As a test, perform a three-dimensional calculation of a hydrogen atom
as follows.
- Using S=[64;64;64], R=diag([16; 16; 16])), X=[0 0 0], Z=1 in ``setup.m''
and sigma1=0.25 in ``ewald.m'', compute the Ewald energy of a single proton
in a cell of size (16 bohr)
. You should find a value for the Ewald
energy of -0.0885440851224253 H.
- Then, run ``Hatoms.m''
for a single state with one electron (gbl_f=1, Ns=1) using the same
parameters in ``setup.m'' as you used for the ewald calculation. You should
find a value (to within about 0.001 H due to possibly different levels of convergence
between you and me) for the electronic energy of -0.356725655980680 H.
When your code functions properly (and you add the Ewald energy), you
should find a total energy within 1 millihartree of the ``official'' LDA
total energy published on the NIST web site1for hydrogen, -0.445671 H.
Tomas Arias
2007-04-26