The Lagrangian density for QED is

$$ \mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\bar{\psi}(i\gamma^{\mu}D_{\mu}-m)\psi $$


$$F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} $$ $$ \bar{\psi}=\gamma^{0}\psi^{\dagger}$$ $$D_{\mu}=\partial_{\mu}+ieA_{\mu}$$

$U(1)$ local gauge transformations are

$$\psi\rightarrow\psi^{'}=e^{-i\alpha(x)}\psi$$ $$A_{\mu}^{'}=A_{\mu}+\frac{1}{e}\partial_{\mu}\alpha(x)$$

I'm trying to see that $\mathcal{L}$ it is invariant under those $U(1)$ transformations, but I finish my calculation with

$$\mathcal{L}^{'}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+e^{-i\alpha(x)}\bar{\psi}[i\gamma^{\mu}D_{\mu}-m-\gamma^{\mu}\partial_{\mu}\alpha(x)]e^{i\alpha(x)}\psi $$

Any hint will be appreciated thanks

  • 2
    $\begingroup$ Hint: Show first that $D_{\mu}\psi$ transforms in a particular simple manner. $\endgroup$
    – Qmechanic
    Commented Feb 27, 2013 at 8:09
  • $\begingroup$ @Qmechanic May it be $D_{\mu}\psi \rightarrow D^{'}_{\mu}\psi^{'}=e^{-i\alpha(x)}D_{\mu}\psi$ ? $\endgroup$
    – J L
    Commented Feb 27, 2013 at 8:15
  • 1
    $\begingroup$ Yes. $\uparrow$ $\endgroup$
    – Qmechanic
    Commented Feb 27, 2013 at 8:23
  • $\begingroup$ The spinor term still troubles me, because $\bar{\psi}(i\gamma^{\mu}D_{\mu}-m)\psi$ will go to $e^{i\alpha(x)}\bar{\psi}(i\gamma^{\mu}D^{'}_{\mu}-m)\psi e^{-i\alpha(x)}=e^{i\alpha(x)}\bar{\psi}(i\gamma^{\mu}e^{-i\alpha(x)}D_{\mu}\psi-m\psi)e^{-i\alpha(x)}=\bar{\psi}(i\gamma^{\mu}D_{\mu}\psi e^{-i\alpha(x)}-m\psi)$ $\endgroup$
    – J L
    Commented Feb 27, 2013 at 8:30
  • 1
    $\begingroup$ if $D_{\mu}\psi$ goes to $e^{-i\alpha}D_{\mu}\psi$ and $-m\psi$ goes to $-me^{-i\alpha}\psi$ then you're done aren't you? $\endgroup$
    – twistor59
    Commented Feb 27, 2013 at 9:04

1 Answer 1


I think you're missing that the partial derivative in $D_\mu$ acts on the phase factor. If you start only with the phase transformation of the Dirac term:

$$\mathcal{L}_D' = \bar \psi e^{i\alpha(x)} \left[ i\gamma^\mu(\partial_\mu + ieA_\mu) - m \right] \psi e^{-i\alpha(x)} $$

you can almost pull the $e^{i\alpha(x)} $ through and cancel them. However, the $\partial_\mu=\partial/\partial x^\mu$ acts on $ e^{i\alpha(x)} $ and gives you an additional term (product, then chain rule). So after canceling, $\partial_\mu \rightarrow \partial_\mu + $ something with $\alpha(x)$. Then perform the $A_\mu \rightarrow A_\mu'$ substitution, and the terms should exactly cancel out.

  • $\begingroup$ I was double counting an exponential factor, I had your answer in mind, however as it is the only one I will accept it $\endgroup$
    – J L
    Commented Feb 27, 2013 at 10:24
  • 1
    $\begingroup$ @Nivalth: Thanks a lot, but you didn't have to accept it if it didn't help you. Qmechanic and twistor59 also deserve some credit, but unfortunately you don't get points for comments. $\endgroup$
    – jdm
    Commented Feb 27, 2013 at 10:53
  • $\begingroup$ Hi, I've been thinking a little about this. Does the recipe goes as follows? 1) $\mathcal{L}_{QED}$ then I impose the local $U(1)$ gauge transformations on the spinors and 2)this tells me how exactly should $A_{\mu}$ transform in order to keep the Lagrangian density invariant? $\endgroup$
    – J L
    Commented Mar 1, 2013 at 13:12

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