
Advanced Quantum Mechanics course sequence · Interactive lesson
The A coefficient from the golden rule over empty modes
Quantum Mechanics II got the spontaneous rate by borrowing Planck's law and running Einstein's argument backwards. Here it comes straight from the field: one golden-rule term per empty mode, summed over frequencies, polarisations and directions. Track where the 3π comes from and the same integrand hands you each line's angular pattern.
UAQM-12.02 · Lesson 2 of 9 in this unit. Position is not a completion score.Build the model
Sum the golden rule over every empty mode and the box cancels, leaving A.
The key idea
The A coefficient is the rate at which an excited atom alone in the photon vacuum emits, obtained here as the golden-rule rate into every empty mode of the field: for an electric-dipole line, A = ω₀³|d|²/(3πε₀ħc³).
Example. Hydrogen 2p → 1s: ħω₀ = 10.20 eV and |⟨1s|r|2p, m⟩|² = (2¹⁵/3¹⁰)a₀² = 0.555a₀² give A = 6.27 × 10⁸ s⁻¹, so the 2p level lives 1.60 ns and Lyman α at 121.6 nm has a natural width A/2π of 99.8 MHz.
The complete physical model and assumptions
Topic 12.01 gave the amplitude for an excited atom in the photon vacuum to put one photon into mode k, s: proportional to √(ħωₖ/(2ε₀V)) êₖₛ*⋅d, and nonzero because of the 1 in √(n + 1). One mode cannot make a rate. Its coupling falls as 1/√V and its frequency is sharp, so an atom facing it alone only trades the excitation back and forth. A rate needs a continuum, and a large box supplies one: Vω²/(8π³c³) modes per unit angular frequency, per steradian and per polarisation, so dense that 5.7 × 10⁸ of them in a 1 cm³ box lie inside the natural width of Lyman α. The golden rule multiplies that density by the squared coupling and the two factors of V cancel: the box was bookkeeping. The polarisation sum leaves |d|² − |k̂⋅d|², the part of the dipole transverse to the photon's direction; the integral over directions gives (8π/3)|d|²; and A = ω₀³|d|²/(3πε₀ħc³). No temperature, Boltzmann factor or Planck spectrum entered, so Einstein's relations become a prediction for topic 12.03 rather than a premise. Before the angular integral, the same integrand is the emission pattern of the line. The derivation assumes the dipole approximation, first order in e, a free-space continuum flat across the line, and times long against 1/ω₀ but short against 1/A; topics 12.05 to 12.08 lift the last two.
1One empty mode cannot make a rate
SI units throughout; no four-vector appears, and where one does in this course the metric is (+, −, −, −). Quoted from topic 12.01: in the dipole approximation −d⋅E takes |e, 0⟩ to |g, 1ₖₛ⟩ with squared matrix element (ħωₖ/(2ε₀V))|êₖₛ*⋅d|², where d = q⟨g|r|e⟩ and êₖₛ, s = 1, 2, are unit polarisation vectors perpendicular to k. Keep only one resonant mode and Unit 11's Jaynes-Cummings solution, quoted, applies: the excitation swings between atom and mode at the vacuum Rabi frequency 2g, with ħg = √(ħω₀/(2ε₀V))|d| for a mode polarised along d, and never leaves for good. For hydrogen 2p in a 1 cm³ box, g = 1.8 × 10⁴ s⁻¹. What turns the swing into a decay is how many modes the atom can reach. The same box holds on average one mode per 1.1 s⁻¹ of angular frequency near Lyman α, so 5.7 × 10⁸ of them lie inside the natural width. Each is coupled feebly, their amplitudes dephase long before any one of them could swing back, and probability leaves at a constant rate: the golden rule's regime, in which the mode density, not any single coupling, sets A.
2Counting modes by direction and frequency
In a periodic box of side L the allowed wavevectors are k = (2π/L)(n₁, n₂, n₃), one per volume (2π)³/V of k-space. A shell of radius k and thickness dk, opened to solid angle dΩ, has volume k² dk dΩ, so it holds Vk² dk dΩ/(2π)³ wavevectors for each polarisation. With k = ω/c that is dN = Vω² dω dΩ/(8π³c³): derived, and a pure number, as Vω² dω/c³ must be. The golden rule's energy δ-function, δ(ħω₀ − ħωₖ), equals δ(ω₀ − ωₖ)/ħ, which turns 2π/ħ into 2π/ħ² and consumes the frequency integral, leaving dΓ = (2π/ħ²)(ħω₀/(2ε₀V))(Vω₀²/(8π³c³)) Σₛ|êₖₛ*⋅d|² dΩ. The V in the mode count meets the 1/V in the squared field per photon and cancels: the rate belongs to the atom and to free space, not to the box. Nothing thermal has appeared either. The vacuum supplied the 1 in n + 1 for every mode, no mode needed a temperature, and Planck's law was never used.
