Work out the matrix elements of ${\mathbf{H}}_{\mathbf{Z}}^{\mathbf{\text{'}}}\mathbf{}{\mathbf{and}\mathbf{}{\mathbf{H}}_{\mathbf{fs}}}^{\mathbf{\text{'}}}$construct the W matrix given in the text, for n = 2.

We get the complete Fine-Structure Formula is,

$E_{fs}^1=\frac{E_n^2}{2m{c}^2}\left(3-\frac{4n}{j+1/2}\right)$

Using Equation 6:66,

$E_{fs}^1=\frac{E_2^2}{2m{c}^2}\left(3-\frac{8}{j+1/2}\right)=\frac{E_1^2}{32m{c}^2}\left(3-\frac{8}{j+1/2}\right);\frac{E_1}{m{c}^2}=-\frac{{\alpha }^2}{2}$,

So.

$E_{fs}^1=-\frac{E_1}{32}\left(\frac{{\alpha }^2}{2}\right)\left(3-\frac{8}{j+1/2}\right)=\frac{13.6eV}{64}{\alpha }^2\left(3-\frac{8}{j+1/2}\right)=\gamma \left(3-\frac{8}{j+1/2}\right).$

$E_{fs}^1=\frac{{\left(E_n\right)}^2}{2m{c}^2}\left(3-\frac{4n}{j+1/2}\right)$

For

For$j=3/2\left({\psi }_3,{\psi }_4,{\psi }_5,{\psi }_7\right)$

$$\begin{gathered} H_{fs}^1=\gamma \left(3-\frac{8}{2}\right) \\ {H}_{fs}^1=-\gamma \end{gathered}$$

This confirms all the γ terms in -W (p. 281).

Meanwhile, $H_z^{\text{'}}=(e/2m)B_{ext}\left(L_z+2S_z\right),{\psi }_1,{\psi }_2,{\psi }_3,{\psi }_4$are Eigen states of $L_z\text{and}{S}_z$ for these there are only diagonal elements:

$H_Z^{\text{'}}=\frac{e}{2m}(L+2S)·B_{ext}$

This confirms the upper left corner of -W.

Finally:

$$\begin{gathered} \left.\begin{array}{r}\left(L_z+2S_z\right)\left|{\psi }_5\right.>=+\hslash \sqrt{\frac{2}{3}}|10⟩\left|\frac{1}{2}\frac{1}{2}\right.> \\ \left(L_z+2S_z\right)\left|{\psi }_6\right.>=-\hslash \sqrt{\frac{1}{3}}|10⟩\left|\frac{1}{2}\frac{1}{2}\right.> \\ \left(L_z+2S_z\right)\left|{\psi }_7\right.>=-\hslash \sqrt{\frac{2}{3}}|10⟩\left|\frac{1}{2}-\frac{1}{2}\right.> \\ \left(L_z+2S_z\right)\left|{\psi }_8\right.>=-\hslash \sqrt{\frac{1}{3}}|10⟩\left|\frac{1}{2}-\frac{1}{2}\right.>\\ \end{array}\right\}so \end{gathered}$$

which confirms the remaining elements.