Return to Revisions

Without loss of generality, we can create spherical coordinates (in three-dimensional $\mathbf{p}$-space) such that $\mathbf{r}$ is along the $\theta=0$ axis. Then, we can transform the integral into those coordinates, and perform the trivial integral over $\phi$: \begin{align} D(x-y) &= \int_0^\infty \int_0^\pi \int_0^{2\pi} \frac{1}{(2\pi)^3}\, \frac{1}{2E_{\mathbf{p}}} e^{i\, p\, r\, \cos\theta}\, p^2\, \sin\theta\, d\phi\, d\theta\, dp \\ &= \frac{1}{(2\pi)^2}\, \int_0^\infty \frac{p^2}{2E_{\mathbf{p}}} \left[\int_0^\pi e^{i\, p\, r\, \cos\theta}\, \sin\theta\, d\theta\right]\, dp. \end{align} Now, the integral in brackets is not hard, but is left as an exercise for the reader, who might note that $\sin\theta\, d\theta = -d \cos\theta$ (which simplifies a change of coordinates) and $\frac{ e^{i p r}- e^{-i pr}}{ipr} = \frac{ 2\sin(pr)}{pr}$.