
Advanced Quantum Mechanics course sequence · Interactive lesson
Scattering of light: Kramers-Heisenberg, Rayleigh, Thomson
Most light reaches your eye by scattering: the blue sky, X-ray diffraction, a gas cloud glowing at its resonance line. This lesson derives all three from one amplitude of order e², shows which piece of the coupling does the work in each regime, and why throwing away the A² term wrecks the answer.
UAQM-12.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Build the model
One amplitude of order e², three regimes: ω⁴ below the lines, λ² on them, rₑ² above.
The key idea
The Kramers-Heisenberg formula is the cross-section for an atom to absorb one photon and emit another, built from the A² coupling at first order and the A⋅p coupling at second order through intermediate states in both time orders.
Example. A free electron removes 6.65 × 10⁻²⁹ m² from any beam of photons far below mc² = 511 keV; an N₂ molecule in 532 nm light scatters 5.3 × 10⁻³¹ m², under a hundredth of that, and a hydrogen atom on its 121.6 nm resonance 7.1 × 10⁻¹⁵ m², 10¹⁴ times it.
The complete physical model and assumptions
Scattering is the simplest process in which the quantised field must act twice: one photon disappears and another appears. Unit 11's coupling can do that in two ways at the same order, e². The A² term destroys and creates in a single step, at first order; the A⋅p term changes photon number by one, so it acts twice, at second order, through an intermediate atomic state reached by absorbing first or by emitting first. The three amplitudes, fed to the golden rule, give the Kramers-Heisenberg cross-section: rₑ² times a dimensionless bracket that holds every regime. Far below the atom's lines the three terms are each of order one and cancel, first through [x, p] = iħ and then through [x, y] = 0, leaving an amplitude of order ω² and Rayleigh's ω⁴ law, scaled by the static polarisability. Far above them the A⋅p terms fade as (ωₙₐ/ω)² and A² alone leaves Thomson's free-electron 0.665 b. At a resonance the absorb-first denominator vanishes; restoring the Weisskopf-Wigner width turns the pole into a Lorentzian whose peak, 3λ²/(2π), is set by the wavelength and not by the size of the atom. The model stops where its assumptions do: once the wavelength shrinks to atomic size the dipole approximation fails, near ħω ≈ mc² recoil and spin turn Thomson into Klein-Nishina, and in a dense gas the waves from neighbouring atoms interfere, which a one-atom cross-section cannot describe.
1Three amplitudes, all of order e²
SI units throughout; where a four-vector appears the metric is (+, −, −, −). Quoted from Unit 11: for an electron of charge −e in the Coulomb gauge the coupling is (e/m)A⋅p + (e²/2m)A², with A = Σₖₛ√(ħ/(2ε₀ωₖV))(aₖₛ + aₖₛ†)êₖₛ once the dipole approximation sets e(±ik⋅r) to 1. Scattering takes |a; kê⟩ to |b; k′ê′⟩, destroying one photon and creating another. A² contains aₖ′†aₖ and aₖaₖ′†, so it does this in one step, at first order, with matrix element (e²ħ/(2mε₀V√(ωω′)))⟨b|a⟩ê⋅ê′, derived. A⋅p changes photon number by one and must act twice. It can absorb first, through |n; 0⟩ with energy denominator Eₐ + ħω − Eₙ, or emit first, through |n; kê, k′ê′⟩ with Eₐ − ħω′ − Eₙ. Each second-order amplitude also carries e², so A² is not a higher-order correction, and the emit-first path, whose intermediate state never conserves energy, is not small away from resonance.
2From amplitudes to a cross-section
Add the three amplitudes and apply the golden rule, which Quantum Mechanics II derived for second-order amplitudes too, with Vω′² dΩ/((2π)³ħc³) final photon states per unit energy and an incident flux of one photon per volume V at speed c. Every V cancels, the couplings collect into (e²/(4πε₀mc²))² = rₑ², and dσ/dΩ = rₑ²(ω′/ω)|bracket|², derived, with the bracket of the first formula card. Dimensions: ⟨p⟩²/(m × energy) is (kg m s⁻¹)²/(kg × kg m² s⁻²) = 1, so dσ/dΩ is an area per steradian. When b = a the scattering is elastic, ω′ = ω, and the A² term ⟨b|a⟩ê⋅ê′ is present. When b ≠ a it vanishes and ω′ = ω − (Eb − Eₐ)/ħ: Raman scattering, Stokes when b lies above a and anti-Stokes when below. Checked against measurement: nitrogen's vibrational spacing, 2331 cm⁻¹, moves 532 nm light to 607.3 nm, the wavelength at which Raman lidars detect atmospheric nitrogen.
