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The use of renormalisation constants often puzzles me. A good example is the use of $Z_2$ in the equation (7.58) of Peskin Schroeder. $Z_2$ is defined in equation (7.26). as $Z_2^{-1} = 1-\frac{d\Sigma}{dp}$. Later in equation (7.31) it is said:

$Z_2-1 = d\Sigma/dp$ although this term is supposed to be infinite. But $d\Sigma/dp$ is treated of being smaller than 1. Okay, in this example the Pauli-Villars renormalisation is used where a rather high $\Lambda$ is needed to make $d\Sigma/dp$ larger than 1. But what would be if $Z_2$ were computed with dimensional regularization ?

Shouldn't be at least : $d\Sigma/dp + (d\Sigma/dp)^2 + (d\Sigma/dp)^3 + \dots $

I know that $d\Sigma/dp$ is of order $\alpha$ and there should be counter term to make the sum of both small (to order $\alpha$). On the other hand I am almost sure that when the counter terms of the next order $\alpha^2$ are calculated that it is already forgotten that there was also a term $(d\Sigma/dp)^2$ which also need a counter term and for $\alpha^3$ order again and so on. Could somebody explain it to me ? Thank you.

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    $\begingroup$ Hi Frederic, may I recommend you to write your equations with MathJax, see here for more info. $\endgroup$ – Hunter Jun 8 '14 at 14:54
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    $\begingroup$ @Hunter: He did, but he forgot the dollar signs :) $\endgroup$ – JamalS Jun 8 '14 at 15:14
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    $\begingroup$ @JamalS look at his previous questions, for instance here and here. The OP always seems to "forget" the dollar signs, so I thought someone should make him/here aware of this so that we don't constantly have to improve his/her posts. $\endgroup$ – Hunter Jun 8 '14 at 15:29
  • $\begingroup$ Did you solve this? I was reading this part of text too, but I found $Z_2-1=\frac{d \Sigma}{d p}$ while $Z_2^{-1}=1-\frac{d \Sigma}{d p}$ not so convincing. Do you have some explaination? $\endgroup$ – Turgon Feb 7 '17 at 3:35
  • $\begingroup$ $Z^{-1}_2 = 1-\frac{d\Sigma}{dp}|_{\not p=m}$ is correct. The propagator of interacting QFT is $\frac{1}{p-m_0-\Sigma(p)}=\frac{1}{\not p-m_0-\Sigma(m)-\frac{d\Sigma}{dp}|_{\not p=m}(\not p-m)}=\frac{1}{\not p-m-\frac{d\Sigma}{dp}(\not p -m)}= \frac{1}{(1-\frac{d\Sigma}{dp})(\not p-m) }=\frac{Z_2}{\not p-m}$. But P&S develops $Z_2$ in a geometrical series as if $\frac{d\Sigma}{dp}$ would be small. A priori possible but each term has to be corrected by a counterterm, I wonder if this ever happens for the higher orders. $\endgroup$ – Frederic Thomas Feb 14 '17 at 15:20

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