Dear Experts,

 I have created a Zr15Nb1 cell containing 15 Zr atoms and 1 Nb atom using
the space group P 63/m m c, (space group number 194).

 But, when I try to calculate the elastic constant of it using the
thermo_pw, I find, pwscf recognizes wrong space group (No. 187). Please
find below my input:

cat > thermo_control << EOF
 &INPUT_THERMO
  what='mur_lc_elastic_constants',
  frozen_ions=.FALSE.
 /
EOF

cat > zr.elastic.in << EOF
 &control
    calculation = 'scf'
    restart_mode='from_scratch',
    prefix='zr',
    verbosity='high',
    tstress = .true.,
   tprnfor = .true.,
    pseudo_dir = '$PSEUDO_DIR/',
    outdir='$TMP_DIR/'
 /
 &system
    ibrav=  4,
    celldm(1) =12.241644950000000E00,
    celldm(3) = 1.591850000000000E00,
    nat= 16,
    ntyp= 2,
    ecutwfc=50.0,
   ecutrho = 180,
   nr1=90,
   nr2=90,
   nr3=144,
   occupations='smearing',
   smearing='marzari-vanderbilt',
   degauss=0.02
   starting_magnetization(1) = 0.7,
   use_all_frac = .true.
 /
 &electrons
    conv_thr =  1.0d-9
 /
ATOMIC_SPECIES
 Zr  91.22  Zr.pz-spn-kjpaw_psl.1.0.0.UPF
 Nb  92.906 Nb.pz-spn-kjpaw_psl.1.0.0.UPF
ATOMIC_POSITIONS (crystal)
Zr     0.166666666666667E+00         0.333333333333333E+00
0.125000000000000E+00
Zr     0.833333333333333E+00         0.666666666666667E+00
0.875000000000000E+00
Zr     0.666666666666667E+00         0.833333333333333E+00
0.125000000000000E+00
Zr     0.333333333333333E+00         0.166666666666667E+00
0.875000000000000E+00
Zr     0.166666666666667E+00         0.833333333333333E+00
0.125000000000000E+00
Zr     0.833333333333333E+00         0.166666666666667E+00
0.875000000000000E+00
Zr     0.833333333333333E+00         0.666666666666667E+00
0.375000000000000E+00
Zr     0.333333333333333E+00         0.166666666666667E+00
0.375000000000000E+00
Zr     0.833333333333333E+00         0.166666666666667E+00
0.375000000000000E+00
Nb     0.166666666666667E+00         0.333333333333333E+00
0.625000000000000E+00
Zr     0.666666666666667E+00         0.833333333333333E+00
0.625000000000000E+00
Zr     0.166666666666667E+00         0.833333333333333E+00
0.625000000000000E+00
Zr     0.666666666666667E+00         0.333333333333333E+00
0.125000000000000E+00
Zr     0.333333333333333E+00         0.666666666666667E+00
0.875000000000000E+00
Zr     0.333333333333333E+00         0.666666666666667E+00
0.375000000000000E+00
Zr     0.666666666666667E+00         0.333333333333333E+00
0.625000000000000E+00
K_POINTS AUTOMATIC
5 5 3 0 0 0

*********************************************************************************************************

Below I have included some part of the output:


 Starting atomic positions in crystallographic axes:

