7
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.


Further reading:

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?.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Liberty Jan 2 '17 at 13:36
  • $\begingroup$ @Liberty Hi. There are no counter-terms in the effective action. Maybe the wikipedia article is not the best reference for this; see instead, e.g., Itzykson & Zuber, Quantum field theory, chapter 6.2.2 (in particular, equations 6-73 to 6.79). $\endgroup$ – AccidentalFourierTransform Jan 2 '17 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.