Nonrelativistic free particle propagators

[Theage]

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?

 

[Einj]

To solve this integrale you first note that the integrand is even under ##p\to-p## and therefore you can extend the boundaries up to ##-\infty##, but paying an addition factor of 1/2. Then you recall that ##\sin(p\Delta x)=\text{Im} e^{ip\Delta x}## and once you have done that you're left if an ordinary gaussian integral (complete the square and so on and so forth). Just remember to take the imaginary part at the end of the gaussian integration.

 

[Theage]

Thanks! Amazingly the [itex]2\pi[/itex]'s work out using Gaussian integral methods.

 

[vanhees71]

And there should be a minus sign in the exponential with the time!