6
$\begingroup$

I am trying to analyze the density of electronic states of a $\ce{TiO2}$ slab, but the result I got from VASP shows the fermi energy is lower than the valence band maximum (VBM) by a small energy difference ($\pu{0.05eV}$).

I am wondering if this is possible for semiconductors?

Band structure with Fermi energy below VBM

$\endgroup$
4
  • 1
    $\begingroup$ I doubt this is good enough for an answer, so posting as a comment. In principle, nothing physical prevents this from happening in the real world. Acceptor-doped semiconductors can do this. Surfaces (referring to the "slab" in the question) are generally weird and can exhibit surface states of hardly predictable energetics. So generally speaking this result is within the realm of possibility. Whether this result is physical in the particular system being studied is a different question. $\endgroup$ Commented Jan 16, 2022 at 6:05
  • 3
    $\begingroup$ An alternative to the answer by @AndreyPoletayev is that the density of states is not a good way to quantitatively determine the position of the VBM as DoS plots typically involve some degree of non-empirical smearing that widen the bands. $\endgroup$
    – ProfM
    Commented Jan 16, 2022 at 9:29
  • $\begingroup$ I agree with @ProfM about the empirical smearing. However, I am more worried about the reference to "slab" in the question. Does the use of this term imply that there is a broken periodicity in the system being computed, or is the original question referring to a fully 3D-periodic computation? $\endgroup$ Commented Jan 17, 2022 at 7:13
  • $\begingroup$ @AndreyPoletayev, Yes, the slab is a surface model, as used in the following tutorial.vasp.at/wiki/index.php/Ni_100_surface_relaxation $\endgroup$
    – Jack
    Commented Jan 17, 2022 at 7:22

1 Answer 1

7
$\begingroup$

No, this isn't possible for an undoped semiconductor. The reason it's happening is because the Fermi level is not well defined for your semiconductor at finite temperature, and you're trying to read the valence band maximum (VBM) from the graph of your broadened DOS. I recommend taking the VBM directly from the band energies, instead.

In a DFT program, the states are computed at a discrete set of k-points, and have well-defined eigenvalues ("band energies") at those points. The DOS for this system is a set of spikes (delta-functions) at those energies, but this doesn't give you a smooth, meaningful DOS. The problem is that the DOS should be integrated over the whole Brillouin zone, but all you have are discrete sampling points.

The common solution is to replace each spike with a narrow Gaussian, or similar "smearing function", which has the effect of smoothing the states out in energy. The states are almost unchanged by any reasonable choice of the width of the Gaussian (the "smearing width"). If you increase the smearing width, the DOS gets smoother and the tail of the smearing distribution stretches to higher energies, which naturally pushes some densities-of-states higher in energy. The way you're reading the DOS, this would raise the VBM, even though the states are the same as they were before. Even worse, strictly-speaking the smearing function never goes to zero, so if you "zoom in" on the DOS, you'll never find somewhere that it decays to zero, until you hit the limits of the calculations finite precision.

The actual VBM is the band-energy of the topmost occupied band, so I recommend you just use that instead. If the precise value is important to you, you will need to ensure that you've sampled the Brillouin zone at the correct k-point (or as close to it as possible). You can check this visually by computing and plotting the band-structure (particularly along the high-symmetry directions).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .