When we set up Hartree-Fock equations for atoms we work with the following Hamiltonian:
$\displaystyle{H=\sum_{i,j}a_i^\dagger \langle i|T|j\rangle a_j+\sum_{i,j} a_i^\dagger \langle i|U|j\rangle a_j +\frac{1}{2} \sum_{i,j,k,m}\langle i,j|V|k,m\rangle a_i^\dagger a_j^\dagger a_m a_k}$
Where Schwabl (Advanced Quantum Mechanics) says that $U$ is the potential the electrons feel due to the nucleus and $V$ is the Coulomb interaction. And here my question arises because well, both $U$ and $V$ are described by Coulomb interaction so, why $U$ term (electron-nucleus interaction) is treated as a single-particle operator and not as a two-particle operator as the case of $V$?
