With the definitions $$ \mathbf{K}=\sum_{i=1}^A\mathbf{k}_i, $$ and $$ \mathbf{k}_{ij}=\frac{1}{2}(\mathbf{k}_i-\mathbf{k}_j). $$ we can rewrite the two-particle kinetic energy (note that we use \( \hbar=c=1 \) as $$ \frac{\mathbf{k}_1^2}{2m_n}+\frac{\mathbf{k}_2^2}{2m_n}=\frac{\mathbf{k}^2}{m_n}+\frac{\mathbf{K}^2}{4m_n}, $$ where \( m_n \) is the average of the proton and the neutron masses.