Introduction

The effect of a small magnetic field is smaller than the fine-structure splitting, so we can solve the problem in two steps:

1. Find the fine structure.

2. 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.

Setting up the perturbation

The orbital magnetic moment of the electron is: $${\mu }_{\text{orb}}=-{\mu }_0\hat{\mathbf{L}},$$ where ${\mu }_0=\frac{e\hslash }{2mc}$ is the Bohr magneton and has the value of $0.579\times {10}^{-8}\text{eV/gauss}$. The spin magnetic moment is: $${\mu }_{\text{spin}}=-2{\mu }_0\hat{\mathbf{S}}.$$ 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.

Symmetry analysis

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 $2P_{3/2}(m=\pm \frac{1}{2}$ and $2P_{1/2}(m=\pm \frac{1}{2})$ couple (positive to positive and negative to negative).

Hence, we reduce from an $8\times 8$ subspace to four $1\times 1$ subspaces: $$2P_{3/2}(m=\frac{3}{2}),2P_{3/2}(m=-\frac{3}{2}),2S_{1/2}(m=\frac{1}{2}),2S_{1/2}(m=-\frac{1}{2})$$ and two $2\times 2$ subspaces: $$\begin{array}{rl}& 2P_{3/2}(m=\frac{1}{2}),2P_{1/2}(m=\frac{1}{2})\\ & 2P_{3/2}(m=-\frac{1}{2}),2P_{1/2}(m=-\frac{1}{2}).\end{array}$$

Calculate the perturbative corrections

First, examine the $1\times 1$ subspaces, which can be analyzed with non-degenerate perturbation theory. $$\mathrm{\Delta }{E}_{\text{mag}}=⟨nljm|({\hat{J}}_z+{\hat{S}}_z)|nljm⟩{\mu }_{0}H={\mu }_{0}H(m+⟨nljm|{\hat{S}}_z|nljm⟩).$$ The radial part of the overlap is $1$. The angular momentum is tricky—need to change the basis from $jmsl$ to $l{m}_{l}s{m}_s$: $$⟨slm|s_z{l}_{z}s{m}_s⟩=\sum _{m_l{m}_s}⟨sl|l{m}_{l}s{m}_s⟩⟨l{m}_{l}s{m}_s|s_z{l}_{z}s{m}_s⟩=\sum _{m_l{m}_s}|⟨slm|l{m}_{l}s{m}_s⟩{|}^2$$ where $⟨slm|l{m}_{l}s{m}_s⟩$ are your Clebsch-Gordon coefficients.

