Calculating Green's functions

TODO

The scheme for these calculations starts from the Lehmann-representation of the Green's function.

$$\begin{equation} G \left(\pmb{r}+\pmb{R}^\alpha,\pmb{r^\prime}+\pmb{R}^{\alpha^\prime},E+i\eta\right) \ =\ \sum_\nu \int_{V_{BZ}}\!\frac{\mathrm{d^3k}}{V_{BZ}} \frac{\Psi_{\pmb{k}\nu}(\pmb{r}+\pmb{R}^\alpha)\Psi^\ast_{\pmb{k}\nu}(\pmb{r^\prime}+\pmb{R}^{\alpha^\prime})}{E-E_{\pmb{k}\nu}+i\eta} \end{equation}$$

Here $\alpha$ and $\alpha^\prime$ are atom sites and $\Psi_{\mathbf{k}\nu}$ and $E_{\mathbf{k}\nu}$ are the eigenstates and eigenenergies of the Kohn-Sham system respectively.

By utilizing the following decomposition of the denominator

$$\begin{equation} \begin{aligned} \lim_{\eta \to 0^+}\ \frac{1}{x-i\eta}\ &=\ \frac{x}{x^2+\eta^2} + i \frac{\eta}{x^2+\eta^2} \\ &= P\left(\frac{1}{x}\right) + i\pi \delta\left(x\right) \end{aligned} \label{eq:principalvaluedecomp} \end{equation}$$

the imaginary part of the expression above is much easier to compute. The complete Green's function can be calculated from the imaginary part via the Kramers-Kronig-Transformation.

$$\begin{equation} G(z)\ =\ -\frac{1}{\pi} \int_{-\infty}^{\infty}\mathrm{dE}\ \frac{1}{z-E}\ \mathrm{Im}[G(E)] \label{eq:kramerkronigcomplex} \end{equation}$$

Here we also deform the energy grid into the complex plane to reduce the number of necessary points on the contour to achieve accurate results.