1.8 × 10⁻¹⁹ s; it puts only about A/(2πω₀) = 6 × 10⁻⁹ of the population at stake.
HardExample 3
An emitter sees a continuum with J(ω) = (γ₀/2π)λ²/((ω − ω₀)² + λ²), the Lorentzian a single leaky cavity mode presents (topic 12.08), so the golden rule would give the rate 2πJ(ω₀) = γ₀. Solve the exact equation for cₑ. For γ₀ = 1.0 × 10¹¹ s⁻¹ and λ = 2.5 × 10¹⁰ s⁻¹, find when Pₑ first reaches zero and how far it revives, and compare with 1/γ₀.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Kernel: with x = ω − ω₀, and the lower limit taken to −∞ since λ ≪ ω₀, the Fourier transform of λ²/(x² + λ²) is πλe(−λ|τ|), so K(τ) = (γ₀λ/2)e(−λτ): the continuum remembers for 1/λ = 40 ps.
- Let I(t) = ∫₀ᵗ e(−λ(t−t′))cₑ(t′) dt′, so dI/dt = cₑ − λI and dcₑ/dt = −(γ₀λ/2)I. Differentiating once more: d²cₑ/dt² + λ dcₑ/dt + (γ₀λ/2)cₑ = 0, with cₑ(0) = 1 and dcₑ/dt(0) = 0, a damped oscillator with g² = γ₀λ/2 in s⁻².
- The roots are −λ/2 ± ½√(λ² − 2γ₀λ). Here λ² − 2γ₀λ = −4.375 × 10²¹ s⁻², so with Ω′ = √(2γ₀λ − λ²) = 6.614 × 10¹⁰ s⁻¹, cₑ = e(−λt/2)[cos(Ω′t/2) + (λ/Ω′)sin(Ω′t/2)], oscillatory whenever λ is below 2γ₀.
- cₑ = 0 where tan(Ω′t/2) = −Ω′/λ = −2.646, so Ω′t/2 = π − 1.209 = 1.932 rad and t = 58.4 ps. Since dcₑ/dt is proportional to −sin(Ω′t/2), the next extremum is at Ω′t/2 = π, t = 95.0 ps, where Pₑ = e(−2πλ/Ω′) = 0.093.
- The golden rule predicted 1/γ₀ = 10.0 ps and no return. With a memory of 40 ps, four times that, the photon is reabsorbed before it can leave.
- Limits: for λ ≫ γ₀ the roots approach −γ₀/2 and −λ, so Pₑ → e(−γ₀t), memoryless; for λ → 0 at fixed g, cₑ → cos(gt), topic 11.09. Units: γ₀λ is in s⁻², like K.
Answer. K(τ) = (γ₀λ/2)e(−λτ) and cₑ = e(−λt/2)[cos(Ω′t/2) + (λ/Ω′)sin(Ω′t/2)] with Ω′ = 6.61 × 10¹⁰ s⁻¹: Pₑ first vanishes at 58.4 ps and revives to 0.093 at 95.0 ps, against a golden-rule lifetime of 10.0 ps.
Quick check
Select an answer and explain it before checking.
Review & continue
The Weisskopf-Wigner ansatz writes the state of an initially excited two-level atom as cₑ(t)|e, 0⟩ plus a sum over modes of cₖ(t)|g, 1ₖ⟩; because the rotating-wave coupling conserves excitation number, this is the general state, and its amplitudes obey closed, exact equations.
- 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.