quantum mechanics - Free particle propagation amplitude calculation

I have a quick calculational question.

In Peskin and Schroeder, Chapter 2, they want to look at the amplitude for a particle to propagate between two arbitrary points, $x$ and $x_0$, in an arbitrary amount of time t given the Hamiltonian $\sqrt{p^2c^2+m^2c^4}$. To do this, we look at the inner product of the time-evolved particle that was at $x_0$ with the $x$ eigenket: $\langle \vec{x}|\exp(-i\hat{H}t/\hbar)|\vec{x_0}\rangle$, correct?

If we insert two complete sets of momenta, the integral becomes (give or take factors of $\hbar$):

$\frac 1 {(2\pi)^3}\int d^3p\; \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar)\exp(i\vec{p}\cdot(\vec{x}-\vec{x_0}))$

I'm not quite sure where to proceed from here... Peskin and Schroeder somehow reach:

$\frac 1 {2\pi^2 |x-x_0|}\int_0^\infty dp\;p\sin(p|x-x_0|)\exp((-it\sqrt{p^2c^2+m^2c^4}/\hbar)$.

To get here, however, it seems that you would have to assume that $p$ points in the same direction as $x-x_0$ to get spherical symmetry or something. Why can we do this? Aren't we summing over all possible momenta?

Thanks!

Answer

Work with spherical coordinates in momentum space and choose $z$-axis in $p$-space along $\vec{x}-\vec{x_0}$. Then $$\vec{p}\cdot(\vec{x}-\vec{x_0})=p|x-x_0| \cos\theta$$

Then your integral $$ \frac 1 {(2\pi)^3}\int d^3p \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar)\exp(i\vec{p}\cdot(\vec{x}-\vec{x_0})) \\ = \frac 1 {(2\pi)^3}\int_0^{\infty} p^2 dp \int_0^{2\pi} d\phi \int_{-1}^{1} d(\cos\theta) \exp(ip|x-x_0| \cos\theta) \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar) $$

Now do the $\phi$ and $\cos\theta$ integrals to get $$ \frac 1 {4\pi^2 \, i \, |x-x_0|}\int_0^{\infty} p dp \Big[ \exp(ip|x-x_0|) - \exp(-ip|x-x_0|) \Big] \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar) $$

Recognize that the quantity in the square brackets is $2i\sin(p|x-x_0|)$ to get your answer.