# Local $U(1)$ gauge invariance of QED

The Lagrangian density for QED is

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

with

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

$U(1)$ 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

• Hint: Show first that $D_{\mu}\psi$ transforms in a particular simple manner. Commented Feb 27, 2013 at 8:09
• @Qmechanic May it be $D_{\mu}\psi \rightarrow D^{'}_{\mu}\psi^{'}=e^{-i\alpha(x)}D_{\mu}\psi$ ?
– J L
Commented Feb 27, 2013 at 8:15
• Yes. $\uparrow$ Commented Feb 27, 2013 at 8:23
• The spinor term still troubles me, because $\bar{\psi}(i\gamma^{\mu}D_{\mu}-m)\psi$ will go to $e^{i\alpha(x)}\bar{\psi}(i\gamma^{\mu}D^{'}_{\mu}-m)\psi e^{-i\alpha(x)}=e^{i\alpha(x)}\bar{\psi}(i\gamma^{\mu}e^{-i\alpha(x)}D_{\mu}\psi-m\psi)e^{-i\alpha(x)}=\bar{\psi}(i\gamma^{\mu}D_{\mu}\psi e^{-i\alpha(x)}-m\psi)$
– J L
Commented Feb 27, 2013 at 8:30
• 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? Commented Feb 27, 2013 at 9:04

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.

• I was double counting an exponential factor, I had your answer in mind, however as it is the only one I will accept it
– J L
Commented Feb 27, 2013 at 10:24
• @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.
– jdm
Commented Feb 27, 2013 at 10:53
• 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?
– J L
Commented Mar 1, 2013 at 13:12