Local $U(1)$ gauge invariance of QED

Question

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

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.