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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 in Theoretical Physics by coolcty (125 points) [ revision history ]
edited May 23, 2014 by coolcty
Most voted comments show all comments

Hi coolcty, this looks like a promising new question :-)

To write centered LaTex equations and also get a preview of the maths, you can use the TEX-button of the editor. Applying this button, you do not need the $s.

You need to consider \(\mathscr{L}_0\) as the free Lagrangian and \(\mathscr{L}_1\) as the interaction, then can you derive the Feynman rules for the interaction term \(\mathscr{L}_1\)?

@JiaYiyang

Thanks for your hint. The tree is just a single line, but I thought it has branches.

Hi @coolcty, as you have figured out the answer, it would be nice if you could write a short answer to your own question (which is completely allowed) if you have time. (just a piece of advice, its completely fine if you don't). 

There is no $m_B$ in this formula, only physical $m$. Also, the third line is $\mathscr{L}_1$.

Most recent comments show all comments

Vladimir Kalitvianski: I was simply considering $\mathcal{L}_1$ as a function of the variable $\Phi$. When you say that $\Phi$ depends of $Z$, I guess you are talking about some specific $\Phi$, which one? A solution of the equation of motion? 

I agree that loops contribute, I have written cancellation of "some" loop contributions.

@40227:  Yes, a solution of the equation of motion. Instead of $\mathcal{L}_1$ we can better consider the equation for $\Phi$ where contribution of $\mathcal{L}_1$ is also $Z$-independent.

In fact, $\mathcal{L}_1$ is somewhat ambiguous since it is determined up to a full derivative of something. We should keep it in mind.

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