Taking the Fourier Transform of a potential

[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.

 

[strangerep]

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

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.)

 

[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

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

 

[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.

 

[The_Duck]

I'm wondering, though, how the form of the cutoff was actually achieved.

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.

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

essentially means that

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

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.

 

[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):
$$F(\vec{x})=F(r)=\frac{1}{r} \mathrm{erf} \left (\frac{r}{\sqrt{2} a} \right ).$$