= Solution
Write $U(x)=e^{i\alpha^a(x)T^a}$ and $A_\mu=A_\mu^aT^a$. With the convention
$$
D_\mu\psi=(\partial_\mu-igA_\mu)\psi,
$$
the <gauge field> must transform as
$$
\boxed{A'_\mu=UA_\mu U^{-1}-\frac{i}{g}(\partial_\mu U)U^{-1}}.
$$
Then $D'_\mu\psi'=U D_\mu\psi$. The conjugate field has $D_\mu\bar\psi=\partial_\mu\bar\psi+ig\bar\psi A_\mu$. For the adjoint scalar,
$$
\boxed{D_\mu\phi=\partial_\mu\phi-ig[A_\mu,\phi]},
$$
or $(D_\mu\phi)^a=\partial_\mu\phi^a+g\epsilon^{abc}A_\mu^b\phi^c$. It transforms as $D'_\mu\phi'=U(D_\mu\phi)U^{-1}$. The <gauge field strength>
$$
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu-ig[A_\mu,A_\nu]
$$
similarly transforms by conjugation.
Back to article page
