Question

Analyze the Zeeman effect for the $n=3$states of hydrogen, in the weak, strong, and intermediate field regimes. Construct a table of energies (analogous to Table 6.2), plot them as functions of the external field (as in Figure 6.12), and check that the intermediate-field results reduce properly in the two limiting cases.

Short answer

Energies in intermediate field are as follows,

${ϵ}_1=E_3-9\gamma +\beta$

${ϵ}_2=E_3-3\gamma +2\beta$

${ϵ}_3=E_3-\gamma +3\beta$

${ϵ}_4=E_3-6\gamma +\frac{\beta }{2}+\sqrt{9{\gamma }^2+\beta \gamma +\frac{{\beta }^2}{4}}$

${ϵ}_5=E_3-6\gamma +\frac{\beta }{2}-\sqrt{9{\gamma }^2+\beta \gamma +\frac{{\beta }^2}{4}}$

${ϵ}_6=E_3-2\gamma +\frac{3}{2}\beta +\sqrt{{\gamma }^2+\frac{3}{5}+\frac{{\beta }^2}{4}}$

${ϵ}_7=E_3-2\gamma +\frac{3}{2}\beta -\sqrt{{\gamma }^2+\frac{3}{5}+\frac{{\beta }^2}{4}}$

${ϵ}_8=E_3-2\gamma +\frac{\beta }{2}+\sqrt{{\gamma }^2+\frac{1}{5}\beta \gamma +\frac{{\beta }^2}{4}}$

.${ϵ}_9=E_3-2\gamma +\frac{\beta }{2}-\sqrt{{\gamma }^2+\frac{1}{5}\beta \gamma +\frac{{\beta }^2}{4}}$

Other nine energies can be obtained by changing the sign of$\beta$ .For$\beta <<\gamma$ weak field limit is obtained and for$\beta ⩾>\gamma$ strong field limit is obtained.

Step-by-step solution

Given. Step 2: Determination of the Weak field

For weak field, from Equation 6.67.

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

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

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

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

$\begin{array}{c}E=E_{3,1/2}+E_Z^1\\ =-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{3}(1-\frac{1}{4})]\pm {\mu }_B{B}_{\text{ext}}\\ =-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{4}]\pm {\mu }_B{B}_{\text{ext}}\end{array}$

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

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

$\begin{array}{c}{g}_{1/2}=1+\frac{\frac{1}{2}\cdot \frac{3}{2}-2+\frac{3}{4}}{2\cdot \frac{1}{2}\cdot \frac{3}{2}}\\ =\frac{2}{3}\end{array}$

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{4}]\pm \frac{1}{3}{\mu }_B{B}_{\text{ext}}$

For$j=\frac{3}{2}$,$m_j=\pm \frac{1}{2},\pm \frac{3}{2}$,

$\begin{array}{c}{g}_{3/2}=1+\frac{\frac{3}{2}\cdot \frac{5}{2}-2+\frac{3}{4}}{2\cdot \frac{3}{2}\cdot \frac{5}{2}}\\ =\frac{4}{3}\end{array}$

$\begin{array}{c}E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{3}(\frac{1}{2}-\frac{1}{4})]\pm {\mu }_B{B}_{\text{ext}}\cdot \{\begin{array}{ll}\frac{4}{3}\cdot \frac{1}{2},& m_j=\pm \frac{1}{2}\\ \frac{4}{3}\cdot \frac{3}{2},& m_j=\pm \frac{3}{2}\end{array}\\ =-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{12}]\pm {\mu }_B{B}_{\text{ext}}\cdot \{\begin{array}{ll}\frac{2}{3},& m_j=\pm \frac{1}{2}\\ 2,& m_j=\pm \frac{3}{2}\end{array}\end{array}$

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

For $j=\frac{3}{2}$, $m_j=\pm \frac{1}{2},\pm \frac{3}{2}$

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

$\begin{array}{c}E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{12}]\pm {\mu }_B{B}_{\text{ext}}\cdot \{\begin{array}{ll}\frac{4}{5}\cdot \frac{1}{2},& m_j=\pm \frac{1}{2}\\ \frac{4}{5}\cdot \frac{3}{2},& m_j=\pm \frac{3}{2}\end{array}\\ =-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{12}]\pm {\mu }_B{B}_{\text{ext}}\cdot \{\begin{array}{ll}\frac{2}{5},& m_j=\pm \frac{1}{2}\\ \frac{6}{5},& m_j=\pm \frac{3}{2}\end{array}\end{array}$

For $j=\frac{5}{2}$, $m_j=\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2}$

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

$\begin{array}{c}E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{3}(\frac{1}{3}-\frac{1}{4})]\pm {\mu }_B{B}_{\text{ext}}\cdot \{\begin{array}{ll}\frac{6}{5}\cdot \frac{1}{2},& m_j=\pm \frac{1}{2}\\ \frac{6}{5}\cdot \frac{3}{2},& m_j=\pm \frac{3}{2}\\ \frac{6}{5}\cdot \frac{5}{2},& m_j=\pm \frac{5}{2}\end{array}\\ =-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{36}]\pm {\mu }_B{B}_{\text{ext}}\cdot \{\begin{array}{ll}\frac{3}{5},& m_j=\pm \frac{1}{2}\\ \frac{9}{5},& m_j=\pm \frac{3}{2}\\ 3,& m_j=\pm \frac{5}{2}\end{array}\end{array}$

