# Problems relating to K-G propagator and path integrals

1. Apr 18, 2013

1) On page the author goes from $\int \frac{d^{3}p}{(2\pi)^3}\left[\frac{1}{2E_P}e^{-ip.(x-y)}+\frac{1}{-2E_P}e^{ip.(x-y)}\right]=\int\frac{d^{3}p}{(2\pi)^3}\int \frac{d^{0}p}{(i2\pi)}\frac{-1}{p^{2}-m^{2}}e^{-ip.(x-y)}$ evaluated at $E_{p}=\pm p^{0}$ (+ for the plus E and minus for minus E) I think the operator $\partial^{2} + m^{2}$ has been applied but surely that would give $p^{2} + m^{2}$ rather than 1 over it and I also don't understand how it it turned into another integral. The book says it has used contour integration which explains the 2pi i term I realize that but I am not sure what it has integrated to get that form, is it that they have used $1/(E_{P}=p^{0} dE_{p} = 2\pi i$ and then substituted E in from there?
2) In a discretized form of an action why is the espilon in the denominator for T and numberator for V, is it because T contains a derivative, so $S=\int dx \left(\frac{m}{2}\dot{x}^{2}-V\right)\rightarrow \sum \left[\frac{T}{\epsilon}-\epsilon V\right]$
3) The book often mentions cfomplete the square and evaluate an integral and to be honest I am not really sure what this means, an example is; $\int\mathcal{D}\phi\mathcal{D}\pi\exp\left[i \int d^{4}x (\pi\dot{\phi}-0.5\pi^{2}=0.5(\nabla\phi)^{2}-V)\right]=\int \mathcal{D}\phi \exp \left[i \int d^{4}x (0.5(\partial_{\mu}\phi)^{2}-V)\right]$ I understand that basically the pi terms are quadratic and so somehow all the pi terms including the integral; cancel, I am not sure why though.