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


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?


1 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.

  • $\begingroup$ Thank you! I see... One thing that that I just realised is that the eoms that I am imposing are obtained from $S$ but not from $S'$. So let's assume that one is interested in on-shell invariance: is it legit to impose the Dirac equation, shouldn't one impose the new equations of motion which would have an extra term proportional to $\partial_\mu\theta$? $\endgroup$
    – TopoLynch
    Sep 4, 2022 at 14:44
  • $\begingroup$ @TopoLynch $\theta$ isn't a field, but a parameter in the transformation. So it does not have an associated equation of motion (nor should it appear in the equations of motion). $\endgroup$
    – Andrew
    Sep 4, 2022 at 17:07
  • $\begingroup$ @TopoLynch Right, so you're finding different Lagrangians related by a $U(1)$ gauge transformation give you different equations of motion, because the Lagrangian isn't gauge invariant. That's not surprising and the solution is to work with a gauge invariant Lagrangian. $\endgroup$
    – Andrew
    Sep 4, 2022 at 23:41
  • $\begingroup$ But, when applying the E-L. equations to the transformed Lagrangain, i.e. $\mathcal{L}’=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi-(\partial\mu\theta)\bar\psi\gamma^\mu\psi$, varying w.r.t. $\bar\psi$, one obtains $(i\gamma^\mu\partial_\mu-m-\gamma^\mu(\partial_\mu\theta))\psi$, which is not the original Dirac equation (due to de extra term) and thus is not the equations that I imposed to cancel the extra term when integrating in the action. Am I mistaken? $\endgroup$
    – TopoLynch
    Sep 4, 2022 at 23:48
  • $\begingroup$ Right! So the Dirac Theory is not even gauge invariant on-shell? (The OP’s reasoning was thus wrong?) $\endgroup$
    – TopoLynch
    Sep 4, 2022 at 23:49

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