
Optional unit · institution-dependent · Interactive lesson
Resonance fluorescence: Mollow triplet and antibunching
This is where the optical Bloch equations meet a photodetector. One atom and one resonant laser give two measurements: a spectrometer finds three lines where a classical scatterer gives one, and a pair of counters never clicks together. Both follow from the atom's lowering operator, and the second is light no classical wave can imitate.
UAQM-14.07 · Lesson 7 of 9 in this unit. Position is not a completion score.Optional unit · institution-dependent. Units 14 and 15 are the course’s optional extension into open systems and quantum gases. Unit 14 builds on radiation and decay; Unit 15 builds on second quantisation. Keep the fifteen-unit numbering and follow your institution’s required syllabus. Optional work is not a required completion gate.
Build the model
One driven atom scatters light into three lines, and never two photons at once.
The key idea
Resonance fluorescence is the light a single two-level atom scatters while a laser near its transition drives it, and every measurable property of that light is a correlation function of the atom's lowering operator σ₋.
Example. A sodium atom driven on resonance at Ω = 2π × 50 MHz, about 5.1Γ, scatters 3.0 × 10⁷ photons s⁻¹: 1.9% elastically at ωL, the rest into lines at ωL and ωL ± 2π × 49.9 MHz, and never two photons at the same instant.
The complete physical model and assumptions
Drive one two-level atom with a resonant laser and ask what light comes out. The answer is already in the atom: in the far zone the positive-frequency scattered field is a classical dipole field with d e(−iω₀t) replaced by d σ₋ at the retarded time, a result outlined from Unit 11's mode equations, so every normally ordered photodetection signal is an atomic correlation function. The mean ⟨σ₋⟩ radiates coherently, an elastic delta function at the laser frequency ωL carrying a fraction 1/(1 + s) of the light, with s = 2Ω²/Γ². The fluctuations radiate the rest, and once the quantum regression theorem is quoted their two-time correlation obeys the optical Bloch equations: the decay constants −Γ/2 and −3Γ/4 ± iΩ′ become a line at ωL of full width Γ and two at ωL ± Ω′ of full width 3Γ/2, which for Ω ≫ Γ carry weights 1:2:1 and stand in heights 1:3:1, the Mollow triplet. Read by two detectors, the same field gives g⁽²⁾(τ), and a detection leaves the atom in |g⟩, so g⁽²⁾(τ) is the excited population regrowing from zero and g⁽²⁾(0) = 0, which no classical intensity allows. The model stops where its approximations do: the Born-Markov treatment of the vacuum needs Γ and Ω far below ω₀, the laser must be narrower than Γ, atomic motion and other atoms blur both the lines and the dip, and in a cavity whose coupling beats its losses the photon comes back and the master equation fails.
1The detector reads the atom's lowering operator
Units are SI with ħ explicit; nothing here is relativistic, so the course's metric signature (+, −, −, −) never enters. Unit 11's mode equations, integrated with the atom as a source, split each mode operator into a free part and a part driven by σ₋ at earlier times. Summing the driven parts over modes, with the Markov step that treats σ₋e(iω₀t) as slow, gives in the far zone the classical dipole field with d e(−iω₀t) replaced by d σ₋(t − r/c): outlined here, not derived in full. The free part annihilates the vacuum, so a detector, which measures normally ordered products, sees only the atom: the intensity is 2ε₀c|ψ|²ρₑₑ, with |ψ| = ω₀²d sin θ/(4πε₀c²r) in V m⁻¹. Over a sphere the sin²θ pattern integrates to 8π/3 and the power to ω₀⁴d²ρₑₑ/(3πε₀c³) = ħω₀Γρₑₑ, so the photon flux is Γρₑₑ with Unit 12's Γ, a check on the normalisation. For sodium's D2 line the measured 16.25 ns lifetime gives Γ = 6.15 × 10⁷ s⁻¹, so one atom scatters at most Γ/2 = 3.1 × 10⁷ photons s⁻¹.
