# Why the sum of one-particle-irreducible graphs $\Pi^*$ can be written in this form?

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I am studying the basics of renormalisation. In 10.3 of Weinberg QFT book, motivated by the renomalisation of field, he express the bare Lagrangian as two part

$\mathscr{L=L}_0+\mathscr{L}_1$,

$\mathscr{L}_0=-\frac{1}{2}\partial_{\mu}\Phi\partial^{\mu}\Phi-\frac{1}{2}m^2\Phi^2$,

$\mathscr{L}_1=-\frac{1}{2}(Z-1)(\partial_{\mu}\Phi\partial^{\mu}\Phi+m^2\Phi^2)+\frac{1}{2}Z\delta m^2\Phi^2-V(\Phi)$,

where

$V(\Phi)\equiv V(\sqrt{Z}\Phi_B)$

After expressing the corrected propagator as a geometric series of uncorrected propagator times the sum of one-particle-irreducible graphs $\Pi^*$, the only thing need to calculate is the $\Pi^*$, but he states:

In calculating $\Pi^*$, we encounter a tree graph arising from a single insertion of vertices corresponding to the terms in $\mathscr{L}_1$ proportional to $\partial_{\mu}\Phi\partial^{\mu}\Phi$ and $\Phi^2$, plus a term $\Pi^*_{LOOP}$ arising from loop graphs like that in Figure 10.4(a):

$\Pi^*(q^2)=-(Z-1)[q^2+m^2]+Z\delta m^2+\Pi^*_{LOOP}(q^2).$

What's the tree graph arising from a single insertion of vertices that corresponds to the terms in $\mathscr{L}_1$ ? And why they are proportional?

asked May 15, 2014
edited May 23, 2014
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Hi coolcty, this looks like a promising new question :-)