Hello Patrizio
yes it is correct for F.C.C. lattice you need to specify ibrav=2. An
celldm(1) is the cube's edge length.
As a rule the celldm and also the A,B,C and cosAB, cosBC, cosAC are
always those of the "conventional" l cell. It means the cube for cubic
lattices , the tetragonal paralleliped for the
tetragonal lattices and so on. See also the documentation on ibrav
https://www.quantum-espresso.org/Doc/INPUT_PW.html#idm199
regards - Pietro
On 11/23/21 3:08 PM, [email protected] wrote:
Hello everybody,
I'd like to have from you a proof/disproof of my understanding on the
use of celldm for F space groups.
I read in the instructions that QE uses the primitive cell, in the
case of Silicon I mean for that the "slanted" cell having parameter
around 0.387 nm and 60° angles, say like here
https://www.iue.tuwien.ac.at/phd/ungersboeck/node27.html .
However, in an example for silicon, here
https://www.quantum-espresso.org/resources/tutorials/shanghai-2013/getting-started/lecture1.pdf
, I see that in the case of silicon with ibrav = 2 the entered celldm
correspond to 0.543 nm, the side of the "cubic cell".
Thus, if I'm working with a F space group, ibrav = 2, I must use the
cube side as celldm, ~ 0.54 nm for silicon. Is this correct?
If I'd like to work with the "slanted primitive cell" with angles of
60°, which in silicon has lattice constant ~ 0.387 nm, how can I do? I
apologize but I couldn't find this in any tutorial I found.
Thanks for your time,
Patrizio
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