- #1

- 11

- 1

## Main Question or Discussion Point

This type of integration is a special case of something that occurs over and over in QM and QFT (it's everywhere in Peskin and Schroeder), but I am having a bit of trouble working out the details. Set [itex]\hbar=1[/itex] and consider the propagation amplitude for a free, nonrelativistic particle to move from x to y in time t, given by [itex]U(x,y,t)=\langle y\vert e^{i(p^2/2m)t}\vert x\rangle[/itex]. Peskin and Schroeder evaluate this as a 3-dimensional momentum space integral, and it's easy to manipulate and find [tex]U(x,y,t)=\int\frac{d^3p}{(2\pi)^3}e^{i(p^2/2m)t}e^{i\vec{p}\cdot\Delta x}=\int\frac{d^3p}{(2\pi)^3}e^{i(p^2/2m)t}e^{ip\Delta x\cos\theta}[/tex] where [itex]\Delta \vec x=\vec y -\vec x[/itex]. The next step is to convert to spherical coordinates, which I don't think I have a problem with: [tex]U(x,y,t)=\frac 1{(2\pi)^3}\int_0^\infty dp\, p^2\int_0^{2\pi}d\varphi\int_{-1}^1d\cos\theta e^{ip^2t/2m}e^{ip\Delta x\cos\theta}.[/tex] After performing the angular integrations (which are both trivial) I find [tex]U(x,y,t)=\frac 2{\Delta x(2\pi)^2}\int_0^\infty dp\,p e^{ip^2t/2m}\sin(p\Delta x)[/tex] which not only looks completely intractable but also is nothing like the answer one is supposed to find for a free propagator. Have I screwed this up or is there just some last touch I'm not seeing?