# Scattering amplitude Green's function integral

On page 208 of Weinberg's QM book, he calculates the following integral

\begin{align} G_k (\vec{x}-\vec{y}) =& \int \frac{d^3 q}{(2\pi \hbar)^3} \frac{e^{i\vec{q} \cdot (\vec{x}-\vec{y})}} {E(k)-E(q)+i\epsilon} \\ =& \frac{4\pi}{(2\pi)^3}\int_0^{\infty} q^2 dq \frac{\sin(q|\vec{x}-\vec{y}|)}{q|\vec{x}-\vec{y}|}\frac{2m/\hbar^2}{k^2-q^2+i\epsilon} \, . \end{align}

It is clear that he is integrating in spherical coordinates. However, I don't see how where the

$$\frac{\sin(q|\vec{x}-\vec{y}|)}{q|\vec{x}-\vec{y}|}$$

comes from. Can someone explain?

• I think you should just try to work out the integral yourself to see where it comes from. Have your tried that? – Danu Feb 6 '16 at 11:47

Write $d^3 q = dq q^2 d\theta d\phi \sin\theta$ and integrate over the angular variables. The only angular dependence in the integrand is in $e^{i \vec{q} \cdot ( \vec{x}-\vec{y}) } = e^{i q r \cos\theta}$ where I've defined $r = | \vec{x} - \vec{y} |$. Then, we have $$\int_0^{2\pi} d\phi \int_0^\pi d\theta \sin\theta e^{i q r \cos\theta}$$ There is no $\phi$ dependence so that just gives $2\pi$. For the $\theta$ integral defined new integration variable $t = \cos\theta$. Then, the above becomes $$2\pi \int_{-1}^{1} dt e^{i q r t} = \frac{2\pi}{i q r } e^{i q r t} \bigg|_{-1}^1 = \frac{2\pi}{ i q r } \left[ e^{i q r } - e^{- i q r } \right] = \frac{4\pi }{ q r } \sin(qr) = \frac{4 \pi \sin \left( q | \vec{x} - \vec{y} | \right) }{ q | \vec{x} - \vec{y} | }$$