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For weak field, from Equation 6.67.

$\begin{array}{l}{E}_{3\text{j}}=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{9}(\frac{3}{j+1/2}鈭�\frac{3}{4})]\\ E_{3\text{j}}=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{3}(\frac{1}{j+1/2}鈭�\frac{1}{4})]\end{array}$

$g_j=1+\frac{j(j+1)鈭�l(l+1)+\frac{3}{4}}{2j(j+1)}$

$E_Z^1={渭}_B{g}_j{B}_{\text{ext}}{m}_j$

It is known that$l=0$, $j=\frac{1}{2}$, $m_j=卤\frac{1}{2}$, and $g_j=2$. Substitute these values.

$\begin{array}{c}E=E_{3,1/2}+E_Z^1\\ =鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{3}(1鈭�\frac{1}{4})]卤{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{4}]卤{渭}_B{B}_{\text{ext}}\end{array}$

It is known that $l=1$, $j=\frac{1}{2}\text{or}\frac{3}{2}$. Substitute these values.

For$j=\frac{1}{2}$,$m_j=卤\frac{1}{2}$,

$\begin{array}{c}{g}_{1/2}=1+\frac{\frac{1}{2}鈰�\frac{3}{2}鈭�2+\frac{3}{4}}{2鈰�\frac{1}{2}鈰�\frac{3}{2}}\\ =\frac{2}{3}\end{array}$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{4}]卤\frac{1}{3}{渭}_B{B}_{\text{ext}}$

For$j=\frac{3}{2}$,$m_j=卤\frac{1}{2},\text{鈥�}卤\frac{3}{2}$,

$\begin{array}{c}{g}_{3/2}=1+\frac{\frac{3}{2}鈰�\frac{5}{2}鈭�2+\frac{3}{4}}{2鈰�\frac{3}{2}鈰�\frac{5}{2}}\\ =\frac{4}{3}\end{array}$

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{3}(\frac{1}{2}鈭�\frac{1}{4})]卤{渭}_B{B}_{\text{ext}}鈰�\{\begin{array}{ll}\frac{4}{3}鈰�\frac{1}{2},& m_j=卤\frac{1}{2}\\ \frac{4}{3}鈰�\frac{3}{2},& m_j=卤\frac{3}{2}\end{array}\\ =鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{12}]卤{渭}_B{B}_{\text{ext}}鈰�\{\begin{array}{ll}\frac{2}{3},& m_j=卤\frac{1}{2}\\ 2,& m_j=卤\frac{3}{2}\end{array}\end{array}$

It is known that$l=2,$$j=\frac{3}{2}\text{or}\frac{5}{2}$. Substitute these values.

For $j=\frac{3}{2}$, $m_j=卤\frac{1}{2},\text{鈥�}卤\frac{3}{2}$

$\begin{array}{c}{g}_{3/2}=1+\frac{\frac{3}{2}鈰�\frac{5}{2}鈭�6+\frac{3}{4}}{2鈰�\frac{3}{2}鈰�\frac{5}{2}}\\ =\frac{4}{5}\end{array}$

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{12}]卤{渭}_B{B}_{\text{ext}}鈰�\{\begin{array}{ll}\frac{4}{5}鈰�\frac{1}{2},& m_j=卤\frac{1}{2}\\ \frac{4}{5}鈰�\frac{3}{2},& m_j=卤\frac{3}{2}\end{array}\\ =鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{12}]卤{渭}_B{B}_{\text{ext}}鈰�\{\begin{array}{ll}\frac{2}{5},& m_j=卤\frac{1}{2}\\ \frac{6}{5},& m_j=卤\frac{3}{2}\end{array}\end{array}$

For $j=\frac{5}{2}$, $m_j=卤\frac{1}{2},\text{鈥�}卤\frac{3}{2},\text{鈥�}卤\frac{5}{2}$

$\begin{array}{c}{g}_{5/2}=1+\frac{\frac{5}{2}鈰�\frac{7}{2}鈭�6+\frac{3}{4}}{2鈰�\frac{5}{2}鈰�\frac{7}{2}}\\ =\frac{6}{5}\end{array}$