3Two polarisations: only the transverse dipole radiates
The two êₖₛ and k̂ form an orthonormal triad, so Σₛ (êₖₛ)ᵢ(êₖₛ*)ⱼ = δᵢⱼ − k̂ᵢk̂ⱼ: derived from completeness, with the transversality k̂⋅êₖₛ = 0 that the Coulomb gauge built into Unit 11's field. Contracting with dᵢ*dⱼ gives Σₛ|êₖₛ*⋅d|² = |d|² − |k̂⋅d|², the squared length of the part of d perpendicular to the photon's direction; a photon has no longitudinal polarisation to receive the rest. For a Δm = 0 line, d = d ẑ is real and the sum is |d|² sin²θ, with θ the angle between k and z: nothing leaves along the dipole, most leaves at right angles, and the light is linearly polarised in the plane of z and k. For Δm = ±1, d ∝ (x̂ ± iŷ)/√2 rotates in the xy plane; then |k̂⋅d|² = |d|² sin²θ/2 and the sum is |d|²(1 + cos²θ)/2. It is largest along z, where the light is circularly polarised, and half as large at θ = 90°, where only the component perpendicular to both k and z survives and the light is linearly polarised.
4The angular integral, and where the 3π comes from
Integrate over directions with ∫k̂ᵢk̂ⱼ dΩ = (4π/3)δᵢⱼ, fixed by rotational symmetry and the trace ∫dΩ = 4π. Then ∫(|d|² − |k̂⋅d|²) dΩ = (4π − 4π/3)|d|² = (8π/3)|d|² for any d, linear or circular. Collecting factors, A = (2π/ħ²)(ħω₀/(2ε₀V))(Vω₀²/(8π³c³))(8π/3)|d|² = ω₀³|d|²/(3πε₀ħc³): derived. Dimensions: ω₀³|d|² is s⁻³ C² m², and ε₀ħc³ is (C² J⁻¹ m⁻¹)(J s)(m³ s⁻³) = C² m² s⁻², so A is in s⁻¹. Correspondence check: ħω₀A = ω₀⁴(2|d|)²/(12πε₀c³), exactly the Larmor power of a dipole p₀ cos ω₀t with p₀ = 2|d|, as it must be, since d is the coefficient of e(−iω₀t), half of a real oscillation. Free-field limit: as e → 0, A → 0 as |d|² ∝ e², the first order assumed. The ω₀³ is worth reading: two powers from the mode density and one from the field per photon. At fixed |d|, moving a line from 10.2 eV down to 10.2 µeV multiplies its lifetime by 10¹⁸, from 1.6 ns to about 50 years.
5Hydrogen 2p: 6.27 × 10⁸ s⁻¹ from the field
For 2p → 1s, d = −e⟨1s|r|2p, m⟩, and Quantum Mechanics II's matrix element, quoted, gives |⟨1s|r|2p, m⟩|² = (2¹⁵/3¹⁰)a₀² = 0.5549a₀² for every m, so |d| = 6.316 × 10⁻³⁰ C m. With ħω₀ = (3/4) × 13.606 eV = 10.204 eV, ω₀ = 1.5503 × 10¹⁶ s⁻¹, and the derived formula gives A = 6.27 × 10⁸ s⁻¹ and τ = 1/A = 1.60 ns: the numbers Quantum Mechanics II reached through Einstein's detailed balance, now with no thermodynamics in the argument. In closed form A = (2/3)⁸α⁵mₑc²/ħ, and the reduced mass multiplies it by μ/mₑ = 0.99946, giving 6.265 × 10⁸ s⁻¹ and 1.596 ns. Beam-foil timing of fast hydrogen atoms measures 1.6 ns. Each approximation can now be priced. The dipole approximation needs ka₀ = (3/8)α = 2.7 × 10⁻³ ≪ 1. The golden rule needs A ≪ ω₀, and A/ω₀ = 4.0 × 10⁻⁸: the line is 2.5 × 10⁷ times narrower than its frequency. Relativistic corrections from the Dirac equation enter at order (Zα)² = 5 × 10⁻⁵.
