# Confution on UV cut-off in the calculation of effective action and Beta function

I am reading David Tong's gauge theory notes and meet some difficulties.

In section 2.4.2, he uses background field to calculate effective action $$S_{eff}$$ and Beta function. Simply like follows: Writting gauge field $$A_{\mu}$$ as $$A_{\mu}=\bar{A}_{\mu}+\delta A_{\mu}$$ for $$\bar{A}_{\mu}$$ being a fixed field, $$\delta A_{\mu}$$ being fluctuation. Also we introduce Faddeev-Popov ghost fields $$c$$ and $$c^{\dagger}$$. The path integral of gauge field $$A_{\mu}$$ goes like $$$$e^{-S_{eff}}=Z=\int \mathcal{D}A \mathcal{D}c \mathcal{D}c^{\dagger}e^{S(A,c,c^{\dagger})}=(\det[\Delta_{gauge}])^{-1/2} \det[\Delta_{ghost}]e^{(-1/2g^2) S_{YM}(\bar{A})}$$$$ where $$$$S_{YM}(A)=\int d^4x tr(F^{\mu\nu}F_{\mu \nu}) .$$$$ So the effective action $$S_{eff}$$ is then $$$$S_{eff}=\frac{1}{2g^2}S_{YM}(\bar{A})+\frac{1}{2}Tr\log\Delta_{gauge}-Tr\log\Delta_{ghost}$$$$ See eq(2.63) in the notes.

Now he gets the contribution of $$-Tr\log\Delta_{ghost}$$ (see page 71 in the notes) as $$$$-Tr\log\Delta_{ghost}=constant \times \int \frac{d^4k}{(2\pi)^4}tr(\bar{A}_{\mu}(k)\bar{A}_{\nu}(-k))(k^{\mu}k^{\nu}-k^2\delta^{\mu\nu})\log\bigg(\frac{\Lambda^2}{k^2}\bigg).$$$$

Here is what I don't understand: how does this $$\log\bigg(\frac{\Lambda^2}{k^2}\bigg)$$ appear in our integral in terms of $$\log$$? Is it a regularization? If it is, then the integrand cannot turn back to $$tr(\bar{A}_{\mu}(k)\bar{A}_{\nu}(-k))(k^{\mu}k^{\nu}-k^2\delta^{\mu\nu})$$ under limit $$\Lambda=\infty$$.

## 1 Answer

Tong is using a cutoff regularization in this calculation, the cutoff being this $$\Lambda$$. Specifically this factor appears because he has just completed the loop momenta integral (integral over the momenta $$p$$). This integral is divergent without a regularization scheme, so of course taking $$\Lambda\rightarrow\infty$$ (removing the cutoff) will cause the what you have written to diverge.