Integrating over all orbitals: Difference between revisions

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  \sum_n \frac{1}{\Omega_{\mathrm{BZ}}} \int_{\Omega_{\mathrm{BZ}}}
  \sum_n \frac{1}{\Omega_{\mathrm{BZ}}} \int_{\Omega_{\mathrm{BZ}}}
   \eps_{n\bold{k}} \, \Theta(\eps_{n\bold{k}}-\mu) \, d \bold{k},
   \epsilon_{n\bold{k}} \, \Theta(\epsilon_{n\bold{k}}-\mu) \, d \bold{k},
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   \sum_{\bold{k}} w_{\bold{k}}  \eps_{n\bold{k}} \, \Theta(\eps_{n\bold{k}}-\mu),
   \sum_{\bold{k}} w_{\bold{k}}  \epsilon_{n\bold{k}} \, \Theta(\epsilon_{n\bold{k}}-\mu),
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Revision as of 10:39, 19 March 2019

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).