1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I think you should just try to work out the integral yourself to see where it comes from. Have your tried that? $\endgroup$ – Danu Feb 6 '16 at 11:47
1
$\begingroup$

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 $$ 2\pi \int_{-1}^{1} dt e^{i q r t} = \frac{2\pi}{i q r } e^{i q r t} \bigg|_{-1}^1 = \frac{2\pi}{ i q r } \left[ e^{i q r } - e^{- i q r } \right] = \frac{4\pi }{ q r } \sin(qr) = \frac{4 \pi \sin \left( q | \vec{x} - \vec{y} | \right) }{ q | \vec{x} - \vec{y} | } $$

$\endgroup$
  • $\begingroup$ Isn't this an explicit solution to a homework-type question? $\endgroup$ – Danu Feb 6 '16 at 11:45
  • $\begingroup$ Oh of course! Sorry I didn't mean for it to be a homework question. I was just trying to understand the derivation from the book. $\endgroup$ – Alex Wang Feb 6 '16 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.