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I wanna clarify some issues about renormalization in the $\bar{MS}$ scheme that I glossed over when I first learnt about this stuff. I am following http://arxiv.org/abs/1411.7853 section 3.1. The gluon part of the QCD Lagrangian is considered and the renormalized coupling and gluon field are written

$$g=\bar{\mu}^{\epsilon}Z_gg_R\qquad{}A_{\mu}=\sqrt{Z_A}A_{\mu}^R\tag{10}$$

where $\bar{\mu}=\frac{\mu}{\sqrt{2\pi}}e^{\gamma_E/2}$. It is immediately stated that the renormalization constant takes the form

$$Z_g=1+\frac{\alpha_s(\mu)}{4\pi}\frac{Z_{11}}{\epsilon}+\bigg(\frac{\alpha_s(\mu)}{4\pi}\bigg)^2\bigg(\frac{Z_{22}}{\epsilon^2}+\frac{Z_{21}}{\epsilon}\bigg)\\+\bigg(\frac{\alpha_s(\mu)}{4\pi}\bigg)^3\bigg(\frac{Z_{33}}{\epsilon^3}+\frac{Z_{32}}{\epsilon^2}+\frac{Z_{31}}{\epsilon}\bigg)+\ldots{}\tag{13}$$ I don't see why it should be obvious that $Z_g$ should take this form. What justifies this?

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  • $\begingroup$ They explain it directly after that equation in a footnote: "Physical quantities can be made finite even if we include terms of order $\epsilon^0,\epsilon^1,\epsilon^2$ in $Z_i$. This corresponds to taking other renormalization prescription." and "The reason why the MS scheme is used more often than other schemes is that empirically perturbative series for various physical quantities exhibit good convergence behaviors.", i.e. this form of the constant is the definition of the MS scheme. $\endgroup$ – ACuriousMind Jan 6 '16 at 13:35
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    $\begingroup$ @ACuriousMind but can we define it this way? I mean, can we choose whatever we like for the $Z$? $\endgroup$ – Yossarian Jan 7 '16 at 12:59
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    $\begingroup$ @AleksandrSokolov That's why you need to pick a scheme to fix the ambiguity in what you choose for $Z$. The point of MSbar is that you only subtract the divergent piece (so there are no finite pieces in $Z$ as you send $\epsilon\rightarrow 0$). This is a choice. The scheme dependence drops out of any physical quantity though, which is guaranteed by setting up RG equations. Also, you are right the form of $Z$ is not obvious without doing some work, they have omitted the calculation of the relevant loop diagrams (which is perfectly acceptable since there are no new ideas in that calculation). $\endgroup$ – Andrew Jan 10 '16 at 3:46
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Regularization is nothing more than a deformation in the theory in the UV scale as well as in IR scale, depending on what divergence you are facing. Don't matter how precisely is the deformation, the requirement is that the physical predictions that connect the events in your usual scale need to be insensitive under this deformations (renormalizable theory). So, if your choice of regularization give you finite predictions in your scale of interest this is great and the regularization is fine. (see this lecture notes)

This $Z_g$ is one of the choices that clearly give you finite results by limiting the loop integrals by an dimensional regularization (see this historical paper too of G,'t Hooft) such that each loop contribute like: $$ \sim\bigg(\frac{\alpha_s(\mu)}{4\pi}\bigg)^n\frac{1}{\epsilon^m} $$ where $n$ is the number of vertices and $m$ the number of internal internal lines in the corresponding feynamn diagrams that you want to regularize.

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  • $\begingroup$ Why is $m\le n $? $\endgroup$ – innisfree Sep 15 '16 at 9:43
  • $\begingroup$ Internal lines are always coming out of vertices, and at least two of them. $\endgroup$ – Nogueira Sep 15 '16 at 20:22
  • $\begingroup$ @innisfree, you can see this from the Euler's characteristic of the feynman diagrams. Faces are loops and edges are lines. $\endgroup$ – Nogueira Sep 15 '16 at 20:27
  • $\begingroup$ So $1-l=n-m $ so $m\ge n $?! $\endgroup$ – innisfree Sep 15 '16 at 23:15
  • $\begingroup$ Isn't the power of epsilon the number of loops? $\endgroup$ – innisfree Sep 15 '16 at 23:19

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