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In the context of Renormalized Pertubation Theory Peskin Schröder says:
The Lagrangian $$ \mathcal{L}=\frac{1}{2} (\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2-\frac{\lambda}{4!}\phi_r^4 + \frac{1}{2} \delta_Z(\partial_\mu\phi_r)^2 -\frac{1}{2}\delta_m^2\phi_r^2-\frac{\delta_\lambda}{4!}\phi_r^4 $$ gives the following set of Feynman rules:
------------>------------ = $\frac{i}{p^2-m^2+i\epsilon}$
------------X------------ = $i(p^2\delta_Z-\delta_m)$
and the two 4-vertices.

The question is: Why look the Feynman rules for the first and the fourth term of the Lagrangian look so different? I believe the answer is connected to the fact that one has to bring the kinetic term of the Lagrangian to its canonical form $\frac{1}{2} (\partial_\mu\phi_r)^2$ and has to interpret everything else as (possibly momentum dependent) vertices. How does this look in formulae?

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marked as duplicate by AccidentalFourierTransform, Jon Custer, Kyle Kanos, knzhou, Cosmas Zachos Jan 15 '18 at 22:15

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    $\begingroup$ This issue is explained here, the trick is to properly define what is the perturbed and what is the unperturbed part of the Lagrangian when doing perturbation calculations. $\endgroup$ – Dilaton Jun 8 '14 at 8:05