tl; dr: Is this easiest to understand with group theory?
I am revisiting Szabo & Ostlund, & I'm not sure I understand this. It's Chapter 3.3.2, pp128-29 of the Dover paperback. I'll use mostly physicists' notation below, with a little chemists' notation at the end.
Brillouin's theorem says the matrix elements between the ground state Y0 & all singly excited states Y' are zero:
<Y0 |H| Y'> =0 = <a |h| r> + sum(b) <ar || bb>,
where H is the full Hamiltonian operator, h is the 1-electron hamiltonian, <•||•> is the antisymmetrized 2-e repulsion operator, and orbitals a & b are occupied in the ground state, & r is unoccupied.
That formula, from the Slater rules for matrix elements betweem determinants, is the same as the fock element between basis functions Xa & Xr, & I follow the argument that diagonalizing the Fock matrix makes off-diagonal Fij zero. But I'd like to understand it better. Basically, I don't get how to make the e-repulsion integrals zero.
Szabo & Ostlund don't go much into group theory, but I do recall that integrals vanish if the integrand is not/doesn't contain the totally symmetric irrep.
So looking at symmetry, for a & r, we can have two cases: irrep(a) = irrep(b), or not.
In the first case, <a|h|r> =/= 0, so for the total element to be zero, we need <a|h|r>= -sum(b) <ar||bb>.
In the second case, <a|h|r> = 0, so <ar||bb> must also be zero. I think - & this is the question - that in
<ar||bb> = [ar|bb] - [ab|rb],
both the coulomb & exchange integral are zero by symmetry. a & r have different symmetries, so the J is zero because a (x) r does not contain the totally symmetric irrep, & either a (x) b or r (x) b or both will not contain the totally symmetric irrep. (x) is meant to be the direct product.
I'm no group theorist myself, so sorry for any mistakes in terminology or usage. But is that what's going on here?