# 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?

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.
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.