= Solution
For the <chiral transformation> $\psi\mapsto e^{i\alpha\gamma^5}\psi$, one also has $\bar\psi\mapsto\bar\psi e^{i\alpha\gamma^5}$. The massless kinetic and vector-current terms are invariant because $\{\gamma^5,\gamma^\mu\}=0$. Since $\gamma^5$ commutes with $[\gamma^\mu,\gamma^\nu]$, the Pauli bilinear transforms with $e^{2i\alpha\gamma^5}$ and is not invariant. Thus the continuous classical axial symmetry requires $\lambda=0$.
A fermion mass also breaks the symmetry, so in the massive theory one additionally needs $m=0$, which contradicts a genuinely massive fermion. Making $\alpha$ local produces $(\partial_\mu\alpha)\bar\psi\gamma^\mu\gamma^5\psi$. No choice of the vector coupling $e$ or Pauli coupling $\lambda$ cancels this term; gauging it requires an axial gauge field, and at the quantum level one must also address the <chiral anomaly>.
Back to article page
