# Is the Dirac Lagrangian locally gauge invariant without gauge field $A$?

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 $$\bar{\psi}\longrightarrow e^{i\alpha(x)} \bar{\psi})$$:

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

yields:

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

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

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

• 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 ? – Frederic Thomas Dec 28 '18 at 19:47
• The gauge-covariant Dirac Lagrangian is of course gauge invariant, but the term $A_{\mu}J^{\mu}$ therein is not. – Qmechanic Dec 28 '18 at 20:11