Obtaining the Slater determinant of a HF calculation

Hi,

I was wondering whether there is an option to retrieve the matrix of molecular orbitals (which is then reduced by a determinant). From my understanding, I should be able to retrieve a NxN matrix for a specific N-electron system.

I thought the following code would result in the molecular orbital matrix of oxygen with a random electron configuration:

import numpy as np
from pyscf import gto, scf

mol = gto.Mole(atom=[
  ['O', (0, 0, 0)],
], basis='sto-6g', unit='bohr', spin=0, charge=0)
mol.build()

mean_field = scf.RHF(mol)
mean_field.kernel()

electrons = np.random.randn(8, 3)

ao = mol.eval_gto('GTOval_sph', electrons)
mo = ao @ mean_field.mo_coeff
print(mo.shape)

Result:

(8, 5)

However, isn’t the matrix supposed to be a NxN matrix? I don’t quite understand where the 5 comes from.

Thank you!

[tberkel]

He doesn’t want the matrix of MOs coefficients in the basis of AOs, but rather the matrix (at a specified set of electron coordinates) whose determinant gives the value of the N-electron wavefunction, i.e. the matrix here: https://wikimedia.org/api/rest_v1/media/math/render/svg/343779d82465137a7292f331e1e8f178307ca60d

On Tue, May 18, 2021 at 6:18 PM Susi Lehtola @.***> wrote:

import numpy as np from pyscf import gto, scf

mol = gto.Mole(atom=[ [‘O’, (0, 0, 0)], ], basis=‘sto-6g’, unit=‘bohr’, spin=0, charge=0) mol.build()

mean_field = scf.RHF(mol) mean_field.kernel() print(mean_field.mo_coeff.shape)

mo_coeff is a 5x5 matrix.

STO-6G is 2s1p for oxygen. This gives 5 spatial orbitals, or 10 spin-orbitals. Oxygen has 8 electrons; this means that you have 2 spin-orbitals that will be unoccupied; however, the 8 occupied spin-orbitals (4 spin-up and 4 spin-down in the present calculation) are each expanded in terms of the 5 basis functions.

Minimal basis sets are unreliable even at the Hartree-Fock level of theory.

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[tberkel]

I think you don’t want the np.concatenate() step. For the wavefunction, you could use the product of the determinants of spin up and spin down. So you could make an (N/2)x(N/2) matrix for spin up and another one for spin down.

On Thu, May 20, 2021 at 11:12 AM Susi Lehtola @.***> wrote:

Well the spin functions are missing from the matrix. I’m not sure how they should be evaluated.

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