Integrating over all orbitals

In this section we discuss partial occupancies. A must for all readers.

First there is the question why to use partial occupancies at all. The answer is: partial occupancies help to decrease the number of k-points necessary to calculate an accurate band-structure energy. This answer might be strange at first sight. What we want to calculate is, the integral over the filled parts of the bands

[math]\displaystyle{ \sum_n \frac{1}{\Omega_{\mathrm{BZ}}} \int_{\Omega_{\mathrm{BZ}}} \epsilon_{n\bold{k}} \, \Theta(\epsilon_{n\bold{k}}-\mu) \, d \bold{k}, }[/math]

where [math]\displaystyle{ \Theta(x) }[/math] is the Dirac step function. Due to our finite computer resources this integral has to be evaluated using a discrete set of k-points\cite{bal73}:

[math]\displaystyle{ \frac{1}{\Omega_{\mathrm{BZ}}} \int_{\Omega_{\mathrm{BZ}}} \to \sum_{\bold{k}} w_{\bold{k}}. }[/math]

Keeping the step function we get a sum

[math]\displaystyle{ \sum_{\bold{k}} w_{\bold{k}} \epsilon_{n\bold{k}} \, \Theta(\epsilon_{n\bold{k}}-\mu), }[/math]

which converges exceedingly slow with the number of k-points included. This slow convergence speed arises only from the fact that the occupancies jump form 1 to 0 at the Fermi-level. If a band is completely filled the integral can be calculated accurately using a low number of k-points (this is the case for semiconductors and insulators).

For metals the trick is now to replace the step function [math]\displaystyle{ \Theta(\epsilon_{n\bold{k}}-\mu) }[/math] by a (smooth) function [math]\displaystyle{ f(\{\epsilon_{n\bold{k}}\}) }[/math] resulting in a much faster convergence speed without destroying the accuracy of the sum. Several methods have been proposed to solve this dazzling problem.