The one-body operators ${\cal O}(\lambda)$ represent a sum over the operators for the individual nucleon degrees of freedom $i$ $${\cal O}(\lambda ) = \sum_{i} O(\lambda ,i).$$

The electric transition operator is given by $$O(E\lambda ) = r^{\lambda } \; Y^{\lambda }_{\mu }(\hat{r}) \; e_{q} e,$$ were $Y^{\lambda }_{\mu }$ are the spherical harmonics and $q$ stands for proton $q=p$ or neutron $q=n$.