3Below every resonance: two sum rules leave ω⁴
Take b = a and use ⟨n|p|a⟩ = imωₙₐ⟨n|r|a⟩, quoted from Quantum Mechanics II. With X₁ = ⟨a|ê′⋅r|n⟩⟨n|ê⋅r|a⟩ and X₂ the other order, the A⋅p part of the bracket is −(m/ħ)Σₙωₙₐ²[X₁/(ωₙₐ − ω) + X₂/(ωₙₐ + ω)]. Split each fraction: ωₙₐ²/(ωₙₐ ∓ ω) = ωₙₐ ± ω + ω²/(ωₙₐ ∓ ω). The first pieces give −(m/ħ)Σₙωₙₐ(X₁ + X₂) = −ê⋅ê′ by the commutator sum rule, cancelling the A² term exactly. The second give −(mω/ħ)⟨a|[ê′⋅r, ê⋅r]|a⟩ = 0, since positions commute. What survives, with no approximation, is the bracket −(mω²/ħ)Σₙ[X₁/(ωₙₐ − ω) + X₂/(ωₙₐ + ω)]: an amplitude of order ω², so σ ∝ ω⁴, derived. For an S state, X₁ and X₂ summed over each level's sublevels are (ê⋅ê′)|⟨n|z|a⟩|², and σ = σT ω⁴[Σₙfₙₐ/(ωₙₐ² − ω²)]². As ω → 0 the sum is m/e² times the static polarisability α, which gives Rayleigh's (8π/3)(ω/c)⁴(α/4πε₀)².
4Rayleigh in numbers: hydrogen, nitrogen, the sky
For hydrogen's ground state Quantum Mechanics II gave α/4πε₀ = 4.5a₀³, quoted here. With Σf = 1 the Rayleigh limit reads σR = σT(ħω/12.83 eV)⁴: the atom scatters like one oscillator of unit strength at 12.83 eV, above Lyman-α's 10.20 eV because the 2p line carries f = 0.416 and supplies only 66% of α. At 1064 nm that is 6.8 × 10⁻⁵ σT, and the next term, of relative order ω², adds 2%. Dimensions: (ω/c)⁴ is m⁻⁴ and (α/4πε₀)² is m⁶, an area. Limit: ω → 0 gives zero, as it must for a bound charge that a static field can only displace. Measurement: nitrogen's refractive index at 532 nm, n − 1 = 2.99 × 10⁻⁴ at 0 °C and 1 atm, gives α/4πε₀ = (n − 1)/(2πN) = 1.77 × 10⁻³⁰ m³ and σR = 5.1 × 10⁻³¹ m²; the molecule's anisotropy adds 3%, quoted, and cavity ring-down measurements of the scattering itself agree within a few per cent. The λ⁻⁴ makes 450 nm light scatter 4.4 times more strongly than 650 nm: the blue of the sky.
5Above every resonance: A² alone gives Thomson
When ħω is far above every Eₙ − Eₐ that carries oscillator strength, each A⋅p denominator is close to ∓ħω, the two time orders cancel because ⟨a|[ê′⋅p, ê⋅p]|a⟩ = 0, and what remains falls as (ωₙₐ/ω)². In the S-state form Σₙfₙₐ/(ωₙₐ² − ω²) → −1/ω² by the Thomas-Reiche-Kuhn rule, so σ → σT. Only A² survives: dσ/dΩ = rₑ²|ê⋅ê′|², which summed over ê′ is rₑ²sin²Θ, Θ the angle between ê and the scattered direction, and over all directions gives (8π/3)rₑ² = 6.652 × 10⁻²⁹ m², derived; rₑ² is an area, as it must be. Unpolarised light scatters as rₑ²(1 + cos²θ)/2 and is fully polarised at 90°. A free electron at rest has no A⋅p term in the dipole approximation, so this is its entire cross-section. Measurement: Thomson's value gives 0.200 cm² g⁻¹ for carbon, six electrons per 12.0 g mol⁻¹, the mass scattering coefficient Barkla measured for hard X-rays in light elements and used to count about half the mass number of electrons per atom. The window is binding ≪ ħω ≪ mc²; beyond it recoil and the Klein-Nishina formula take over, named here only.
