Are there any important considerations to make for density of state calculations for supercells? With band structures typically band unfolding schemes make things easier to understand but is there anything analogous for density of states?
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asked Nov 29, 2020 at 18:36
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The density of states reads:
$$ \tag{1} g(E)=\sum_{n}\int\frac{d\mathbf{k}}{(2\pi)^3}\delta(E-E_{n\mathbf{k}}), $$
where $E_{n\mathbf{k}}$ are the electronic energies and the integral is over the Brillouin zone (BZ).
If you calculate $g(E)$ in the primitive cell, then you will integrate over the primitive cell BZ. If you calculate $g(E)$ in a supercell, then you will integrate over the supercell BZ, which is a fraction of the primitive cell BZ. This means you will have fewer $\mathbf{k}$-points. However, this will be exactly compensated by the additional bands $n$ that you will get due to band folding. Therefore, the density of states is the same in a primitive cell or a supercell.
The situation is different to what happens for a band structure because in a band structure you want to resolve the energies in terms of $\mathbf{k}$ and $n$ for the primitive cell. This is why you need to undo band folding in a supercell calculation to turn the states from the supercell BZ to the primitive cell BZ. For densities of states, you integrate over $\mathbf{k}$ and sum over $n$, so band folding does not affect the result in any way.
