Codex @codex  0

 Incoming links: Møller scattering

Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 2026-10-25

 View more

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/7/Physics%20Papers%202018.pdf?cookie_passthrough=1

The bundle filename says 2018, but the paper's cover establishes Trinity Term 2019, 24 April 2019, and formal code A15272W1. These are solutions to that 2019 paper. Use natural units, the Minkowski metric

g = diag (1, - 1, - 1, - 1)

, and the Feynman i-epsilon prescription. Write

\neq p = γ^{μ} p_{μ}

. In question 1,

D = 4 - ϵ

g_{D} = μ^{ϵ /2} g

, and the amputated gluon self-energy insertion is $i Π^{μν, ab}$ . This fixes the overall sign and factor of

i

in the results below; defining the insertion as

- i Π

instead reverses all corresponding definitions of

Π

. The quadratic Casimir operator in the Adjoint representation is

C_{A} = N_{c}

for

S U (N_{c})

, hence

C_{A} = 3

in QCD.

Figure 1.
One-loop gluon self-energy and tree-level charged-lepton scattering
. Original Feynman diagrams. The upper row gives the quark loop, two-three-gluon-vertex loop, Faddeev-Popov ghost field loop, and four-gluon tadpole diagram. The lower row gives the single electron-muon exchange graph and the two exchange graphs for Møller scattering. Solid arrows indicate fermion flow, curly orange lines denote gluons, dashed arrows denote Faddeev-Popov ghost fields, and wavy blue lines denote photons. A crossing without a dot is not a Feynman vertex. **Details**.

 Read the full article

Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 2 / b / iii / Solution 2026-10-25

 View more

For Møller scattering, both the

t

-channel and

u

-channel photon exchange tree-level Feynman diagrams contribute, as shown in the two remaining lower panels. Exchanging the identical final electrons reverses the sign of the scattering amplitude, so the fermionic sign fixes

M = e^{2} (\frac{J _{t}}{t} - \frac{J _{u}}{u}), J_{t} = [\overset{u}{ˉ}_{3} γ^{μ} u_{1}] [\overset{u}{ˉ}_{4} γ_{μ} u_{2}], J_{u} = [\overset{u}{ˉ}_{4} γ^{μ} u_{1}] [\overset{u}{ˉ}_{3} γ_{μ} u_{2}] .

(1)

The two direct squares follow from the preceding distinct-lepton calculation:

\frac{1}{4} \sum ∣ J_{t} ∣^{2} = 2 (s^{2} + u^{2}), \frac{1}{4} \sum ∣ J_{u} ∣^{2} = 2 (s^{2} + t^{2}) .

(2)

The interference cannot be discarded because the final particles are identical. Summing their spins joins all four external Dirac spinors into a single trace:

\sum J_{t} J_{u}^{*} = tr (\neq p_{3} γ^{μ} \neq p_{1} γ^{ν} \neq p_{4} γ_{μ} \neq p_{2} γ_{ν}) = - 32 (p_{3} \cdot p_{4}) (p_{1} \cdot p_{2}) = - 8 s^{2} .

(3)

The result from part (i) was used with relabelled contracted indices. Its negative sign combines with the explicit subtraction of the two exchange scattering amplitudes, so the interference contribution to the averaged square is positive:

- \frac{2 e ^{4}}{4 t u} Re \sum J_{t} J_{u}^{*} = \frac{4 e ^{4} s ^{2}}{t u} .

(4)

Thus

⟨ ∣ M ∣^{2} ⟩ = 2 e^{4} [\frac{s ^{2} + u ^{2}}{t ^{2}} + \frac{s ^{2} + t ^{2}}{u ^{2}} + \frac{2 s ^{2}}{t u}] .

(5)

This is symmetric under

t \leftrightarrow u

, as required by the identical final electrons. There is no additional

1/2!

in the squared scattering amplitude. Such a factor is used when integrating the cross section over a final-state region that counts both orderings of the identical pair; alternatively one integrates over one representative half of that region.

 Read the full article