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?

Brillouin's theorem is the direct consequence of the so-called Slater–Condon rules which for the case of the HF reference determinant $\Phi_{0}$ and a singly excited determinant $\Phi_{a}^{r}$ show that the Hamiltonian matrix element between these determinants reduces to the Fock matrix element between the corresponding orbitals $$\newcommand{\braket}{\langle{#1}\vert{#2}\rangle} \newcommand{\bracket}{\langle{#1}\vert{#2}\vert{#3}\rangle} \newcommand{\op}{\hat{#1}} \newcommand{\el}{_{\mathrm{e}}} \newcommand{\core}{^{\mathrm{core}}} \newcommand{\oneeintphys}{\bracket{#1}{#2}{#3}} \newcommand{\twoeintphysrantisymm}{\langle{#1}\,||\,{#2}\rangle} \bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} } = \oneeintphys{a}{\op{H}\core}{r} + \sum\limits_{j=1}^{n} \twoeintphysrantisymm{aj}{rj} = \bracket{ \psi_{a} }{ \op{F} }{ \psi_{r} } \, .$$ Now taking the HF equations and orthogonality of spin orbitals into account we indeed get zero for $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }$ since $$\bracket{ \psi_{a} }{ \op{F} }{ \psi_{r} } = \varepsilon_{r} \braket{ \psi_{a} }{ \psi_{r} } = 0 \, .$$ Now try to use the Slater–Condon rules for the case of the HF reference determinant $\Phi_{0}$ and a double excited determinant $\Phi_{ab}^{rs}$ to convince yourself that $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{ab}^{rs} }$ unlike $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }$ won't vanish. This is a very well-known result which manifests itself in many different ways. For instance, this is the reason why in order to improve the HF energies in configuration interaction scheme we include doubly excited determinants first (CID) rather than singly excited once (CIS): the later won't have any effect on electronic energy.
• 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? Mar 5 '16 at 16:56
• 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! 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. Mar 5 '16 at 18:26