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I am performing vc relax calculations to find the effect of pressure on the cell. Both tensile and compression pressure. But I am confused about the lattice parameters it is giving me. Its a cubic FCC structure with ibrav = 2

CELL_PARAMETERS (alat= 11.14000000)
  -0.499839154   0.000000000   0.499839154
   0.000000000   0.499839154   0.499839154
  -0.499839154   0.499839154   0.000000000
these are at 0 pressure
CELL_PARAMETERS (alat= 11.14000000)
  -0.498989421   0.000000000   0.498989421
   0.000000000   0.498989421   0.498989421
  -0.498989421   0.498989421   0.000000000
these are at 10 Kbar pressure

I wanted to clarify, does the 0 pressure give the actual cell parameters? Or are they different than what you get from volume optimization?

Are the parameters = 11.14*(0.499839154)? I have to do further calculations after this to see their effects on properties.

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    $\begingroup$ alat is in bohr (11.14*0.499839154/0.5) that is 11.13641 bohr or 5.893 Angstrom. It will be better if you share your input file. $\endgroup$
    – Pranav kumar
    Commented Jan 1, 2022 at 11:31
  • $\begingroup$ @pranavkumar What you calculated is correct with lattice parameter of the structure. please help me understand why did you divide it with 0.5? that conversion part which changes wrt to different lattices is still confusing to me (previously done on hexagonal structure too). Feel free to ask me any input parameter you want to know. As above i got parameters at different pressures, so I have to use these in further calculations to check change in properties. $\endgroup$
    – epsilon02fft
    Commented Jan 1, 2022 at 13:45

1 Answer 1

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From documentation

ibrav      structure                   celldm(2)-celldm(6)
                                     or: b,c,cosbc,cosac,cosab
  0          free
      crystal axis provided in input: see card CELL_PARAMETERS

  1          cubic P (sc)
      v1 = a(1,0,0),  v2 = a(0,1,0),  v3 = a(0,0,1)

  2          cubic F (fcc)
      v1 = (a/2)(-1,0,1),  v2 = (a/2)(0,1,1), v3 = (a/2)(-1,1,0)

  3          cubic I (bcc)
      v1 = (a/2)(1,1,1),  v2 = (a/2)(-1,1,1),  v3 = (a/2)(-1,-1,1)
 -3          cubic I (bcc), more symmetric axis:
      v1 = (a/2)(-1,1,1), v2 = (a/2)(1,-1,1),  v3 = (a/2)(1,1,-1)

  4          Hexagonal and Trigonal P        celldm(3)=c/a
      v1 = a(1,0,0),  v2 = a(-1/2,sqrt(3)/2,0),  v3 = a(0,0,c/a)

When ibrav=2 and celldm(1)=11.44 is chosen

Following size of simulation box

-0.5 0.0 0.5
 0.0 0.5 0.5
-0.5 0.5 0.0

Here celldm(1) act as scaling factor which changes simulation box. After relaxation scaling parameter remain same where as simulation cell changed to

CELL_PARAMETERS (alat= 11.14000000)
  -0.499839154   0.000000000   0.499839154
   0.000000000   0.499839154   0.499839154
  -0.499839154   0.499839154   0.000000000

Hence you have to find cell change that happen relative to 0.5 of original cell vector
Hence output 11.14x0.499839154/0.5=11.1364163511 bohr

You can redo whole exercise instead of ibrav=2, use ibrav=0 and define CELL_PARAMETERS along with ATOMIC_POSITIONS

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  • $\begingroup$ Thank you so much. I am still not well versed in understanding all this in this way. I will do the calculations again to see it. Where can i read about simulation box and scaling factor? the only thing i have read it the quantum espresso documentation. by using ibrav = 0 mean that it is necessary to use cell_parameters? $\endgroup$
    – epsilon02fft
    Commented Jan 1, 2022 at 15:07
  • $\begingroup$ Pranav might be able to give brief answers in a comment, but in general, each question should be posted separately so that everyone can see it and you can get a quicker answer. $\endgroup$
    – Nike Dattani - No Free Time
    Commented Jan 1, 2022 at 17:52
  • $\begingroup$ just read about primitive cell vector and conventional cell vectors $\endgroup$
    – Pranav kumar
    Commented Jan 2, 2022 at 5:16
  • $\begingroup$ @NikeDattani ok Sir. Any further questions I will ask in new question post. Thank you. $\endgroup$
    – epsilon02fft
    Commented Jan 2, 2022 at 8:16
  • $\begingroup$ @pranavkumar Ok I will read and try to understand further. $\endgroup$
    – epsilon02fft
    Commented Jan 2, 2022 at 8:17

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