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 -- Patrizio Graziosi, PhD Research Scientist CNR - ISMN Institute for the Study of Nanostructured Materials _______________________________________________ Quantum ESPRESSO is supported by MaX (www.max-centre.eu) users mailing list [email protected] https://lists.quantum-espresso.org/mailman/listinfo/users
