1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – Frederic Thomas Dec 28 '18 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 '18 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.