Taking the Fourier Transform of a potential

  • #1
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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.
 

Answers and Replies

  • #2
strangerep
Science Advisor
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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.)
 
  • #3
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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:
  • #4
692
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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
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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.
 
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
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Well' let's see. The Fourier transform is
[tex]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}).[/tex]
Introducing spherical coordinates with the polar axis in direction of ##x## leads to
[tex]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}.[/tex]
The latter integral gives indeed the desired result (according to Mathematica):
[tex]F(\vec{x})=F(r)=\frac{1}{r} \mathrm{erf} \left (\frac{r}{\sqrt{2} a} \right ).[/tex]
 

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