- #1

- 1,344

- 32

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.