     site n.     atom                  positions (cryst. coord.)
         1           Zr  tau(   1) = (  0.1666667  0.3333333  0.1250000  )
         2           Zr  tau(   2) = (  0.8333333  0.6666667  0.8750000  )
         3           Zr  tau(   3) = (  0.6666667  0.8333333  0.1250000  )
         4           Zr  tau(   4) = (  0.3333333  0.1666667  0.8750000  )
         5           Zr  tau(   5) = (  0.1666667  0.8333333  0.1250000  )
         6           Zr  tau(   6) = (  0.8333333  0.1666667  0.8750000  )
         7           Zr  tau(   7) = (  0.8333333  0.6666667  0.3750000  )
         8           Zr  tau(   8) = (  0.3333333  0.1666667  0.3750000  )
         9           Zr  tau(   9) = (  0.8333333  0.1666667  0.3750000  )
        10           Nb  tau(  10) = (  0.1666667  0.3333333  0.6250000  )
        11           Zr  tau(  11) = (  0.6666667  0.8333333  0.6250000  )
        12           Zr  tau(  12) = (  0.1666667  0.8333333  0.6250000  )
        13           Zr  tau(  13) = (  0.6666667  0.3333333  0.1250000  )
        14           Zr  tau(  14) = (  0.3333333  0.6666667  0.8750000  )
        15           Zr  tau(  15) = (  0.3333333  0.6666667  0.3750000  )
        16           Zr  tau(  16) = (  0.6666667  0.3333333  0.6250000  )

     The energy minimization will require  9 scf calculations

     The point group 107 D_3h (-62m) is compatible with the Bravais lattice.

     The rotation matrices with the order used inside thermo_pw are:

     12 Sym. Ops. (no inversion) found (10 have fractional translation)


                          s                  frac. trans.

      isym =  1     identity

 cryst.   s( 1) = (  1    0    0   )
                  (  0    1    0   )
                  (  0    0    1   )

 cart.    s( 1) = (  1.000  0.000  0.000 )
                  (  0.000  1.000  0.000 )
                  (  0.000  0.000  1.000 )


      isym =  2     180 deg rotation - cart. axis [0,1,0]

 cryst.   s( 2) = ( -1    0    0   )
                  (  1    1    0   )
                  (  0    0   -1   )

 cart.    s( 2) = ( -1.000  0.000  0.000 )
                  (  0.000  1.000  0.000 )
                  (  0.000  0.000 -1.000 )


      isym =  3     120 deg rotation - cryst. axis [0,0,1]

 cryst.   s( 3) = (  0    1    0   )
                  ( -1   -1    0   )
                  (  0    0    1   )

 cart.    s( 3) = ( -0.500 -0.866  0.000 )
                  (  0.866 -0.500  0.000 )
                  (  0.000  0.000  1.000 )


      isym =  4     120 deg rotation - cryst. axis [0,0,-1]

 cryst.   s( 4) = ( -1   -1    0   )
                  (  1    0    0   )
                  (  0    0    1   )

 cart.    s( 4) = ( -0.500  0.866  0.000 )
                  ( -0.866 -0.500  0.000 )
                  (  0.000  0.000  1.000 )


      isym =  5     180 deg rotation - cryst. axis [1,-1,0]

 cryst.   s( 5) = (  0   -1    0   )
                  ( -1    0    0   )
                  (  0    0   -1   )

 cart.    s( 5) = (  0.500 -0.866  0.000 )
                  ( -0.866 -0.500  0.000 )
                  (  0.000  0.000 -1.000 )


      isym =  6     180 deg rotation - cryst. axis [2,1,0]

 cryst.   s( 6) = (  1    1    0   )
                  (  0   -1    0   )
                  (  0    0   -1   )

 cart.    s( 6) = (  0.500  0.866  0.000 )
                  (  0.866 -0.500  0.000 )
                  (  0.000  0.000 -1.000 )


      isym =  7     inv. 180 deg rotation - cart. axis [0,0,1]

 cryst.   s( 7) = (  1    0    0   )
                  (  0    1    0   )
                  (  0    0   -1   )

 cart.    s( 7) = (  1.000  0.000  0.000 )
                  (  0.000  1.000  0.000 )
                  (  0.000  0.000 -1.000 )


      isym =  8     inv. 180 deg rotation - cart. axis [1,0,0]

 cryst.   s( 8) = ( -1    0    0   )
                  (  1    1    0   )
                  (  0    0    1   )

 cart.    s( 8) = ( -1.000  0.000  0.000 )
                  (  0.000  1.000  0.000 )
                  (  0.000  0.000  1.000 )


