[Pw_forum] ibrav7 doesn't show the structure of BaNi2As2 correctly

[David Foster]

Dear Tone and Giovanni

thank you for your helps. So, how can I correct crystal positions by using v1, v2, and v3?
Regards

David Foster

Ph.D. Student of Chemistry

--------------------------------------------
On Tue, 5/27/14, Giovanni Pizzi <giovanni.pizzi at epfl.ch> wrote:

 Subject: Re: [Pw_forum] ibrav7 doesn't show the structure of BaNi2As2 correctly
 To: "PWSCF Forum" <pw_forum at pwscf.org>
 Date: Tuesday, May 27, 2014, 7:28 AM
 
 
   
 
     
   
   
     Dear David,
 
       I think that the relative (crystal) coordinates that
 you write for
       the 5-atom cell are still expressed in terms of the
 conventional
       cell (the one with vectors (a,0,0), (0,a,0), (0,0,c).
 This is the
       cell used in the case ibrav=14, reason for which you
 get the
       correct structure.
 
       
 
       Instead, the cell used in ibrav=7 is the primitive one
 - the docs
       say:
 
       
 
       v1=(a/2)(1,-1,c/a),  v2=(a/2)(1,1,c/a), 
 v3=(a/2)(-1,-1,c/a)
 
       
 
       Therefore you have to express the relative (crystal)
 coordinates
       with respect to this basis set, and not in terms of
 the lattice
       vectors of the conventional cell.
 
       
 
       Best,
 
       
 
       Giovanni Pizzi
 
       
 
       
 
       
 
       On 05/27/2014 03:48 PM, David Foster wrote:
 
     
     
       Dear users
 
 The atomic positions for BaNi2As2 crystal with space group
 of I4/mmm (139) are:
 
 Ba (2a): 0, 0,   0 
 As (4e): 0, 0,   0.3471 
 Ni(4d):   0, 0.5, 0.25
 
 I used MS Visualizer to build the conventional tetragonal
 cell (a = b = 4.1474  c = 11.619).
 
 After seeing the structure with MS, I convert it to
 primitive and finding all atoms positions. Only 5 atoms
 there are in primitive (If I am right):
 
 Ba 0.000000 0.000000  0.000000
 As 0.347100 0.347100  0.000000
 As 0.652900 0.652900  0.000000
 Ni 0.750000 0.250000  0.500000
 Ni 0.250000 0.750000  0.500000
 
 I prepared two input file for QE5.0.3, one with ibrav=7 and
 5 atoms, and the other with ibrav=14 and all atoms in
 conventional cell instead of its primitive:
 with P1 symmetry, the cell has 10 atoms:
 Ba 0.0000 0.0000  0.0000
 Ba 0.5000 0.5000  0.5000
 As 0.000 0.000  0.3471
 As 0.5000 0.5000  0.8471
 As 0.000 0.000  0.6529
 As 0.5000 0.5000  0.1529
 Ni 0.000 0.5000  0.25000
 Ni 0.500 0.0000  0.250000
 Ni 0.000 0.5000  0.75000
 Ni 0.500 0.0000  0.750000
 
 I have attached two inputs and two pictures that xcrysden
 shows. While ibrav=14 shows the crystal correctly, ibrav=7
 has a problem and doesn't show the crystal properly. I
 am confused!!
 Any help will be appreciated.
 
 
 
 Then, I 
  
 Regards
 
 David Foster
 
 Ph.D. Student of Chemistry
       
 
       
       
 
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     -- 
 Giovanni Pizzi
 Post-doctoral Research Scientist
 EPFL STI IMX THEOS
 MXC 340 (Bâtiment MXC)
 Station 12
 CH-1015 Lausanne (Switzerland)
 Phone: +41 21 69 31124
   
 
 
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