Slater-Koster Table for CH

[Anshuman5]

Dear pekkosk,
The Slater-Koster file for CH_parameterization in the example is not symmetry ( C-H table is not the same as H-C table ). If we compare this with mio sk CH file C-H table is same as the H-C table.
I am wondering why is this the case for Hotbit. We also modified the def select_orbitals in slako.py as this was not permuting the orbitals correctly for Hamiltonian integral(this is what we think). We modified the select_orbitals function to fetch all the orbitals for C-H and H-C and then obtain the C-H_no_repulsion.par file.

def select_orbitals(val1,val2,integral):
"""
Select orbitals from given valences to calculate given integral.
e.g. ['2s','2p'],['4s','3d'],'sds' --> '2s' & '3d'
"""
nl1=None
#for nl in val1:
# if nl[1]==integral[0]: nl1=nl
nl2=None
#for nl in val2:
# if nl[1]==integral[1]: nl2=nl
for pair in ((a,b) for a in val1 for b in val2):
x = pair[0]
y = pair[1]
if x[1]==integral[0] and y[1]==integral[1]:
nl1 = x
nl2 = y
if x[1]==integral[1] and y[1]==integral[0]:
nl1 = x
nl2 = y
print (nl1,nl2)
return nl1,nl2

Best Regards,
Anshuman
C_H_norepulsion_par.zip

[pekkosk]

KS-tables are so that CH-table means <C-orbitals|operator|H-orbitals> and HC-table means <H-orbitals|operator|C-orbital>. For example, H_sp-sigma-tables then are not symmetric, because H has no p-orbitals. H_ss, H_pp, and H_dd tables, on the other hand, should be symmetric.

[Anshuman5]

Thanks for clarification. So to match mio-1-1 sk file: C-H.skf and H-C.skf (which are identical file), do we need to copy H-C.skf to C-H.skf ?
I am baffled when I compare the C-H.skf with H-C.skf in the mio library. Why are they exactly the same?

[pekkosk]

That's a good question; I don't know. Perhaps you should take a look at the exact description of .skf file format.