[Wien] Bi almost cubic

Fri Jul 8 13:07:53 CEST 2022

Plot the bands and you will "understand" the minimum at EF in the 2-atom 
structure when you consider how the tetrahedron method works (i.e. by 
connecting bands according to energy (and not character). Due to the 
finite k-mesh you get a lot of "avoided" crossings and a pseudogap in 
the DOS.

Try a mesh of 100**3 or even 200**3 k-points for the DOS (probably you 
still get some artefacts at EF, but they should be "smaller".

It is a nice example of Fermi surface nesting and a resulting Peierls 
distortion.

PS: I don't think anything has changed in sgroup. It finds the identical 
SG for both structures, but with c/2. And in addition, it shifts the 
origin, so that the Bi atom sits at the origin.

PPS: The 2 k-meshes should NOT be identical for the 1 and 2 atom cells, 
but as close as possible be of similar k-point density.

Am 08.07.2022 um 12:22 schrieb 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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