To see this, we look at the contributions arising from $$ \langle \Phi_H^P | = \langle \Phi_0| $$ in Eq. (1), that is we multiply with \( \langle \Phi_0 | \) from the left in $$ (\hat{H} -E)\sum_{P'H'}C_{H'}^{P'}|\Phi_{H'}^{P'} \rangle=0. $$ If we assume that we have a two-body operator at most, Slater's rule gives then an equation for the correlation energy in terms of \( C_i^a \) and \( C_{ij}^{ab} \) only. We get then $$ \langle \Phi_0 | \hat{H} -E| \Phi_0\rangle + \sum_{ai}\langle \Phi_0 | \hat{H} -E|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0, $$ or $$ E-E_0 =\Delta E=\sum_{ai}\langle \Phi_0 | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{abij}\langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}, $$ where the energy \( E_0 \) is the reference energy and \( \Delta E \) defines the so-called correlation energy. The single-particle basis functions could be the results of a Hartree-Fock calculation or just the eigenstates of the non-interacting part of the Hamiltonian.