Substitution of the gauge transformations and $UU^\dagger=I$ gives
$$
D_\mu'\phi'=U(D_\mu\phi)U^\dagger.
$$
A direct commutator calculation gives
$$
[D_\mu,D_\nu]\phi=-ig[F_{\mu\nu},\phi].
$$
Covariance of the left side then implies $F'_{\mu\nu}=UF_{\mu\nu}U^\dagger$. The <cyclic property of the trace> makes the traces of $F_{\mu\nu}F^{\mu\nu}$, $D_\mu\phi D^\mu\phi$, and $\phi^3$ invariant, so the Lagrangian is <gauge invariance>[gauge invariant].

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-304/4/a/solution