Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 305 / 3 / c

Table of contents
- Solution c

Solution

A Fierz rearrangement puts the charged-current operator in the same current ordering as the neutral-current operator. Accounting for the interchange of fermionic fields, the combined amplitude is

M = \frac{G _{F}}{2} [\overset{u}{ˉ} (k^{'}) γ^{α} (1 - γ^{5}) u (k)] [\overset{u}{ˉ} (p^{'}) γ_{α} (C_{V} - C_{A} γ^{5}) u (p)], C_{V} = c_{V} + 1, C_{A} = c_{A} + 1.

Sum over final spins and average over the initial electron spin. The fermion spin sum and gamma-matrix trace identities give

\overline{∣ M ∣^{2}} = 4 G_{F}^{2} [(C_{V} + C_{A})^{2} s^{2} + (C_{V} - C_{A})^{2} u^{2}] .

For massless two-body scattering in the centre-of-momentum frame,

d σ / d t = \overline{∣ M ∣^{2}} / (16 π s^{2})

and

u = - s - t

. Integrating

- s \leq t \leq 0

therefore yields

σ = \frac{G _{F}^{2} s}{4 π} [(c_{V} + c_{A} + 2)^{2} + \frac{1}{3} (c_{V} - c_{A})^{2}] = \frac{G _{F}^{2} s}{3 π} (c_{V}^{2} + c_{A}^{2} + c_{V} c_{A} + 3 c_{V} + 3 c_{A} + 3) .

Consequently

H (s) = \frac{s}{3 π}, B = 1, C = 1, D = 3, E = 3, F = 3.

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