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{3}(\frac{1}{3}鈭�\frac{1}{4})]卤{渭}_B{B}_{\text{ext}}鈰�\{\begin{array}{ll}\frac{6}{5}鈰�\frac{1}{2},& m_j=卤\frac{1}{2}\\ \frac{6}{5}鈰�\frac{3}{2},& m_j=卤\frac{3}{2}\\ \frac{6}{5}鈰�\frac{5}{2},& m_j=卤\frac{5}{2}\end{array}\\ =鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{36}]卤{渭}_B{B}_{\text{ext}}鈰�\{\begin{array}{ll}\frac{3}{5},& m_j=卤\frac{1}{2}\\ \frac{9}{5},& m_j=卤\frac{3}{2}\\ 3,& m_j=卤\frac{5}{2}\end{array}\end{array}$

Energies in weak field from lowest state to highest are determined as follows,

For,$l=0,\text{鈥�}j=1/2,\text{鈥�}{m}_j=鈭�1/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{4}]鈭�{渭}_B{B}_{\text{ext}}$

For,$l=0,j=1/2,m_j=+1/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{4}]+{渭}_B{B}_{\text{ext}}$

For,$l=1,j=1/2,m_j=鈭�1/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{4}]鈭�\frac{1}{3}{渭}_B{B}_{\text{ext}}$

For,$l=1,j=1/2,m_j=+1/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{4}]+\frac{1}{3}{渭}_B{B}_{\text{ext}}$

For,$l=1,j=3/2,m_j=鈭�3/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{12}]鈭�2{渭}_B{B}_{\text{ext}}$

For $l=1,j=3/2,m_j=鈭�1/2$,

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{12}]鈭�\frac{2}{3}{渭}_B{B}_{\text{ext}}$

For,$l=1,j=3/2,m_j=+1/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{12}]+\frac{2}{3}{渭}_B{B}_{\text{ext}}$

For,$l=1,j=3/2,m_j=+3/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{12}]+2{渭}_B{B}_{\text{ext}}$

For ,$l=2,j=3/2,m_j=鈭�3/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{12}]鈭�\frac{6}{5}{渭}_B{B}_{\text{ext}}$

For,$l=2,j=3/2,m_j=鈭�1/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{12}]鈭�\frac{2}{5}{渭}_B{B}_{\text{ext}}$

For,$l=2,j=3/2,m_j=+1/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{12}]+\frac{2}{5}{渭}_B{B}_{\text{ext}}$

For,$l=2,j=3/2,m_j=+3/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{12}]+\frac{6}{5}{渭}_B{B}_{\text{ext}}$

For,$l=2,j=5/2,m_j=鈭�5/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{36}]鈭�3{渭}_B{B}_{\text{ext}}$

For,$l=2,j=5/2,m_j=鈭�3/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{36}]鈭�\frac{9}{5}{渭}_B{B}_{\text{ext}}$

For,$l=2,j=5/2,m_j=鈭�1/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{36}]鈭�\frac{3}{5}{渭}_B{B}_{\text{ext}}$

For,$l=2,j=5/2,m_j=+1/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{36}]+\frac{3}{5}{渭}_B{B}_{\text{ext}}$

For,$l=2,j=5/2,m_j=+3/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{36}]+\frac{9}{5}{渭}_B{B}_{\text{ext}}$

For,$l=2,j=5/2,m_j=+5/2$

$E=鈭�\frac{13.6\text{eV}}{9}[1+\frac{{伪}^2}{36}]+3{渭}_B{B}_{\text{ext}}$

For Strong field,the total energy is for $n=3$.

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{n^2}+{渭}_B{B}_{\text{ext}}(m_l+2m_s)+\frac{13.6\text{eV}}{n^3}{伪}^2[\frac{3}{4n}鈭�\frac{l(l+1)鈭�m_l{m}_s}{l(l+1)(l+\frac{1}{2})}]\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{l(l+1)鈭�m_l{m}_s}{l(l+1)(l+\frac{1}{2})}鈭�\frac{1}{4}]\}+{渭}_B{B}_{\text{ext}}(m_l+2m_s)\end{array}$

For$l=0,m_l=0,m_s=鈭�1/2$,

$E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{4}\}鈭�{渭}_B{B}_{\text{ext}}$

For$l=0,m_l=0,m_s=+1/2$,

$E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{4}\}+{渭}_B{B}_{\text{ext}}$