6Δm = 0 and Δm = ±1: the pattern before the integral
Divide dΓ/dΩ by A and a Δm = 0 line sends a fraction (3/(8π))sin²θ of its photons into each steradian, a Δm = ±1 line (3/(16π))(1 + cos²θ): derived, and each integrates to 1, because the 8π/3 does not care which way d points. That is why all three 2p sublevels decay at the same A. It also fixes what a level with its sublevels equally filled must do: sin²θ + 2 × (1 + cos²θ)/2 = 2 at every θ, so it radiates isotropically and unpolarised, as it must with no axis to prefer. Single out an axis with a weak magnetic field B ∥ z and the patterns become visible. Viewed along B, only the two σ lines appear, circularly polarised in opposite senses. Viewed across B, the π line is linearly polarised along B and the σ lines across it, and with equal populations the π line is twice as bright as each σ line, since sin²90° = 1 against (1 + 0)/2. That 2:1:1 transverse triplet with those polarisations is the normal Zeeman pattern measured on singlet lines such as cadmium's 643.8 nm, and the quantised-field sum reproduces it.
Key equations
Read the units, assumptions and limits before substituting.
Golden rule summed over empty modes
Γ = (2π/ħ²) Σₖₛ (ħωₖ/(2ε₀V)) |êₖₛ*⋅d|² δ(ω₀ − ωₖ)
The squared vacuum matrix element of −d⋅E is quoted from topic 12.01, with n + 1 = 1. Every factor of V is now in view, ready to cancel.
d = q⟨g|r|e⟩ in C m; êₖₛ (s = 1, 2) unit polarisations ⊥ k; V the box volume in m³; ħω₀ the energy of e above g, ω₀ in s⁻¹.Modes per polarisation in dω dΩ
dN = Vω² dω dΩ/(8π³c³)
Derived by counting k-points in the shell k² dk dΩ. Its V cancels the 1/V of the field per photon: the box was bookkeeping.
Periodic box of volume V in m³, one k per (2π)³/V; ω = ck in s⁻¹; dΩ in sr. dN is a pure number.Sum over the two polarisations
Σₛ |êₖₛ*⋅d|² = |d|² − |k̂⋅d|²
Derived: only the part of d transverse to the photon's direction radiates, so no photon leaves along a linearly oscillating dipole.
From Σₛ (êₖₛ)ᵢ(êₖₛ*)ⱼ = δᵢⱼ − k̂ᵢk̂ⱼ, since k̂⋅êₖₛ = 0; d may be complex, with |d|² = d⋅d* in C² m².Emission per unit solid angle
dΓ/dΩ = ω₀³(|d|² − |k̂⋅d|²)/(8π²ε₀ħc³)
Derived: the integrand before the angular integral is the line's emission pattern, sin²θ for Δm = 0 and (1 + cos²θ)/2 for Δm = ±1.
In s⁻¹ sr⁻¹, with k̂ the photon direction. d ∥ z gives |d|² sin²θ; d ∝ x̂ ± iŷ gives |d|²(1 + cos²θ)/2.The A coefficient, derived
A = ω₀³|d|²/(3πε₀ħc³)
V has cancelled and no temperature or Planck spectrum entered. ħω₀A is the Larmor power of a classical dipole of amplitude 2|d|.
|d|² = q²|⟨g|r|e⟩|² in C² m², ω₀ in s⁻¹, A in s⁻¹; ε₀ in F m⁻¹. ∫(|d|² − |k̂⋅d|²) dΩ = (8π/3)|d|² for any d.Hydrogen 2p → 1s in closed form
A = (2/3)⁸ α⁵ mₑc²/ħ
6.27 × 10⁸ s⁻¹ and 1.60 ns; the reduced mass multiplies it by μ/mₑ = 0.99946, giving 6.265 × 10⁸ s⁻¹. Beam-foil timing measures 1.6 ns.
α = 1/137.036, mₑc²/ħ = 7.763 × 10²⁰ s⁻¹; uses ħω₀ = (3/8)α²mₑc² and |⟨1s|r|2p, m⟩|² = (2¹⁵/3¹⁰)a₀².Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set n₁ = 0: no photon leaves along z, where each dipole points. Set n₀ = 0: the light is brightest along z. Make n₀ equal to n₁ and the solid curve lies on the thin isotropic line and the polarisation reads 0 at every θ; for any mix the all-directions readout is (n₀ + 2n₁)A, since every sublevel decays at the same A.