Step 3: Determination of energies in weak field

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

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{4}]-{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{4}]+{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{4}]-\frac{1}{3}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{4}]+\frac{1}{3}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{12}]-2{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{12}]-\frac{2}{3}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{12}]+\frac{2}{3}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{12}]+2{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{12}]-\frac{6}{5}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{12}]-\frac{2}{5}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{12}]+\frac{2}{5}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{12}]+\frac{6}{5}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{36}]-3{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{36}]-\frac{9}{5}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{36}]-\frac{3}{5}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{36}]+\frac{3}{5}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{36}]+\frac{9}{5}{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}[1+\frac{{\alpha }^2}{36}]+3{\mu }_B{B}_{\text{ext}}$

Determination of the Strong field 

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

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

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

$E=-\frac{13.6\text{eV}}{9}\{1+\frac{{\alpha }^2}{4}\}-{\mu }_B{B}_{\text{ext}}$

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

$E=-\frac{13.6\text{eV}}{9}\{1+\frac{{\alpha }^2}{4}\}+{\mu }_B{B}_{\text{ext}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Step 5:Determination of the Intermediate field

For Intermediate fieldthe fine structure energy is given as follows,

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

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

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

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

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

$H_Z^{\text{'}}=\frac{1}{\hslash }\beta (L_Z+2S_Z)$

Here, $\beta ={\mu }_B{B}_{\text{ext}}$.

Step 6:Determination of non-zero blocks of matrix

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

$9\gamma -\beta ,9\gamma +\beta ,3\gamma -2\beta ,3\gamma +2\beta ,\left(\begin{array}{cc}3\gamma -\frac{2}{3}\beta & \frac{\sqrt{2}}{3}\beta \\ \frac{\sqrt{2}}{3}\beta & 9\gamma -\frac{1}{3}\beta \end{array}\right),\left(\begin{array}{cc}3\gamma +\frac{2}{3}\beta & \frac{\sqrt{2}}{3}\beta \\ \frac{\sqrt{2}}{3}\beta & 9\gamma +\frac{1}{3}\beta \end{array}\right)$

Step 7:Use ofClebsch-Gordan coefficients

For,$l=2$

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

$|\frac{5}{2}-\frac{5}{2}⟩=|2-2⟩|\frac{1}{2}-\frac{1}{2}⟩$

$\begin{array}{c}5/2,-5/2{⟨H_Z^{\text{'}}⟩}_{5/2,-5/2}=⟨\frac{5}{2}-\frac{5}{2}\left|\frac{\beta }{\hslash }(L_Z+2S_Z)\right|\frac{5}{2}-\frac{5}{2}⟩A\\ =\frac{\beta }{\hslash }⟨2-2\left|⟨\frac{1}{2}-\frac{1}{2}\left|(L_Z+2S_Z)\right|2-2⟩\right|\frac{1}{2}-\frac{1}{2}⟩\\ =\beta ⟨2-2\left|⟨\frac{1}{2}-\frac{1}{2}\left|\right(-2-1\left)\right|2-2⟩\right|\frac{1}{2}-\frac{1}{2}⟩\\ =-3\beta \end{array}$

For ,$j=3/2$

$|\frac{5}{2}-\frac{3}{2}⟩=\sqrt{\frac{4}{5}}|2-1⟩|\frac{1}{2}-\frac{1}{2}⟩+\sqrt{\frac{1}{5}}|2-2⟩|\frac{1}{2}\frac{1}{2}⟩$

$\begin{array}{c}5/2,-3/2{⟨H_Z^{\text{'}}⟩}_{5/2,-3/2}=⟨\frac{5}{2}-\frac{3}{2}\left|\frac{\beta }{\hslash }(L_Z+2S_Z)\right|\frac{5}{2}-\frac{3}{2}⟩\\ =\frac{\beta }{\hslash }\frac{5}{2}-\frac{3}{2}|(L_Z+2S_Z)\left\{\sqrt{\frac{4}{5}}|2-1⟩|\frac{1}{2}-\frac{1}{2}⟩+\sqrt{\frac{1}{5}}|2-2⟩|\frac{1}{2}\frac{1}{2}⟩\right\}⟩\\ =\beta \frac{5}{2}-\frac{3}{2}|\left\{(-1-1)\sqrt{\frac{4}{5}}|2-1⟩|\frac{1}{2}-\frac{1}{2}⟩+(-2+1\left)\sqrt{\frac{1}{5}}\right|2-2⟩|\frac{1}{2}\frac{1}{2}⟩\right\}⟩\\ =\beta \{\sqrt{\frac{4}{5}}2-1|\frac{1}{2}-\frac{1}{2}|+\sqrt{\frac{1}{5}}2-2|⟨\frac{1}{2}\frac{1}{2}|\}\left\{(-2)\sqrt{\frac{4}{5}}|2-1⟩|\frac{1}{2}-\frac{1}{2}⟩-\sqrt{\frac{1}{5}}|2-2⟩|\frac{1}{2}\frac{1}{2}⟩\right\}⟩⟩⟩\\ =\beta (\frac{-8}{5}-\frac{1}{5})\\ =-\frac{9}{5}\beta \end{array}$

For$j=-1/2$,

.

