University Quantum Field Theory I · QED at Tree Level: Moller, Bhabha and Compton Scattering and Pair Annihilation · 10.2
Møller scattering at low energy: Mott's formula recovered
Topic 10.01 put a minus sign between Møller's two diagrams. This lesson makes that sign earn its keep: slow the electrons down and field theory must hand back the Mott formula of Advanced QM, singlet and triplet weights included. It does, and the way it does marks where tree-level QED stops being the right tool.
Build the model
Connect the measurement to the mechanism.
Slow electrons give field theory nothing new to say: a Coulomb potential and the Pauli principle already fix how they scatter. That makes the low-speed limit of Møller's amplitude a test, and this lesson runs it. In the centre-of-mass frame the energy transfer is exactly zero, so each photon line is the Coulomb 1/|q|²; topic 9.03's reduction keeps only the charge density 2m at each vertex, and topic 5.06's dictionary f = ℳ/(8π√s) turns each diagram into a first Born amplitude for V = α/r with reduced mass m/2.
What field theory adds is bookkeeping that the Schrödinger equation imposes by hand. The t channel keeps each electron's spin, the u channel swaps them, and the relative minus sign from anticommuting fields makes the spin dependence f − f′Pσ, with Pσ the spin-exchange operator. Triplet pairs therefore scatter with f − f′ and singlet pairs with f + f′, and the unpolarised average is the Born-level Mott formula, now derived.
The match holds in a window, α ≪ v ≪ 1. Above it, spatial currents and spinor normalisation add corrections of order v², and the interference itself vanishes at s = 6m² before turning constructive. Below it, every further photon in the ladder of repeated exchanges costs α/v rather than α, so the tree fails as v falls toward α, and the logarithmic Coulomb phase it cannot supply takes over everywhere except at exactly 90°, where that phase cancels.
- Simple definition
- Low-energy Møller scattering is tree-level e⁻e⁻ → e⁻e⁻ at α ≪ v ≪ 1: each diagram becomes a Coulomb Born amplitude, the t channel keeping both spins and the u channel swapping them with a minus sign, and the spin average is Mott's formula at first Born order.
- Example
- Two electrons of 500 eV each in their centre-of-mass frame move at v = 0.044c, and α/(4mv²) becomes αħc/(8T) = 360 fm. At 90° the unpolarised dσ/dΩ = 4 × (360 fm)² = 5.18 × 10³ b sr⁻¹, half the 1.04 × 10⁴ b sr⁻¹ that distinguishable particles would give.
The t channel keeps each spin and the u channel swaps them; the minus is topic 7.05's fermion sign. With q⁰ = 0 in the CM frame, each photon line is exactly the Coulomb 1/|q|².
ℳ dimensionless; f in GeV⁻¹, or fm after × ħc; δᵢⱼ = 1 when electrons i and j carry the same spin label; derived, up to corrections of order v²
From ℳₜ → −4m²e²/|q|² and topic 5.06's f = ℳ/(8π√s) at √s = 2m. It is UQMII-15.08's Born amplitude for V = +α/r: negative, because like charges repel.
|q| = 2mv sin(θ/2) in GeV; v each electron's CM speed, a fraction of c; α = 1/137.036; f in GeV⁻¹, fm after × ħc; derived
Derived: δ₄₁δ₃₂ is the matrix element of Pσ. Triplets scatter with f − f′ and the singlet with f + f′, with nothing antisymmetrised by hand.
f′ = f(π − θ); Pσ swaps the two spin labels, +1 on the three triplet states and −1 on the singlet; dimensionless
It is |f|² + |f′|² − Re(f f′*) = (1/4)|f + f′|² + (3/4)|f − f′|², UAQM-02.06's weights. At 90° it is half the distinguishable sum.
dσ/dΩ in GeV⁻² sr⁻¹, × 0.3894 mb GeV²; α/(4mv²) = αħc/(8T), T = ½mv² each; counts either electron; derived
Relativity weakens the destructive interference: R(90°) is 0.504 at 1 keV and 0.541 at 10 keV. The interference vanishes at s = 6m², 115 keV each, and is constructive above.