2Mean and fluctuation: elastic and inelastic light
Write σ₋ = ⟨σ₋⟩ + δσ₋. The correlation splits into |⟨σ₋⟩|², which never decays and so transforms into a delta function at ωL, and ⟨δσ₊(t + τ)δσ₋(t)⟩, which decays and gives lines of finite width. The first is elastic scattering by the atom's mean dipole, as sharp as the laser; the second is fluorescence proper. Topic 14.06's resonant steady state, used here, has ρₑₑ = Ω²/(Γ² + 2Ω²) and |⟨σ₋⟩|² = Ω²Γ²/(Γ² + 2Ω²)², so the elastic fraction is 1/(1 + s), derived, with s = 2Ω²/Γ² = I/Iₛₐₜ and Iₛₐₜ = πhcΓ/(3λ³) = 6.26 mW cm⁻² for sodium, computed from the measured Γ and λ. The limits check it. As Ω → 0 all the light is elastic: the atom responds linearly, like a driven classical dipole, and this is Rayleigh scattering. As s grows the atom scatters faster, ρₑₑ → ½, yet the Bloch vector shrinks to the centre of the ball and the mean dipole vanishes, so the elastic fraction falls as 1/s, to 1.9% at Ω = 5.1Γ. Inelastic light is the price of saturation, a nonlinearity no linear oscillator has.
3Regression: the Bloch equations run in τ
To evaluate ⟨δσ₊(t + τ)δσ₋(t)⟩ we quote the quantum regression theorem: under the Born-Markov approximations a two-time average ⟨A(t + τ)B(t)⟩ obeys, in τ, the same linear equations as ⟨A(t + τ)⟩, starting from ⟨A(t)B(t)⟩. For fluctuations these are the homogeneous Bloch equations. On resonance, in Bloch components, du/dt = −Γu/2, dv/dt = −Γv/2 − Ωw and dw/dt = −Γw + Ωv, so the decay constants are −Γ/2 and the roots of (λ + Γ/2)(λ + Γ) + Ω² = 0, which are λ = −3Γ/4 ± iΩ′ with Ω′ = √(Ω² − Γ²/16), derived. Each constant becomes a line centred by its imaginary part, with half-width minus its real part: a Lorentzian at ωL of full width Γ and a pair at ωL ± Ω′ of full width 3Γ/2, which also carry a dispersive part, of relative size about 5Γ/(4Ω′) for Ω ≫ Γ, that pulls their maxima slightly inside ±Ω′. The initial values ⟨σᵢσ₋⟩ − ⟨σᵢ⟩⟨σ₋⟩ fix the weights, and the inelastic area is ρₑₑ − |⟨σ₋⟩|², as it must be. Carried through exactly, this gives the closed form the figure draws at every Ω, and a direct numerical integration of the master equation reproduces it to five decimal places.
4Dressed states count the weights 1:2:1
Why three lines, and why 1:2:1, is clearest for Ω ≫ Γ with dressed states. In the frame rotating at ωL the resonant Hamiltonian (ħΩ/2)(σ₊ + σ₋) has eigenstates |±⟩ = (|g⟩ ± |e⟩)/√2 split by ħΩ; counting laser photons, each manifold of N photons is such a doublet, the next ħωL lower. Emission takes the atom from manifold N to N − 1 through σ₋, whose four elements ⟨±|σ₋|±⟩ and ⟨∓|σ₋|±⟩ all have modulus ½, so each transition runs at Γ/4 per unit population: |+⟩ → |+⟩ and |−⟩ → |−⟩ at ωL, |+⟩ → |−⟩ at ωL + Ω, |−⟩ → |+⟩ at ωL − Ω. The cross rates balance, so the populations are ½ each and the lines carry Γ/8, Γ/4 and Γ/8: weights 1:2:1 summing to Γρₑₑ = Γ/2, derived. Divided by the widths 3Γ/2, Γ and 3Γ/2 they give heights 1:3:1. For sodium at Ω = 2π × 100 MHz the lines lie 100 MHz apart, with full widths 9.8 MHz and 14.7 MHz. Measured, quoted: Schuda, Stroud and Hercher saw the triplet in a sodium beam in 1974, and Wu, Grove and Ezekiel in 1975, crossing laser and beam at right angles to remove the Doppler width, found its spacing and heights in good agreement with Mollow's theory.
