The effect of a small magnetic field is smaller than the fine-structure splitting, so we can solve the problem in two steps:
Find the fine structure.
Perturb the fine structure due to the field.
This weak field regime is called the Zeeman regime. When \(H\) is large, the fine structure is small compared to the energy shifts due to the field (called the Paschen-Back regime). We will solve the general case and then extract the limiting behavior.
The orbital magnetic moment of the electron is: \[\mathbf{\mu}_{\text{orb}} = -\mu_0 \hat{\mathbf{L}},\] where \(\mu_0 = \frac{e\hbar}{2mc}\) is the Bohr magneton and has the value of \(0.579 \times 10^{-8} \, \text{eV/gauss}\). The spin magnetic moment is: \[\mathbf{\mu}_{\text{spin}} = -2\mu_0 \hat{\mathbf{S}}.\] It is the extra factor of 2 that makes life difficult. \[\hat{V}_{\text{mag}} = \mu_0 \mathbf{H} \cdot (\hat{\mathbf{L}} + 2\hat{\mathbf{S}}) = \mu_0 \mathbf{H} \cdot (\hat{\mathbf{J}} + \hat{\mathbf{S}}).\] Choose the \(z\)-direction along \(\mathbf{H}\), so \(\mathbf{H} = H \mathbf{e}_z\). Then \(\hat{S}^2\), \(\hat{L}^2\), \(\hat{L}_z\), and \(\hat{S}_z\) commute with \(\hat{V}_{\text{mag}}\). But \(\hat{L}_z\) and \(\hat{S}_z\) do not separately commute with \(\hat{V}_{\text{fine structure}}\), only the sum does. This implies the field will mix states, and we do not know the parallel directions in the degenerate subspace.
One important note: \(\hat{L}_z + 2\hat{S}_z\) is an even parity operator, so it cannot connect states with different parity. Therefore \(\ell\) must be the same or differ by a multiple of 2, as \(\ell+1\) is different parity from \(\ell\). This reduces a lot of our work.
We have eight degenerate energy levels \(2P_{3/2}, 2P_{1/2}, 2S_{1/2}\) with degeneracies of \(4\), \(2\), and \(2\) respectively. Because of the parity argument, the \(2S_{1/2}\) state cannot connect to \(2P_{3/2}\) or \(2P_{1/2}\). Therefore \(S\) is a parallel direction. Similarly, \(2P_{3/2} m = \pm 3/2\) cannot couple, since \(\hat{J}_z\) is a good quantum number, as \(\hat{L}_z + \hat{S}_z\) commutes with \(\hat{H}\). So only \(2P_{3/2} (m{=} \pm \tfrac{1}{2}\) and \(2P_{1/2} (m{=} \pm \tfrac{1}{2})\) couple (positive to positive and negative to negative).
Hence, we reduce from an \(8 \times 8\) subspace to four \(1 \times 1\) subspaces: \[2P_{3/2} (m{=}\tfrac{3}{2}), \quad 2P_{3/2} (m{=}-\tfrac{3}{2}), \quad 2S_{1/2} (m{=}\tfrac{1}{2}),\quad 2S_{1/2} (m{=}- \tfrac{1}{2})\] and two \(2\times 2\) subspaces: \[\begin{aligned} &2P_{3/2}( m{=}\tfrac{1}{2}), \quad 2P_{1/2} (m{=}\tfrac{1}{2}) \\ &2P_{3/2} (m{=} -\tfrac{1}{2}), \quad 2P_{1/2} (m{=}-\tfrac{1}{2}). \end{aligned}\]
First, examine the \(1 \times 1\) subspaces, which can be analyzed with non-degenerate perturbation theory. \[\Delta E_{\text{mag}} = \langle n l j m \lvert (\hat{J}_z + \hat{S}_z )\lvert n l j m \rangle \, \mu_0 H= \mu_0 H \left(m + \langle n l j m \lvert \hat{S}_z \lvert n l j m \rangle \right).\] The radial part of the overlap is \(1\). The angular momentum is tricky—need to change the basis from \(jms l\) to \(l m_l s m_s\): \[\langle s l m \lvert s_z l_z s m_s \rangle = \sum_{m_l m_s} \langle s l \lvert l m_l s m_s \rangle \langle l m_l s m_s \lvert s_z l_z s m_s \rangle= \sum_{m_l m_s} \lvert \langle s l m \lvert l m_l s m_s \rangle \rvert^2\] where \(\langle s l m \lvert l m_l s m_s \rangle\) are your Clebsch-Gordon coefficients.