For$l=1,m_l=鈭�1,m_s=鈭�1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{1}{2}鈭�\frac{1}{4}]\}鈭�2{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{12}\}鈭�2{渭}_B{B}_{\text{ext}}\end{array}$

For$l=1,m_l=鈭�1,m_s=+1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{5}{6}鈭�\frac{1}{4}]\}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{7{伪}^2}{36}\}\end{array}$

For$l=1,m_l=0,m_s=鈭�1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{2}{3}鈭�\frac{1}{4}]\}鈭�{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{5{伪}^2}{36}\}鈭�{渭}_B{B}_{\text{ext}}\end{array}$

For$l=1,m_l=0,m_s=+1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{2}{3}鈭�\frac{1}{4}]\}+{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{5{伪}^2}{36}\}+{渭}_B{B}_{\text{ext}}\end{array}$

For$l=1,m_l=+1,m_s=鈭�1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{5}{6}鈭�\frac{1}{4}]\}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{7{伪}^2}{36}\}\end{array}$

For$l=1,m_l=+1,m_s=+1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{1}{2}鈭�\frac{1}{4}]\}+2{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{12}\}+2{渭}_B{B}_{\text{ext}}\end{array}$

For$l=2,m_l=鈭�2,m_s=鈭�1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{1}{3}鈭�\frac{1}{4}]\}鈭�3{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{12}\}鈭�3{渭}_B{B}_{\text{ext}}\end{array}$

For$l=2,m_l=鈭�2,m_s=+1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{7}{15}鈭�\frac{1}{4}]\}鈭�{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{13{伪}^2}{180}\}鈭�{渭}_B{B}_{\text{ext}}\end{array}$

For$l=2,m_l=鈭�1,m_s=鈭�1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{11}{30}鈭�\frac{1}{4}]\}鈭�2{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{7{伪}^2}{180}\}鈭�2{渭}_B{B}_{\text{ext}}\end{array}$

For$l=2,m_l=鈭�1,m_s=+1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{13}{30}鈭�\frac{1}{4}]\}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{11{伪}^2}{180}\}\end{array}$

For$l=2,m_l=0,m_s=鈭�1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{2}{5}鈭�\frac{1}{4}]\}鈭�{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{20}\}鈭�{渭}_B{B}_{\text{ext}}\end{array}$

For$l=2,m_l=0,m_s=+1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{2}{5}鈭�\frac{1}{4}]\}+{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{20}\}+{渭}_B{B}_{\text{ext}}\end{array}$

For$l=2,m_l=+1,m_s=鈭�1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{13}{30}鈭�\frac{1}{4}]\}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{11{伪}^2}{180}\}\end{array}$

For$l=2,m_l=+1,m_s=+1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{11}{30}鈭�\frac{1}{4}]\}+2{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{7{伪}^2}{180}\}鈭�2{渭}_B{B}_{\text{ext}}\end{array}$

For$l=2,m_l=+2,m_s=鈭�1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{7}{15}鈭�\frac{1}{4}]\}+{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{13{伪}^2}{180}\}+{渭}_B{B}_{\text{ext}}\end{array}$

For$l=2,m_l=+2,m_s=+1/2$,

$\begin{array}{c}E=鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{3}[\frac{1}{3}鈭�\frac{1}{4}]\}+3{渭}_B{B}_{\text{ext}}\\ =鈭�\frac{13.6\text{eV}}{9}\{1+\frac{{伪}^2}{12}\}+3{渭}_B{B}_{\text{ext}}\end{array}$

For Intermediate fieldthe fine structure energy is given as follows,

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

For $E_n=\left(\frac{E_1}{9}\right)$,

$\begin{array}{c}{E}_{fs}^1=\frac{{\left(\frac{E_1}{9}\right)}^2}{2m{c}^2}(3鈭�\frac{12}{j+1/2})\\ =\frac{3E_1^2}{162m{c}^2}(1鈭�\frac{4}{j+1/2})\\ =\frac{E_1^2}{54m{c}^2}(1鈭�\frac{4}{j+1/2})\\ =鈭�\frac{E_1{伪}^2}{108}(1鈭�\frac{4}{j+1/2})\\ =3纬(1鈭�\frac{4}{j+1/2})\end{array}$

It is known that $纬=\frac{13.6\text{eV}}{324}{伪}^2$.

