Past exam of the quantum field theory course of the University of Oxford / 2020 / 1 / b

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.

iii

Let

p_{1}, p_{3}

label the incoming and outgoing fermion, and

p_{2}, p_{4}

the incoming and outgoing antifermion. The bottom row of the figure gives the two tree-level Feynman diagrams:

The $t$ channel has an open fermion line $1 \to 3$ and an open antifermion line with fermion flow $4 \to 2$ . The scalar propagator carries $q_{t} = p_{1} - p_{3} = p_{4} - p_{2}$ .
The $s$ channel has initial particle-antiparticle annihilation into a scalar propagator with $q_{s} = p_{1} + p_{2} = p_{3} + p_{4}$ , followed by creation of the final pair.