      isym =  9     inv.  60 deg rotation - cryst. axis [0,0,1]

 cryst.   s( 9) = ( -1   -1    0   )
                  (  1    0    0   )
                  (  0    0   -1   )

 cart.    s( 9) = ( -0.500  0.866  0.000 )
                  ( -0.866 -0.500  0.000 )
                  (  0.000  0.000 -1.000 )


      isym = 10     inv.  60 deg rotation - cryst. axis [0,0,-1]

 cryst.   s(10) = (  0    1    0   )
                  ( -1   -1    0   )
                  (  0    0   -1   )

 cart.    s(10) = ( -0.500 -0.866  0.000 )
                  (  0.866 -0.500  0.000 )
                  (  0.000  0.000 -1.000 )


      isym = 11     inv. 180 deg rotation - cryst. axis [0,1,0]

 cryst.   s(11) = (  1    1    0   )
                  (  0   -1    0   )
                  (  0    0    1   )

 cart.    s(11) = (  0.500  0.866  0.000 )
                  (  0.866 -0.500  0.000 )
                  (  0.000  0.000  1.000 )


      isym = 12     inv. 180 deg rotation - cryst. axis [1,1,0]

 cryst.   s(12) = (  0   -1    0   )
                  ( -1    0    0   )
                  (  0    0    1   )

 cart.    s(12) = (  0.500 -0.866  0.000 )
                  ( -0.866 -0.500  0.000 )
                  (  0.000  0.000  1.000 )


     point group D_3h (-62m)
     there are  6 classes
     the character table:

       E     2C3   3C2   s_h   2S3   3s_v
A'_1   1.00  1.00  1.00  1.00  1.00  1.00
A'_2   1.00  1.00 -1.00  1.00  1.00 -1.00
E'     2.00 -1.00  0.00  2.00 -1.00  0.00
A''1   1.00  1.00  1.00 -1.00 -1.00 -1.00
A''2   1.00  1.00 -1.00 -1.00 -1.00  1.00
E''    2.00 -1.00  0.00 -2.00  1.00  0.00

     the symmetry operations in each class and the name of the first
element:

     E        1
          identity
     2C3      3    4
          120 deg rotation - cryst. axis [0,0,1]
     3C2      2    6    5
          180 deg rotation - cart. axis [0,1,0]
     s_h      7
          inv. 180 deg rotation - cart. axis [0,0,1]
     2S3      9   10
          inv.  60 deg rotation - cryst. axis [0,0,1]
     3s_v     8   11   12
          inv. 180 deg rotation - cart. axis [1,0,0]

     Space group identification,  12 symmetries:

     Bravais lattice   4  hexagonal
     Point group number  21 / 107  D_3h (-62m)

     Nonsymmorphic operations not found: All fractional translations vanish
     Symmetries of the point group in standard order

        1       E   1
        2     i6z  57
        3      3z  27
        4     i2z  34
        5     3-z  28
        6    i6-z  58
        7     i2x  36
        8    2210  30
        9   i2110  64
       10      2y   3
       11   i2010  63
       12   21-10  29


     Space group nymber  187

     Space group P-6m2   (group number 187).
     The origin coincides with the ITA tables.

     The Laue class is D_6h(6/mmm)

     In this class the elastic tensor is

     ( c11  c12  c13   .    .    . )
     ( c12  c11  c13   .    .    . )
     ( c13  c13  c33   .    .    . )
     (  .    .    .   c44   .    . )
     (  .    .    .    .   c44   . )
     (  .    .    .    .    .    X )
     X=(c11-c12)/2

     It requires three strains: e1, e3, and e4
     for a total of 12 scf calculations

*****************************************************************************************************

It will be great if you can help me in solving this problem,

Thanks,
Best regards,
Krishnendu

-- 
Dr. Krishnendu Mukherjee,

Principal Scientist,
CSIR-NML,
Jamshedpur.
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