[Wien] Bi almost cubic

[Fecher, Gerhard]

Hallo Peter,

b) I did that for all tested structures P1 and the one after sgroup/symmetry

the Fermi energy is slightly different (0.4092709547 or 0.4104773792) if one compares the different structures,
and, if one plots the density of states as Pablo did they are also clearly different
In the original structure with 2 atoms in the cell, one has a very clear minimum at the Fermi energy,
in the reduced structure with only one atom and c/2, one has a high density of states at the Fermi energy.
I performed both calculations with 25x25x25 initial k-mesh and otherwise also identical parameters

For 1 atom c/2 one has 2 half filled bands crossing the Fermi energy==> clearly metallic
For 2 atoms c, the bands are nearly filled up to Ef or empty (within 0.01 electrons) at the used parameters ==> clearly semimetallic

I do not remember that sgroup asked to change the structure (warning: !!! Unit cell has been reduced. sgroup found: 166 (R -3 m) )
in the possibly very old version that I used in the past for my calculations with x=1/4

Ciao
Gerhard

DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy:
"I think the problem, to be quite honest with you,
is that you have never actually known what the question is."

====================================
Dr. Gerhard H. Fecher
Institut of Physics
Johannes Gutenberg - University
55099 Mainz
________________________________________
Von: Wien [wien-bounces at zeus.theochem.tuwien.ac.at] im Auftrag von Peter Blaha [peter.blaha at tuwien.ac.at]
Gesendet: Freitag, 8. Juli 2022 11:50
An: wien at zeus.theochem.tuwien.ac.at
Betreff: Re: [Wien] Bi almost cubic

This is the effect of:


a) a complicated "folding" of the BZ together with a shift of origin.

b) Plot the band structures (use xcrysden in the hexagonal setting to get "identical k-mehes".

c) you will see that some bands at some k-points agree, others do not and they are very complicated to trace by backfolding.

d) In any case it should become obvious what the DOS at EF looks so different. It is most likely a k-mesh problem (you should use enormous unshifted meshes), but still, the Tetrahedron method has no problems for the DOS at EF in the small one-atom cell since all bands go through EF in a straight line and you get a metal.

For the doubled cell, there are many "pseudo gaps" and the Tetrahedron method will make a completely different interpolation for the bands and if your k-mesh is not VERY dense, give you a semimetal (or "gap").


Otherwise, the results are "identical" as they should, but you have to be careful with the interpretation.


When A is changed to 0.25;
----------------
R   LATTICE,NONEQUIV.ATOMS:  1 166_R-3m
MODE OF CALC=RELA unit=ang
  8.591340  8.591340 22.415740 90.000000 90.000000120.000000
ATOM   1: X=0.25000000 Y=0.25000000 Z=0.25000000
ATOM   1:X= 0.75000000 Y=0.75000000 Z=0.75000000
Bi         NPT=  781  R0=0.00000500 RMT=    2.5000   Z: 83.000
-------------------
sgroup gives this warning;

warning: !!! Unit cell has been reduced.
sgroup found: 166 (R -3 m)

and the cell is reduced to;
--------------------
R   LATTICE,NONEQUIV.ATOMS:  1 166 R-3m
MODE OF CALC=RELA unit=ang
  8.591340  8.591340 11.207870 90.000000 90.000000120.000000
ATOM   1: X=0.00000000 Y=0.00000000 Z=0.00000000
Bi1        NPT=  781  R0=0.00000500 RMT=    2.5000   Z: 83.0
---------------------
which is "semicubic" with an angle;

Angle is 87.539 deg
and only one Bi atom in the cell, now in the corners

And in this reduced Bi structure the "gap" at Ef in DOS disappears.

What I see is that with A=0.249 (0.25-0.001) and A=0.2499 (0.25-0.0001)
the DOS have a "gap" and they are quite symilar, but with the addition of the 0.0001 (A=0.25) the cel is reduced and the "gap" disappears.

I hope that this answers your questions and becomes clear what I am trying to show.
Saludos

Pablo
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