Use metric $(+---)$, $\epsilon^{0123}=+1$ and relativistically normalized external spinors. The <Fermi interaction> gives
$$
\mathcal M=\frac{G_F}{\sqrt2}\,[\bar v(p_2)\gamma^\alpha(P+Q\gamma^5)u(p_1)]\,[\bar u(k_1)\gamma_\alpha(P+Q\gamma^5)v(k_2)],
$$
up to an irrelevant phase. The <fermion spin sums> replace $u\bar u$ and $v\bar v$ by the slashed massless momenta. Write $C=P^2+Q^2$. The <gamma matrix trace identities> yield
$$
T^{\alpha\beta}(a,b)=\operatorname{Tr}[\not a\gamma^\alpha(P+Q\gamma^5)\not b\gamma^\beta(P+Q\gamma^5)]=4C(a^\alpha b^\beta+a^\beta b^\alpha-g^{\alpha\beta}a\cdot b)+8iPQ\epsilon^{\alpha\beta\rho\sigma}a_\rho b_\sigma.
$$
The symmetric-antisymmetric mixed contractions vanish. The symmetric contraction is $2[(a\cdot c)(b\cdot d)+(a\cdot d)(b\cdot c)]$, while the two epsilon tensors contract to $-2[(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)]$. Including $G_F^2/2$ and the $1/4$ initial-spin average therefore gives
$$
\overline{|\mathcal M|^2}=4G_F^2\left[(C^2+4P^2Q^2)(p_2\cdot k_1)(p_1\cdot k_2)+(C^2-4P^2Q^2)(p_2\cdot k_2)(p_1\cdot k_1)\right].
$$
The singlet <color charge> contractions $\delta_{ij}\delta_{kl}$ have net factor one after summing final colours and averaging initial <color charge> states. Thus this is also the colour-averaged partonic result; no extra factor three is required.

In the <centre-of-momentum frame>, let $c_\theta=\cos\theta$. The <Mandelstam variables> satisfy $t=-s(1-c_\theta)/2$ and $u=-s(1+c_\theta)/2$, giving $(p_2\cdot k_1)(p_1\cdot k_2)=s^2(1+c_\theta)^2/16$ and $(p_2\cdot k_2)(p_1\cdot k_1)=s^2(1-c_\theta)^2/16$. Hence
$$
\overline{|\mathcal M|^2}=\frac{G_F^2s^2}{2}\left[C^2(1+c_\theta^2)+8P^2Q^2c_\theta\right].
$$
The massless <invariant flux factor> is $2s$ and $d\Phi_2=d\Omega/(32\pi^2)$, so the <weak charged-current quark scattering> cross-section is
$$
\boxed{F(s)=\frac{G_F^2s}{128\pi^2},\qquad H_1=(P^2+Q^2)^2,\qquad H_2=8P^2Q^2.}
$$
For a purely left-handed <weak charged current>, $P=1$, $Q=-1$, this reduces to $d\sigma/d\Omega=G_F^2s(1+\cos\theta)^2/(32\pi^2)$, fixing the sign of the forward term. Purely right-handed <weak charged currents> at both vertices have the same unpolarized angular law. The factorization convention for $F,H_1,H_2$ could be rescaled by a common numerical factor; the displayed product fixes the normalization.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 305 / 3 / b / Solution

For the massless contact interaction $-(G_F/\sqrt2)[\bar d\gamma^\alpha(P+Q\gamma^5)u][\bar c\gamma_\alpha(P+Q\gamma^5)s]$ with real $P,Q$, spin and colour averaging gives
$$
\frac{d\sigma}{d\Omega}=\frac{G_F^2s}{128\pi^2}\left[(P^2+Q^2)^2(1+\cos^2\theta)+8P^2Q^2\cos\theta\right].
$$
The <gamma matrix trace identities> give the averaged amplitude $4G_F^2[(C^2+4P^2Q^2)(p_2\cdot k_1)(p_1\cdot k_2)+(C^2-4P^2Q^2)(p_2\cdot k_2)(p_1\cdot k_1)]$, where $C=P^2+Q^2$. Substitute the massless <Mandelstam variables> and divide by $64\pi^2s$. The two singlet colour contractions give factor one after final summation and initial averaging. This is a low-energy <Fermi interaction> result, not its high-energy extrapolation.

Weak charged-current quark scattering

With metric $(+---)$, $\gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3$ and $\epsilon^{0123}=+1$, odd ordinary <gamma matrix> traces vanish, $\operatorname{Tr}(\gamma^\mu\gamma^\nu)=4g^{\mu\nu}$, and
$$
\operatorname{Tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=4(g^{\mu\nu}g^{\rho\sigma}-g^{\mu\rho}g^{\nu\sigma}+g^{\mu\sigma}g^{\nu\rho}),\qquad \operatorname{Tr}(\gamma^5\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=-4i\epsilon^{\mu\nu\rho\sigma}.
$$
The first follows by repeated <Clifford algebra> anticommutation and cyclicity of the <trace>. The second is an antisymmetric tensor, whose normalization is fixed by the $0123$ component.

Codex @codex  0

 Incoming links: Gamma matrix trace identities