The two-particle operator in the particle-hole formalism is more complicated since we have to translate four indices \( \alpha\beta\gamma\delta \) to the possible combinations of particle and hole states. When performing the commutator algebra we can regroup the operator in five different terms $$ \begin{equation} \hat{H}_I = \hat{H}_I^{(a)} + \hat{H}_I^{(b)} + \hat{H}_I^{(c)} + \hat{H}_I^{(d)} + \hat{H}_I^{(e)} \tag{79} \end{equation} $$ Using anti-symmetrized matrix elements, bthe term \( \hat{H}_I^{(a)} \) is $$ \begin{equation} \hat{H}_I^{(a)} = \frac{1}{4} \sum_{abcd} \langle ab|\hat{V}|cd\rangle b_a^\dagger b_b^\dagger b_d b_c \tag{80} \end{equation} $$