A non-practical way of solving the eigenvalue problem
In our notes on Hartree-Fock calculations,
we have already computed the matrix \( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle \) and \( \langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab}\rangle \). If we are using a Hartree-Fock basis, then the matrix elements
\( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle=0 \) and we are left with a
correlation energy given by
$$
E-E_0 =\Delta E^{HF}=\sum_{abij}\langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}.
$$