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Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2018 / 1 / b / i / Solution by

Codex  0 2026-10-25

The tree-level Feynman diagrams are the two ordered quark-exchange graphs and the

s

-channel gluon-exchange graph with a three-gluon vertex. There is no two-gluon-two-quark contact term in the ordinary QCD Lagrangian density. Let

q = k_{1} + k_{2} = p_{1} + p_{2}

,

r_{t} = p_{1} - k_{1}

,

r_{u} = p_{1} - k_{2}

, and

s = q^{2}

,

t = r_{t}^{2}

,

u = r_{u}^{2}

i M

as the product of the stated Feynman rules.

Figure 1.
The three gluon-fusion tree diagrams and the Yukawa fermion-loop correction
. The first three panels show ordered quark exchange and three-gluon exchange. The last panel shows a directed closed fermion loop between two scalar insertions. Crossings without dots are not vertices. Details.

Suppressing external colour indices, the quark graphs are

M_{t}^{μν} = - g^{2} \overset{u}{ˉ} (p_{1}) T^{a} T^{b} γ^{μ} \frac{\neq r _{t}}{t} γ^{ν} v (p_{2}), M_{u}^{μν} = - g^{2} \overset{u}{ˉ} (p_{1}) T^{b} T^{a} γ^{ν} \frac{\neq r _{u}}{u} γ^{μ} v (p_{2}) .

The massless Dirac equations give

\overset{u}{ˉ} \neq p_{1} = 0

and

\neq p_{2} v = 0

. Consequently

\frac{u ˉ \neq k _{1} \neq r _{t} \neq k _{2} v}{t} = - \overset{u}{ˉ} \neq k_{2} v = \overset{u}{ˉ} \neq k_{1} v, \frac{u ˉ \neq k _{2} \neq r _{u} \neq k _{1} v}{u} = - \overset{u}{ˉ} \neq k_{1} v .

The last equality uses

\overset{u}{ˉ} \neq q v = 0

. The quark-exchange sum is therefore

- g^{2} \overset{u}{ˉ} \neq k_{1} [T^{a}, T^{b}] v

.

For the gluon graph, use all incoming momenta

(k_{1}, k_{2}, - q)

in the three-gluon tensor:

V^{μν ρ} = (k_{1} - k_{2})^{ρ} g^{μν} + (k_{2} + q)^{μ} g^{ν ρ} - (q + k_{1})^{ν} g^{ρ μ} .

Its double contraction is

k_{1 μ} k_{2 ν} V^{μν ρ} = s (k_{2} - k_{1})^{ρ} /2

. The vertex and gluon propagator give

M_{s}^{μν} = - \frac{i g ^{2} f ^{ab c}}{s} \overset{u}{ˉ} γ_{ρ} T^{c} v V^{μν ρ}, k_{1 μ} k_{2 ν} M_{s}^{μν} = i g^{2} f^{ab c} \overset{u}{ˉ} \neq k_{1} T^{c} v .

Combining the graphs yields

k_{1}^{μ} k_{2}^{ν} M_{μν} = - g^{2} \overset{u}{ˉ} (p_{1}) \neq k_{1} ([T^{a}, T^{b}] - i f^{ab c} T^{c}) v (p_{2}) .

Thus the requested spinor proportionality factor is

- g^{2} \overset{u}{ˉ} (p_{1}) \neq k_{1} v (p_{2})

when the colour matrix is displayed separately; retaining it inside the spinor bilinear makes its external colour indices explicit. For a convention calling the Feynman-rule product itself

M

, rather than

i M

, the entire result gains one factor of

i

.

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