Hamiltonian matrix without the bit representation, one and two-body operators

For two determinants which differ only by the substitution of single-particle states \( i \) with a single-particle state \( j \): $$ \begin{eqnarray} \langle SD | {\cal O}_1 | SD_i^j \rangle & = & \langle \phi_i | {\cal O}_1 | \phi_j \rangle \\ \langle SD | {\cal O}_2 | SD_i^j \rangle & = & \sum_{k \in SD} \langle \phi_i \phi_k | {\cal O}_2 | \phi_j \phi_k \rangle - \langle \phi_i \phi_k | {\cal O}_2 | \phi_k \phi_j \rangle \nonumber \end{eqnarray} $$ For two determinants which differ by two single-particle states $$ \begin{eqnarray} \langle SD | {\cal O}_1 | SD_{ik}^{jl} \rangle & = & 0 \\ \langle SD | {\cal O}_2 | SD_{ik}^{jl} \rangle & = & \langle \phi_i \phi_k | {\cal O}_2 | \phi_j \phi_l \rangle - \langle \phi_i \phi_k | {\cal O}_2 | \phi_l \phi_j \rangle \nonumber \end{eqnarray} $$ All other matrix elements involving determinants with more than two substitutions are zero.