What's wrong with my attempt to test Brillouin's Theorem?

1

Dear Sirs,
I am trying to test the validity of Brillouin’s theorem by computing the coupling of the ground configuration <a,b| with the singly excited configuration, |r,b>, of LiH, i.e., <a,b|H|b,r> = [a|T + V|r] + Sum_b {[ar|bb] - [ab|br]. I first run a Hartree-Fock calculation on LiH [(1\sigma)^2 (2\sigma)^2) at the STO-3G basis set, and got the core Hamiltonian (T + V) and the matrix of the molecular-orbital coefficients, C. For testing purposes, I truncated C to the third column, containing the two occupied molecular orbitals and the lowest-lying virtual. I have first transformed the core Hamiltonian from the AO basis to the MO basis using the truncated C matrix, thus getting a 3x3 core Hamiltonian matrix. The matrix looks like the following:

[[-4.75302645 -0.11033346 -0.16833039]
[-0.11033346 -1.53858794 0.03604979]
[-0.16833039 0.03604979 -1.13369892]]

Then I have computed the two-electron integrals coupling the ground configuration with the 2\sigma → 3\sigma excitation:
<2\sigma b|| 3\sigma b> = \Sum_b [2\sigma 3\sigma|bb] - [2\sigma b|b 3\sigma], with b standing for the two paired electrons in 1\sigma and the uncoupled electron in 2\sigma. The one-electron integral analogue for the coupling of the ground configuration with the single excitation should be the [2,3] element of the core Hamiltonian matrix (0.03604979). However, upon computing the two-electron integrals using mints.mo_eri(2\sigma, 3\sigma,b,b) - mints.mo_eri(2\sigma,b,b, 3\sigma), I get a number which is negative but one order of magnitude smaller in modulus compared to the core Hamiltonian element, thus making <a,b|H|r,b> non-zero.
I cannot understand where the mistake is, as the coupling of the ground configuration with itself, \Sum_a [a|T+V|a] + 2J - K, yields the expected result. Is there perhaps some missing coefficient multiplying the J and K part of the two-electron integrals that arise when using a spatial-MO basis? I looked into several textbooks, but I could not find anything on that, except for the case of the Slater-Condon rule, case I, applied to a closed shell.
I am really stuck on that, so your suggestions and/or help would be greatly appreciated.

Thanks for your patience and understanding!
Best regards,
Giorgio Visentin

2
  1. Just so you know, you haven’t supplied enough information to reproduce your computation. You need a bond length for that.
  2. The most likely issue is that you did not spin-integrate correctly. I work in spin-orbital formalisms as a matter of habit. In that formalism, there are a total of 3 b. Two exchange terms vanish for a total of four integrals. Spin-integration makes two of them identical.
3

Thanks a lot for your reply! These are the charge, spin multiplicity and Z matrix of LiH:
0 1
Li -0.665300 -0.293129 -0.148122
H 0.665300 0.293129 0.148122

Moreover, I have just followed your advice. In the spin-orbital formalism, b should correspond to 1/sigma(down), 1\sigma(up) and 2/sigma(up) if the two configurations abovementioned differ for the 2\sigma(down) and 3/sigma(down) spin-MOs, where “down” and “up” refer to the spin orientation. For the sake of simplicity, I will hereinafter label the spin-MOs n\sigma(spin) as simply n(spin).
Spin integration should lead me to the following expression:
J = [2(down) 3(down) | 1(down) 1(down)] + [2(down) 3(down) | 1(up) 1(up)] + [2(down) 3(down) | 2(up) 2(up)];
K = [2(down) 1(down)| 1(down) 3(down)],
while [2(down) 1(up)| 1(up) 3(down)], [2(down) 2(up)| 2(up) 3(down)] vanish due to the zero-exchange among electrons with different spins.
Following your suggestions, J - K thus computed yields -0.03604938, which is, indeed, the inverse of the [2,3] element of the core Hamiltonian matrix.
Thanks for your patience and kindness.
Kind regards,
Giorgio Visentin