For$j=1/2$,$E_{fs}^1=鈭�9纬$, for$j=3/2$,and $E_{fs}^1=鈭�3纬$for$j=5/2$,$E_{fs}^1=鈭�纬$. These are diagonal elements in$鈭�W$ matrixZeeman Write the expression for the Hamiltonian.

$H_Z^{\text{'}}=\frac{1}{\mathrm{鈩�}}尾(L_Z+2S_Z)$

Here, $尾={渭}_B{B}_{\text{ext}}$.

Non-zero blocks of$鈭�W$matrix are as follows,

$9纬鈭�尾,9纬+尾,3纬鈭�2尾,3纬+2尾,\text{鈥�}\left(\begin{array}{cc}3纬鈭�\frac{2}{3}尾& \frac{\sqrt{2}}{3}尾\\ \frac{\sqrt{2}}{3}尾& 9纬鈭�\frac{1}{3}尾\end{array}\right),\text{鈥�}\left(\begin{array}{cc}3纬+\frac{2}{3}尾& \frac{\sqrt{2}}{3}尾\\ \frac{\sqrt{2}}{3}尾& 9纬+\frac{1}{3}尾\end{array}\right)$

For,$l=2$

Use Clebsch-Gordan coefficients (in this case$j=5/2$or$j=3/2$).

$|\frac{5}{2}鈭�\frac{5}{2}鉄�=|2鈭�2鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�$

$\begin{array}{c}5/2,鈭�5/2{鉄�H_Z^{\text{'}}鉄�}_{5/2,鈭�5/2}=鉄�\frac{5}{2}鈭�\frac{5}{2}\left|\frac{尾}{\mathrm{鈩�}}(L_Z+2S_Z)\right|\frac{5}{2}鈭�\frac{5}{2}鉄�A\\ =\frac{尾}{\mathrm{鈩�}}鉄�2鈭�2\left|鉄�\frac{1}{2}鈭�\frac{1}{2}\left|(L_Z+2S_Z)\right|2鈭�2鉄�\right|\frac{1}{2}鈭�\frac{1}{2}鉄�\\ =尾鉄�2鈭�2\left|鉄�\frac{1}{2}鈭�\frac{1}{2}\left|\right(鈭�2鈭�1\left)\right|2鈭�2鉄�\right|\frac{1}{2}鈭�\frac{1}{2}鉄�\\ =鈭�3尾\end{array}$

For ,$j=3/2$

$|\frac{5}{2}鈭�\frac{3}{2}鉄�=\sqrt{\frac{4}{5}}|2鈭�1鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�+\sqrt{\frac{1}{5}}|2鈭�2鉄�|\frac{1}{2}\frac{1}{2}鉄�$

$\begin{array}{c}5/2,鈭�3/2{鉄�H_Z^{\text{'}}鉄�}_{5/2,鈭�3/2}=鉄�\frac{5}{2}鈭�\frac{3}{2}\left|\frac{尾}{\mathrm{鈩�}}(L_Z+2S_Z)\right|\frac{5}{2}鈭�\frac{3}{2}鉄�\\ =\frac{尾}{\mathrm{鈩�}}\frac{5}{2}鈭�\frac{3}{2}|(L_Z+2S_Z)\left\{\sqrt{\frac{4}{5}}|2鈭�1鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�+\sqrt{\frac{1}{5}}|2鈭�2鉄�|\frac{1}{2}\frac{1}{2}鉄�\right\}鉄�\\ =尾\frac{5}{2}鈭�\frac{3}{2}|\left\{(鈭�1鈭�1)\sqrt{\frac{4}{5}}|2鈭�1鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�+(鈭�2+1\left)\sqrt{\frac{1}{5}}\right|2鈭�2鉄�|\frac{1}{2}\frac{1}{2}鉄�\right\}鉄�\\ =尾\{\sqrt{\frac{4}{5}}2鈭�1|\frac{1}{2}鈭�\frac{1}{2}|+\sqrt{\frac{1}{5}}2鈭�2|鉄�\frac{1}{2}\frac{1}{2}|\}\left\{(鈭�2)\sqrt{\frac{4}{5}}|2鈭�1鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�鈭�\sqrt{\frac{1}{5}}|2鈭�2鉄�|\frac{1}{2}\frac{1}{2}鉄�\right\}鉄�鉄�鉄�\\ =尾(\frac{鈭�8}{5}鈭�\frac{1}{5})\\ =鈭�\frac{9}{5}尾\end{array}$

For$j=鈭�1/2$,

.

