# Poles bit in a propagator

Hi I am trying to derive the K-G propagator and am stuck on the bit where Cauchy's Integral formula is needed i.e evaluating from $$\int \frac{d^{3}p}{(2\pi)^3}\left\lbrace\frac{1}{2E_{p}}e^{-ip.(x-y)}|_{p^{o}=E_{p}}+\frac{1}{-2E_{p}}e^{-ip.(x-y)}|_{p^{o}=-E_{p}}\right\rbrace$$ to $$\int \frac{d^{3}p}{(2\pi)^3}\int \frac{dp^{0}}{i2\pi}\frac{-1}{P^{2}-m^{2}}e^{-ip.(x-y)}$$

I understand that the formula $g(z_{0})=\frac{1}{2\pi i}\int \frac{g(z)}{z-z_{0}}dz$ must be used but I just don't see how the solution can be found,

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I would advise going the other way round first. If you perform the $dp^0$ integration you will end up with a 3 dimensional integral that matches what you start with. So you basically have to 'reverse' a complex integration which is a bit of a pain the first time you see it. – alexarvanitakis Apr 19 '13 at 13:28
This question is very incomplete but potentially interesting if it gets clarified. So can you elaborate it a bit? – Dilaton Jul 23 '13 at 23:40
@ChrisWhite thanks for this information, this version looks now good and clear enough to be left open. – Dilaton Jul 24 '13 at 11:01

For Feynman prescription, the poles are located at $p^0=\pm(E_p-i\epsilon)$. When $x^0>y^0$, we close the counter below the positive pole such that $\Re(-ip^0(x^0-y^0))<0$; When $x^0<y^0$, we close the counter above the negative pole such that $\Re(-ip^0(x^0-y^0))<0$. According to Jodan lemma, we know that

$$\int_{|p^0|=+\infty}\frac{dp^0}{2\pi i}\,\frac{-1}{P^2-m^2}e^{-ip^0(x^0-y^0)}=0$$

Notice that $(p^0)^2-E^2_p=(p^0-E_p)(p^0+E_p)$ and the counter we choose only has one pole. For $x^0>y^0$, we have

$$z_0=E_p,\quad g(z)=\frac{-1}{p^0+E_p}e^{-ip^0(x^0-y^0)}$$

and for $x^0<y^0$, we have

$$z_0=-E_p,\quad g(z)=\frac{-1}{p^0-E_p}e^{-ip^0(x^0-y^0)}$$

Then, with the residue theorem

$$\frac{1}{2\pi i}\int dp^0\,\frac{g(z)}{z-z_0}=g(z_0)$$

we can obtian that

$$\frac{1}{2\pi i}\int dp^0\,\frac{-1}{(p^0)^2-E^2_p+i\epsilon}e^{-ip^0(x^0-y^0)}=\frac{1}{2E_p}e^{-iE_p(x^0-y^0)}\theta(x^0-y^0)+\frac{1}{-2E_p}e^{iE_p(x^0-y^0)}\theta(y^0-x^0)$$

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sorry I am still a bit confused, I dont see how you did the last integral. I realise that p can be split into $p^{0}$ and its spatial terms but i stil cant see how to do the last integral – user21119 Apr 19 '13 at 13:34
Sorry I am getting muddled with conventions firstly is $E_{p}^{2}=m^{2}+p^{2}_{i}$, also shouldn't the exponentials have a $p$ not an $E_{p}$, I don't really see why there is a $p^{2}-m^{2}$, sorry for being awkwrd – user21119 Apr 19 '13 at 14:14
$p^2=E^2_p-p^2_i=m^2$ is the on-shell condition. Please notice that $e^{-ip_i(x_i-y_i)}$ appears in the three-dimentional integral $\int d^3\vec{p}$ – soliton Apr 19 '13 at 14:34
Thanks I get it now – user21119 Apr 19 '13 at 14:50