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I have performed the DFT calculation for Silicon in Quantum ESPRESO. I have tried to calculate the electronic band structure, the shape of wiggles of bands are mostly matching but the energy values are not matching with the referenced literatures. I have attached the figure of the calculated band structure by me.enter image description here

It's valance bands are going above the '0' value, which is not correct.
Please, suggest me how I can resolve this energy value issue. It will be helpful in my learning process.
Thanks!

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The y-axis corresponds to the direct eigenvalues from your DFT calculation. You need to set the valence band maximum to '0' manually. The DFT output of your calculation will give the highest occupied state energy (or the Fermi energy, if you use occupations smearing). So whatever that energy is, subtract that from the energies in your band plot.

A note on the Fermi energy value outputted in DFT codes: it's only really useful in the case of metals, or when likewise there are occupied states at the Fermi level, such as in heterostructures, slabs etc. In insulators and semiconductors, its position within the band gap depends on the smearing and some other computational factors, and isn't physically meaningful.

When you do align to the Fermi energy (by defining $E_F=0$) in these cases, you should use the value calculated from the scf calculation. Prof. Stefano de Gironcoli answered this question in the Quantum Espresso mailing list (pw_forum):

the ONLY Fermi energy that is meaningful is the one corresponding to a scf calculation... which is calculated consistently with the corresponding charge density. All nscf or band calculations are performed on grids which might give non-regular sampling of the BZ... you should not use the corresponding fermi energy .

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    $\begingroup$ Thanks for the answer, but how could I know, by what amount I have to shift the eigenvalues? Is this shifting needs to be done in this particular case or we have to do it manually in all the DFT calculation? Wouldn't it be weird if we do it manually(I mean manually we can shift it by whatever we want)? $\endgroup$
    – UJM
    Sep 26, 2020 at 18:02
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    $\begingroup$ @UjjawalModanwal I will try to reword it in my answer, but the code outputs the highest occupied state energy. This is what you want to define as zero. It's not weird, I don't think the absolute values of eigenvalue energies have any physical meaning with pseudopotentials. You can shift it to anything you want as long as everything shifts equally (e.g. it's common shift to the Fermi energy in metals). And yes, you will always have to do this when plotting. Quantum espresso has plotband.x which will ask for the Fermi energy and do this for you, however. Other codes may have similar tools. $\endgroup$
    – Kevin J. May
    Sep 26, 2020 at 19:13
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    $\begingroup$ @Kevin J. M., Thanks, I understood your answer now. But there is one more doubt I have, the indirect band gap calculated for Si by me using DFT is 0.6551 eV. I want to know if it is at room temp. or at 0K. If it is at room temp. It's values should be around 1.1 eV. So, is there any 'Energy correction factor' used in DFT or I have performed the calculation wrongly. Please, clarify my doubt! $\endgroup$
    – UJM
    Sep 27, 2020 at 9:21
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    $\begingroup$ @Kevin J. M.: I am not a QE user, but if we generalize from Si, the choice of the Fermi energy to be used for the correction is also related to the k-point set. For band structure calculations one uses a k-point path along high symmetry lines while for obtaining the self-consistent density one samples the Brillouin zone by some mesh. The question now is which k-point set is the best choice to select the Fermi energy for the correction. Typically this depends on the material. Is this something the user has to decide when using QE? Could you extend your answer with respect to this aspect? $\endgroup$
    – Gregor Michalicek
    Sep 27, 2020 at 14:12
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    $\begingroup$ The eigenvalues have meaning if you use energy-consistent pseudopotentials. Solid state physicists tend to employ only shape-consistent pseudopotentials, which are less accurate than the energy-consistent ones: energy-consistent potentials are also shape consistent, whereas shape-consistent ones are not energy consistent. The shape consistent pseudo induces an error in the orbital energies. $\endgroup$
    – Susi Lehtola
    Sep 28, 2020 at 10:58

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