It is however, from a computational point of view, convenient to introduce an external potential \( \hat{u}_{\mathrm{ext}}(x_i) \) by adding and substracting it to the original Hamiltonian. This means that our Hamiltonian can be rewritten as $$ \hat{H} = \hat{H}_0 + \hat{H}_I = \sum_{i=1}^A \hat{h}_0(x_i) + \sum_{i < j=1}^A \hat{v}(x_{ij})-\sum_{i=1}^A\hat{u}_{\mathrm{ext}}(x_i), $$ with $$ \hat{H}_0=\sum_{i=1}^A \hat{h}_0(x_i) = \sum_{i=1}^A\left(\hat{t}(x_i) + \hat{u}_{\mathrm{ext}}(x_i)\right). $$

The interaction (or potential energy term) reads now $$ \hat{H}_I= \sum_{i < j=1}^A \hat{v}(x_{ij})-\sum_{i=1}^A\hat{u}_{\mathrm{ext}}(x_i). $$

In nuclear physics the one-body part \( u_{\mathrm{ext}}(x_i) \) is often approximated by a harmonic oscillator potential or a Woods-Saxon potential. However, this is not fully correct, because as we have discussed, nuclei are self-bound systems and there is no external confining potential. As we will see later, the \( \hat{H}_0 \) part of the hamiltonian cannot be used to compute the binding energy of a nucleus since it is not based on a model for the nuclear forces. That is, the binding energy is not the sum of the individual single-particle energies.