Past exam of the advanced quantum field theory course of the University of Oxford / 2022 / 2 / b / i

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- Solution i

Solution

There is one tree-level Feynman diagram: annihilation of the up quark and its antiquark into an

s

-channel gluon, followed by production of the down quark and its antiquark. A gluon interaction vertex preserves quark flavor, so a

t

-channel line joining the incoming up quark to the outgoing down quark is impossible.

Figure 1.
Annihilation and exchange channels with labelled external momenta
. Original Feynman diagrams, with arrows denoting fermion flow. The left panel with $V = g$ and $f^{'} = d$ is the distinct-quark flavor process. Both panels with $V = γ$ describe Bhabha scattering; both with $V = g$ and $f^{'} = f = u$ describe identical-quark flavor scattering. The wavy internal line stands for the indicated vector mediator. Details.

Let

i, j, k, l

label the initial quark, initial antiquark, final quark, final antiquark color charges. Up to an irrelevant common sign, the scattering amplitude is

M_{s} = \frac{g ^{2}}{s} (T^{a})_{ji} (T^{a})_{k l} [\overset{v}{ˉ} (p_{2}) γ^{μ} u (p_{1})] [\overset{u}{ˉ} (p_{3}) γ_{μ} v (p_{4})] .

The Mandelstam variables are

s = (p_{1} + p_{2})^{2}

,

t = (p_{1} - p_{3})^{2}

,

u = (p_{1} - p_{4})^{2}

, with

s + t + u = 0

in the zero-mass limit. The spin sum uses

\sum u \overset{u}{ˉ} = \sum v \overset{v}{ˉ} = \neq p

, and the initial spin average contributes

1/4

. The resulting two gamma matrix traces contract to

\frac{1}{4} tr (\neq p_{2} γ^{μ} \neq p_{1} γ^{ν}) tr (\neq p_{3} γ_{μ} \neq p_{4} γ_{ν}) = 8 [(p_{2} \cdot p_{3}) (p_{1} \cdot p_{4}) + (p_{2} \cdot p_{4}) (p_{1} \cdot p_{3})] = 2 (t^{2} + u^{2}) .

The initial color average contributes

1/ N_{c}^{2}

. Summing final color charges gives the distinct-flavor quark scattering color factor

K_{ss} = \frac{1}{N _{c}^{2}} \sum_{a, b} tr (T^{a} T^{b}) tr (T^{a} T^{b}) = \frac{N _{c}^{2} - 1}{4 N _{c}^{2}} .

Therefore

\overline{∣ M_{s} ∣^{2}} = g^{4} \frac{N _{c}^{2} - 1}{2 N _{c}^{2}} \frac{t ^{2} + u ^{2}}{s ^{2}} = \frac{4 g ^{4}}{9} \frac{t ^{2} + u ^{2}}{s ^{2}}, C = \frac{4 g ^{4}}{9} .

Equivalently

C = 64 π^{2} α_{s}^{2} /9

for the QCD coupling

α_{s} = g^{2} / (4 π)

. There is no initial gluon spin average: the gluon is internal.

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