6On resonance: restore the width
As ω → ωₙₐ the absorb-first denominator Eₙ − Eₐ − ħω vanishes and the bracket diverges, because second-order perturbation theory has let the intermediate state live forever. Topics 12.06 and 12.07 showed that it decays, its amplitude carrying Eₙ − iħΓ/2, quoted here. Near an S to P line at ω₀ the third card becomes σ = (8π/3)rₑ²(fω₀/2)²/((ω − ω₀)² + Γ²/4), a Lorentzian of full width Γ peaking at (8π/3)rₑ²(fω₀/Γ)². If the upper level decays only back to a, Γ is topic 12.02's A, which written through f is 2frₑω₀²/(3c). Then f cancels: σ₀ = 3λ²/(2π), derived, for every such line. Lyman-α at 121.6 nm gives 7.06 × 10⁻¹⁵ m², 10¹⁴ times σT and 8 × 10⁵ times πa₀². Dimensions: λ² is an area. Limit: weaken the line and the peak does not fall, because the width narrows with it. Measurement: the same A gives the 2p lifetime, 1.60 ns, as measured. Over frequency the line holds ∫σ dν = σ₀Γ/4 = πrₑcf, the classical absorption area, since with Γ = A every photon absorbed is re-emitted: it is set by f alone, and broadening that conserves photons, such as the Doppler shifts of a warm gas, spreads it without changing it.
Key equations
Read the units, assumptions and limits before substituting.
Kramers-Heisenberg cross-section
dσ/dΩ = rₑ²(ω′/ω)|⟨b|a⟩ê⋅ê′ − (1/m)Σₙ[⟨b|p⋅ê′|n⟩⟨n|p⋅ê|a⟩/(Eₙ − Eₐ − ħω) + ⟨b|p⋅ê|n⟩⟨n|p⋅ê′|a⟩/(Eₙ − Eₐ + ħω′)]|²
Derived: A² at first order, then A⋅p at second order, absorbing first and emitting first. For b ≠ a the A² term drops out and the lines are Raman, shifted by (Eb − Eₐ)/ħ.
rₑ = e²/(4πε₀mc²) in m; ê, ê′ real unit polarisations; p in kg m s⁻¹; E in J; ħω′ = ħω + Eₐ − Eb. The bracket is dimensionless.Commutator sum rule that cancels A²
ê⋅ê′ = (m/ħ)Σₙ ωₙₐ[⟨a|ê′⋅r|n⟩⟨n|ê⋅r|a⟩ + ⟨a|ê⋅r|n⟩⟨n|ê′⋅r|a⟩]
Derived; with ê = ê′ it is the Thomas-Reiche-Kuhn rule Σₙfₙₐ = 1. It makes the A² amplitude cancel the frequency-independent part of second-order A⋅p exactly.
ωₙₐ = (Eₙ − Eₐ)/ħ in s⁻¹, r in m, m in kg; from ⟨a|[ê′⋅r, ê⋅p]|a⟩ = iħê⋅ê′ and ⟨n|p|a⟩ = imωₙₐ⟨n|r|a⟩.Elastic scattering from an S state, widths neglected
σ(ω) = σT ω⁴[Σₙ fₙₐ/(ωₙₐ² − ω²)]²
Derived from the first card with both sum rules, exact away from resonances. As ω → 0 it is Rayleigh; as ω → ∞, (Σf)² = 1 gives Thomson; at each ωₙₐ it has a pole.
fₙₐ = (2mωₙₐ/ħ)|⟨n|z|a⟩|², summed over the sublevels of n, is dimensionless; ω in s⁻¹; σT in m².Rayleigh limit, far below every line
σR = (8π/3)(ω/c)⁴(α/4πε₀)²
Derived: the surviving amplitude is of order ω², so σ ∝ λ⁻⁴. Hydrogen acts as one f = 1 oscillator at 12.83 eV, so σR = σT(ħω/12.83 eV)⁴.
α = (e²/m)Σₙfₙₐ/ωₙₐ², the static polarisability in C m² V⁻¹; α/4πε₀ in m³, 4.5a₀³ = 6.67 × 10⁻³¹ m³ for hydrogen 1s.Thomson limit, far above every line
σT = (8π/3)rₑ² = 6.652 × 10⁻²⁹ m²
Derived from A² alone: Σ over ê′ of |ê⋅ê′|² is sin²Θ, whose solid-angle integral is 8π/3. Needs binding ≪ ħω ≪ mc²; the Klein-Nishina regime beyond is named only.
rₑ = e²/(4πε₀mc²) = 2.818 × 10⁻¹⁵ m; 0.6652 b, 1 b = 10⁻²⁸ m²; unpolarised: dσ/dΩ = rₑ²(1 + cos²θ)/2.Resonant peak with the width restored
σ₀ = 3λ²/(2π)
Derived from (8π/3)rₑ²(fω₀/Γ)²: f cancels, because the coupling that makes the atom scatter strongly also empties the level fast. About 10⁶ times πa₀².
