# Schwartz loop result in $\bar{c}b\rightarrow \bar{u}d$ EFT

I am trying to reproduce the result of equation $$(31.97)$$ of Schwart's book Quantum field theory and the Standard Model. The results involves the calculation of the following diagram

where the external quark momenta are set to zero. Schwartz finds the following result

$$\mathcal{A} = \frac{G_Fg_s^2}{32\pi^2}\left(\frac{1}{\varepsilon_\mathrm{IR}}+\frac{3}{4}+\frac{1}{2}\log\frac{\tilde{\mu}^2}{m_W^2}\right)(\bar{c}_L\gamma^\nu T^a\gamma^\alpha\gamma^\mu b_L)(\bar{d}_L\gamma_\nu T^a\gamma_\alpha\gamma_\mu u_L)$$

Beside the spinor structure which is trivial, I cannot get the same result from the loop integral. In what follows I'll highlight some steps in my evaluation so that we can find what I'm doing wrong.

The relevant integral to be solved in dimensional regularization is the following

$$\frac{g_2^2 g_s^2}{d} \left(\mu^2\right)^\frac{4-d}{2}\int\frac{\mathrm{d}^d k}{(2\pi)^d}\frac{1}{(k^2)^2(k^2-m_W^2)}$$

Using the Feynman parametrisation we get

$$\frac{1}{(k^2)^2(k^2-m_W^2)} = \int_0^1 \mathrm{d}x\mathrm{d}y\,\delta(x+y-1)\frac{2y}{[x(k^2-m_W^2)+yk^2]^3}=\int_0^1\mathrm{d}x\frac{2(1-x)}{(k^2-xm_W^2)^3}$$

Swapping the integrals and using the following result given by Schwartz

$$\int\frac{\mathrm{d}^d k}{(2\pi)^d}\frac{1}{(k^2-\Delta)^3} = \frac{-i}{2(4\pi)^{d/2}}\frac{1}{\Delta^{3-d/2}}\Gamma\left(\frac{6-d}{2}\right)$$

where in our case $$\Delta = xm_W^2$$, then we get

$$\frac{-i}{2(4\pi)^{d/2}}\frac{g_2^2 g_s^2}{d} \left(\mu^2\right)^\frac{4-d}{2}\Gamma\left(\frac{6-d}{2}\right)\int_0^1 \mathrm{d}x\;2(1-x)\frac{1}{(xm_W^2)^{3-d/2}}$$

Now, given that $$\epsilon = \frac{4-d}{2}$$, then $$3-d/2 = 1+\epsilon$$ so that

$$\frac{-i}{(4\pi)^2}\frac{g_2^2}{m_W^2}\frac{g_s^2}{d}\Gamma\left(\epsilon+1\right) \left(\frac{4\pi\mu^2}{m_W^2}\right)^\epsilon\int_0^1\mathrm{d}x\;\frac{1-x}{x^{1+\epsilon}}$$

In the limit $$d\rightarrow 4$$, or $$\epsilon\rightarrow 0$$, the Gamma function is finite and gives $$\Gamma(1)=1$$. The integral becomes

$$\int_0^1\mathrm{d}x\;\frac{1-x}{x^{1+\epsilon}} = \frac{1}{\epsilon(\epsilon-1)} = \frac{1}{\epsilon-1}-\frac{1}{\epsilon}$$

where the first part is finite and the second divergent. We also expand the factor in parenthesis

$$\left(\frac{4\pi\mu^2}{m_W^2}\right)^\epsilon = 1+\epsilon\log\left(\frac{4\pi\mu^2}{m_W^2}\right)+\mathcal{O}(\epsilon^2)$$

What I get in the end is the following (factors of $$i$$ aside)

$$\frac{G_Fg_s^2}{8\pi^2}\left[1+\epsilon\log\left(\frac{4\pi\mu^2}{m_W^2}\right)+\mathcal{O}(\epsilon^2)\right]\left(1+\frac{1}{\epsilon}\right) = \frac{G_Fg_s^2}{8\pi^2}\left[1+\frac{1}{\epsilon}+\log\left(\frac{4\pi\mu^2}{m_W^2}\right)\right]$$

Which is not at all the result given by Schwartz! What bugs me the most is the additional factor of $$3/4$$ which Schwartz get which seems to me to be linked to some color factor which I cannot find since we did not touch the Dirac structure at all.

What am I doing wrong?

Two things. First, it looks to me that you did not expand the 1/d factor. That is, the $$\epsilon$$ piece in $$1/d=1/4+\epsilon/8$$ combines with the pole and gives the puzzling 3/4 factor.
Second, when you expand the $$\Gamma$$ function you need to include higher-order terms which give you some Eulergamma rubbish (which is then included in the RG-scale).
• Oh, the $1/d$ factor, of course! In light of this I think that I can make it work. I left out the $\gamma_E$ factor since it's going to be absorbed in the renorm. constant, but yes, my bad not explicitly saying it! Commented Jul 12, 2021 at 12:20