# Taking the Fourier Transform of a potential

1. Dec 6, 2014

### spaghetti3451

Hi, I've been reading a paper on renormalisation theory as applied to a simple one-particle Coulombic system with a short-range potential.

In the process of renormalisation, the authors introduce an ultraviolet cutoff into the Coulomb potential through its Fourier transform:

$\frac{1}{r} \xrightarrow{\text{F.T.}} \frac{4\pi}{q^{2}} \xrightarrow{\text{cutoff}} \frac{4\pi}{q^{2}} e^{-q^{2}a^{2}/2} \xrightarrow{\text{F.T.}} \frac{erf(r/\sqrt{2}a)}{r}$

I don't understand how $\frac{1}{r}$ becomes $\frac{4\pi}{q^2}$ when Fourier transformed.

2. Dec 6, 2014

### strangerep

That's just a standard 3D Fourier transform. You can perform the 3D Fourier transform in maybe 3-4 lines by changing the cartesian integration variables to spherical polar.

(If you need more detail, you should probably post this in one of the homework forums.)

3. Dec 6, 2014

### spaghetti3451

I see.

It won't be necessary to post this on the homework forums as I can work out the details of the fourier transforming by myself, I think.

I'm wondering, though, how the form of the cutoff was actually achieved. I understand that the cutoff scale length was arbitrary, but

$\frac{4\pi}{q^{2}} \xrightarrow{\text{cutoff}} \frac{4\pi}{q^{2}} e^{-q^{2}a^{2}/2}$

essentially means that

$q \xrightarrow{\text{cutoff}} qe^-frac{qa}{sqrt(2)}$

Last edited: Dec 6, 2014
4. Dec 6, 2014

### HomogenousCow

I'd like to know this as well, is there a specific reason a gaussian was chosen?
Other than, you know because an elementary inverse fourier transform exists.

5. Dec 7, 2014

### The_Duck

Without seeing this paper, probably the form of the cutoff is just chosen to be mathematically convenient. There's a great deal of freedom in choosing the form of the cutoff, because the whole point of renormalization theory is that physical results end up being independent of the cutoff.

You shouldn't think of this procedure as modifying q; we're modifying the potential. We just want to make the potential go to zero above some given finite momentum scale.

6. Dec 8, 2014

### vanhees71

Well' let's see. The Fourier transform is
$$F(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{q} \frac{1}{(2 \pi)^3} \frac{4 \pi}{\vec{q}^2} \exp(-\vec{q}^2 a^2/2+\mathrm{i} \vec{q} \cdot \vec{x}).$$
Introducing spherical coordinates with the polar axis in direction of $x$ leads to
$$F(\vec{x})=\frac{1}{\pi} \int_0^{\infty} \mathrm{d} q \int_{-1}^{1} \mathrm{d} u \exp(-q^2 a^2/2+\mathrm{i} r q u)=\frac{2}{\pi r} \int_0^{\infty} \mathrm{d} q \exp(-\vec{q}^2 a^2/2)\frac{\sin(q r)}{q}.$$
The latter integral gives indeed the desired result (according to Mathematica):
$$F(\vec{x})=F(r)=\frac{1}{r} \mathrm{erf} \left (\frac{r}{\sqrt{2} a} \right ).$$