5A detection resets the atom: g⁽²⁾(τ) derived
Two detectors measure Glauber's coincidence rate G⁽²⁾(τ) ∝ ⟨E⁽⁻⁾(t)E⁽⁻⁾(t + τ)E⁽⁺⁾(t + τ)E⁽⁺⁾(t)⟩, which the source field turns into ⟨σ₊(t)σ₊(t + τ)σ₋(t + τ)σ₋(t)⟩. Read it as a measurement. The first detection applies σ₋, and σ₋ρₛₛσ₊ = ρₑₑ|g⟩⟨g| whatever ρₛₛ was: a photon has just left, so the atom is certainly in its ground state. In the Markov approximation that photon never returns, so the master equation evolves |g⟩⟨g| for a time τ and the second detection comes at rate Γρₑₑ(τ), with ρₑₑ(0) = 0. So G⁽²⁾ ∝ Γρₑₑ × Γρₑₑ(τ), and dividing by (Γρₑₑ)², the rate of uncorrelated pairs, gives g⁽²⁾(τ) = ρₑₑ(τ)/ρₑₑ(∞), derived; the regression theorem returns the same expression. On resonance the Bloch equations solved from w = −1 give the closed form 1 − e(−3Γτ/4)[cos Ω′τ + 3Γ/(4Ω′) sin Ω′τ]. Checks: g⁽²⁾(0) = 0 at every Ω; g⁽²⁾ → 1 as τ → ∞, independent detections; for Ω ≪ Γ the roots turn real and g⁽²⁾ = (1 − e(−Γτ/2))²; for Ω ≫ Γ it rings, first peaking at Ω′τ = π at 1 + e(−3πΓ/(4Ω′)), which is 1.45 at Ω = 3Γ.
6No classical intensity can do this
A classical field gives the detectors a non-negative intensity I(t). Then g⁽²⁾(0) = ⟨I²⟩/⟨I⟩² = 1 + Var(I)/⟨I⟩² ≥ 1, shown, and for stationary light Cauchy-Schwarz, ⟨I(t)I(t + τ)⟩ ≤ √(⟨I(t)²⟩⟨I(t + τ)²⟩) = ⟨I²⟩, gives g⁽²⁾(τ) ≤ g⁽²⁾(0). Thermal light sits at 2 and an ideal laser at 1. One atom breaks both bounds: g⁽²⁾(0) = 0 is sub-Poissonian, and g⁽²⁾ rising from it is antibunching, photons spaced by the time to re-excite, about half a Rabi period under strong drive and a few lifetimes under weak. Measured, quoted: in 1977 Kimble, Dagenais and Mandel crossed a sodium beam with a resonant laser and saw coincidences rise from a minimum at τ = 0 over roughly the 16 ns lifetime, the first light shown to need a quantum description. Their beam held a fluctuating number of atoms, whose independent photons pull g⁽²⁾(0) up towards 1, so sub-Poissonian counts waited for Short and Mandel in 1983, and a single stored ion in 1987 removed the averaging over atoms altogether.
Key equations
Read the units, assumptions and limits before substituting.
Source field of the atom (outlined)
E⁽⁺⁾(r, t) = E⁽⁺⁾free(r, t) + [ω₀²/(4πε₀c²r)] d⊥ σ₋(t − r/c)
Outlined from Unit 11, not derived in full. Detector signals become atomic correlations; integrating 2ε₀c⟨E⁽⁻⁾⋅E⁽⁺⁾⟩ over a sphere returns ħω₀Γρₑₑ, Unit 12's photon rate.
far zone, r ≫ λ; d = ⟨g|d|e⟩ in C m; d⊥ = (r̂ × d) × r̂, of size d sin θ; ω₀ in rad s⁻¹; ε₀ in F m⁻¹; E in V m⁻¹Spectrum from a two-time correlation
S(ω) = (1/π) Re ∫₀(∞) dτ e(−iωτ) ⟨σ₊(t + τ) σ₋(t)⟩ₛₛ
The Wiener-Khinchin theorem applied to the source field, quoted. Evaluating the correlation needs the regression theorem, also quoted: in τ it obeys the Bloch equations.
