 Solution

For a nucleus of spin $I>0$ and electronic $J=1/2$, the magnetic-dipole hyperfine interaction is $A\mathbf{I}\cdot \mathbf{J}$, using dimensionless angular-momentum operators. The electric-quadrupole hyperfine contribution vanishes for $J=1/2$. In zero field, $F=I\pm 1/2$ labels the levels:Their signed separation is $\Delta =A(I+1/2)$. For $A>0$, the larger-$F$ multiplet lies higher; negative $A$ reverses this order. If $I=0$ there is no such hyperfine splitting.

Let the field define $z$ and writeHere $g_I$ is the signed nuclear Zeeman coefficient expressed in Bohr-magneton units, including its nuclear-to-Bohr-magneton factor; it is not an unconverted nuclear $g$ factor. Total projection $m=m_I+m_J$ is conserved, while $F$ need not be. In the basis $|m-1/2,+1/2⟩$, $|m+1/2,-1/2⟩$, the two-state block isFor $A>0$, define $x=(g_J-g_I){\mu }_{B}B/\Delta$. Diagonalization gives the Breit-Rabi formula,For either sign of $A$, the block itself remains valid; its ordered eigenvalues are $-A/4+g_I{\mu }_{B}Bm\pm \frac{1}{2}\sqrt{{\Delta }^2+2Am(g_J-g_I){\mu }_{B}B+(g_J-g_I{)}^2{\mu }_B^2{B}^2}$.

For a stretched hyperfine Zeeman state at projection $m=\pm (I+1/2)$ there is only one state each. Their energies are exactlyAssigning two branches to each stretched projection would double-count states.

In weak fields the shift is $g_F{\mu }_{B}mB$, with $g_F=g_I\pm (g_J-g_I)/(2I+1)$ for $F=I\pm 1/2>0$; an $F=0$ state has no first-order Zeeman shift. At strong hyperfine-decoupling fields the eigenstates approach product states withThis is the hyperfine regime of the Paschen-Back effect, still assuming the field is weak enough not to invalidate the electronic $J=1/2$ manifold. States of the same $m$ mix and avoid crossing when their off-diagonal coupling is nonzero; different-$m$ levels can cross.