 Solution

Take dimensionless spin operators and $H=D{S}_z^2+g{\mu }_{B}B{S}_z$, with $g>0$. The four wavefunctions are $|m⟩$, $m=-3/2,-1/2,1/2,3/2$, withAt zero field they form doublets at $D/4$ and $9D/4$. A transverse microwave magnetic field drives $\Delta m=\pm 1$ transitions. When $h\nu >2D$, their positive resonance fields in ascending order areThey connect $(1/2,3/2)$, $(-1/2,1/2)$, and $(-3/2,-1/2)$, respectively. ThusHere $D/h=4.25\mathrm{GHz}$ and the zero-field doublet gap is $2D$.

At lower frequency it is essential to use absolute energy gaps, not discard a transition after its levels cross. For $0<h{\nu }^{\prime }<2D$, the distinct candidate positive fields areThe first two coincide whenThis axial spin-three-halves resonance coincidence leaves two distinct resonance positions, at $0.1275$ and $0.3825\mathrm{T}$, with two transitions superposed at the lower position. At ${\nu }^{\prime }=2D/h$ there are also only two strictly positive-field lines, but an additional zero-field resonance remains; this is a different counting convention. Observable intensities also require thermal occupation of the relevant lower levels.