Live results
- SAMPLE dΓ/dΩ AT θ
- 5.98 10⁸ s⁻¹ sr⁻¹
- ALL DIRECTIONS, N × A
- 62.65 10⁸ s⁻¹
- POLARISATION, + IN z–k PLANE
- 0.50
- SHARE WITHIN θ OF +z
- 0.500
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. Atoms in m = 0, n₀: 6; Atoms in each of m = ±1, n₁: 2; Viewing angle θ from z: 90 °
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting picture is that spontaneous emission is ordinary stimulated emission, driven by the vacuum's zero-point field of ½ħω per mode. Put numbers on it and it fails twice. A classical random field holding ½ħω per mode is half a photon per mode, n̄ = ½, and stimulated emission runs at n̄A, so the picture predicts A/2. The same field would drive absorption too, so a ground-state hydrogen atom in a cold, dark box would be pumped into 2p within nanoseconds; that never happens. In the derivation the 1 is ⟨1ₖₛ|aₖₛ†|0⟩ = 1, from aa† = a†a + 1, not a zero-point energy, and a ground-state atom absorbs nothing because aₖₛ|0⟩ = 0. Splitting the rate symmetrically, Dalibard, Dupont-Roc and Cohen-Tannoudji (1982) found that vacuum fluctuations and radiation reaction each supply A/2 to an excited atom and cancel for a ground-state one: the vacuum field is half of the story, never all of it.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Hydrogen-like ions keep hydrogen's wavefunctions with a₀ replaced by a₀/Z and energies multiplied by Z². Using the derived A = ω₀³|d|²/(3πε₀ħc³) and hydrogen's 2p rate of 6.268 × 10⁸ s⁻¹ (infinite nuclear mass), find the 2p → 1s rate, lifetime and wavelength for He⁺ (Z = 2) and Ne⁹⁺ (Z = 10), and check that the dipole approximation and the golden rule still hold.
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.
- Scaling: ħω₀ = (3/8)Z²α²mₑc² grows as Z², and |d| = e|⟨1s|r|2p, m⟩| shrinks as 1/Z, so A ∝ ω₀³|d|² ∝ Z⁶ × Z⁻² = Z⁴. Nothing else in A knows about the atom: the mode density and the polarisation and angular sums belong to free space.
- He⁺: A = 16 × 6.268 × 10⁸ s⁻¹ = 1.003 × 10¹⁰ s⁻¹ and τ = 1/A = 99.7 ps; ħω₀ = 4 × 10.204 eV = 40.82 eV, so λ = 121.50 nm/4 = 30.38 nm.
- Ne⁹⁺: A = 10⁴ × 6.268 × 10⁸ s⁻¹ = 6.27 × 10¹² s⁻¹ and τ = 0.160 ps; ħω₀ = 100 × 10.204 eV = 1.020 keV, so λ = 1.215 nm, a soft X-ray line.
- Dipole approximation: k times the orbit size is (ω₀/c)(a₀/Z) = (3/8)αZ, 5.5 × 10⁻³ for He⁺ and 2.7 × 10⁻² for Ne⁹⁺, still small. Golden rule: A/ω₀ = 4.04 × 10⁻⁸ Z², 1.6 × 10⁻⁷ and 4.0 × 10⁻⁶, still tiny.
- Limits: Z = 1 returns hydrogen. What does grow is relativity, corrections of order (Zα)² = 5.3 × 10⁻³ for neon, which come from the Dirac equation and not from this first-order, non-relativistic calculation.
Answer. A ∝ Z⁴. He⁺: A = 1.00 × 10¹⁰ s⁻¹, τ = 99.7 ps, λ = 30.4 nm. Ne⁹⁺: A = 6.27 × 10¹² s⁻¹, τ = 0.160 ps, λ = 1.215 nm. Both stay dipole-like (ka ≤ 2.7 × 10⁻²) with A/ω₀ ≤ 4.0 × 10⁻⁶.
MediumExample 2
A hydrogen atom in 2p, m = 0, sits in a periodic cubic box of side 1.00 mm. Find the vacuum field per photon of a Lyman-α mode, the coupling g to a mode polarised along z, the number of modes per unit angular frequency at ω₀ and how many lie inside the natural width. Show that 2π(g²/3)ρ reproduces A, and find the box side below which the continuum picture fails.
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.
- From above, ω₀ = 1.5503 × 10¹⁶ s⁻¹, ħω₀ = 1.6349 × 10⁻¹⁸ J and |d| = 6.316 × 10⁻³⁰ C m. Field per photon: √(ħω₀/(2ε₀V)) = √(1.6349 × 10⁻¹⁸ J ÷ (2 × 8.854 × 10⁻¹² F m⁻¹ × 1.00 × 10⁻⁹ m³)) = 9.61 V m⁻¹.
- With ê ∥ d, ħg = 9.61 V m⁻¹ × 6.316 × 10⁻³⁰ C m = 6.07 × 10⁻²⁹ J, so g = 5.75 × 10⁵ s⁻¹. Alone with this mode the atom would swing at 2g = 1.15 × 10⁶ s⁻¹, about 550 times slower than the decay itself: no single mode has time to matter.
