Analytic Integrals

Consider a cell periodic function, $f(\mathbf{r})$, which may be written in terms of its discrete Fourier series

$$f(\mathbf{r}) = \frac{1}{V}\sum_{l=-\infty}^{\infty} \sum_{m... ...B}_{3})e^{i(l\mathbf{B}_{1}+m\mathbf{B}_{2}+n\mathbf{B}_{3})}.$$ (50)

Consider also the bandwidth limited version of this same function, $f_{D}(\mathbf{r})$, which has only the same frequency components as $D(\mathbf{r})$:

$$f_{D}(\mathbf{r}) = \frac{1}{V}\sum_{l=-J_{1}}^{J_{1}+1} \su... ...B}_{3})e^{i(l\mathbf{B}_{1}+m\mathbf{B}_{2}+n\mathbf{B}_{3})}.$$ (51)

It can be shown that the projection of $f(\mathbf{r})$ onto a particular basis function is exactly equal to that of $f_{D}(\mathbf{r})$, and that furthermore, replacing the integral by a discrete sum over grid points leads to exactly the same answer:

$$\int_{V} d\mathbf{r} \: f^{\ast}(\mathbf{r})D_{KLM}(\mathbf{r}) = \int_{V} d\mathbf{r} \: f^{\ast}_{D}(\mathbf{r})D_{KLM}(\mathbf{r})$$

$$= \Omega \sum_{X=0}^{N_{1}-1}\sum_{Y=0}^{N_{2}-1}\sum_{Z=0}^{N_{3}-1}f^{\ast}_{D}(\mathbf{r}_{FGH})D_{KLM}(\mathbf{r}_{FGH})$$

$$= \Omega f^{\ast}_{\mathrm{D}}(\mathbf{r}_{KLM}).$$ (52)

This result is very useful for our purposes as it tells us that the overlap integral of any cell periodic function with a function that is represented by our basis set, can be evaluated exactly as a summation over grid points.