# How can we use geometric series in calculating Green's function sum over loop effects

I found Greens function summing over repeated insertion of 1PI in Schwartz p.330: \begin{aligned} iG(\not p)&=\frac{i}{\not p-m}(i\Sigma(\not p))\frac{i}{\not p-m}+\frac{i}{\not p-m}(i\Sigma(\not p))\frac{i}{\not p-m}(i\Sigma(\not p))\frac{i}{\not p-m}+\cdots\\ &=\frac{i}{\not p-m}\left[1+\frac{-\Sigma(\not p)}{\not p-m}+\left(\frac{-\Sigma(\not p)}{\not p-m}\right)^2+...\right]\\ &=\frac{i}{\not p-m}\frac{1}{1+\frac{\Sigma(\not p)}{\not p-m}}\\ &=\frac{i}{\not p -m+\Sigma(\not p)} \end{aligned}

It seems to use geometric series assuming $|\frac{-\Sigma(\not p)}{\not p-m}|<1$ from 2nd line to 3rd line. I can't understand how the assumption $\left|\frac{-\Sigma(\not p)}{\not p-m}\right|<1$ is valid for all cases.

If I understand correctly, field renormalization and mass renormalization are done in next page, such that $\Sigma(\not p)$ is 1PI of self-energy before field renormalization. For example, electron self energy up to $e^2$ order in dimensional regularization is $\Sigma(\not p)=\frac{\alpha}{2\pi}\frac{\not p-4m}{\epsilon}$. Obviously as $\epsilon \rightarrow 0$, $|\frac{-\Sigma(\not p)}{\not p-m}|<1$ doesn't seem to hold.

Can anyone explain how I can use geometric series in calculating Green's function? Or is it just divergent and can't I use this?

Also I don't understand the idea of 'repeated insertion of 1PI' because it looks like we can count on the loop effect in a very ordered manner, but I think we need proof to legitimate this counting.

Some remarks:

• Recall that $\Sigma(\not p)$ is given by the sum of all 1PI diagrams, and therefore $\Sigma(\not p)=\frac{\alpha}{2\pi}\frac{\not p-4m}{\epsilon}$ cannot be right: it is an incomplete result. In fact, you have to take into account the counter-term diagrams as well, which makes $\Sigma(\not p)$ a finite function, independent of $\epsilon$ (in the limit $\epsilon\to 0$). This function satisfies $$\Sigma(m)=\Sigma'(m)=0$$ that is, $$\Sigma(\not p)=\mathcal O(\not p-m)^2$$

This in turns implies that $\frac{\Sigma(\not p)}{\not p-m}<1$ at least in a neighbourhood of $\not p=m$. The Dyson resummation is understood in the sense of analytical continuation to a larger region in the formal complex variable $\not p$.

• The equality $G(\not p)=\frac{1}{\not p-m+\Sigma(\not p)}$ can be justified non-perturbatively, that is, without resummation of a divergent series. See Itzykson & Zuber, Quantum field theory, chapter 6-2-2 (in particular, equations 6.73 to 6.79) for the details. See also the Wikipedia entry Effective action.

• If you insist on defining $\Sigma(\not p)$ through the Dyson resummation, then you are correct in your scepticism: in fact, as you are summing a divergent series, the ordering of the terms affects the sum itself. One may admit that such an ordering prescription is a part of the definition of the QFT (in the same way that a regularisation prescription is another fundamental ingredient of the theory). An interesting reference for this is On Laplace–Borel Resummation of Dyson–Schwinger Equations.

For the use of $\not p$ as a formal complex variable, see Spinor field normalisation from poles in the propagator. For the asymptotic behaviour of $\Sigma(\not p)$ for $\not p\to\infty$, instead of $\not p\to m$, see Is there any known bound to the growth of interacting correlation functions?.
• Although effective action doesn't use resumation of divergences, the counter term of lagrangian($\delta L$) in exponent of effective action contains perterbative expansion of some coupling constant, as far as I know. Maybe you can correct me. Thanks. – Liberty Jan 2 '17 at 13:36