Write $\Psi=\Psi^aT_a$. The infinitesimal form of $\Psi\mapsto U\Psi U^\dagger$ is
$$
\delta\Psi=i[\alpha^bT_b,\Psi^cT_c],
\qquad
\boxed{\delta\Psi^a=-f_{bc}{}^a\alpha^b\Psi^c}.
$$
The adjoint <gauge covariant derivative> is
$$
D_\mu\Psi=\partial_\mu\Psi-ig[A_\mu,\Psi],
$$
which transforms as $D_\mu\Psi\mapsto U(D_\mu\Psi)U^\dagger$. A gauge-invariant Lagrangian is therefore
$$
\boxed{\mathcal L=-\frac14\operatorname{Tr}(F_{\mu\nu}F^{\mu\nu})
+\operatorname{Tr}\!\left[\bar\Psi(i\gamma^\mu D_\mu-m)\Psi\right]}.
$$
The trace and cyclicity make each term invariant under conjugation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-304/4/c/solution