R and E dimensionless; v each electron's CM speed, a fraction of c; E is 1/2 at 90°, 0.3 at 60°; derived from topic 10.01's trace
Tree level is its η → 0 limit. At 10 eV and 30°, η = 0.58 and RC = 0.998 against Born's 0.929; η = 1 at T = 3.40 eV, a quarter of the Rydberg.
quoted from UAQM-02.08, non-relativistic, all orders in α/v; E as in the card above; η dimensionless; the phase vanishes at 90°
Two Coulomb exchanges, read in the centre-of-mass frame
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout; ħc = 197.327 MeV fm restores lengths. Topic 10.01's amplitude is ℳ = ℳₜ − ℳᵤ, with ℳₜ = e²[ū₃ γμ u₁][ū₄ γμ u₂]/t, e² = 4πα, and ℳᵤ the same with electrons 3 and 4 interchanged. In the centre-of-mass frame the electrons arrive with momenta ±p and leave with ±p′ at angle θ. Each keeps its energy, so the exchanged photon has q⁰ = 0 exactly, t = −|q|² = −4p² sin²(θ/2) and u = −4p² cos²(θ/2). The propagator is therefore the Coulomb 1/|q|² at every speed, and everything that depends on speed sits in the spinor currents. Topic 9.03 reduced those, quoted: ū(p′, s′) γ⁰ u(p, s) tends to 2m times a Kronecker delta in spin, while the spatial currents are of order mv. The contracted product of two currents is then 4m² times two spin deltas, plus magnetic and convection terms of relative order v², which at 1 keV per electron are below 1%.
Each diagram is a Coulomb Born amplitude
Insert the reduction. With t = −|q|², ℳₜ → −4m²e² δ₃₁δ₄₂/|q|², where δᵢⱼ is 1 when electrons i and j carry the same spin label. Topic 5.06's dictionary, quoted, gives the non-relativistic amplitude whose square is dσ/dΩ: f = ℳ/(8π√s) with √s → 2m. So the t channel gives f(θ) = −4m²(4πα)/(16πm|q|²) = −mα/|q|², and with |q| = 2mv sin(θ/2), f(θ) = −α/(4mv² sin²(θ/2)), derived. Three checks. Dimensions: α is a pure number and mv² a mass, so f has mass dimension −1, a length. Physics: it is UQMII-15.08's first Born amplitude −(2μ/4π)Ṽ(q) for V = +α/r with μ = m/2, negative because like charges repel. Number: restoring units, α/(4mv²) = αħc/(8T), with T = ½mv² each electron's kinetic energy, is 180.0 fm at T = 1.00 keV. The u channel is the same calculation with p′ → −p′: |q|² becomes 4m²v² cos²(θ/2), giving f(π − θ), but its deltas pair electron 4 with 1 and 3 with 2.
The fermion sign becomes the spin-exchange operator
Assemble the channels: ℳ/(16πm) = f(θ) δ₃₁δ₄₂ − f(π − θ) δ₄₁δ₃₂, derived. Read the spin factors as matrix elements ⟨s₃s₄|O|s₁s₂⟩ of an operator on the two spins. The first is the identity; the second swaps the labels, so it is the spin-exchange operator Pσ = (1 + σ₁⋅σ₂)/2, and O = f − f′Pσ with f′ = f(π − θ). Because σ₁⋅σ₂ is +1 on the three triplet states and −3 on the singlet, Pσ is +1 and −1 there: triplet pairs scatter with f − f′, the singlet with f + f′. UAQM-02.06 reached the same pattern by antisymmetrising a space-times-spin wavefunction by hand. Here nothing was imposed; it came from topic 7.05's ordering of anticommuting fields. Reverse that sign, as if electrons were spin-half bosons, and the triplet would get f + f′ and the singlet f − f′, interchanging the weights. The Coulomb force does not act on spin, so the total spin is conserved and the two channels never mix.