We already showed: $$|\ell +\frac{1}{2},m⟩=\sqrt{\frac{\ell +m+\frac{1}{2}}{2\ell +1}}|m_{\ell }=m-\frac{1}{2},m_s=+\frac{1}{2}⟩+\sqrt{\frac{\ell -m+\frac{1}{2}}{2\ell +1}}|m_{\ell }=m+\frac{1}{2},m_s=-\frac{1}{2}⟩$$ and $$|\ell -\frac{1}{2},m⟩=-\sqrt{\frac{l-m+\frac{1}{2}}{2l+1}}|m_{\ell }=m-\frac{1}{2},m_s=+\frac{1}{2}⟩+\sqrt{\frac{\ell +m+\frac{1}{2}}{2\ell +1}}|m_{\ell }=m+\frac{1}{2},m_s=-\frac{1}{2}⟩$$ 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: $$⟨s\ell jm|s_z|s\ell jm⟩=\pm \frac{m}{2\ell +1}$$ for $j=\ell \pm \frac{1}{2}$, and: $$\mathrm{\Delta }{E}_{\text{mag}}={\mu }_{0}H(m+\frac{m}{2\ell +1})={\mu }_{0}Hm\left(\frac{2\ell +1\pm 1}{2\ell +1}\right).$$ Now recall: $$\mathrm{\Delta }{E}_{\text{FS}}=E_2^0\frac{{\alpha }^2}{n^2}[\frac{1}{j+\frac{1}{2}}-\frac{3}{4}].$$ So: $$\{\begin{array}{l}\mathrm{\Delta }E(2P_{3/2},m=\pm \frac{3}{2})=E_2^0\frac{{\alpha }^2}{4}\left[\frac{1}{4}\right]+{\mu }_{0}H\frac{3}{2}\cdot \frac{4}{3}=E_2^0\frac{{\alpha }^2}{16}+2{\mu }_{0}H\\ \mathrm{\Delta }E(2S_{1/2},m=\pm \frac{1}{2})=E_2^0\frac{{\alpha }^2}{4}\left[\frac{5}{4}\right]+{\mu }_{0}H\frac{1}{2}\cdot 2=E_2^0\frac{5{\alpha }^2}{16}+{\mu }_{0}H.\end{array}$$ Now onto the $2\times 2$ cases. The diagonal fine-structure matrix elements are: $$\{\begin{array}{l}{\mathrm{\Delta }}^2{E}_2^0\frac{{\alpha }^2}{4}[\frac{2}{2}-\frac{3}{4}]=\frac{E_2^0{\alpha }^2}{16},j=\frac{3}{2},\\ {\mathrm{\Delta }}^2{E}_2^0\frac{{\alpha }^2}{4}\left[\frac{5}{16}\right]=\frac{5E_2^0{\alpha }^2}{16},j=\frac{1}{2}.\end{array}$$ The diagonal magnetic matrix elements are: $${\mu }_{0}Hm\frac{2\ell +1\pm 1}{2\ell +1}=\{\begin{array}{l}\pm \frac{2}{3}{\mu }_{0}Hj=\frac{3}{2},\text{ since }\pm \frac{1}{2}\times \frac{4}{3}\\ \pm \frac{1}{3}{\mu }_{0}Hj=\frac{1}{2},\text{ since }\pm \frac{1}{2}\times \frac{2}{3}\end{array}$$ The off-diagonal elements are: $$\begin{array}{rl}⟨P_{1/2},m|S_z|P_{3/2},m⟩& =\sum _{m_{\ell },m_s}⟨s=\frac{1}{2},\ell =1,j=\frac{1}{2},m|s=\frac{1}{2},\ell =1,m_{\ell },m_s⟩\\ & \times ⟨s=\frac{1}{2},\ell =1,m_{\ell },m_s|s=\frac{1}{2},\ell =1,j=\frac{3}{2},m⟩\\ & =\sum _{m_{\ell },m_s}⟨s=\frac{1}{2},\ell =1,j=\frac{1}{2},m|s=\frac{1}{2},\ell =1,m_{\ell },m_s⟩\\ & \times ⟨s=\frac{1}{2},\ell =1,m_{\ell },m_s|s=\frac{1}{2},\ell =1,j=\frac{3}{2},m⟩\\ & \text{Since only the two }{m}_{\ell }\text{ values contribute, we get:}\\ & =\frac{1}{2}(-\sqrt{\frac{\ell -m+\frac{1}{2}}{2\ell +1}}\sqrt{\frac{\ell +m+\frac{1}{2}}{2\ell +1}}-\frac{1}{2}\sqrt{\frac{\ell -m+\frac{1}{2}}{2\ell +1}}\sqrt{\frac{\ell +m+\frac{1}{2}}{2\ell +1}})\\ & =-\frac{1}{2}\frac{1}{2\ell +1}\left(\sqrt{(\ell +\frac{1}{2}{)}^2-m^2}\right)2\end{array}$$ 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 $-{\mu }_{0}H\sqrt{2}/3$. We can write the matrix in full, subtract off $ϵ\mathbb{I}$ and take its determinant, setting it to zero: $$\text{det}\left(\begin{array}{cc}\frac{{\alpha }^2{E}_2^0}{16}+\frac{2}{3}{\mu }_{0}H-ϵ& -{\mu }_{0}H\frac{\sqrt{2}}{3}\\ -{\mu }_{0}H\frac{\sqrt{2}}{3}& \frac{5{\alpha }^2{E}_2^0}{16}\pm \frac{1}{3}{\mu }_{0}H-ϵ\end{array}\right)=0.$$ to find: $${ϵ}^2-ϵ(\frac{3}{8}{\alpha }^2{E}_2^0+{\mu }_{0}H)+\frac{5}{256}{\alpha }^4(E_2^0{)}^2\pm \frac{11}{48}{\alpha }^2{E}_2^0{\mu }_{0}H=0$$ Solving for $ϵ$ gives: $$ϵ=\frac{3}{16}{\alpha }^2{E}_2^0\pm \frac{1}{2}{\mu }_{0}H\pm \frac{1}{2}\sqrt{{\left(\frac{1}{16}{\alpha }^2{E}_2^0\right)}^2\mp \frac{1}{2}{\alpha }^2{E}_2^0{\mu }_{0}H+{\mu }_0^2{H}^2}.$$ 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: $$\{\begin{array}{l}\mathrm{\Delta }E(2P_{3/2}\mathrm{\&}2P_{1/2},m_j=\frac{1}{2})=\frac{3}{16}{\alpha }^2{E}_2^0\pm \frac{1}{2}{\mu }_{0}H+\frac{1}{2}\sqrt{{\left(\frac{1}{16}{\alpha }^2{E}_2^0\right)}^2-\frac{1}{2}{\alpha }^2{E}_2^0{\mu }_{0}H+{\mu }_0^2{H}^2}\\ \mathrm{\Delta }E(2P_{3/2}\mathrm{\&}2P_{1/2},m_j=-\frac{1}{2})=\frac{3}{16}{\alpha }^2{E}_2^0-\frac{1}{2}{\mu }_{0}H\pm \frac{1}{2}\sqrt{{\left(\frac{1}{16}{\alpha }^2{E}_2^0\right)}^2+\frac{1}{2}{\alpha }^2{E}_2^0{\mu }_{0}H+{\mu }_0^2{H}^2}\end{array}$$

Limiting behavior

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 $m_j=+\frac{1}{2}:$ $$\frac{{\mu }_{0}H}{2}\pm \frac{{\mu }_{0}H}{2}=\{\begin{array}{l}{\mu }_{0}H\\ 0\end{array}$$ and for $m_j=-\frac{1}{2}$: $$-\frac{{\mu }_{0}H}{2}\pm \frac{{\mu }_{0}H}{2}=\{\begin{array}{l}-{\mu }_{0}H\\ 0\end{array}$$ This is ${\hat{J}}_z\pm {\hat{S}}_z$ when we think of ${\hat{J}}_z$ and ${\hat{S}}_z$ as independent.