I am trying to follow this tutorial from the FLEUR DFT code's official website. The tutorial goes over calculating the ground state energies for the ferromagnetic and antiferromagnetic configurations of fcc Fe. I understand that one needs to have at least two atoms in a primitive cell to have an antiferromagnetic order (for example, for the case of bcc Fe, one converts one atom-bcc to two atom simple cubit lattice structure).

However, here looks like they are converting a two-atom fcc unit cell into a tetragonal cell (Fleur manual says tp refers to simple-tetragonal). Unfortunately I'm not able to see how this can be accomplished. For convenience, I'm attaching the part of the code where they write this.

this image

Any help in understanding how this can be achieved would be greatly appreciated.


1 Answer 1


I found the answer for this (in the form of a picture, from some textbook or other) on Chegg, of all places. Consider two conventional fcc unit cells (the actual cubes, not the Wigner–Seitz cells) side to side. One primitive cell for the two-cell system is tetragonal: it’s turned 45 degrees along the vertical axis and centered on the central atom, while keeping the same vertical height as either of the two original cells.

Answer #11 to this FLEUR forum post clarifies the proper size:

I think if you want to have a 2-atom fcc unit cell, the shape would typically be a tetragonal unit cell ('tP'). The lattice parameter 'a' would be the distance between an atom at a corner and the nearest atom at a face center in a conventional fcc unit cell. The lattice parameter 'c' would be the lattice constant of the fcc unit cell.

In the example you provided, you’ll find the lovely relation $a = c /\sqrt{2}$, as predicted by the answerer above.

  • 2
    $\begingroup$ Wonderful! Thank you so much! $\endgroup$
    – rahman62
    Commented Oct 12, 2023 at 10:44
  • $\begingroup$ If anybody knows the actual textbook from which the Chegg solution (to which we only need the statement, haha) is taken, I'd be happy to edit my answer with the better reference! $\endgroup$
    – elutionary
    Commented Oct 13, 2023 at 12:26

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