Averaging the spins gives Mott's formula at Born level
An unpolarised beam on an unpolarised target puts each of the four initial spin states in with weight 1/4, and a detector blind to spin sums the four final states. Square O and sum: the direct terms give 4|f|² and 4|f′|², and the cross term needs all four labels equal, which happens for two of the sixteen combinations, giving −4 Re(f f′*). Divided by 4, dσ/dΩ = |f|² + |f′|² − Re(f f′*) = (1/4)|f + f′|² + (3/4)|f − f′|², derived: one singlet state in four and three triplet states in four. With f and f′ real and negative this is (α/(4mv²))²[1/sin⁴(θ/2) + 1/cos⁴(θ/2) − 1/(sin²(θ/2) cos²(θ/2))], the Born-level form of Mott's 1930 formula (UAQM-02.08), shown. Limits: as θ → 0 the first term dominates and Rutherford's single-particle formula returns; delete the exchange term and the distinguishable sum remains; at 90°, f = f′, the triplet amplitude vanishes and dσ/dΩ = |f|², half the distinguishable sum at any speed inside the window.
The mass-kept trace: the same limit, and its first correction
Topic 10.01 squared the full amplitude with the mass kept, quoted: ⟨|ℳ|²⟩ = 2e⁴[(s² + u² + 8m²t − 8m⁴)/t² + (s² + t² + 8m²u − 8m⁴)/u² + 2(s − 2m²)(s − 6m²)/(tu)]. As v → 0, s → 4m² while t and u vanish, so the three terms tend to 8m⁴/t², 8m⁴/u² and −8m⁴/(tu). Divided by 64π²s = 256π²m², they become α²m²/|q|⁴ = |f|², |f′|² and −f f′: the spin-averaged result again, now from traces with no spinors written down. Keep the speed finite, with v² = 1 − 4m²/s, and add 1 to the interference divided by the direct terms: the ratio to the distinguishable sum is R = 1 − (1 − 3v²)(1 + v²) sin²θ/[2 + 3v⁴ + (2 + 8v²)cos²θ + v⁴cos⁴θ], derived. At 90° it reads 1/2 + v² + O(v⁴): 0.504 at T = 1 keV and 0.541 at 10 keV. The factor 1 − 3v², which is (s − 6m²) in disguise, vanishes at v = 0.577, T = 115 keV, and beyond it the exchange interference turns constructive, heading for 10.01's 9/5 at 90°.
Where tree level stops: α ≪ v ≪ 1
The upper edge is relativity: the corrections run as v², so the window closes well before the interference changes sign. The lower edge is the Born series. Each extra rung in the ladder of repeated exchanges brings α times a factor that grows as the pair slows, so the true expansion parameter is η = α/(2v), the Sommerfeld parameter for relative speed 2v, not α. The ladders rebuild UAQM-02.07's exact amplitude, whose phase −η ln sin²(θ/2) no single exchange supplies. Its angle-independent part, infrared divergent once the photon is massless, is common to both channels and cancels in f f′*, leaving cos(η ln tan²(θ/2)) on the interference (UAQM-02.08, quoted). At 90°, tan(θ/2) = 1 and the phase drops out, so the 90° value holds at every η. Elsewhere it does not: η = 1 at T = 3.40 eV, a quarter of the measured Rydberg 13.606 eV, and real targets add binding energies of ten eV and more. Møller's full formula, of which this is the limit, is checked against measured electron-electron scattering and calibrates the Møller polarimeters at electron accelerators (quoted).
Change one variable at a time
Make the relationship visible.
