I want to understand the electronic band structure diagram of the following image, corresponding to $\text{MoS}_2$ (TMD):

enter image description here

I read about DFT (density functional theory).

DFT is based on solving the Schrodinger equation for a set of atoms. Through the Born Oppenheimer approach, it is possible to decouple the nuclear and electronic wave functions. We just then need to solve the Schrodinger equation for electrons. Using the two Hohenberg and Kohn theorems, we obtain the Kohn Scham equations: $$\bigg[-\frac{\hbar}{2m}\nabla^2+V(r)+V_H(r)+V_{XC}(r)\bigg]\psi_i(r)=\epsilon_i(r)\psi_i$$

Computationally the problem is solved as follows:

-Define n[r] (random)

-Solve Kohn Scham equation and find $\psi(r)$

-Calculate new n(r): $n(r)=2\sum\psi^{*}(r)\psi(r)$

-If n(r) calculated is equal to the original, the program will stop. If they're different, the program recalculate the Kohn Scham equation with n(r) calculated previously.

According to Hohenberg and Kohn's theorem 1, each electronic density uniquely corresponds to an energy in the ground state, therefore, once the program ends, knowing the electronic density, we know the energy in the ground state.

I don't understand how DFT relates to the various lines obtained from the graph and I don't understand what DFT is related to orbitals. How was the graph obtained based on the DFT?

  • 1
    $\begingroup$ In crystals the solutions to the Kohn-Sham equations actually depend on the k point. What you see in band structure plots are just the Kohn-Sham energy eigenvalues along some k point path. $\endgroup$ Feb 10, 2020 at 19:34
  • $\begingroup$ @GregorMichalicek Thanks for the reply. The points under the graph correspond to the different zones within the Brillouin cell. So, why don't we get just one line for the conduction band and another line for the valence band? Why are there several lines for each of the bands and what is the relationship between this and DFT? $\endgroup$ Feb 10, 2020 at 19:48
  • 2
    $\begingroup$ This question would also fit well on the new Materials Modeling SE $\endgroup$
    – Cody Aldaz
    May 16, 2020 at 22:40
  • $\begingroup$ Was the answer below satisfactory? Or can you copy and paste it on Matter Modeling SE (please)? $\endgroup$ Jul 23, 2020 at 16:43

1 Answer 1


$\epsilon_i$ in the Kohn-Sham equation actually has two indexes: $\epsilon_{nk}$, where $n$ and $k$ respectively, represent, the band and $k$ point indexes. So, for each point in a given line in that band-structure diagram you actually solve a Kohn-Sham equation for $\epsilon_{nk}$.

So, the vertical (y) axis represent $\epsilon$, whereas the horizontal (x) axis represents $k$ and different lines represent the bands $n$.

  • 2
    $\begingroup$ There is a proposal for a Materials Modeling SE which is currently in the commitment phase. It would be great if you could support the proposal by committing to it. There was many related questions in the example questions proposed for the site. $\endgroup$
    – rashid
    Feb 26, 2020 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.