S in s, per unit ω in rad s⁻¹, with ∫S dω = ρₑₑ; σ₋ = |g⟩⟨e| is dimensionless; ss marks the driven steady stateElastic fraction on resonance
Pcoh/P = |⟨σ₋⟩ₛₛ|²/ρₑₑ = Γ²/(Γ² + 2Ω²) = 1/(1 + s)
Derived from the Bloch steady state of topic 14.06. Ω → 0 is pure Rayleigh scattering; at s ≫ 1 the atom scatters fastest while its mean dipole ⟨σ₋⟩ tends to zero.
Γ = A coefficient and Ω = d⋅E₀/ħ, both in s⁻¹; s = 2Ω²/Γ² = I/Iₛₐₜ; the elastic part is a delta function at ωLMollow triplet, Ω ≫ Γ, on resonance
Sinc(ω) ≈ ρₑₑ[¼L(ωL − Ω′, 3Γ/2) + ½L(ωL, Γ) + ¼L(ωL + Ω′, 3Γ/2)]
Derived, given the regression theorem: the Bloch decay constants −Γ/2 and −3Γ/4 ± iΩ′ set centres and widths. Weights 1:2:1 over widths 3:2:3 give peak heights 1:3:1.
L(ωc, w) is a unit-area Lorentzian of full width w centred on ωc, in s; Ω′ = √(Ω² − Γ²/16) in rad s⁻¹Photon correlation of one atom, on resonance
g⁽²⁾(τ) = 1 − e(−3Γτ/4)[cos Ω′τ + 3Γ/(4Ω′) sin Ω′τ]
Derived: a detection leaves σ₋ρσ₊ ∝ |g⟩⟨g|. g⁽²⁾(0) = 0 at every Ω, and the first maximum, at Ω′τ = π, is 1 + e(−3πΓ/(4Ω′)).
the excited population regrowing from |g⟩ over its steady value; dimensionless; below Ω = Γ/4, cos and sin become cosh and sinhClassical bound on intensity correlations
g⁽²⁾(0) = ⟨I²⟩/⟨I⟩² = 1 + Var(I)/⟨I⟩² ≥ 1
Shown in one line. One atom breaks it at τ = 0, and as g⁽²⁾ rises it also breaks the Cauchy-Schwarz bound g⁽²⁾(τ) ≤ g⁽²⁾(0): sub-Poissonian and antibunched.
I(t) ≥ 0 is any classical, stationary intensity in W m⁻²; Var(I) = ⟨I²⟩ − ⟨I⟩²Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at Ω = 0.5Γ: two thirds of the light is in the arrow and the inelastic part is one hump, yet g⁽²⁾(0) is already 0. Drag Ω up: the arrow shrinks, the sidebands walk out to just inside the dashed ±Ω′ and settle near a third of the central height, and g⁽²⁾ rings at Ω′.
Live results
- ELASTIC FRACTION
- 5.3 %
- SIDEBAND OFFSET Ω′
- 2.99 Γ
- g⁽²⁾ AT THE DOT
- 1.449
- FIRST g⁽²⁾ MAXIMUM
- 1.455
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. Rabi frequency Ω: 3.0 Γ; Delay Γτ at the dot: 1.0
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Drive the atom weakly and almost all of its light is elastic: at Ω = 0.2Γ, 93% sits in the delta function at ωL, as narrow as the laser. The tempting conclusion is that this light is laser-like, photons arriving at random with g⁽²⁾(0) = 1, and that antibunching, like the triplet, needs strong driving. It does not. g⁽²⁾(0) = 0 at every Ω, because a detection leaves the atom in |g⟩ whatever state it was in, and at Ω ≪ Γ the dip is g⁽²⁾(τ) = (1 − e(−Γτ/2))², still only 0.15 at τ = 1/Γ. The spectrum is a first-order property, set by ⟨σ₊(t + τ)σ₋(t)⟩; photon statistics are second-order, set by ⟨σ₊σ₊σ₋σ₋⟩, and one does not fix the other. The zero is an interference: σ₋² = 0 forces the fluctuations' pair amplitude ⟨δσ₋δσ₋⟩ to equal −⟨σ₋⟩², cancelling exactly the pairs the elastic part alone would carry.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A sodium atom on its D2 cycling transition, Γ/2π = 9.79 MHz, is driven on resonance with Ω/2π = 50.0 MHz. Find the elastic fraction of the scattered light, the photon rate, and the positions, full widths, shares of the inelastic power and peak-height ratio of the three Mollow lines.
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.
- Ω/Γ = 50.0/9.79 = 5.11, so s = 2Ω²/Γ² = 52.2. The elastic fraction is 1/(1 + s) = 0.0188, about 1.9%, and ρₑₑ = (s/2)/(1 + s) = 0.491.
- Γ = 2π × 9.79 MHz = 6.15 × 10⁷ s⁻¹, so the photon rate is Γρₑₑ = 3.02 × 10⁷ s⁻¹, of which 0.0188 × 3.02 × 10⁷ = 5.7 × 10⁵ s⁻¹ is elastic. Both are rates in s⁻¹, as they must be.
- The sideband lines are centred at ±Ω′ from ωL, with Ω′/2π = √(50.0² − 9.79²/16) MHz = 49.94 MHz, only 0.06 MHz inside Ω. The central line has full width Γ/2π = 9.79 MHz and each sideband 1.5 × 9.79 = 14.7 MHz.
- Because Ω ≫ Γ, the inelastic power divides ¼ : ½ : ¼. Heights go as weight over width: (¼)/1.5 : (½)/1 : (¼)/1.5 = 1/6 : 1/2 : 1/6 = 1:3:1.
- Check against the exact resonant spectrum: its central peak stands 3.02 times each sideband peak, its central line holds 0.510 of the inelastic power, and the sidebands' dispersive part, of relative size about 5Γ/(4Ω′) = 0.25, puts their maxima at ±2π × 49.0 MHz, 0.9 MHz inside ±Ω′: small corrections that vanish as Γ/Ω → 0.
Answer. 1.9% elastic; 3.02 × 10⁷ photons s⁻¹; lines centred at ωL and ωL ± 2π × 49.9 MHz (exact sideband maxima at ±2π × 49.0 MHz), full widths 9.79 MHz and 14.7 MHz, carrying ¼ : ½ : ¼ of the inelastic power, heights 1:3:1 (3.02 in the exact spectrum).
MediumExample 2
A sodium atom (λ = 589.2 nm, Γ = 6.15 × 10⁷ s⁻¹) at the origin has its transition dipole along z and is driven on resonance at Ω = Γ. A photodiode of area 1.0 mm² faces it from r = 2.0 cm in the equatorial plane. Find d from Γ, then from the source field the field scale |ψ|, the intensity and the photon flux at the diode, and check the flux against Γρₑₑ.
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.
- At Ω = Γ, ρₑₑ = Ω²/(Γ² + 2Ω²) = 1/3 and the elastic fraction is 1/(1 + 2) = 1/3. ω₀ = 2πc/λ = 3.20 × 10¹⁵ rad s⁻¹, so ħω₀ = 3.37 × 10⁻¹⁹ J, or 2.10 eV.
- Unit 12's Γ = ω₀³d²/(3πε₀ħc³) inverts to d = √(3πε₀ħc³Γ/ω₀³) = 2.11 × 10⁻²⁹ C m, about 2.5 ea₀.
- At θ = 90°, |ψ| = ω₀²d/(4πε₀c²r) = 1.08 × 10⁻³ V m⁻¹. Units: s⁻² × C m ÷ (F m⁻¹ × m² s⁻² × m) = C F⁻¹ m⁻¹ = V m⁻¹.
- I = 2ε₀c|ψ|²ρₑₑ = 2 × 8.85 × 10⁻¹² × 3.00 × 10⁸ × (1.08 × 10⁻³)² ÷ 3 = 2.06 × 10⁻⁹ W m⁻². Through 1.0 × 10⁻⁶ m² that is 2.06 × 10⁻¹⁵ W, or 2.06 × 10⁻¹⁵ ÷ 3.37 × 10⁻¹⁹ = 6.1 × 10³ photons s⁻¹.
