Is the Dirac action invariant under $U(1)$ local gauge transformations?

Question

I have usually found in books/lectures that the Dirac theory, given by

$$S=\int d^4x\bar\psi(i\gamma^\mu\partial_\mu-m)\psi, $$ is invariant under $U(1)$ global transformations (which is evident) but not invariant under $U(1)$ local (gauge) transformations, due to an extra term appearing in the Lagrangian of the form

$$\mathcal{L}\rightarrow\mathcal{L}’=\mathcal{L}-(\partial_\mu\theta)\bar{\psi}\gamma^\mu\psi.$$

Nevertheless, if one integrates this Lagrangian to obtain the new action, you find when integrating by parts that this extra term vanishes.

$$\int d^4x (\partial_\mu\theta)\bar\psi\gamma^\mu\psi=-\int d^4x \theta \partial_\mu(\bar\psi\gamma^\mu\psi)=0,$$ since $$\partial_\mu(\bar\psi\gamma^\mu\psi)=\partial_\mu\bar\psi\gamma^\mu\psi-im\bar\psi\psi=i(\partial_\mu\bar\psi\gamma^\mu\psi+m\bar\psi\psi)=0,$$ where I used the Dirac equation both for $\psi$ and for $\bar\psi$. So one concludes $S=S’$. Is then the Dirac theory invariant to gauge transformation with no need of introducing the covariant derivative?

Answer 1

Since you used the equations of motion, you can only conclude that the Dirac Lagrangian is gauge invariant on-shell. The covariant derivative is needed to ensure off-shell gauge invariance (ie, when the equations of motion are not satisfied). Generally speaking you will derive incorrect conclusions if you plug the equations of motion back into the Lagrangian, since (a) classically you want to vary the action which requires considering off-shell paths as part of the variation, (b) quantum mechanically in the path integral you integrate over all configurations of the fields, including off-shell ones.