# Fermion self-energy and vertex renormalization in Non-Abelian Gauge Theories

I am currently going through chapter 16 of Peskin and Schroeder and some of the calculations seem very obscure to me. The problems are as follows:

1. On page 528, the authors compute the value of the diagram for the fermion self-energy in the Feynman-'t Hooft gauge (equation 16.75):
$$\int\frac{d^4p}{(2\pi)^4}(ig)^2\gamma^{\mu}t^{a}\frac{i\displaystyle{\not}p+ \displaystyle{\not}k}{(p + k)^2}\gamma_{\mu}t^a\frac{-i}{p^2}.\tag{16.75}$$ After a bit of algebra, using some identities for the $$t^a$$ matrices and employing the Feynman parametrization, we arrive at
$$\frac{ig^2}{(4\pi)^{d/2}}C_2(r)\displaystyle{\not}k\int_0 ^1 dx (1-x)(d-2)\frac{\Gamma(2-\frac{d}{2})}{\Delta^{2-d/2}}$$ with $$C_2(r)$$ being the Casimir operator and $$\Delta = -x(1-x)k^2$$. The authors then conclude that this expression results in equation 16.76:
$$\frac{ig^2}{(4\pi)^2}\displaystyle{\not}kC_2(r)\Gamma(2-\frac{d}{2}).\tag{16.76}$$ I understand that we could write $$d = 4-\epsilon$$ and expand the $$\Gamma$$ and $$\Delta$$ functions as
$$\frac{\Gamma(\epsilon/2)}{\Delta^{\epsilon/2}} = \frac{2}{\epsilon} - \ln\mu^2 - \ln(x(1-x)\kappa) + \mathcal{O}(\epsilon)$$ where we set the renormalization condition at $$k^2 = -\mu^2$$ and $$\kappa$$ is a constant that depends on the way we choose to do our Feynman parametrization. The thing is that we are interested in computing the beta function, so we only consider the $$\ln\mu^2$$ term in the expansion, from what I can gather. However, I don't understand how restricting the expansion to the log term results in equation 16.76, let alone equation 16.77 which is the expression for the counterterm:
$$\delta_2 = -\frac{g^2}{(4\pi)^2}\frac{\Gamma(2-\frac{d}{2})}{(\mu^2)^{2-d/2}}C_2(r).\tag{16.77}$$ note that Peskin & Schroeder use $$M^2$$ where I have used $$\mu^2$$.
Edit: the problem that I am having is that I don't understand how Peskin & Schroeder managed to compute the integral $$\int_0^1 dx(1-x)(d-2)\frac{\Gamma(2-\frac{d}{2})}{\Delta^{2-d/2}} = \frac{\Gamma(2-\frac{d}{2})}{(\mu^2)^{2-\frac{d}{2}}}$$, or if that is even the computation being made after all. I also don't understand how the energy scale shows up in the denominator.

2. The next computation is for the $$\delta_1$$ counterterm, which refers to the vertex renormalization. The starting point is equation 16.78:
$$\int\frac{d^4 p}{(2\pi)^4}g^3 t^b t^a t^b \frac{\gamma^{\nu}(\displaystyle{\not}p + \displaystyle{\not}k\,')\gamma^{\mu}(\displaystyle{\not}p +\displaystyle{\not}k)\gamma_{\nu}}{(p+k\,')^2(p+k)^2p^2}.\tag{16.78}$$

Again, some algebra, identities and a subtle argument about the superficial degree of divergence of this diagram allows us to get to an approximation for this expression, given by equation 16.81:
$$g^3 [C_2(r) - \frac{1}{2}C_2(G)]t^a(2-d)^2\frac{1}{d}\gamma^{\mu}\int\frac{d^4 p}{(2\pi)^4}\frac{1}{(p^2)^2}.\tag{16.81}$$ Where $$C_2(G)$$ is the Casimir operator in the adjoint representation. I don't really understand how the second line is then obtained:
$$\frac{ig^3}{(4\pi)^2}[C_2(r) - \frac{1}{2}C_2(G)]t^a\gamma^\mu(\Gamma(2-\frac{d}{2}) + ...)$$ Doing the integral of $$\frac{1}{(p^2)^2}$$ in $$d$$ dimensions actually diverges once we break the integral in the solid angle part and in the "radial" part (plus, the gamma function actually appears in the denominator, not in the numerator). To illustrate:
$$\int\frac{d^d p}{(2\pi)^d}\frac{1}{p^4} = \int\frac{d\Omega_d}{(2\pi)^d}\int_0^\Lambda dp \frac{p^{d-1}}{p^4} = \frac{2\pi^{d/2}}{(2\pi)^d\Gamma(d/2)}\frac{\Lambda^{d-4}}{d-4}$$

I don't know if the way I proceeded to compute that integral is correct and how to get to the second line of 16.81. The result for the counterterm in 16.84 again has the energy scale in the denominator, which makes me believe that my problem in both questions have similar explanations.
This is my first question on SE, and I apologize if any part of it is confusing or awfully basic. I appreciate any tip or answer.

