Rewriting $\int d \textbf{r} \int d \textbf{R} \rho_1(\textbf{r})\rho_2(\textbf{R})v(|\textbf{r}-\textbf{R}|)$ in terms of Fourier components

Question

I'm trying to reproduce some steps from a paper (https://journals.aps.org/pre/pdf/10.1103/PhysRevE.62.3855, eq. 5 and eq. 17) where the author rewrites the following integral

$$\int d \textbf{r} \int d \textbf{R} \rho_1(\textbf{r})\rho_2(\textbf{R})v(|\textbf{r}-\textbf{R}|),$$

in terms of its Fourier components as follows:

$$\frac{1}{V}\sum_{\textbf{k}}\tilde{v}(k)\tilde{\rho_1}(\textbf{k})\tilde{\rho_2}(-\textbf{k}).$$

where $V$ denotes the volume and the integral is taken over the whole volume of the system. It is not clear for me how this form is obtained, as I would expect, if one writes a general function in terms of a Fourier series, there should always be an exponential term showing up. Would someone be able to clarify, possibly show the steps?

Answer 1

This is an application of the Convolution theorem for Fourier series (or Fourier transform) telling that the Fourier coefficient of $\rho_2*v$ is proportional to $\tilde{\rho}_2(k)\,\tilde{v}(k)$, together with the Parceval–Plancherel Formula. This yields $$ \iint \rho_1(r)\,\rho_2(R)\,v(r-R)\,\mathrm d r\,\mathrm d R = \int \rho_2(R)\,(\rho_1*v)(R)\,\mathrm d R \\ = \frac{1}{V}\sum_k \overline{\tilde{\rho}_2(k)}\,(\tilde{\rho}_1(k)\,\tilde{v}(k)) = \frac{1}{V}\sum_k \tilde{v}(k)\,\tilde{\rho}_1(k)\,\tilde{\rho}_2(-k). $$