I Expansion at first order in QCD counterterm

[Siupa]

What is the meaning of the expansion at first order in $\delta_2$ and $\delta_3$ at the second step in the last line? These quantities are not "small" - on the contrary, the entire point is to then take the $\epsilon \to 0$ limit and the counterterms blow up

 

[Vanadium 50]

I can't read light gray on a less light gray background. Can you use LaTex and maybe post your source?

 

[vanhees71]

The Dyson series is an expansion in the coupling constant $g$. Obviously you are evaluating the quoted 2nd-order loop corrections to the quark and gluon self-energies as well as the gluon 3-vertex to get the counter terms to evaluate $Z_1$, $Z_2$, and $Z_3$, from which $Z_g$ follows through a Slavnov-Taylor identity (which maybe is Eq. (75), which you didn't quote). Of course since you have the said $Z$-factors only to order $g^2$ (or order $\alpha_s=g^2/(4 \pi)$), you can determine only the counterterm contributing to $Z_g$ to this order, and thus you have to expand the expression for it also up to order $\alpha_s$.

In the here obviously applied minimal-subtraction scheme you need to find the coefficients to $1/\epsilon$ order by order perturbation theory. This determines the counter terms order by order. At higher loop order you have to take care of the subdivergences by using the corresponding counter terms of subdiagrams.

 

[Siupa]

I can't read light gray on a less light gray background. Can you use LaTex and maybe post your source?

If you open the imgur link it should be in high res. Anyways the reference is chapter 3, end of subchapter 3.3 of this pdf

 

[Siupa]

The Dyson series is an expansion in the coupling constant $g$. Obviously you are evaluating the quoted 2nd-order loop corrections to the quark and gluon self-energies as well as the gluon 3-vertex to get the counter terms to evaluate $Z_1$, $Z_2$, and $Z_3$, from which $Z_g$ follows through a Slavnov-Taylor identity (which maybe is Eq. (75), which you didn't quote). Of course since you have the said $Z$-factors only to order $g^2$ (or order $\alpha_s=g^2/(4 \pi)$), you can determine only the counterterm contributing to $Z_g$ to this order, and thus you have to expand the expression for it also up to order $\alpha_s$.

In the here obviously applied minimal-subtraction scheme you need to find the coefficients to $1/\epsilon$ order by order perturbation theory. This determines the counter terms order by order. At higher loop order you have to take care of the subdivergences by using the corresponding counter terms of subdiagrams.

Eq. (75) is just the definition of $Z_1 = Z_g Z_2 Z_3^{\frac{1}{2}}$. Anyways thank you I understand now, the expansions of $1/Z_2$ , $1/Z_3$ are obviously at first order in $\alpha_s$, which in turn means first order in the counterterms since they are proportional to $\alpha_s$ because they were computed at 1-loop. I guess the divergence in $1/\epsilon$ doesn't matter since we only take the limit $\epsilon \to 0$ in the end?

 

[vanhees71]

Right! That's the idea!