Set θ to 90° and drag v down to 0.002c, where α/(2v) nears 2: the faint all-orders curve swings away from the dashed Born-Mott curve everywhere except 90°, where the Coulomb phase cancels. Then drag v above 0.3c and watch the solid tree-level curve lift as relativity weakens the exchange dip.
η = α/(2v)0.058
BORN-LEVEL MOTT R(θ)0.700
TREE-LEVEL MØLLER R(θ)0.703
MOTT WITH COULOMB PHASE R(θ)0.701
Live interpretationη = α/(2v): 0.058. BORN-LEVEL MOTT R(θ): 0.700. TREE-LEVEL MØLLER R(θ): 0.703. MOTT WITH COULOMB PHASE R(θ): 0.701
Catch the common trap
Explain before calculating.
Unpolarised electrons scatter at a speed inside α ≪ v ≪ 1. In the centre-of-mass frame at θ = 90°, how does the tree-level dσ/dΩ compare with |f(90°)|², the square of a single Coulomb Born amplitude?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTwo free electrons collide head-on in their centre-of-mass frame, each with kinetic energy T = 1.00 keV. Check that they lie inside the window α ≪ v ≪ 1, then find the Born-level Mott dσ/dΩ at θ = 90° and θ = 30° in b sr⁻¹, with its ratio to the distinguishable sum at each angle.
- Speed: v²/c² = 2T/(mc²) = 2(1.00 keV)/(511.0 keV) = 3.91 × 10⁻³, so v = 0.0626c. Then η = α/(2v) = 0.00730/0.125 = 0.058 ≪ 1 and v² = 0.004 ≪ 1: inside the window.
- Length scale: restoring ħ and c, α/(4mv²) becomes αħc/(4mc² v²/c²) = αħc/(8T) = (1.440 MeV fm)/(8.00 × 10⁻³ MeV) = 180.0 fm. Its square is 3.240 × 10⁴ fm² = 324.0 b, since 1 b = 100 fm².
- At 90°: sin²45° = cos²45° = 1/2, so the bracket is 4 + 4 − 4 = 4 and dσ/dΩ = 4 × 324.0 b sr⁻¹ = 1.30 × 10³ b sr⁻¹. The distinguishable bracket is 4 + 4 = 8, so the ratio is 0.500.
- At 30°: with sin⁴(θ/2) + cos⁴(θ/2) = 1 − 2 sin²(θ/2) cos²(θ/2) and sin²(θ/2) cos²(θ/2) = sin²θ/4, the Mott bracket becomes 16/sin⁴θ − 12/sin²θ = 256 − 48 = 208, so dσ/dΩ = 208 × 324.0 b sr⁻¹ = 6.74 × 10⁴ b sr⁻¹.
- The distinguishable bracket simplifies the same way to 16/sin⁴θ − 8/sin²θ = 224, so the ratio at 30° is 208/224 = 0.929. Exchange removes half the rate at 90° but only 7% at 30°, where the partner's amplitude f′ is small.
Answerdσ/dΩ = 1.30 × 10³ b sr⁻¹ at 90° (ratio 0.500) and 6.74 × 10⁴ b sr⁻¹ at 30° (ratio 0.929); η = 0.058 and v² = 0.004 place the pair inside the window.
MediumElectrons of T = 2.50 keV each meet in their centre-of-mass frame. From ℳ/(16πm) = f δ₃₁δ₄₂ − f′ δ₄₁δ₃₂, find dσ/dΩ at θ = 60° when beam and target are both spin-up along z, and when a spin-up electron meets a spin-down one with final spins unobserved. Average to the unpolarised result and check it against Mott's formula.
- Length: αħc/(8T) = (1.440 MeV fm)/(0.0200 MeV) = 72.0 fm. At 60°, sin²30° = 1/4 and cos²30° = 3/4, so f = −72.0 fm/(1/4) = −288 fm and f′ = −72.0 fm/(3/4) = −96.0 fm.
