Non-invariance of the Interaction term in QED lagrangian The interaction term in the QED Lagrangian $$\mathcal{L}_{int}=e\bar\psi\gamma^\mu A_\mu\psi$$ changes under a gauge transformation $$A_\mu\rightarrow A_\mu+\partial_\mu\chi$$ Doesn’t it affect the QED Feynman rules? Does it mean in different gauges the Feynman rules are different? But if this is so, isn't the cross-section which is a measurable come out to be different for different gauge choices?
 A: The kinetic term $$\mathcal{L}_\text{fermion} = i\overline{\psi}\gamma^\mu \partial_\mu \psi$$
changes too, by precisely the right quantity to cancel the change in the interaction term. Thus, the total Lagrangian is invariant, and this is what matters.
A: As @RobinEkman mentioned, the kinetic term changes as well. This can be easily computed
\begin{equation}
\begin{split}
{\cal L}_D = i {\bar \psi} \gamma^\mu \partial_\mu \psi &\to i {\bar \psi} e^{- i \alpha} \gamma^\mu \partial_\mu \left( e^{i \alpha} \psi \right) =  i {\bar \psi} \gamma^\mu \partial_\mu  \psi  - \partial_\mu \alpha ( {\bar \psi}   \gamma^\mu   \psi  )
\end{split}
\end{equation}
Similarly
$$
{\cal L}_{int} = e{\bar \psi} \gamma^\mu A_\mu \psi \to e {\bar \psi} \gamma^\mu \left( A_\mu + \frac{1}{e} \partial_\mu \alpha \right) \psi = e {\bar \psi} \gamma^\mu  A_\mu   \psi +  \partial_\mu \alpha ( {\bar \psi} \gamma^\mu  \psi )
$$
Adding the two together, we find that under a gauge transformation
$$
\delta \left( {\cal L}_D + {\cal L}_{int} \right) = 0
$$
Thus, the full Lagrangian is gauge invariant. 
