 Solution

ID: past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2020/1/b/ii/solution

With all four external fermions of species 1, both final attachments are possible. The middle row of the figure gives the two tree-level Feynman diagrams:

The $t$ channel has lines $1 \to 3$ and $2 \to 4$ , joined by a scalar propagator with $q_{t} = p_{1} - p_{3} = p_{4} - p_{2}$ .
The $u$ channel has lines $1 \to 4$ and $2 \to 3$ , joined by a scalar propagator with $q_{u} = p_{1} - p_{4} = p_{3} - p_{2}$ .

For a fixed ordering of external fermion states, the fermionic sign from exchanging the final identical fermions gives

M = - g^{2} [\frac{( u ˉ _{3} u _{1} ) ( u ˉ _{4} u _{2} )}{t - m ^{2} + i 0} - \frac{( u ˉ _{4} u _{1} ) ( u ˉ _{3} u _{2} )}{u - m ^{2} + i 0}], u = q_{u}^{2} .

Here

u_{j} = u_{s_{j}} (p_{j})

are Dirac spinors. In particular, the scattering amplitude changes sign when the complete final labels

3

and

4

are exchanged, as required for identical fermions.

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