# Why volume of unit cell is not the same as cubing the lattice constant?

I have a cubic-FCC with a=b=c and $$\alpha$$=$$\beta$$=$$\gamma$$=$$90^o$$. As far as I understand the volume of this cubic unit cell should be a cube of the lattice constant. But this not the case I am seeing in the quantum ESPRESSO, SCF output file:

 bravais-lattice index     =            2
lattice parameter (alat)  =      10.9569  a.u.
unit-cell volume          =     328.8522 (a.u.)^3
number of atoms/cell      =            4
number of atomic types    =            3
number of electrons       =        60.00
number of Kohn-Sham states=           36
kinetic-energy cutoff     =     200.0000  Ry
charge density cutoff     =    2000.0000  Ry


As you can see that unit cell volume is not the cube of the lattice parameter. Why is it so? Am I putting any parameter wrongly?

Thank you!

• I'll leave it to somebody who knows these things properly to answer, but it looks like the volume quoted is for the primitive unit cell, while the lattice constant is for the conventional cell, which is 4 times the size size of the primitve one for a FCC lattice. See en.wikipedia.org/wiki/Unit_cell for an explanation of the difference. May 21, 2021 at 8:11
• @Ian Bush, Could you please confirm if the quantum ESPRESSO gives total energy for the primitive cell and not for the conventional cell at the end of the scf cycle?
– UJM
May 21, 2021 at 12:38
• No idea, I don't use Quantum Espresso, but the documentation should tell you that May 21, 2021 at 13:32
• @UjjawalM. If you use the primitive cell in your input file, QE will give the total energy of the primitive cell at the end of the SCF cycle. On the other hand, if you use the conventional cell in your input file it will give the total energy of the conventional cell. Remember that the total energy is meaningless in the case of QE as it uses the pseudopotential method. Only energy differences are meaningful. Jun 1, 2021 at 9:45

The cell vectors for the primitive cell corresponding to the cubic face-centered Bravais lattice (ibrav=2) are
The cell volume is $$V=|v_1 \times v_2 \cdot v_3| = (a/2)^32=328.85\ \mathrm{a.u.}^3$$