- #1
pleasehelpmeno
- 154
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Hi I am struggling trying to see understand the basic propagator integral trick.
\int \frac{d^{3}p}{(2\pi^{3})}\left\lbrace \frac{1}{2E_{p}}e^{-ip.(x-y)}|_{p_{0}=E_{p}}+\frac{1}{-2E_{p}}e^{ip.(x-y)}|_{p_{0}=-E_{p}}\right\rbrace = \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 know it can be solved by contour integration about poles p^{0}=E_{p} and p^{0}=-E_{p} respectively but I just can't do it. Any help would be appreciated, I can provide working to what I have done but there doesn't seem much point because I am very confused.
\int \frac{d^{3}p}{(2\pi^{3})}\left\lbrace \frac{1}{2E_{p}}e^{-ip.(x-y)}|_{p_{0}=E_{p}}+\frac{1}{-2E_{p}}e^{ip.(x-y)}|_{p_{0}=-E_{p}}\right\rbrace = \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 know it can be solved by contour integration about poles p^{0}=E_{p} and p^{0}=-E_{p} respectively but I just can't do it. Any help would be appreciated, I can provide working to what I have done but there doesn't seem much point because I am very confused.