= Solution
The <Zeeman effect> comes from $-\boldsymbol\mu\cdot\mathbf B$. For an <Electron>, the orbital and <spin> <magnetic moments> are approximately $-\mu_B\mathbf L/\hbar$ and $-2\mu_B\mathbf S/\hbar$, respectively. This explains both the sign and the relative factor two in the term linear in $B$. The second term is the <spin-orbit interaction>: the <Electron>'s motion through the atomic <electric field> produces a magnetic interaction in its rest frame. The relativistic correction, including <Thomas precession>, and the radial <expectation value> are incorporated into $K$. This effective <Hamiltonian> retains the linear <Zeeman splitting> and <spin-orbit coupling>, while omitting diamagnetic and <hyperfine structure> corrections.
Define the <angular momentum ladder operators> by $L_\pm=L_x\pm iL_y$ and similarly for $S_\pm$. Direct expansion gives
$$
\frac12(L_+S_-+L_-S_+)=L_xS_x+L_yS_y,
$$
so adding $L_zS_z$ gives the required scalar product. The off-diagonal terms change $m_\ell$ and $m_s$ in opposite directions. Consequently $J_z=L_z+S_z$ commutes with $H$, and $m_j$ remains a good <quantum number> at every <magnetic field>.
In the given product <eigenbasis>, the diagonal <matrix elements> are
$$
\langle m_\ell,m_s|H|m_\ell,m_s\rangle
=\mu_BB(m_\ell+2m_s)+2Km_\ell m_s.
$$
For $\ell=1$ and $s=1/2$, the only off-diagonal couplings are between $(1,-1/2)$ and $(0,1/2)$, and between $(0,-1/2)$ and $(-1,1/2)$. Each relevant orbital ladder factor is $\hbar\sqrt2$ and each <spin> ladder factor is $\hbar$. Multiplying by $K/\hbar^2$ gives $\sqrt2K$. Therefore
$$
\boxed{a=\mu_BB,\qquad b=K,\qquad c=\sqrt2K}.
$$
All other off-diagonal entries vanish either by the ladder endpoints or by conservation of $m_j$. This derives every entry of the displayed matrix.
The stretched states have no mixing and yield
$$
\boxed{E_{m_j=3/2}=K+2a,\qquad E_{m_j=-3/2}=K-2a}.
$$
The $m_j=1/2$ block is $\begin{pmatrix}-K&\sqrt2K\\\sqrt2K&a\end{pmatrix}$, while the $m_j=-1/2$ block is $\begin{pmatrix}-a&\sqrt2K\\\sqrt2K&-K\end{pmatrix}$. Taking the trace and determinant of each $2\times2$ block gives the remaining four <energy eigenvalues>:
$$
\boxed{E_{+}^{(\pm)}=\frac{a-K}{2}\pm\frac12\sqrt{(a+K)^2+8K^2}},
$$
$$
\boxed{E_{-}^{(\pm)}=\frac{-a-K}{2}\pm\frac12\sqrt{(a-K)^2+8K^2}}.
$$
The subscripts indicate $m_j=\pm1/2$ and the superscripts indicate the upper or lower root. These formulas are the <Zeeman spectrum for orbital angular momentum one> and remain valid for either sign of $K$.
For the usual ordering $K>0$, including sodium, the zero-field states follow from
$$
\mathbf L\cdot\mathbf S=\frac12(J^2-L^2-S^2),\qquad
E_j=K\left[j(j+1)-\ell(\ell+1)-s(s+1)\right].
$$
The $j=3/2$ quartet lies at $K$ and the $j=1/2$ doublet at $-2K$, separated by $3K$. In a weak <magnetic field>, the <Landé g-factors> give
$$
E_{3/2,m_j}=K+\frac43am_j+O(a^2/K),\qquad
E_{1/2,m_j}=-2K+\frac23am_j+O(a^2/K).
$$
These label the curves near $B=0$. In a strong <magnetic field>, the <Paschen-Back effect> makes $(m_\ell,m_s)$ the appropriate approximate labels:
$$
E_{m_\ell,m_s}=a(m_\ell+2m_s)+2Km_\ell m_s+O(K^2/a)
$$
for the mixed states, with the stretched-state formulas already exact. The branch connections for $B>0$ are
$$
\begin{array}{c|c}
\text{zero-field label }(j,m_j)&\text{strong-field label }(m_\ell,m_s)\\\hline
(3/2,3/2)&(1,1/2)\\
(3/2,1/2)&(0,1/2)\\
(1/2,1/2)&(1,-1/2)\\
(3/2,-1/2)&(-1,1/2)\\
(1/2,-1/2)&(0,-1/2)\\
(3/2,-3/2)&(-1,-1/2)
\end{array}
$$
Within a fixed $m_j$, the nonzero coupling prevents a crossing; this is an <avoided crossing>. Curves with different $m_j$ may cross because the <Hamiltonian> does not couple them.
\Image[/past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2017-spin-orbit-zeeman-spectrum.png]
{title=Exact spin-orbit and Zeeman energies for orbital angular momentum one and positive spin-orbit constant}
For the sodium estimate, the zero-field frequency separation fixes
$$
3K=h\delta f,\qquad K=1.14\times10^{-22}\ \mathrm J.
$$
The lower $3s$ doublet has shifts $\pm a$. The <Sodium D-line Zeeman pattern> gives photon-frequency shifts $a/h$ times $\{\pm2/3,\pm4/3\}$ for D$_1$ and $\{\pm1/3,\pm1,\pm5/3\}$ for D$_2$. Thus the highest-frequency D$_1$ component and lowest-frequency D$_2$ component approach each other by $3a/h$. Extrapolating the <weak-field Zeeman splitting> gives
$$
\boxed{B_{\rm cross}\sim\frac{h\delta f}{3\mu_B}=12.3\ \mathrm T},
$$
so the crossover occurs at a field of order $10\ \mathrm T$, where $a\sim K$ and the weak-field approximation is becoming marginal.
The exact <eigenvalues> refine the estimate within this <Hamiltonian>. The relevant photon energies, with a common zero-field reference omitted, are $E_+^{(-)}+a$ and $E_-^{(+)}-a$. The corresponding <electric dipole transitions> have $\Delta m_j=+1$ and $-1$, respectively. Equating them gives
$$
6a=\sqrt{(a+K)^2+8K^2}+\sqrt{(a-K)^2+8K^2}.
$$
Its positive solution is $a/K=\sqrt{10}/3$, hence
$$
\boxed{B_{\rm cross}=\frac{\sqrt{10}\,h\delta f}{9\mu_B}\simeq13.0\ \mathrm T}
$$
in the retained model. The crossing is between optical transition frequencies; the participating upper states themselves need not be degenerate.
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