σ₀ is σ at ω₀, λ in m; S to P line whose upper level decays only to a, so Γ = A = 2frₑω₀²/(3c). Lyman-α: 7.06 × 10⁻¹⁵ m².Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at ω₀/100: the dashed A² line and A⋅p curve both sit at σT, yet their sum, solid, lies 10⁸ below. Pass ω₀ and the A⋅p curve falls as ω⁻⁴ while the sum lands on the A² line. Narrow Γ, exaggerated here since real lines sit near 10⁻⁷ω₀, and only the peak moves, as (ω₀/Γ)².
Live results
- ALL TERMS, log₁₀ σ/σT
- -2.76
- PURE ω⁴ LAW, log₁₀ σ/σT
- -2.80
- A⋅p ALONE, log₁₀ σ/σT
- 0.04
- PEAK AT ω₀, log₁₀ σ/σT
- 3.00
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. Photon frequency log₁₀(ω/ω₀): −0.70; Line width log₁₀(Γ/ω₀): −1.5
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Fresh from the Jaynes-Cummings model, students drop A² as a small diamagnetic correction and keep only the resonant, absorb-first A⋅p term. Both moves fail for scattering. A² is not higher order: its amplitude rₑê⋅ê′ carries e², exactly as second-order A⋅p does, and below resonance it is as large as the A⋅p sum and opposite in sign. Nor is the emit-first term negligible: the rotating-wave argument needs |ω − ω₀| ≪ ω₀ and slow near-resonant dynamics, the opposite of scattering far from a line. Keep only part of the bracket and the cross-section stops falling at low frequency, tending to σT with A² dropped and to σT/4 with only the resonant term, whatever the colour. The sky would be white, and clear air far more opaque. The ω⁴ law needs all three amplitudes and the sum rules that cancel them, the same sum rules that make the A⋅p and d⋅E forms of the coupling agree.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A Thomson-scattering diagnostic sends a 1.00 J pulse of 532 nm light, polarised perpendicular to the scattering plane, through a plasma with nₑ = 1.0 × 10²⁰ m⁻³. Optics at 90° view a 10 mm length of the beam through a solid angle of 0.010 sr. Using the A² term alone, how many scattered photons do they collect?
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.
- A free electron at rest offers A⋅p no other state to reach in the dipole approximation, so A² alone acts: dσ/dΩ = rₑ²|ê⋅ê′|², which summed over the two outgoing polarisations is rₑ²sin²Θ, Θ the angle between ê and the scattered direction.
- Here ê is perpendicular to the scattering plane and the scattered direction lies in it, so Θ = 90° and dσ/dΩ = rₑ² = (2.818 × 10⁻¹⁵ m)² = 7.94 × 10⁻³⁰ m² sr⁻¹.
- Photons in the pulse: hc/λ = 3.734 × 10⁻¹⁹ J each, so N = 1.00 J ÷ 3.734 × 10⁻¹⁹ J = 2.68 × 10¹⁸.
- Electrons per unit beam area in the viewed length: nₑL = 1.0 × 10²⁰ m⁻³ × 0.010 m = 1.0 × 10¹⁸ m⁻². Collected: N nₑL (dσ/dΩ) ΔΩ = 2.68 × 10¹⁸ × 1.0 × 10¹⁸ m⁻² × 7.94 × 10⁻³⁰ m² sr⁻¹ × 0.010 sr = 2.1 × 10⁵.
- Regime check: ħω = 2.33 eV against mc² = 511 keV, so recoil is negligible and Thomson holds; the electrons' thermal motion Doppler-broadens the scattered light, which is how the diagnostic reads the temperature, without changing the count.
Answer. About 2.1 × 10⁵ photons, 8 × 10⁻¹⁴ of those in the pulse, from dσ/dΩ = rₑ² = 7.94 × 10⁻³⁰ m² sr⁻¹ at 90°.
MediumExample 2
A model atom has all its oscillator strength, f = 1, in one line at ħω₀ = 10.0 eV above an S ground state. For elastic scattering of 1.00 eV photons, find the A² amplitude and the two A⋅p amplitudes in units of rₑê⋅ê′, their sum and σ. What would dropping A², or keeping only the resonant term, predict?
