= Solution
Put $U=e^{-i\alpha\gamma_5/2}$. The transformed <Dirac adjoint> is $\bar\psi'=\psi^\dagger U^\dagger\gamma^0=\bar\psi U$, since $\gamma^0$ anticommutes with $\gamma_5$. It is therefore incorrect to transform the two sides of a Yukawa bilinear with mutually inverse matrices.
Write $M(\phi)=\phi_1+i\gamma_5\phi_2$. The transformed <Yukawa interaction> is invariant exactly when
$$
U M(\phi')U=M(\phi).
$$
Since $M$ commutes with $\gamma_5$, this is $M(\phi')=e^{i\alpha\gamma_5}M(\phi)$. Use <3/a/i> and equate the coefficients of $\mathbf1$ and $i\gamma_5$:
$$
\boxed{\begin{pmatrix}\phi_1'\\\phi_2'\end{pmatrix}
=\begin{pmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{pmatrix}
\begin{pmatrix}\phi_1\\\phi_2\end{pmatrix}.}
$$
This is the global <SO(2)> rotation of a <chiral scalar-doublet Yukawa model>. Equivalently, $\Phi=\phi_1+i\phi_2$ transforms as $\Phi'=e^{i\alpha}\Phi$, while the two chiral <spinor> components transform with opposite half-angle phases. The sign of the scalar rotation follows from the negative sign in the specified <fermion> transformation.
Back to article page