- Both up: s₁ = s₂ = ↑, and both terms survive only for s₃ = s₄ = ↑, so the amplitude is f − f′ = −192 fm and dσ/dΩ = (192 fm)² = 3.686 × 10⁴ fm² = 368.6 b sr⁻¹, a pure triplet.
- Up on down: s₁ = ↑, s₂ = ↓. The direct term needs s₃ = ↑, s₄ = ↓ and gives f; the exchange term needs s₃ = ↓, s₄ = ↑ and gives −f′. Different final states do not interfere: dσ/dΩ = (288² + 96.0²) fm² = 9.216 × 10⁴ fm² = 921.6 b sr⁻¹.
- Unpolarised: ↑↑ and ↓↓ behave like the first case and ↑↓ and ↓↑ like the second, each with weight 1/4, so dσ/dΩ = (368.6 + 921.6)/2 b sr⁻¹ = 645.1 b sr⁻¹.
- Check: Mott's bracket at 60° is 16 + 16/9 − 16/3 = 112/9 = 12.44, and (112/9) × (72.0 fm)² = 6.451 × 10⁴ fm² = 645.1 b sr⁻¹. Equally (1/4)|f + f′|² + (3/4)|f − f′|² = (1/4)(1474.6) + (3/4)(368.6) = 645.1 b sr⁻¹, and the ratio to the distinguishable 921.6 b sr⁻¹ is 0.700.
Answer369 b sr⁻¹ for parallel spins, 922 b sr⁻¹ for opposite spins, and 645 b sr⁻¹ unpolarised, which is Mott's formula; η = 0.037 keeps the Coulomb phase negligible.
HardFrom topic 10.01's mass-kept Møller result, find the ratio R(60°) of the tree-level cross-section to the distinguishable sum to first order in v², with v each electron's CM speed, and the CM kinetic energy per electron at which it exceeds the Born-level 0.700 by 1%. Then use UAQM-02.08's Coulomb phase to find the energy below which the all-orders ratio exceeds 0.700 by 1%. State the window at 60° and at 90°.
- With m = 1, B = v² and z = cos θ, the CM kinematics give s = 4/(1 − B), t = −2B(1 − z)/(1 − B), u = −2B(1 + z)/(1 − B). One plus 10.01's interference term over its two direct terms is R = 1 − (1 − 3B)(1 + B) sin²θ/[2 + 3B² + (2 + 8B)z² + B²z⁴].
- To first order in B the numerator is (1 − 2B) sin²θ and the denominator 2(1 + z²) + 8Bz². With E = sin²θ/[2(1 + z²)] and 2 + 4z²/(1 + z²) = 4(1 − E), R = 1 − E + 4B E(1 − E) + O(B²); B → 0 returns the Born-level 1 − E.
- At 60°, E = 0.75/2.5 = 0.300, so R = 0.700 + 0.84v². A rise of 1%, 0.0070, needs v² = 8.33 × 10⁻³; the closed form gives 8.29 × 10⁻³. Then γ = 1/√(1 − v²) = 1.0042 and T = (γ − 1)mc² = 0.0042 × 511.0 keV = 2.1 keV.
- Below: RC = 1 − E cos(η ln tan²30°) = 1 − 0.300 cos(1.099η), since ln(1/3) = −1.099. RC − 0.700 = 0.0070 needs cos(1.099η) = 0.9767, so 1.099η = 0.2164 rad and η = 0.197.
- η = α/(2v) gives v = 0.00730/0.394 = 0.0185c, and T = ½mc²v² = 0.5 × 511.0 keV × 3.43 × 10⁻⁴ = 87.6 eV.
- At 90°, tan 45° = 1 kills the phase, so only the upper edge remains: R = 0.500 + v² reaches 0.505 at v² = 0.005, T = 1.3 keV.
AnswerR(60°) = 0.700 + 0.84v² + …; at the 1% level the window at 60° is about 88 eV to 2.1 keV per electron, while at 90° it has no lower edge and closes near 1.3 keV.