Changing the summation indices \( \alpha \) and \( \beta \) in (51) we obtain $$ \begin{equation} \sum_{\alpha\beta} \langle \alpha\beta|\hat{v}|\gamma\delta\rangle a^{\dagger}_\alpha a^{\dagger}_\beta a_\delta a_\gamma = \sum_{\alpha\beta} \langle \alpha\beta|\hat{v}|\delta\gamma\rangle a^{\dagger}_\alpha a^{\dagger}_\beta a_\gamma a_\delta \tag{52} \end{equation} $$ From this it follows that the restriction on the summation over \( \gamma \) and \( \delta \) can be removed if we multiply with a factor \( \frac{1}{2} \), resulting in $$ \begin{equation} \hat{H}_I = \frac{1}{2} \sum_{\alpha\beta\gamma\delta} \langle \alpha\beta|\hat{v}|\gamma\delta\rangle a^{\dagger}_\alpha a^{\dagger}_\beta a_\delta a_\gamma \tag{53} \end{equation} $$ where we sum freely over all single-particle states \( \alpha \), \( \beta \), \( \gamma \) og \( \delta \).