Question in code performance

[Kee-Wang]

This might be a simple question. Why here we do not use the I_AO in symmetry orbital basis which has smaller dimension? Is it because there is no way to get symmetry orbital version of ERI tensor?

psi4numpy/Self-Consistent-Field/RHF_symmetry.py

Line 178 in 0d2b091

J_ = transform_aotoso(np.einsum("mnls,ls->mn", I_AO, D_AO), transformers_)

[dgasmith]

You can obtain a SO version of the ERI; however, the symmetry rules become much more complex. Interestingly it's often not favorable to transform ERIs in SCF settings as it is more costly to perform the transform than using the AO basis. Naively the AO -> SO ERI transform is N^5 and prevents a number of sparsity considerations. Psi4 for example only considers transforms to the SO basis for correlated methods such as CC.

As Psi4NumPy isn't about performance we would definitely consider a PR which uses the SO basis for ERIs!

[Kee-Wang]

Ah ,that makes sense.
Maybe one more follow up question (or we can move this topic to psi4 forum if you think that's a better place). I was trying to get the ERI under the SO basis just to make sure I understand the conversion behind it. However, I did not find a psi4 function for that. My goal would be to get the J_SO and K_SO using I_SO and D_SO for each sub-symmetry space instead of from I_AO and D_AO as did in the code above. Also I want to get the second quantized Hamiltonian under symmetry, so I_SO is essential to get AO->SO-> MO for each sub-symmetry.

Is there an example of how to do that? I tried to do the following basis transformation which works for AO->MO, but it seems does not do the trick for the AO->SO.
I_SO = [np.einsum("abcd,aA,bB,cC,dD",I_AO, transformer) for transformer in transformers_]