We already showed: \[\bigg\lvert \ell+\frac{1}{2}, m \bigg\rangle = \sqrt{\frac{\ell+m+\frac{1}{2}}{2\ell+1}} \bigg\lvert m_\ell = m-\frac{1}{2}, m_s = +\frac{1}{2} \bigg\rangle + \sqrt{\frac{\ell-m+\frac{1}{2}}{2\ell+1}} \bigg\lvert m_\ell = m+\frac{1}{2}, m_s = -\frac{1}{2} \bigg\rangle\] and \[\bigg\lvert \ell-\frac{1}{2}, m \bigg\rangle = -\sqrt{\frac{l-m+\frac{1}{2}}{2l+1}} \bigg\lvert m_\ell = m-\frac{1}{2}, m_s = +\frac{1}{2} \bigg\rangle + \sqrt{\frac{\ell+m+\frac{1}{2}}{2\ell+1}} \bigg\lvert m_\ell = m+\frac{1}{2}, m_s = -\frac{1}{2} \bigg\rangle\] so for \(j= \ell+1/2\), only two \(m_\ell\) terms contribute to each sum and so we get: \[= \frac{1}{2} \frac{\ell+m+\frac{1}{2}}{2\ell+1} + \frac{1}{2} \frac{\ell-m+\frac{1}{2}}{2\ell+1} = \frac{m}{2\ell+1}\] and for \(j = \ell - \frac{1}{2}\), only two \(m_s\) terms contribute, and we get: \[= \frac{1}{2} \frac{\ell-m+\frac{1}{2}}{2\ell+1} + \frac{1}{2} \frac{\ell+m+\frac{1}{2}}{2\ell+1} = -\frac{m}{2\ell+1}.\] So: \[\langle s \ell j m \lvert s_z \lvert s \ell j m \rangle = \pm \frac{m}{2\ell+1}\] for \(j = \ell \pm \frac{1}{2}\), and: \[\Delta E_{\text{mag}} = \mu_0 H \left(m + \frac{m}{2\ell+1}\right) = \mu_0 H m \left(\frac{2\ell+1 \pm 1}{2\ell+1}\right).\] Now recall: \[\Delta E_{\text{FS}} = E_2^0 \frac{\alpha^2}{n^2} \left[\frac{1}{j + \frac{1}{2}} - \frac{3}{4}\right].\] So: \[\begin{cases} \Delta E (2P_{3/2}, m = \pm \tfrac{3}{2}) = E_2^0 \frac{\alpha^2}{4} \left[\frac{1}{4}\right] + \mu_0 H \frac{3}{2} \cdot \frac{4}{3}= E_2^0 \frac{\alpha^2}{16} + 2\mu_0 H \\ \Delta E (2S_{1/2}, m = \pm \tfrac{1}{2}) = E_2^0 \frac{\alpha^2}{4} \left[\frac{5}{4}\right] + \mu_0 H \frac{1}{2} \cdot 2= E_2^0 \frac{5 \alpha^2}{16} + \mu_0 H. \end{cases}\] Now onto the \(2 \times 2\) cases. The diagonal fine-structure matrix elements are: \[\begin{cases} \Delta^2 E_2^0 \frac{\alpha^2}{4} \left[ \frac{2}{2} - \frac{3}{4} \right] = \frac{ E_2^0 \alpha^2}{16}, \quad j = \tfrac{3}{2}, \\ \Delta^2 E_2^0 \frac{\alpha^2}{4} \left[ \frac{5}{16} \right] = \frac{5 E_2^0 \alpha^2}{16}, \quad j =\tfrac{1}{2}. \end{cases}\] The diagonal magnetic matrix elements are: \[\mu_0 H m \frac{2 \ell + 1\pm1}{2 \ell + 1} =\begin{cases} \pm\frac{2}{3} \mu_0 H \quad j = \tfrac{3}{2}, \text{ since } \pm\frac{1}{2}\times \frac{4}{3}\\ \pm\frac{1}{3} \mu_0 H \quad j=\tfrac{1}{2}, \text{ since } \pm\frac{1}{2}\times \frac{2}{3} \end{cases}\] The off-diagonal elements are: \[\begin{aligned} \langle P_{1/2}, m | S_z | P_{3/2}, m \rangle &= \sum_{m_\ell, m_s} \langle s = \frac{1}{2}, \ell = 1, j = \frac{1}{2}, m | s = \frac{1}{2}, \ell = 1, m_\ell, m_s \rangle \\ &~~~~~~~~~\times \langle s = \frac{1}{2}, \ell = 1, m_\ell, m_s | s = \frac{1}{2}, \ell = 1, j = \frac{3}{2}, m \rangle \\ &= \sum_{m_\ell, m_s} \langle s = \frac{1}{2}, \ell = 1, j = \frac{1}{2}, m | s = \frac{1}{2}, \ell = 1, m_\ell, m_s \rangle \\ &~~~~~~~~~\times \langle s = \frac{1}{2}, \ell = 1, m_\ell, m_s | s = \frac{1}{2}, \ell = 1, j = \frac{3}{2}, m \rangle\\& \text{Since only the two $m_\ell$ values contribute, we get:}\\ &= \frac{1}{2} \left( -\sqrt{\frac{\ell - m + \frac{1}{2}}{2\ell + 1}} \sqrt{\frac{\ell + m + \frac{1}{2}}{2\ell + 1}} - \frac{1}{2} \sqrt{\frac{\ell - m + \frac{1}{2}}{2\ell + 1}} \sqrt{\frac{\ell + m + \frac{1}{2}}{2\ell + 1}} \right) \\ &= -\frac{1}{2} \frac{1}{2\ell+1} \left(\sqrt{(\ell+\tfrac{1}{2})^2 - m^2} \right)2 \end{aligned}\] Hence, for \(m = \pm \frac{1}{2}, \ell = 1\), we get: \[= -\frac{1}{3} \sqrt{\frac{4}{4} - \frac{1}{4}} = -\frac{\sqrt{2}}{3}\] Hence, the off diagonal elements are \(-\mu_0H\sqrt{2}/3\). We can write the matrix in full, subtract off \(\epsilon \mathbb{I}\) and take its determinant, setting it to zero: \[\text{det} \begin{pmatrix} \frac{\alpha^2 E_2^0}{16} + \frac{2}{3} \mu_0 H - \epsilon & -\mu_0 H \frac{\sqrt{2}}{3} \\ -\mu_0 H \frac{\sqrt{2}}{3} & \frac{5 \alpha^2 E_2^0}{16} \pm \frac{1}{3} \mu_0 H - \epsilon \end{pmatrix} = 0.\] to find: \[\epsilon^2 - \epsilon \left( \frac{3}{8} \alpha^2 E_2^0 + \mu_0 H \right) + \frac{5}{256} \alpha^4 ( E_2^0)^2 \pm \frac{11}{48} \alpha^2 E_2^0 \mu_0 H = 0\] Solving for \(\epsilon\) gives: \[\epsilon = \frac{3}{16} \alpha^2 E_2^0 \pm\frac{1}{2} \mu_0 H \pm \frac{1}{2} \sqrt{\left( \frac{1}{16} \alpha^2 E_2^0 \right)^2 \mp \frac{1}{2} \alpha^2 E_2^0 \mu_0 H + \mu_0^2 H^2}.\] The first and third \(\pm\) and \(\mp\) correspond to \(m_j = \pm \frac{1}{2}\). The second \(\pm\) corresponds to the fact that we have two roots. As a result, we have: \[\begin{cases} \Delta E(2P_{3/2} \& 2P_{1/2}, m_j = \tfrac{1}{2}) = \frac{3}{16} \alpha^2 E_2^0 \pm \frac{1}{2} \mu_0 H + \frac{1}{2} \sqrt{\left( \frac{1}{16} \alpha^2 E_2^0 \right)^2 -\frac{1}{2} \alpha^2 E_2^0 \mu_0 H + \mu_0^2 H^2}\\ \Delta E(2P_{3/2} \& 2P_{1/2}, m_j = -\tfrac{1}{2}) = \frac{3}{16} \alpha^2 E_2^0 - \frac{1}{2} \mu_0 H \pm \frac{1}{2} \sqrt{\left( \frac{1}{16} \alpha^2 E_2^0 \right)^2 + \frac{1}{2} \alpha^2 E_2^0 \mu_0 H + \mu_0^2 H^2} \end{cases}\]
In the small \(H\) limit, we get the following simplifications. For \(m_j = \pm \frac{1}{2}\), we have: \[\frac{3}{16} \alpha^2 E_0^2 + \frac{1}{8} \alpha^2 E_0^2 + \frac{1}{3} \mu_0 H + \frac{1}{3} \alpha^2 E_0^2 \left(\frac{16}{6} \frac{\mu_0 H}{\alpha^2 E_0^2}\right)^{\frac{1}{2}}\] \[\frac{5}{16} \alpha^2 E_0^2 + \frac{1}{3} \mu_0 H\] \[\frac{1}{16} \alpha^2 E_0^2 + \frac{2}{3} \mu_0 H.\] And for \(m_j = -\frac{1}{2}\), we have: \[\frac{3}{16} \alpha^2 E_0^2 + \frac{1}{8} \alpha^2 E_0^2 - \frac{1}{3} \mu_0 H + \frac{1}{3} \alpha^2 E_0^2 \left(\frac{16}{6} \frac{\mu_0 H}{\alpha^2 E_0^2}\right)^{\frac{1}{2}}\] \[\frac{5}{16} \alpha^2 E_0^2 - \frac{1}{3} \mu_0 H\] \[\frac{1}{16} \alpha^2 E_0^2 - \frac{2}{3} \mu_0 H.\] which can be read of the above matrix with \(H\) small. In the large \(H\) limit, we instead get for \(m_j = +\frac{1}{2}:\) \[\frac{\mu_0 H}{2} \pm \frac{\mu_0 H}{2} = \begin{cases} \mu_0 H\\ 0 \end{cases}\] and for \(m_j = -\frac{1}{2}\): \[-\frac{\mu_0 H}{2} \pm \frac{\mu_0 H}{2} = \begin{cases} -\mu_0 H\\ 0 \end{cases}\] This is \(\hat{J}_z \pm \hat{S}_z\) when we think of \(\hat{J}_z\) and \(\hat{S}_z\) as independent.