Brillouin's Theorem: How do other determinants mix with the ground state?
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
$$
  \newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
  \newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}
  \newcommand{\op}[1]{\hat{#1}}
  \newcommand{\el}{_{\mathrm{e}}}
  \newcommand{\core}{^{\mathrm{core}}}
  \newcommand{\oneeintphys}[3]{\bracket{#1}{#2}{#3}}
  \newcommand{\twoeintphysrantisymm}[2]{\langle{#1}\,||\,{#2}\rangle}
  \bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }
  =
  \oneeintphys{a}{\op{H}\core}{r} + \sum\limits_{j=1}^{n} \twoeintphysrantisymm{aj}{rj}
  =
  \bracket{ \psi_{a} }{ \op{F} }{ \psi_{r} } \, .
$$
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.