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(This is purely for personal study - the exercise is 20.2(a) from Lancaster and Blundell (2014), Oxford Uni. Press - an excellent textbook btw.)

"Confirm that the Fourier transform of $V(\underline{r})=\frac{-g^2}{4\pi|\underline{r}|}e^{-\mu|\underline{r}|}$ is $\tilde{V}(\underline{q})=\frac{-g^2}{|\underline{q}|^2+\mu^2}$."

The definition of the 3D transform (equation 22) is $\tilde{f}(\underline{k})=\int d^3x\,\,e^{-i\underline{k}.\underline{x}}f(\underline{x})$; therefore using spherical co-ordinates, I can get as far as

$\tilde{V}(\underline{q})=\int_0^\infty dr\,|\underline{r}|^2\frac{-g^2}{4\pi|\underline{r}|}e^{-\mu|\underline{r}|}e^{i|\underline{q}||\underline{r}|}\int_0^\pi \sin\theta d\theta\int_0^{2\pi}d\phi =4\pi\int_0^\infty dr\,|\underline{r}|\frac{-g^2}{4\pi}e^{-\mu|\underline{r}|+i|\underline{q}||\underline{r}|}\\ =-g^2\int_0^\infty dr\,|\underline{r}|e^{-(\mu-i|\underline{q}|)|\underline{r}|}.$

However, this integral appears to yield $\frac{-1}{\mu-i|\underline{q}|}$, for constant $\mu$ and $q$, since (with $Z_q=\mu-i|\underline{q}|$)

$-g^2\int_0^\infty dr\,|\underline{r}|e^{-Z_q|\underline{r}|} =Z_q^{-1}\lim(c\rightarrow\infty)[(1-c)e^{-Z_q c}-1]$,

and the first term vanishes in the limit (due to L'Hopital's rule).

One last comment is that I have had a sneaking suspicion this is somehow related to Laplace transforms, a) because the formula for $\tilde{V}(\underline{q})$ is so very Laplace-looking, and b) the Laplace formula itself containing a factor of $\exp(-st)$ (where $s$ is the integration variable). If someone could just please tell me what I am missing I'd be grateful!

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    $\begingroup$ It should not be $e^{i|q||r|}$ but $e^{i{\bf q}\cdot {\bf r}}$ in the Fourier integral. To proceed choose a coordinate system where ${\bf q}$ points along the $z$-axis. We then have ${\bf q}\cdot {\bf r} = r|q|\cos\theta$ and the $\theta$ integral will give you a different $r$ integral to perform which should give the desired result. $\endgroup$ – Winther Nov 8 '15 at 23:55
  • $\begingroup$ Thanks very much, I shall try this. $\endgroup$ – Pan Daemonium Nov 9 '15 at 2:21
  • $\begingroup$ One more question, if you can see this: is it related to Laplace Transforms? This is more out of curiosity. $\endgroup$ – Pan Daemonium Nov 9 '15 at 2:23
  • $\begingroup$ I've done it, thanks very much for the hint (and especially for not giving me the full solution, it was a good puzzle!) To be clear by 'Laplace transform' I meant how the integral basically comes down to $\int_0^\infty e^{-\mu r}\sin(r|q|)dr$, which you'll notice is the Laplace transform $L[\sin(r|q|)](\mu)$. That'll be why it was familiar to me :) By the way, if you want to Answer the question (i.e. in the box), I'll give you some time, else I might Answer it myself for the aid of myself and others. $\endgroup$ – Pan Daemonium Nov 9 '15 at 3:41
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    $\begingroup$ Yes that's right, the final result is the Laplace transform of $\sin(r|q|)$, but this I think accidental in the sense that you would have no way of knowing that without having solved the problem first. $\endgroup$ – Winther Nov 9 '15 at 3:46
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Carrying on from Winther's hint above, I obtain

$\tilde{V}(q)=\int_0^{2\pi}d\phi\int_0^\pi d(\cos\theta) \int_0^\infty dr\,|r|^2\frac{-g^2}{4\pi|r|}e^{-\mu|r|}e^{iq.r} \\=(2\pi)(\frac{-g^2}{4\pi})\int_0^\pi d(\cos\theta) \int_0^\infty dr\,|r|e^{-\mu|r|}e^{i|q||r|\cos\theta} \\ =\frac{ig^2}{2|q|}\int_0^\infty dr\,e^{-\mu|r|}(e^{-i|q||r|}-e^{i|q||r|}) \\ =\frac{g^2}{|q|}\int_0^\infty dr\,e^{-\mu|r|}\sin(|q||r|) \\ =-\frac{g^2}{|q|}\left[\frac{e^{-\mu|r|}(\mu\sin(|q||r|)+|q|\cos(|q||r|))}{|q|^2+\mu^2}\right]^\infty_0 \\ =\left(\frac{g^2}{|q|}\right)\left(\frac{-|q|}{|q|^2+\mu^2}\right)=\frac{-g^2}{|q|^2+\mu^2}.$

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