Inserting this in (46) gives $$\begin{eqnarray} H_I |\alpha_1\alpha_2\dots\alpha_n\rangle &=& \sum_{\alpha_1', \alpha_2'} \langle \alpha_1'\alpha_2'|\hat{v}|\alpha_1\alpha_2\rangle a_{\alpha_1'}^{\dagger} a_{\alpha_2'}^{\dagger} a_{\alpha_2} a_{\alpha_1} |\alpha_1\alpha_2\dots\alpha_n\rangle \nonumber \\ &+& \dots \nonumber \\ &=& \sum_{\alpha_1', \alpha_n'} \langle \alpha_1'\alpha_n'|\hat{v}|\alpha_1\alpha_n\rangle a_{\alpha_1'}^{\dagger} a_{\alpha_n'}^{\dagger} a_{\alpha_n} a_{\alpha_1} |\alpha_1\alpha_2\dots\alpha_n\rangle \nonumber \\ &+& \dots \nonumber \\ &=& \sum_{\alpha_2', \alpha_n'} \langle \alpha_2'\alpha_n'|\hat{v}|\alpha_2\alpha_n\rangle a_{\alpha_2'}^{\dagger} a_{\alpha_n'}^{\dagger} a_{\alpha_n} a_{\alpha_2} |\alpha_1\alpha_2\dots\alpha_n\rangle \nonumber \\ &+& \dots \nonumber \\ &=& \sum_{\alpha, \beta, \gamma, \delta} ' \langle \alpha\beta|\hat{v}|\gamma\delta\rangle a^{\dagger}_\alpha a^{\dagger}_\beta a_\delta a_\gamma |\alpha_1\alpha_2\dots\alpha_n\rangle \tag{48} \end{eqnarray}$$