$|\frac{5}{2}鈭�\frac{1}{2}鉄�=\sqrt{\frac{3}{5}}|20鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�+\sqrt{\frac{2}{5}}|2鈭�1鉄�|\frac{1}{2}\frac{1}{2}鉄�$

role="math" localid="1659005505710" $\begin{array}{c}5/2,鈭�1/2{鉄�H_Z^{\text{'}}鉄�}_{5/2,鈭�1/2}=鈭�\frac{3}{5}尾|\frac{5}{2}\frac{1}{2}鉄�\\ =\sqrt{\frac{3}{5}}|21鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�鈭�\sqrt{\frac{2}{5}}|20鉄�|\frac{1}{2}\frac{1}{2}鉄�\end{array}$

For$j=1/2$,

$\begin{array}{c}5/2,1/2{鉄�H_Z^{\text{'}}鉄�}_{5/2,1/2}=\frac{3}{5}尾|\frac{5}{2}\frac{3}{2}鉄�\\ =\sqrt{\frac{1}{5}}|22鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�鈭�\sqrt{\frac{1}{5}}|21鉄�|\frac{1}{2}\frac{1}{2}鉄�5/2,3/2{鉄�H_Z^{\text{'}}鉄�}_{5/2,3/2}\\ =\frac{9}{5}尾\end{array}$

$\begin{array}{c}|\frac{5}{2}\frac{5}{2}鉄�=|22鉄�|\frac{1}{2}\frac{1}{2}鉄�5/2,5/2{鉄�H_Z^{\text{'}}鉄�}_{5/2,5/2}\\ =3尾\end{array}$

The Diagonal elements of matrix$鈭�W$for$j=5/2$are as follows,

$纬+3尾,\text{鈥�}纬鈭�3尾,\text{鈥�}纬鈭�\frac{9}{5}尾,\text{鈥�}纬+\frac{9}{5}尾,\text{鈥�}纬鈭�\frac{3}{5}尾,\text{鈥�}纬+\frac{3}{5}尾$

For,$j=\frac{3}{2}$,

$\begin{array}{c}|\frac{3}{2}鈭�\frac{3}{2}鉄�=\sqrt{\frac{1}{5}}|2鈭�1鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�鈭�\sqrt{\frac{4}{5}}|2鈭�2鉄�|\frac{1}{2}\frac{1}{2}鉄�3/2,鈭�3/2{鉄�H_Z^{\text{'}}鉄�}_{3/2,鈭�3/2}\\ =鈭�\frac{6}{5}尾\end{array}$

$\begin{array}{c}|\frac{3}{2}鈭�\frac{1}{2}鉄�=\sqrt{\frac{2}{5}}|20鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�鈭�\sqrt{\frac{3}{5}}|2鈭�1鉄�|\frac{1}{2}\frac{1}{2}鉄�3/2,鈭�1/2{鉄�H_Z^{\text{'}}鉄�}_{3/2,鈭�1/2}\\ =鈭�\frac{2}{5}尾\end{array}$

$\begin{array}{c}|\frac{3}{2}\frac{1}{2}鉄�=\sqrt{\frac{3}{5}}|21鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�鈭�\sqrt{\frac{2}{5}}|20鉄�|\frac{1}{2}\frac{1}{2}鉄�3/2,1/2{鉄�H_Z^{\text{'}}鉄�}_{3/2,1/2}\\ =\frac{2}{5}尾\end{array}$

$\begin{array}{c}|\frac{3}{2}\frac{3}{2}鉄�=\sqrt{\frac{4}{5}}|2鈭�1鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�鈭�\sqrt{\frac{1}{5}}|2鈭�2鉄�|\frac{1}{2}\frac{1}{2}鉄�3/2,3/2{鉄�H_Z^{\text{'}}鉄�}_{3/2,3/2}\\ =\frac{6}{5}尾\end{array}$

The Diagonal elements of matrix$鈭�W$for$j=3/2$are as follows,

$3纬+\frac{6}{5}尾,\text{鈥�}3纬鈭�\frac{6}{5}尾,\text{鈥�}3纬+\frac{2}{5}尾,\text{鈥�}3纬鈭�\frac{2}{5}尾$

