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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?

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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.

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  • $\begingroup$ @ Robin and prahar - That is fine. I know that the total lagrangian is gauge invariant. But I still do not get the point. How does the gauge invariance of the full lagrangian help me out? After all the Feynman rules are dictated by the interaction term only. Isn't it? Therefore, if the Feynman rules change, then the result of all physical calculations will also be different which is not desired. $\endgroup$ – SRS Oct 19 '14 at 15:46
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    $\begingroup$ But to identify the interaction term in the gauge transformed Lagrangian, you must look at the whole gauge transformed Lagrangian. A transformation of fields doesn't have to respect a splitting of the Lagrangian into free and interacting parts. $\endgroup$ – Robin Ekman Oct 19 '14 at 16:07
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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.

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