 Solution

Within a fixed isolated electronic $J$ multiplet, the electronic field at the nucleus is a vector operator. Rotational symmetry and the Wigner-Eckart theorem imply that its projection into this multiplet is proportional to $\mathbf{J}$:Its detailed orbital, spin-dipolar and contact contributions are included in the reduced coefficient $b_J$. Consequently the projected magnetic-dipole hyperfine interaction becomesin the same dimensionless angular-momentum convention as a. The proportionality is a statement about matrix elements within the chosen electronic manifold, not an identity for the electron field operator on the entire atomic Hilbert space.

Define $\mathbf{F}=\mathbf{I}+\mathbf{J}$. In zero external field the quantum states obtained by addition of angular momentum are labelled $|\gamma J,I;F,m_F⟩$, where $\gamma$ specifies the electron configuration and other fixed labels. The good quantum numbers include $I,J,F,m_F$, with $F=|I-J|,\dots ,I+J$ and $m_F=-F,\dots ,F$. In a description using LS coupling the fixed electronic labels can also include $L,S$. The energies depend on $F$ but not $m_F$:This neglects coupling to other electronic manifolds, electric-quadrupole hyperfine interactions and the Zeeman effect from external fields. Those additions can shift the intervals or change which quantum numbers remain useful.