The non-zero off-diagonal elements are as follows,

$\begin{array}{c}5/2,鈭�3/2{鉄�H_Z^{\text{'}}鉄�}_{3/2,鈭�3/2}{=}_{3/2,鈭�3/2}{鉄�H_Z^{\text{'}}鉄�}_{5/2,鈭�3/2}\\ =\frac{尾}{\mathrm{鈩�}}\{\sqrt{\frac{4}{5}}2鈭�1|\frac{1}{2}鈭�\frac{1}{2}|+\sqrt{\frac{1}{5}}2鈭�2|鉄�\frac{1}{2}\frac{1}{2}|\}(L_Z+2S_Z)\left\{\sqrt{\frac{1}{5}}|2鈭�1鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�鈭�\sqrt{\frac{4}{5}}|2鈭�2鉄�|\frac{1}{2}\frac{1}{2}鉄�\right\}鉄�鉄�鉄�\\ =尾\left\{\sqrt{\frac{4}{5}}2鈭�1|\frac{1}{2}鈭�\frac{1}{2}|+\sqrt{\frac{1}{5}}2鈭�2|鉄�\frac{1}{2}\frac{1}{2}|\left\}\right\{(鈭�1鈭�1)\sqrt{\frac{1}{5}}|2鈭�1鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�鈭�(鈭�2+1\left)\sqrt{\frac{4}{5}}\right|2鈭�2鉄�|\frac{1}{2}\frac{1}{2}鉄�鉄�鉄�鉄�鉄�\right\}\\ =尾\{\sqrt{\frac{4}{5}}2鈭�1|\frac{1}{2}鈭�\frac{1}{2}|+\sqrt{\frac{1}{5}}2鈭�2|鉄�\frac{1}{2}\frac{1}{2}|\}\left\{(鈭�2)\sqrt{\frac{1}{5}}|2鈭�1鉄�|\frac{1}{2}鈭�\frac{1}{2}鉄�+\sqrt{\frac{4}{5}}|2鈭�2鉄�|\frac{1}{2}\frac{1}{2}鉄�\right\}鉄�鉄�鉄�\\ =尾(鈭�\frac{4}{5}+\frac{2}{5})\\ =鈭�\frac{2}{5}尾\end{array}$

For $j=鈭�\frac{1}{2}$,

$\begin{array}{c}5/2,鈭�1/2{鉄�H_Z^{\text{'}}鉄�}_{3/2,鈭�1/2}{=}_{3/2,鈭�1/2}{鉄�H_Z^{\text{'}}鉄�}_{5/2,鈭�1/2}\\ =鈭�\frac{\sqrt{6}}{5}尾\end{array}$

$\begin{array}{c}5/2,1/2{鉄�H_Z^{\text{'}}鉄�}_{3/2,1/2}{=}_{3/2,1/2}{鉄�H_Z^{\text{'}}鉄�}_{5/2,1/2}\\ =鈭�\frac{\sqrt{6}}{5}尾\end{array}$

$\begin{array}{c}5/2,3/2{鉄�H_Z^{\text{'}}鉄�}_{3/2,3/2}{=}_{3/2,3/2}{鉄�H_Z^{\text{'}}鉄�}_{5/2,3/2}\\ =鈭�\frac{2}{5}尾\end{array}$

Non-zero off-diagonal elements of matrix$鈭�W$ are as follows,

$\frac{2}{5}尾,\text{鈥�}\frac{\sqrt{6}}{5}尾$

Matrix$鈭�W$(which has$18脳18$elements) has six$1脳1$ blocksand six$2脳2$ blocks:

$9纬鈭�尾,\text{鈥�}9纬+尾,\text{鈥�}3纬鈭�2尾,\text{鈥�}3纬+2尾,\text{鈥�}纬+3尾,\text{鈥�}纬鈭�3尾$

$\left(\begin{array}{cc}3纬鈭�\frac{2}{3}尾& \frac{\sqrt{2}}{3}尾\\ \frac{\sqrt{2}}{3}尾& 9纬鈭�\frac{1}{3}尾\end{array}\right),\text{鈥�}\left(\begin{array}{cc}3纬+\frac{2}{3}尾& \frac{\sqrt{2}}{3}尾\\ \frac{\sqrt{2}}{3}尾& 9纬+\frac{1}{3}尾\end{array}\right)$

