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?


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