# Particle Physics Integral Calculation

1. May 12, 2010

### Sistine

1. The problem statement, all variables and given/known data
In calculating the quantum mechanical amplitude for the Coulomb potential (scattering of say $$\alpha$$ particle off a massive particle of charge $$Ze$$), I came across a fourier transform which I could not calculate.

If
$$U(r)=\frac{2Ze^2}{4\pi\epsilon_0 r}$$
then
$$\tilde{U}(k)=\int U(\mathbf{x})e^{i\mathbf{k}\cdot\mathbf{x}}d^3x=\frac{Ze^2}{\epsilon_0 k^2}$$

2. Relevant equations

3. The attempt at a solution

2. May 12, 2010

### gabbagabbahey

It looks pretty straightforward, just choose your coordinate system so that $\textbf{k}$ points along the z-axis and perform the integration in spherical coordinates.

Edit: It turns out that the integral doesn't converge directly, so instead you may wish to transform the Yukawa potential $\frac{1}{r}e^{-ur}$, and then take the limit as $u\to 0$. Strictly speaking, this is only valid if you treat the Coulomb potential as a distribution and not a mathematical function (otherwise, the limit of the integral need not be the same as the integral of the limit).

Last edited: May 12, 2010
3. May 12, 2010

### Sistine

Thanks for your reply, I tried integrating over $$\mathbb{R}^3$$ but my integral does not converge for $$0\leq r<\infty$$ i.e. integrating

$$\int_0^{2\pi}\int_0^{\pi}\int_0^{\infty}r e^{ikr \cos\theta} \sin\theta dr d\theta d\phi$$

Doing the $$\theta$$ integration first I get

$$\int_0^{2\pi}\int_0^{\infty}\frac{2\sin kr}{k}dr d\phi$$

But the r integration does not converge. Am I using the wrong limits

4. May 12, 2010

### gabbagabbahey

Did you read the edit part of my post?

5. May 13, 2010

### Sistine

Yes I read the edit part of your post, but in my lecture notes there was nothing said about approximation by the Yukawa potential, the integral was just given as I stated it above. Is the integral an approximation? It partially works out when integrated. (also is it related to the first born approximation?)