Use Minkowski reduced unit cell

[legoses]

I am investigating the affect of using a non-reduced unit cell. When I say non-reduced I mean the lattice vectors have not been reduced to the shortest possible choices. Algorithms suggested by Minkowski and others give an easy way to accomplish this.

i want to know how bad the error in the total energy is if I use a unit cell that is not reduced. So I chose an fcc crystal and used the lattice vectors:

.5 1.0 .5
0 .5 .5
.5 0 .5

instead of the common choice

.5 .5 0
0 .5 .5
.5 0 .5

Indeed the energy per atom was off by ~800 meV. In the calculation I also notice the following warning in the OUTCAR file:

WARNING: grid for Broyden might be to small

Then I increased NGX by a factor of 1.3 or something and I notice that the energy is closer to the correct value. It appears that the error is associated with the FFT grid, but I can't see how.

From my understanding, NGX,Y,Z help define the density of the FFT grid by dividing the lattice vectors up by NGX, NGY and NGZ. However, if you draw out several choices of lattice vectors for a given lattice and then divide all of those choices up you will soon start to notice that it doesn't matter what your choice of lattice vectors are... the FFT density is the same in all cases.

Anyway, my question comes down to this: How does choosing a different set of lattice vectors affect the calculation in VASP? Any insight would be greatly appreciated.

[legoses]

Just another note. I am not looking to be convinced that the cell shape matters... I know that it does. I just need help understanding the details of why it matters.

[admin]

please check how the reciprocal basis set vectors are constructed by vasp in both cases, and whether the PW-basis sets are really equivalent (NPLWV)