2D Fourier transform of Coulomb potenial

[realtrip]

The result is well known, but i need more details about the integral below

\int \mathrm{d}^2x \frac{1}{|\mathbf{x}|} e^{- \mathrm{i} \mathbf{q} \cdot \mathbf{x}} = \frac{2 \pi}{q}

I've done the Fourier transform of the Coulomb potential in 3D. But failed to get the right answer in 2D.

I did only a few practice about 2D integrals. Will anyone show me more details about it?

Thanks in advance!

[gabbagabbahey]

The result is well known, but i need more details about the integral below

\int \mathrm{d}^2x \frac{1}{|\mathbf{x}|} e^{- \mathrm{i} \mathbf{q} \cdot \mathbf{x}} = \frac{2 \pi}{q}

I've done the Fourier transform of the Coulomb potential in 3D. But failed to get the right answer in 2D.



Well, what did you try? Show us your attempt.

[realtrip]

A little too long, and no latex in this computer.

So, i upload a screenshot in the attachment.

Thanks!

[gabbagabbahey]

:yuck:Yuck!:wink:

I wouldn't write the complex exponential in terms of sines and cosines if I were you. Instead, just switch to polar coordinates right away (with your coordinate system chosen so that \textbf{q} points in the positive x-direction) to get:

\oint d^2 x \frac{e^{-i\textbf{q}\cdot\textbf{x}}}{|\textbf{x}|}=\int_0^ \infty dr \left(\int_0^{2\pi} e^{-iqr\cos\theta}d\theta \right)

If you don't immediately recognize the angular integral, try defining \overline{r}=qr and compute the first two derivatives of the integral w.r.t \overline{q} to show that it satisfies a well known differential equation....:wink:

[gabbagabbahey]

Also, unlike the 3D case, the integral converges without treating the coulomb potential as a limiting case of the Yukawa potential.

[realtrip]

Thanks for your reply.

I'm not sure whether you want me to do the second derivatives of the angular integral w.r.t \overline{r}=qr or something else. If i havent misunderstood, i'm sorry to tell you that i really dont know what the well known differential equation is.....

Would you give me a little more details? Thanks a lot!

[gabbagabbahey]

Well, I'll tell you that \int_0^{2\pi} e^{-i\overline{r}\cos\theta}d\theta=2\pi J_0(\overline{r})....as for proving it, I'll leave that to you (Hint: What differential equation does the n=0 Bessel function of the 1st kind satisfy?)...

[realtrip]

Thank you very much!!!!