When it comes to the check of the invariance of the Lagrangian of the Dirac equation under local $U(1)$-transformations I have made the following observation:

$$L = \bar{\psi} (i\gamma^{\mu}\partial_\mu \psi -m)\psi.$$

I assume the following local gauge transformation: $\psi \longrightarrow e^{-i\alpha(x)} \psi$ ( and $\bar{\psi}\longrightarrow e^{i\alpha(x)} \bar{\psi})$:

$$L' =e^{i\alpha(x)}\bar{\psi}e^{-i\alpha(x)}(i\gamma^\mu\partial_\mu\psi+e\gamma^{\mu}(\partial_\mu\alpha) \psi) -m\bar{\psi}{\psi}$$


$$L' = L + e \bar{\psi}\gamma^\mu\psi \partial_\mu\alpha= L + e J^\mu \partial_\mu\alpha$$

where $J^{\mu}$ represents the 4-current density of the Dirac field. Now instead of introducing the gauge field $A_\mu$ I apply a partial integration and get:

$$\delta L = L'-L = e\partial_\mu(J^\mu \alpha) - e\partial_\mu J^\mu \alpha.$$

Applying two arguments:

  1. under the action integral the partial derivative of the first term can be transformed into a surface integral at whose borders the current density $J^{\mu}$ vanishes.

  2. due to current conservation the second term also vanishes. Therefore the Lagrangian (or at least the action) remains invariant under the local $U(1)$ transformation. What is the flaw of this conclusion?

BTW: I considered a similar argument for checking the gauge invariance of the coupling term $J^\mu A_\mu$ in the Maxwell equations. If $A_\mu$ is changed under gauge transformations one would get an additional term $J^\mu \partial_\mu \alpha$ which can be eliminated by the same strategy.

So if the first (invariance of the Dirac-equation without gauge field) does not work out, so how can work the second? Or would it mean that the Maxwell equations are not really gauge-invariant?


1 Answer 1


Concerning OP's first question the problem is that one is not allowed to use on-shell relations (like the continuity equation $d_{\mu}J^{\mu}\approx 0$ for the Dirac matter current $J^{\mu}$) when proving that the off-shell Dirac Lagrangian is gauge invariant.

  • $\begingroup$ Thank you for the answer. $\partial_\mu J^\mu=0$ is based on the field equations. On the other hand how can $A_\mu J^\mu$ be gauge-invariant ? Or is it not necessary to be invariant ? $\endgroup$ Dec 28, 2018 at 19:47
  • $\begingroup$ The gauge-covariant Dirac Lagrangian is of course gauge invariant, but the term $A_{\mu}J^{\mu}$ therein is not. $\endgroup$
    – Qmechanic
    Dec 28, 2018 at 20:11

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