The translationally invariant one- and two-body Hamiltonian reads for an A-nucleon system, $$ \tag{5} \hat{H}=\left[\sum_{i=1}^A\frac{\boldsymbol{p}_i^2}{2m} -\frac{\boldsymbol{P}^2}{2mA}\right] +\sum_{i < j}^A V_{ij} \; , $$ where \( V_{ij} \) is the nucleon-nucleon interaction. Adding zero as here $$ \sum_{i=1}^A\frac{1}{2}m\omega^2\boldsymbol{r}_i^2- \frac{m\omega^2}{2A}\left[\boldsymbol{R}^2+\sum_{i < j}(\boldsymbol{r}_i-\boldsymbol{r}_j)^2\right]=0. $$ we can then rewrite the Hamiltonian as $$ \hat{H}=\sum_{i=1}^A \left[ \frac{\boldsymbol{p}_i^2}{2m} +\frac{1}{2}m\omega^2 \boldsymbol{r}^2_i \right] + \sum_{i < j}^A \left[ V_{ij}-\frac{m\omega^2}{2A} (\boldsymbol{r}_i-\boldsymbol{r}_j)^2 \right]-H_{\mathrm{CoM}}. $$