I am calculating the band structure of a crystal along high-symmetry lines in the BZ, using ICHARG=11 after a calculation with a Monkhors-Pack grid. I have two questions:
The Fermi energy in the MP and bandline runs differ by 1eV. I assume the energy from the Monkhorst-Pack run (Auto/100) is more exact. Does the large difference mean there is something wrong?
The second run takes 27 iterations to converge. Should it be iterative at all if it calculates a band structure from existing files? In otherwords, again, am I doing something wrong?
Thanks in advance,
Herbert
Fermi energy in band calculation
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hfruchtl
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Fermi energy in band calculation
Last edited by hfruchtl on Wed Mar 17, 2010 12:35 am, edited 1 time in total.
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Fermi energy in band calculation
please note that the MP k-mesh probably includes a completely different set of k-points (not including the k-points with the maxima of the KS-eigenvalues) therefore it is easily possible that the maximum KS values (and hence the Fermi level) differ. Please also note that in 3D periodic systems the E=0 is defined only within a constant offset.
2) by using ICHARG=11 you start with the converged,self consistent potential of the BZ. However, when changing the k-krid from an equally distributed grid to a grid spanned along the symmetry lines at the edges of the BZ, the wavefunction corresponding to these kpoints has to be determined self-consistently (the run starts with a random wavefunction, because the k-points have been changed). This will take a few electronic steps, therefore your run seems ok.
Of course the ionic geometries must not be relaxed in this bandstructure run anymore (but this is certainly not what you have done)
2) by using ICHARG=11 you start with the converged,self consistent potential of the BZ. However, when changing the k-krid from an equally distributed grid to a grid spanned along the symmetry lines at the edges of the BZ, the wavefunction corresponding to these kpoints has to be determined self-consistently (the run starts with a random wavefunction, because the k-points have been changed). This will take a few electronic steps, therefore your run seems ok.
Of course the ionic geometries must not be relaxed in this bandstructure run anymore (but this is certainly not what you have done)
Last edited by admin on Mon Mar 22, 2010 3:53 pm, edited 1 time in total.