= Solution
Use the uncoupled <angular momentum eigenstates> $|m_L,\uparrow\rangle$ and $|m_L,\downarrow\rangle$, with $m_L=-1,0,1$. For the <Hamiltonian operator> actually written in the paper, these are the <energy eigenstates>, with
$$
E=\mu_BB_z(m_L+2m_S),\qquad m_S=\pm\frac12.
$$
The initial $J=3/2$ states are not all <energy eigenstates> of this operator. To find their energy populations, first perform the <Clebsch-Gordan decomposition>. The extremal states are $|3/2,3/2\rangle=|1,\uparrow\rangle$ and $|3/2,-3/2\rangle=|-1,\downarrow\rangle$. Applying the total <angular momentum lowering operator> to the first gives
$$
\sqrt3\,|3/2,1/2\rangle=\sqrt2\,|0,\uparrow\rangle+|1,\downarrow\rangle.
$$
Similarly,
$$
|3/2,-1/2\rangle=\sqrt{\frac23}|0,\downarrow\rangle+\sqrt{\frac13}|-1,\uparrow\rangle.
$$
The initial preparation is an equal incoherent mixture of the four coupled states. Each extremal state contributes population $1/4$ to its one <energy eigenvalue>, while each nonextremal state contributes $2/3$ and $1/3$ of its own $1/4$ population to the two displayed components. The resulting <Zeeman splitting> is therefore
$$
\boxed{\begin{array}{c|ccccc}
E/(\mu_BB_z)&-2&-1&0&1&2\\ \hline
\text{fraction of atoms}&1/4&1/6&1/6&1/6&1/4\\
\text{energy multiplicity}&1&1&2&1&1
\end{array}}
$$
for nonzero $B_z$. The relative populations are $\boxed{3:2:2:2:3}$ in this order; the two zero-energy states each have population $1/12$. These <Zeeman populations from an unpolarized angular-momentum multiplet> remain constant under subsequent evolution by the stated static <Hamiltonian operator>, although off-diagonal <quantum coherence> can evolve.
There is a distinct weak-field model in which a retained <spin-orbit interaction> keeps $J=3/2$ isolated and one projects the magnetic perturbation into that quartet. The <Landé g-factor> is then $4/3$, giving four shifts $\mu_BB_z(\pm2,\pm2/3)$ with equal populations. That calculation requires the additional projection or dominant <spin-orbit interaction>; it is not diagonalization of the full magnetic <Hamiltonian operator> specified here, which also mixes the $J=1/2$ sector.
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