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Given a Lagrangian: $$L = \frac{1}{2}(\partial_u\phi)^2 + g (\partial_u\phi)^4.$$ Does anyone have idea how to write down the 1-loop correction? The derivative coupling is the part that confuses me.

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    $\begingroup$ The 1-loop correction to what? The self-energy? The vertex function? $\endgroup$ – Qmechanic May 22 '20 at 1:07
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Diagrammatically, the diagram will look the same as if the derivatives weren't there. However, when writing down the Feynman rules, the vertices will carry extra factors of momentum that are brought down by the derivatives (one way to see this is by decomposing the fields into creation and annihilation operators, where the factors of $e^{ipx}$ are explicit - the derivative hits this exponential and does nothing but just bring down a factor of $i p_\mu$). Check out section 5.4.3 of this reference for a more careful derivation of the Feynman rules. In summary though, the one-loop integral will look pretty similar to what it would look like with just a $\phi^4$ interaction, but with some additional factors of momentum in the numerator.

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  • $\begingroup$ Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic Sep 21 '20 at 11:33

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