- Modes of both polarisations in all directions: ρ = Vω₀²/(π²c³) = 1.00 × 10⁻⁹ m³ × 2.403 × 10³² s⁻² ÷ (9.870 × 2.694 × 10²⁵ m³ s⁻³) = 9.04 × 10⁻⁴ s. That is one mode per 1.11 × 10³ s⁻¹ on average, and ρA = 5.67 × 10⁵ modes inside the natural width.
- Averaged over directions and polarisations a mode catches |êₖₛ*⋅d|² = |d|²/3, so the golden rule gives 2π(g²/3)ρ = 2π × (5.755 × 10⁵ s⁻¹)² ÷ 3 × 9.04 × 10⁻⁴ s = 6.27 × 10⁸ s⁻¹ = A. Since g² ∝ 1/V and ρ ∝ V, every box gives the same number.
- The continuum picture needs the average spacing 1/ρ ≪ A. The two are equal at V = π²c³/(ω₀²A) = 1.77 × 10⁻¹⁵ m³, a box side of 12.1 µm, about a hundred Lyman-α wavelengths. Well below that the atom sees discrete modes, the regime of the Purcell factor in topic 12.08.
Answer. Field per photon 9.61 V m⁻¹ and g = 5.75 × 10⁵ s⁻¹; ρ = 9.04 × 10⁻⁴ s, with 5.67 × 10⁵ modes inside the natural width; 2π(g²/3)ρ = 6.27 × 10⁸ s⁻¹ = A in any box much larger than 12 µm.
HardExample 3
Hydrogen atoms in 2p, with the electron treated as spinless so that m is the orbital quantum number, sit in a weak field B ∥ z. A lens collecting a cone of half-angle 30° is placed either on the z axis or on the x axis. For a π decay (m = 0 → 1s) and a σ⁺ decay (m = +1 → 1s), find the fraction of photons each lens collects and the polarisation on its axis. Then check the result for a sample with all three sublevels equally filled.
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.
- Normalised patterns from dΓ/dΩ ÷ A: π gives (3/(8π))sin²θ and σ⁺ gives (3/(16π))(1 + cos²θ), with d ∝ ẑ and d ∝ x̂ + iŷ. Each integrates to 1 over the sphere, since both angular integrals give 8π/3.
- Lens on z, with u = cos θ and dΩ = 2π du over cos 30° ≤ u ≤ 1: π collects (3/4)∫(1 − u²)du = 1/2 − (3/4)cos 30° + (1/4)cos³30° = 0.0129, and σ⁺ collects (3/8)∫(1 + u²)du = (3/8)(1 − cos 30°) + (1/8)(1 − cos³30°) = 0.0941. An isotropic source would give (1 − cos 30°)/2 = 0.0670.
- Lens on x: inside a cone about x, symmetry gives ⟨(k̂⋅ẑ)²⟩ = ⟨(k̂⋅ŷ)²⟩ = (1 − ⟨(k̂⋅x̂)²⟩)/2. The π pattern 1 − (k̂⋅ẑ)² then averages like (1 + (k̂⋅x̂)²)/2, the σ pattern about its own axis, so π collects 0.0941. The σ⁺ pattern (1 + (k̂⋅ẑ)²)/2 averages like (3 − (k̂⋅x̂)²)/4, and its integral over the cone gives 0.0535.
- Polarisation on each lens axis is the transverse part of d. Along z, σ⁺ keeps all of (x̂ + iŷ): circular. Along x, π keeps ẑ, linear along z (B), while σ⁺ keeps only iŷ, linear along y, with half the π intensity per steradian at the axis.
- Equal sublevels: lens on z collects (0.0129 + 2 × 0.0941)/3 = 0.0670, lens on x (0.0941 + 2 × 0.0535)/3 = 0.0670, both the isotropic value, as a level with no preferred axis must give. Limit check: as the half-angle reaches 180° every fraction goes to 1.
Answer. Lens on z: π 1.29%, σ⁺ 9.41%, circularly polarised. Lens on x: π 9.41%, linear along z; σ⁺ 5.35%, linear along y. Equal sublevels: 6.70% at either lens, the isotropic (1 − cos 30°)/2.
Quick check
Select an answer and explain it before checking.
Review & continue
The A coefficient is the rate at which an excited atom alone in the photon vacuum emits, obtained here as the golden-rule rate into every empty mode of the field: for an electric-dipole line, A = ω₀³|d|²/(3πε₀ħc³).
- 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.