# Brillouin's Theorem: How do other determinants mix with the ground state?

I am a bit confused about Brillouin's Theorem. I get that a singly excited determinant can not mix with the ground state wavefunction because that translates to an off-diagonal matrix element ($\langle \chi_a \chi_b|\mathcal H |\chi_r \chi_b \rangle$) and the Hartree-Fock eigenvalue problem requires off-diagonal elements to equal 0, but how about when a doubly excited determinant mixes with the ground state ($\langle \chi_a \chi_b|\mathcal H |\chi_r \chi_s \rangle$)? Both of them do not have $i=j$ thus both should be off-diagonal entries, yet a doubly excited determinant can mix with the ground state so what am I missing?


• Thanks for your answer but I still do not understand. I have used the Slater-Condon rules for the mixture between the HF reference determinant and the doubly excited determinant which yields <-ab||rs-> for the matrix element . But this is also an off-diagonal matrix element so why does it not equate to 0 as in the above case? I also know that the singly excited determinant and the doubly excited determinant can mix resulting in the matrix element <-Xb|f|Xs-> = <-Xb|h|Xs-> + sum_r <-rb || rs-> which also does not equate to 0 but it is an off-diagonal matrix element so it should equal 0 no? – Stuff Mar 5 '16 at 16:15
• Sorry, but it sounds like you have a mishmash in your head. :| Indeed, you get $\langle ab||rs \rangle$ for the case of reference and double excited determinants, but what do you mean by saying that this is an off-diagonal matrix element? To start from, which matrix you think $\langle ab||rs \rangle$ belongs to? – Wildcat Mar 5 '16 at 16:56
• No need to apologize I think you are right. I think <-ab||rs-> belongs to the HF matrix no? – Stuff Mar 5 '16 at 17:06
• There is no HF matrix. However, there is a Fock one. It is composed of the elements $F_{ij} = \langle i | \hat{F} | j \rangle$ and is indeed diagonal. Now look at $\langle a b || rs \rangle$. First, it certainly not an element of a Fock matrix: it has totally different form. Secondly, it is not even an element of a square (or, in general, rectangular, or two-dimensional) matrix! – Wildcat Mar 5 '16 at 17:35
• My another mistake in comments above is that $\langle \Phi_{a}^{r} | \hat{H} | \Phi_{bc}^{st} \rangle$ will actually vanish because there are more than two different orbitals in $\Phi_{a}^{r}$ and $\Phi_{bc}^{st}$. The non-zero Hamiltonian matrix element between singly and doubly excited determinant would be, say, $\langle \Phi_{a}^{r} | \hat{H} | \Phi_{ab}^{st} \rangle$ since these determinants differ by two orbitals. – Wildcat Mar 5 '16 at 18:26