K-point integration

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

$\sum_n \frac{1}{\Omega_{\mathrm{BZ}}} \int_{\Omega_{\mathrm{BZ}}} \epsilon_{n\bold{k}} \, \Theta(\epsilon_{n\bold{k}}-\mu) \, d \bold{k},$

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:

$\frac{1}{\Omega_{\mathrm{BZ}}} \int_{\Omega_{\mathrm{BZ}}} \to \sum_{\bold{k}} w_{\bold{k}}.$

Keeping the step function we get a sum

$\sum_{\bold{k}} w_{\bold{k}} \epsilon_{n\bold{k}} \, \Theta(\epsilon_{n\bold{k}}-\mu),$

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

Linear tetrahedron methods

Within the linear tetrahedron method, the term $\epsilon_{n\bold{k}}$ is interpolated linearly between two k-points. Bloechl has recently revised the tetrahedron method to give effective weights $f(\{\epsilon_{n\bold{k}}\})$ for each band and k-point. In addition Bloechel was able to derive a correction formula which removes the quadratic error inherent in the linear tetrahedron method (linear tetrahedron method with Bloechel corrections). The linear tetrahedron is more or less fool proof and requires a minimal interference by the user.

The main drawback is that the Bloechels method is not variational with respect to the partial occupancies if the correction terms are included, therefore the calculated forces might be wrong by a few percent. If accurate forces are required we recommend a finite temperature method.

Finite temperature approaches - smearing methods

In this case the step function is simply replaced by a smooth function, for example the Fermi-Dirac function

$f(\frac{\epsilon-\mu}{\sigma}) = \frac{1}{\exp(\frac{\epsilon-\mu}{\sigma})+1}.$

or a Gauss like function

$f(\frac{\epsilon-\mu}{\sigma}) = \frac{1}{2} \left ( 1- {\mathrm{erf}} \left [ \frac{\epsilon - \mu}{\sigma} \right ] \right ).$

is one used quite frequently in the context of solid state calculations. Nevertheless, it turns out that the total energy is no longer variational (or minimal) in this case. It is necessary to replace the total energy by some generalized free energy

$F =E - \sum_{n\bold{k}} w_{\bold{k}} \sigma S(f_{n\bold{k}}).$ The calculated forces are now the derivatives of this free energy $F$ (see Forces). In conjunction with Fermi-Dirac statistics the free energy might be interpreted as the free energy of the electrons at some finite temperature $\sigma=k_{\mathrm{B}} T$, but the physical significance remains unclear in the case of Gaussian smearing. Despite this problem, it is possible to obtain an accurate extrapolation for $\sigma \to 0$ from results at finite $\sigma$ using the formula

$E(\sigma \to 0) =E_0= \frac{1}{2} ( F +E) .$

In this way we get a "physical" quantity from a finite temperature calculation, and the Gaussian smearing method serves as an mathematical tool to obtain faster convergence with respect to the number of k-points. For Al this method converges even faster than the linear tetrahedron method with Bloechel corrections.

Improved functional form by Methfessel and Paxton

Typical convenient settings for $\sigma$ for different metals: Aluminium possesses an extremely simple DOS, Lithium and Tellurium are also simple nearly free electron metals, therefore $\sigma$ might be large. For Copper $\sigma$ is restricted by the fact that the d-band lies approximately 0.5 eV beneath the Fermi-level. Rhodium and Vanadium posses a fairly complex structure in the DOS at the Fermi-level, $\sigma$ must be small.

	forces	Stress tensor
0	yes	no
1	yes	trace only
2	yes	yes
3	yes	yes
4	yes	yes
5	yes	yes
6	yes	yes
7	yes	yes

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