I am so sorry for posting this long question. But I've been confused and frustrated by perturbation series in Quantum Field Theory for years. I hope someone can help me. Thank you so much!

In the bare perturbation, the exact (bare) 2-point function is $$ D^{\prime}=D+D\left(-i\Sigma\right)D+D\left(-i\Sigma\right)D\left(-i\Sigma\right)D+\cdots, $$ where $D=\frac{i}{p^{2}-m^{2}}$ and $-i\Sigma$ is the 1PI. Obviously, the above series does not converge, although I can obtain the 'correct' result by multiplying the equation by $D^{-1}$ from the left hand side and by $D^{\prime-1}$ from the right hand side,

However, people claim that it equals to $$ \frac{i}{p^{2}-m^{2}-\Sigma}, $$ because of the identity $\frac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots$. Since this identity is valid only if $|x|<1$, I once hoped that there would be a cure in renormalization theory. Unfortunately, I finally realized that it is hopeless to expect that $\Sigma$ is very small even after adding counter terms. I once asked this question to some postdocs and they said that I should not worry about this since perturbative expansions in QFT are asymptotic. Well, I admit that perturbation series like $$ F(g)=a_{0}+a_{1}g+a_{2}g^{2}+a_{3}g^{3}+\cdots $$
is an asymptotic expansion, which should stop at some high order in $g$ and the renormalization can only make the coefficients $a_{i}$s finite, even though the theory is still divergent.

In the above case of the 2-point function, the expansion is clearly not dependent on the coupling $g$ since each 1PI contains arbitrary order of $g$.

An example in renormalization is in the following picture enter image description here

We expanded the factor $\frac{1}{1+\delta_{2}}\simeq 1-\delta_{2}$, where $\delta_{2}$ contains a high momentum cutoff which is divergent in the limit $\Lambda\rightarrow\infty$.

Another similar example in renormalization is the following enter image description here

where we replaced $m$ by $m_{1}$ because $$ m_{1}^{2}=m^{2}\left(1-\frac{g}{16\pi^{2}\epsilon}\right)\\ m^{2}=m_{1}^{2}\frac{1}{1-\frac{g}{16\pi^{2}\epsilon}}\simeq m_{1}^{2}\left(1+\frac{g}{16\pi^{2}\epsilon}\right) $$ How can one do an expansion like this $1/(1-\infty)\simeq 1+\infty$?

So far the best mathematical theory for divergent series I can find are the following:

  1. If $n\neq 1$ is an integer, then $[1+n+n^{2}+\cdots]_{reg}=\frac{1}{1-n}$. This was first 'discovered' by Euler.

  2. If $\Re{z}<1$, then $[1+z+z^{2}+\cdots]_{reg}=\frac{1}{1-z}$ in the sense of Borel summation.

But in QFT, I don't think the above mathematics make things clearer. How should I think about those expansions in QFT? Can anyone help me?

Thank you very much!

  • 2
    $\begingroup$ Hi, and welcome to Physics Stack Exchange! You have a good question in here somewhere, but the post is pretty long and I think it's a little too easy to lose track of what you're asking. I don't see why the images you included are necessary. It might make the question clearer to take them out. (If you really do want to include them, then you will need to transcribe the text into Markdown/MathJax, quote it, and reference the source.) $\endgroup$ – David Z Dec 11 '14 at 9:52
  • $\begingroup$ What @David Z said. Tip concerning the title (v1) An annoying question about perturbative quantum field theory. Try to pick a title that tells what the question is. Avoid irrelevant and redundant information. E.g. we already know that it is QFT from the tag, so no need to repeat that in the title. $\endgroup$ – Qmechanic Dec 11 '14 at 10:09
  • $\begingroup$ Hi David and Qmechanic. Thank you so much for telling me this! But I am sorry that now I have to go to pick up my new ID card otherwise I would be kicked out of this country. I will be back a few hours later and I'll do as you said. $\endgroup$ – Xiaoyi Jing Dec 11 '14 at 10:18
  • $\begingroup$ Hi David and Qmechanic. I suddenly realized that there is no need to use the invalid expansion in field theory. I think I just asked the wrong question. I am sorry. $\endgroup$ – Xiaoyi Jing Dec 12 '14 at 21:42

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