# Peskin QFT Contour Integral — Chapter 6

On page 178 of Peskin's QFT, they have the vector potential $$A^\mu(x)=\int\frac{d^4k}{(2\pi)^4}e^{-ik\cdot x}\frac{-ie}{k^2}\left(\frac{p'^\mu}{k\cdot p'+i\epsilon}-\frac{p^\mu}{k\cdot p-i\epsilon}\right)\tag{6.5}$$ and they set the poles to be $$k^0=\pm|\mathbf{k}|$$ and $$k\cdot p=0$$ and $$k'\cdot p =0$$. For $$t<0$$ they close the contour upwards they pick up the pole at $$k\cdot p=0$$ and they get $$A^\mu(x)=\int\frac{d^3k}{(2\pi)^3}e^{i\mathbf{k}\cdot \mathbf{x}}e^{-i(\mathbf{k}\cdot\mathbf{p}/p^0)t}\frac{(2\pi i) (+ie)}{(2\pi)k^2}\frac{p^\mu}{p^0}$$

I understand every step except for how they obtain the $$p^0$$ in the denominator.

Given that the pole we pick is $$k\cdot p =0$$ then we will have $$\begin{split}\int_{\text{above}} dk^0 \left[\int\frac{d^3k}{(2\pi)^4}e^{-ik\cdot x}\frac{-ie}{k^2}\left(\frac{p'^\mu}{k\cdot p'+i\epsilon}-\frac{p^\mu}{k\cdot p-i\epsilon}\right)\right]\end{split}\\ =2\pi i\ \text{Res}\bigg[\int\frac{d^3k}{(2\pi)^4}e^{-ik\cdot x}\frac{-ie}{k^2}\left(\frac{p'^\mu}{k\cdot p'+i\epsilon}-\frac{p^\mu}{k\cdot p-i\epsilon}\right)\bigg]_{k^0=\mathbf{k}\cdot\mathbf{p}/p^0}\\ =\int\frac{d^3k}{(2\pi)^3}e^{i\mathbf{k}\cdot \mathbf{x}}e^{-i(\mathbf{k}\cdot\mathbf{p}/p^0)t}\frac{(2\pi i) (+ie)}{(2\pi)k^2}p^\mu$$ I am clearly missing the step that brings the $$p^0$$ in the denominator. I also assume that we are implying that $$k^2=-\frac{(\mathbf{k}\cdot\mathbf{p})^2}{{p^0}^2}+|\mathbf{k}|^2$$. What am I missing?

• The $k^{2}=k^{\mu}k_{\mu}$ in the denominator includes a $k_{0}^{2}$, which produces poles that need to be taken into account. – Buzz Dec 6 '19 at 0:07

$$\frac{1}{k\cdot p}=\frac{1}{k^0p^0-\mathbf{k}\cdot\mathbf{p}}=\frac{1/p^0}{k_0-\frac{\mathbf{k}\cdot\mathbf{p}}{p^0}}$$
$$k^0=\frac{\mathbf{k}\cdot\mathbf{p}}{p^0}$$
$$\frac{1}{p^0}.$$