 Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2017/2/b/ii/solution

Continue with

η = - g

and dot products in that auxiliary metric. Odd gamma traces vanish. The two current tensors are

H_{μα} L^{μα} = 4 [p_{2 μ} p_{1 α} + p_{2 α} p_{1 μ} - η_{μα} (p_{1} \cdot p_{2} + M^{2})], = 4 [k_{1}^{μ} k_{2}^{α} + k_{1}^{α} k_{2}^{μ} - η^{μα} (k_{1} \cdot k_{2})] .

Contracting gives

H_{μα} L^{μα} = 32 [(p_{1} \cdot k_{1}) (p_{2} \cdot k_{2}) + (p_{1} \cdot k_{2}) (p_{2} \cdot k_{1}) + M^{2} (k_{1} \cdot k_{2})] .

Let

u = 2 M^{2} - s - t

p_{1} \cdot k_{1} = p_{2} \cdot k_{2} = (M^{2} - t) /2

p_{1} \cdot k_{2} = p_{2} \cdot k_{1} = (M^{2} - u) /2

, and

k_{1} \cdot k_{2} = s /2

. Substitution yields

\overline{∣ T ∣^{2}} = \frac{4 g _{s}^{4}}{9 s ^{2}} [(M^{2} - t)^{2} + (M^{2} - u)^{2} + 2 M^{2} s] = \frac{64 π ^{2} α _{s}^{2}}{9 s ^{2}} [(M^{2} - t)^{2} + (s + t - M^{2})^{2} + 2 M^{2} s] .

The result is already averaged over incoming colours and spins and summed over outgoing ones. For

M \to 0

, it becomes

4 g_{s}^{4} (t^{2} + u^{2}) / (9 s^{2})

An additional check uses the centre-of-mass angle and velocity

β_{t} = 1 - 4 M^{2} / s

\overline{∣ T ∣^{2}} = \frac{2 g _{s}^{4}}{9} [1 + cos^{2} θ + \frac{4 M ^{2}}{s} sin^{2} θ] .

It tends to

4 g_{s}^{4} /9

at threshold and is positive throughout the physical region. Keeping both printed minus signs would instead give

4 g_{s}^{4} [(M^{2} - t)^{2} + (M^{2} - u)^{2} - 2 M^{2} s] / (9 s^{2})

, displaying the incorrect mass contribution explicitly.

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