 Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2016/1/b/ii/solution

Use the fundamental SU(3) normalization

tr (T^{a} T^{b}) = δ^{ab} /2

, with Hermitian generators. The average over the

3 \times 3

incoming colours and sum over the outgoing colours gives

C_{col} = \frac{1}{N _{c}^{2}} i, j, k, l \sum a \sum (T^{a})_{ji} (T^{a})_{k l}^{2} = \frac{1}{N _{c}^{2}} a, b \sum tr (T^{a} T^{b}) tr (T^{a} T^{b}) = \frac{N _{c}^{2} - 1}{4 N _{c}^{2}} = \frac{2}{9} (N_{c} = 3) .

The complete spin and colour averaged result is consequently

\overline{∣ M ∣^{2}} = \frac{4 g _{s}^{4}}{9} \frac{t ^{2} + u ^{2}}{s ^{2}} = \frac{64 π ^{2} α _{s}^{2}}{9} \frac{t ^{2} + u ^{2}}{s ^{2}}, α_{s} = \frac{g _{s}^{2}}{4 π} .

For massless two-body centre-of-momentum kinematics,

t = - s (1 - cos θ) /2

and

u = - s (1 + cos θ) /2

, so the same answer is

2 g_{s}^{4} (1 + cos^{2} θ) /9

. This checks positivity, the symmetry under exchanging the outgoing directions, and the absence of a forward

t

-channel singularity in this particular process.

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