[Wien] Bi almost cubic

[Fecher, Gerhard]

Hallo Pablo,
you are right, something strange happens if one accepts the suggestion of sgroup and/or symmetry.
Unfortunately I can not clearly find out anymore which version of Wien2k I used some years ago (only hint, the .struct files did not contain the part "Precise positions").

Peter (or someone else who knows the subroutines well) should check what is wrong with the new structure or symmetry operations.

When accepting the suggested changes the result is a metal as you observed.

I checked by using the structure in P1 (struct is below), in that case the semimetallic character is kept if I strictly use P1 (note in hexagonal setup one has 6 atoms in the cell, if interested I can send it in P1, too).
(Note one has to ignore some warnings during initialisation. Even if a calculation in P1 may be in general a bad idea with respect to calculation time, I do not see why I "MUST" shift the atomic positions while using P1 ????)

I also played another -- possibly nasty --  trick, I accepted everything of sgroup and symmetry (resulting in .struct with only 1 atom and half c), then I inserted the symmetry operations into the initial structure file (with 2 atoms and original c)
and continued with the initialisation with this .struct file, in that case, the semimetallicity of Bi was kept.

PS.: the program findsym (see https://stokes.byu.edu/iso/findsym.php)  reduces the structure with z=1/4 in the same way as sgroup (just with the atom at 3b and not 3a, what should principally not matter),
however, it uses consistently hexagonal set up and not a mixture of hexagonal lattice parameters with rhombohedral atomic positions, which is indeed an unlucky choice.

PSS.: The symmetry checker of Eneavor (a commercial crystallographic program) did not reduce to a one atom structure, for unknown reason.

Note the following structure has experimenta a and c from the Pearson database.

blebleble                                                                      
P   LATTICE,NONEQUIV.ATOMS:  2 1_P1                                            
MODE OF CALC=RELA unit=bohr                                                    
  8.966184  8.966184  8.966184 57.227000 57.227000 57.227000                   
ATOM  -1: X=0.25000000 Y=0.25000000 Z=0.25000000
          MULT= 1          ISPLIT= 8
Bi1        NPT=  781  R0=0.00000500 RMT=    2.5000   Z: 83.000                 
LOCAL ROT MATRIX:    1.0000000 0.0000000 0.0000000
                     0.0000000 1.0000000 0.0000000
                     0.0000000 0.0000000 1.0000000
ATOM  -2: X=0.75000000 Y=0.75000000 Z=0.75000000
          MULT= 1          ISPLIT= 8
Bi2        NPT=  781  R0=0.00000500 RMT=    2.5000   Z: 83.000                 
LOCAL ROT MATRIX:    1.0000000 0.0000000 0.0000000
                     0.0000000 1.0000000 0.0000000
                     0.0000000 0.0000000 1.0000000
   2      NUMBER OF SYMMETRY OPERATIONS
 1 0 0 0.00000000
 0 1 0-0.00000000
 0 0 1-0.00000000
       1
-1 0 0 0.50000000
 0-1 0 0.50000000
 0 0-1 0.50000000
       2
Precise positions
   0.250000000000000   0.250000000000000   0.250000000000000
   0.750000000000000   0.750000000000000   0.750000000000000


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 delamora [delamora at unam.mx]
Gesendet: Freitag, 8. Juli 2022 01:15
An: A Mailing list for WIEN2k users
Betreff: [Wien] Bi almost cubic

Dear Gerhard,

The optimized cell is;
---------------
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=0.23592070 Y=0.23592070 Z=0.23592070
ATOM   1:X= 0.76407930 Y=0.76407930 Z=0.76407930
Bi         NPT=  781  R0=0.00000500 RMT=    2.5000   Z: 83.000
----------------
with two atoms in the cell
XYZ = A and 1-A
A=0.23592070 and 1-A=0.76407930

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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