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Partial occupancies, different methods

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

$\displaystyle \vspace*{1mm}
\sum_n \frac{1}{\Omega_{BZ}} \int_{\Omega_{BZ}}
\epsilon _{n{\bf k}}\, \Theta(\epsilon _{n{\bf k}}-\mu) \, {\rm d}{\bf k},
$


where $ \Theta(x)$ is the Dirac step function. Due to our finite computer resources this integral has to be evaluated using a discrete set of k-points[37]:

$\displaystyle \vspace*{1mm} \frac{1}{\Omega_{BZ}} \int_{\Omega_{BZ}} \to \sum_{\bf k}w_{\bf k}.$ (7.7)


Keeping the step function we get a sum

$\displaystyle \vspace*{1mm}
\sum_{\bf k}w_{\bf k}\epsilon _{n{\bf k}}\, \Theta(\epsilon _{n{\bf k}}-\mu),
$


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 $ \Theta(\epsilon _{n{\bf k}}-\mu)$ by a (smooth) function $ f(\{\epsilon _{n{\bf k}}\})$ resulting in a much faster convergence speed without destroying the accuracy of the sum. Several methods have been proposed to solve this dazzling problem.



Subsections
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