- Check: the atom emits Γρₑₑ = 2.05 × 10⁷ s⁻¹ with angular density (3/(8π)) sin²θ per steradian, and the diode subtends 1.0 × 10⁻⁶/(0.020)² = 2.5 × 10⁻³ sr, so 2.05 × 10⁷ × 0.119 × 2.5 × 10⁻³ = 6.1 × 10³ s⁻¹, the same number. A third of it, 2.0 × 10³ s⁻¹, is elastic.
Answer. d = 2.11 × 10⁻²⁹ C m, |ψ| = 1.08 × 10⁻³ V m⁻¹, I = 2.06 × 10⁻⁹ W m⁻² and 6.1 × 10³ photons s⁻¹ at the diode, equal to Γρₑₑ × (3/(8π)) × 2.5 × 10⁻³ sr; one third of them elastic.
HardExample 3
Sodium atoms (Γ = 6.15 × 10⁷ s⁻¹) are driven on resonance at Ω = 3.0Γ. For one atom, find the delay and height of the first maximum of g⁽²⁾(τ). A beam then holds a Poisson-distributed number of atoms, mean ⟨N⟩ = 1.5, in view for about 1 μs. Treating the atoms as independent, with position-dependent cross terms averaging to zero, find g⁽²⁾ at τ = 0, at that maximum and on the plateau 1/Γ ≪ τ ≪ 1 μs, and compare a fixed N = 2.
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.
- Ω′ = Γ√(3.0² − 1/16) = 2.990Γ. The derivative of g⁽²⁾ = 1 − e(−3Γτ/4)[cos Ω′τ + 3Γ/(4Ω′) sin Ω′τ] is e(−3Γτ/4)(9Γ²/16 + Ω′²)(sin Ω′τ)/Ω′, so the first maximum is at Ω′τ = π.
- That is τ = π/(2.990Γ) = 1.051/Γ = 17.1 ns, where g⁽²⁾ = 1 + e(−3π/(4 × 2.990)) = 1 + e(−0.788) = 1.455.
- With N independent atoms the coincidences are N single-atom terms plus N(N − 1) uncorrelated pairs from different atoms: G⁽²⁾ = ⟨N⟩R²g₁(τ) + ⟨N(N − 1)⟩R², with R one atom's rate and g₁ the single-atom function. Dividing by (⟨N⟩R)², g⁽²⁾(τ) = g₁(τ)/⟨N⟩ + ⟨N(N − 1)⟩/⟨N⟩².
- Poisson statistics give ⟨N(N − 1)⟩ = ⟨N⟩², so g⁽²⁾ = 1 + g₁(τ)/1.5: 1.00 at τ = 0, 1 + 1.455/1.5 = 1.97 at 17.1 ns, and 1 + 1/1.5 = 1.67 on the plateau, which falls to 1 only as the atoms leave, over about 1 μs.
- For fixed N = 2, ⟨N(N − 1)⟩/⟨N⟩² = 2/4 = 0.5: g⁽²⁾(0) = 0.50, g⁽²⁾ = 1.455/2 + 0.5 = 1.23 at the maximum, and 1.00 on the plateau.
- So the beam's light is still antibunched, rising from τ = 0, but g⁽²⁾(0) = 1 is not sub-Poissonian: number fluctuations hide the zero. Limits check it: ⟨N⟩ → ∞ gives 1, uncorrelated light, and a fixed N = 1 returns the single atom.
Answer. One atom: first maximum at 17.1 ns with g⁽²⁾ = 1.455. Poisson beam, ⟨N⟩ = 1.5: g⁽²⁾ = 1.00 at τ = 0, 1.97 at 17.1 ns, 1.67 on the plateau, antibunched but not sub-Poissonian. Fixed N = 2: 0.50, 1.23 and 1.00.
Quick check
Select an answer and explain it before checking.
Review & continue
Resonance fluorescence is the light a single two-level atom scatters while a laser near its transition drives it, and every measurable property of that light is a correlation function of the atom's lowering operator σ₋.
- 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.