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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?

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  • $\begingroup$ 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. $\endgroup$ – Buzz Dec 6 '19 at 0:07
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The residue of

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

at

$$k^0=\frac{\mathbf{k}\cdot\mathbf{p}}{p^0}$$

is

$$\frac{1}{p^0}.$$

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