$\begin{array}{l}\left(\begin{array}{cc}纬+\frac{9}{5}尾& \frac{2}{5}尾\\ \frac{2}{5}尾& 3纬+\frac{6}{5}尾\end{array}\right),\text{鈥�}\left(\begin{array}{cc}纬鈭�\frac{9}{5}尾& \frac{2}{5}尾\\ \frac{2}{5}尾& 3纬鈭�\frac{6}{5}尾\end{array}\right)\\ \left(\begin{array}{cc}纬鈭�\frac{3}{5}尾& \frac{\sqrt{6}}{5}尾\\ \frac{\sqrt{6}}{5}尾& 3纬鈭�\frac{2}{5}尾\end{array}\right),\text{鈥�}\left(\begin{array}{cc}纬+\frac{3}{5}尾& \frac{\sqrt{6}}{5}尾\\ \frac{\sqrt{6}}{5}尾& 3纬+\frac{2}{5}尾\end{array}\right)\end{array}$

Eigen values of$2脳2$block is need to be determined. It is required to calculate for$3$of them as eigen values for other three will be obtained by changing the sign of.

$\begin{array}{c}\left|\begin{array}{cc}3纬鈭�\frac{2}{3}尾鈭�位& \frac{\sqrt{2}}{3}尾\\ \frac{\sqrt{2}}{3}尾& 9纬鈭�\frac{1}{3}尾鈭�位\end{array}\right|=0\\ {位}^2+位(尾鈭�12纬)+纬(27纬鈭�7尾)=0\\ 位=6纬鈭�\frac{尾}{2}卤\sqrt{9{纬}^2+尾纬+\frac{{尾}^2}{4}}\end{array}$

$\begin{array}{c}\left|\begin{array}{cc}纬鈭�\frac{9}{5}尾鈭�位& \frac{2}{5}尾\\ \frac{2}{5}尾& 3纬鈭�\frac{6}{5}尾鈭�位\end{array}\right|=0\\ {位}^2+位(3尾鈭�4纬)+3{纬}^2鈭�\frac{33}{5}尾纬+2{尾}^2=0\\ 位=2纬鈭�\frac{3}{2}尾卤\sqrt{{纬}^2+\frac{3}{5}+\frac{{尾}^2}{4}}\end{array}$

$\begin{array}{c}\left|\begin{array}{cc}纬鈭�\frac{3}{5}尾鈭�位& \frac{\sqrt{6}}{5}尾\\ \frac{\sqrt{6}}{5}尾& 3纬鈭�\frac{2}{5}尾鈭�位\end{array}\right|=0\\ {位}^2+位(尾鈭�4纬)+纬(3纬鈭�\frac{11}{5}尾)=0\\ 位=2纬鈭�\frac{尾}{2}卤\sqrt{{纬}^2+\frac{1}{5}尾纬+\frac{{尾}^2}{4}}\end{array}$

Finally, energies in intermediate field are as follows,

${系}_1=E_3鈭�9纬+尾$

${系}_2=E_3鈭�3纬+2尾$

${系}_3=E_3鈭�纬+3尾$

${系}_4=E_3鈭�6纬+\frac{尾}{2}+\sqrt{9{纬}^2+尾纬+\frac{{尾}^2}{4}}$

${系}_5=E_3鈭�6纬+\frac{尾}{2}鈭�\sqrt{9{纬}^2+尾纬+\frac{{尾}^2}{4}}$

${系}_6=E_3鈭�2纬+\frac{3}{2}尾+\sqrt{{纬}^2+\frac{3}{5}+\frac{{尾}^2}{4}}$

${系}_7=E_3鈭�2纬+\frac{3}{2}尾鈭�\sqrt{{纬}^2+\frac{3}{5}+\frac{{尾}^2}{4}}$

${系}_8=E_3鈭�2纬+\frac{尾}{2}+\sqrt{{纬}^2+\frac{1}{5}尾纬+\frac{{尾}^2}{4}}$

${系}_9=E_3鈭�2纬+\frac{尾}{2}鈭�\sqrt{{纬}^2+\frac{1}{5}尾纬+\frac{{尾}^2}{4}}$

Other nine energies can be obtained by changing the sign of$尾<<纬$ .For weak field limit is obtained and for strong $尾猢�>纬$field limit is obtained.