Does integrating PDOS give total charge of a system?

Question

Suppose you have a two atom cubic cell, and you have calculated the atomic PDOS. Does integrating the atomic PDOS and summing them to Fermi energy give exactly the total charge of a system?

Answer 1

You shouldn't integrate to the Fermi energy, unless you want the T = 0 K result and that was what you computed with your DFT program (which is actually quite difficult to get).

The correct expression is the integral over all energies of the DOS multiplied by the broadening function (the occupation of the states), and if you want to get the total number of electrons then you should use exactly the same function and smearing width as the DFT program used.

$$ N = \int_{-\infty}^\infty f\left(\frac{E-E_F}{\sigma}\right)g(E)dE \tag{1} $$ where $N$ is the number of electrons, $f$ is the broadening function with width $\sigma$ and Fermi energy $E_F$ and $g$ is the DOS.

Note that, in practice, you don't actually have to integrate to $\infty$ since the broadening function decays to zero for $E>>E_F$.

If you integrate the PDOS rather than the DOS, then you will usually get a slightly smaller value from the integral, because of the spilling factor; i.e. not all of the states of a real material or molecule can be represented as a sum of a few atomic-like (usually hydrogenic or pseudoatomic) orbitals.