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 the first card with b = a and ⟨n|p|a⟩ = imω₀⟨n|r|a⟩, and using (m/ħ)ω₀²|⟨n|z|a⟩|² = fω₀/2: the A² amplitude is +1, the absorb-first A⋅p amplitude −(f/2)ω₀/(ω₀ − ω) and the emit-first one −(f/2)ω₀/(ω₀ + ω).
- With ω/ω₀ = 0.100: absorb-first −0.5/0.900 = −0.5556, emit-first −0.5/1.100 = −0.4545.
- Sum: 1 − 0.5556 − 0.4545 = −0.0101, equal to −(ω/ω₀)²/(1 − (ω/ω₀)²) = −0.01/0.99 = −0.010101: the sum rule has removed every order-one part.
- σ = σT × 0.010101² = 1.020 × 10⁻⁴ σT = 6.79 × 10⁻³³ m². The pure ω⁴ law, 1.000 × 10⁻⁴ σT, is 2% low, because the line is near enough to raise the polarisability.
- Dropping A² leaves |−1.0101|² = 1.020 σT, 1.0 × 10⁴ times too large. Keeping only the resonant term leaves 0.5556² = 0.309 σT, 3.0 × 10³ times too large.
Answer. Amplitudes +1, −0.5556 and −0.4545 summing to −0.0101; σ = 1.02 × 10⁻⁴ σT = 6.8 × 10⁻³³ m². Without A² the answer is 10⁴ times too big; with only the resonant term, 3 × 10³ times.
HardExample 3
Hydrogen in its ground state is lit at Lyman-α, λ = 121.57 nm, where f = 0.4162 for 1s to 2p and 2p decays only to 1s. Ignoring fine structure, find Γ, the natural peak cross-section and the line's frequency-integrated cross-section, then the peak in gas at 1.0 × 10⁴ K, where Doppler broadening dominates. What column of atoms is optically thick at line centre?
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.
- ω₀ = 2πc/λ = 1.5494 × 10¹⁶ s⁻¹. Topic 12.02's A written through f: Γ = A = 2frₑω₀²/(3c) = 2 × 0.4162 × 2.8179 × 10⁻¹⁵ m × (1.5494 × 10¹⁶ s⁻¹)² ÷ (3 × 2.9979 × 10⁸ m s⁻¹) = 6.26 × 10⁸ s⁻¹, a lifetime of 1.60 ns, as measured.
- Peak with the width restored: (8π/3)rₑ²(fω₀/Γ)² = 3λ²/(2π) = 3 × (1.2157 × 10⁻⁷ m)² ÷ 2π = 7.06 × 10⁻¹⁵ m², with full width Γ/2π = 99.7 MHz.
- Area under the Lorentzian: ∫σ dν = σ₀ × Γ/4 = 7.06 × 10⁻¹⁵ m² × 6.26 × 10⁸ s⁻¹ ÷ 4 = 1.10 × 10⁻⁶ m² s⁻¹, the same as πrₑcf: it is fixed by f, so broadening that conserves photons can spread it but not change it.
- Doppler full width: ΔνD = (ν₀/c)√(8kT ln 2/mH), with ν₀ = 2.466 × 10¹⁵ Hz and √(8kT ln 2/mH) = 2.139 × 10⁴ m s⁻¹, so ΔνD = 1.76 × 10¹¹ Hz: 1.8 × 10³ times Γ/2π and 16 times the 10.9 GHz fine-structure splitting.
- The area survives but is spread over a Gaussian: peak = (2/ΔνD)√(ln 2/π) × 1.10 × 10⁻⁶ m² s⁻¹ = 0.9394 × 1.10 × 10⁻⁶ ÷ 1.76 × 10¹¹ m² = 5.9 × 10⁻¹⁸ m², 1.2 × 10³ times below the natural peak.
- Optical depth one at line centre needs N = 1/σ = 1.7 × 10¹⁷ m⁻², or 1.7 × 10¹³ cm⁻²: clouds this thin already take Lyman-α out of a quasar's light, the absorbers of the Lyman-α forest.
Answer. Γ = 6.26 × 10⁸ s⁻¹; natural peak 7.06 × 10⁻¹⁵ m²; ∫σ dν = 1.10 × 10⁻⁶ m² s⁻¹; Doppler peak at 10⁴ K 5.9 × 10⁻¹⁸ m²; optically thick beyond N ≈ 1.7 × 10¹⁷ m⁻².
Quick check
Select an answer and explain it before checking.
Review & continue
The Kramers-Heisenberg formula is the cross-section for an atom to absorb one photon and emit another, built from the A² coupling at first order and the A⋅p coupling at second order through intermediate states in both time orders.
- 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.