Answer to the first question: I think I see the confusion now. Actually, you are right in saying that P&S go from the third line of Eq. (16.76) $$\frac{ig^2}{(4\pi)^{d/2}}C_2(r)\not{k}\int_0^1 dx(1-x)(d-2) \frac{\Gamma(2-\frac{d}{2})}{\Delta^{2-d/2}}$$ to the fourth one $$\frac{ig^2}{(4\pi)^2}\not{k}C_2(r)\Gamma(2-\frac{d}{2})+...$$ just by computing the integral, i.e. using $$\int_0^1 dx(1-x)=\frac{1}{2}$$ and that $$(d-2)=2$$ and that the remaining terms are constant in $$x$$, since the authors substitute $$\Delta$$ with $$M^2$$ (or $$\mu^2$$ according to your question/post). This is discussed in the previous page, page 527.

So, given that the counter term $$\delta_2$$ is given by the diagram of the form of the fifth diagram in Figure 10.4 (without the mass counterterm, since we have negleceted that, and with the momentum carried by the external legs being $$\not{k}$$ instead of $$\not{p}$$), then we can write that $$i\delta_2\not{k}+\frac{ig^2}{(4\pi)^{d/2}}C_2(r)\not{k}\int_0^1 dx(1-x)(d-2) \frac{\Gamma(2-\frac{d}{2})}{\Delta^{2-d/2}}=0$$ at the energy scale $$M^2$$. From that, and by making the substitution $$\Delta=M^2$$, we conclude that $$\delta_2=-\frac{g^2}{(4\pi)^2}C_2(r)\frac{\Gamma(2-\frac{d}{2})}{(M^2)^{2-d/2}}+\text{finite}$$ So, the renormalization condition of the 2 point function is the equation that determines $$\delta_2$$, just as in the QED case. This is implied by P&S. Now, the only loose end to my answer, I think, is the reason why we can set $$\Delta=M^2$$ (and hence neglecting any $$x-$$ dependence of $$\Delta$$), but this is discussed, as I mentioned earlier, in the previous page in P&S.

Answer to the second question: You say that you are comfortable when P&S arrive at the first line of Eq. (16.81) $$g^3[C_2(r)-\frac{1}{2}C_2(G)] t^a(2-d)^2\frac{1}{d}\gamma^{\mu} \int\frac{d^4p}{(2\pi)^4}\frac{1}{(p^2)^2} \tag{16.81}$$ There are two ways to proceed in deriving the second line of Eq. (16.81), i.e. $$\frac{ig^3}{(4\pi)^2}[C_2(r)-\frac{1}{2}C_2(G)] t^a\gamma^{\mu}\Big(\Gamma(2-\frac{d}{2}+...)\Big) \tag{16.81}$$

I will simply describe both of these ways and if something does not make sense, you can always comment.

1. You can add on the denominator a small parameter $$\epsilon$$, i.e. $$p^2\rightarrow p^2-\delta$$ and hence use the formulas in the appendix: $$\int\frac{d^dp}{(2\pi)^d}\frac{1}{(p^2-\delta)^2}= \frac{(-1)^2i}{(4\pi)^{d/2}}\frac{\Gamma(2-\frac{d}{2})}{\Gamma(2)}\Bigg(\frac{1}{\delta}\Bigg)^{2-\frac{d}{2}}$$ and after expanding everything in powers of $$\epsilon$$, where $$\epsilon$$ is defined by $$d=4-2\epsilon$$, you will see that in the $$d\rightarrow4$$ limit, the divergence does not depend on the artificially introduced parameter $$\delta$$, but it is coming from the poles of the $$\Gamma$$ function (i.e. the infinitesimal parameter $$\epsilon$$).
2. Go back to evaluating the vertex quantum correction, i.e. Eq. (16.78) and evaluate the integrals without neglecting the external momenta (at least from the denominator, if I remember correctly). This would give you the desired dependence on the $$\Gamma$$ function.

I hope this helps. If there are more questions, please comment.

• Wow, that was a nice way to solve the second question! Thanks a lot! About the first question, I have edited that part for clarity :) Nov 15, 2022 at 18:55
• Also, in your answer the denominator in the integral is supposed to be $p^2 - \delta$, right? You are reserving the $\epsilon$ for the shift in the dimension, correct? Nov 15, 2022 at 18:58
• Yes, this is a different infinitesimal variable! I will look at the revised question and edit my answer (hopefully soon). Sorry, there was a mistake. Now the anwer to the seqond question is correct. I will take a look on the first one Nov 16, 2022 at 7:16