$|\frac{5}{2}-\frac{1}{2}⟩=\sqrt{\frac{3}{5}}|20⟩|\frac{1}{2}-\frac{1}{2}⟩+\sqrt{\frac{2}{5}}|2-1⟩|\frac{1}{2}\frac{1}{2}⟩$

role="math" localid="1659005505710" $\begin{array}{c}5/2,-1/2{⟨H_Z^{\text{'}}⟩}_{5/2,-1/2}=-\frac{3}{5}\beta |\frac{5}{2}\frac{1}{2}⟩\\ =\sqrt{\frac{3}{5}}|21⟩|\frac{1}{2}-\frac{1}{2}⟩-\sqrt{\frac{2}{5}}|20⟩|\frac{1}{2}\frac{1}{2}⟩\end{array}$

For$j=1/2$,

$\begin{array}{c}5/2,1/2{⟨H_Z^{\text{'}}⟩}_{5/2,1/2}=\frac{3}{5}\beta |\frac{5}{2}\frac{3}{2}⟩\\ =\sqrt{\frac{1}{5}}|22⟩|\frac{1}{2}-\frac{1}{2}⟩-\sqrt{\frac{1}{5}}|21⟩|\frac{1}{2}\frac{1}{2}⟩5/2,3/2{⟨H_Z^{\text{'}}⟩}_{5/2,3/2}\\ =\frac{9}{5}\beta \end{array}$

$\begin{array}{c}|\frac{5}{2}\frac{5}{2}⟩=|22⟩|\frac{1}{2}\frac{1}{2}⟩5/2,5/2{⟨H_Z^{\text{'}}⟩}_{5/2,5/2}\\ =3\beta \end{array}$

Step 8:Determination of diagonal elements of matrix -W

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

$\gamma +3\beta ,\gamma -3\beta ,\gamma -\frac{9}{5}\beta ,\gamma +\frac{9}{5}\beta ,\gamma -\frac{3}{5}\beta ,\gamma +\frac{3}{5}\beta$

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

$\begin{array}{c}|\frac{3}{2}-\frac{3}{2}⟩=\sqrt{\frac{1}{5}}|2-1⟩|\frac{1}{2}-\frac{1}{2}⟩-\sqrt{\frac{4}{5}}|2-2⟩|\frac{1}{2}\frac{1}{2}⟩3/2,-3/2{⟨H_Z^{\text{'}}⟩}_{3/2,-3/2}\\ =-\frac{6}{5}\beta \end{array}$

$\begin{array}{c}|\frac{3}{2}-\frac{1}{2}⟩=\sqrt{\frac{2}{5}}|20⟩|\frac{1}{2}-\frac{1}{2}⟩-\sqrt{\frac{3}{5}}|2-1⟩|\frac{1}{2}\frac{1}{2}⟩3/2,-1/2{⟨H_Z^{\text{'}}⟩}_{3/2,-1/2}\\ =-\frac{2}{5}\beta \end{array}$

$\begin{array}{c}|\frac{3}{2}\frac{1}{2}⟩=\sqrt{\frac{3}{5}}|21⟩|\frac{1}{2}-\frac{1}{2}⟩-\sqrt{\frac{2}{5}}|20⟩|\frac{1}{2}\frac{1}{2}⟩3/2,1/2{⟨H_Z^{\text{'}}⟩}_{3/2,1/2}\\ =\frac{2}{5}\beta \end{array}$

$\begin{array}{c}|\frac{3}{2}\frac{3}{2}⟩=\sqrt{\frac{4}{5}}|2-1⟩|\frac{1}{2}-\frac{1}{2}⟩-\sqrt{\frac{1}{5}}|2-2⟩|\frac{1}{2}\frac{1}{2}⟩3/2,3/2{⟨H_Z^{\text{'}}⟩}_{3/2,3/2}\\ =\frac{6}{5}\beta \end{array}$

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

$3\gamma +\frac{6}{5}\beta ,3\gamma -\frac{6}{5}\beta ,3\gamma +\frac{2}{5}\beta ,3\gamma -\frac{2}{5}\beta$

Step 10:Determination of non-zero off-diagonal elements

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

$\begin{array}{c}5/2,-3/2{⟨H_Z^{\text{'}}⟩}_{3/2,-3/2}{=}_{3/2,-3/2}{⟨H_Z^{\text{'}}⟩}_{5/2,-3/2}\\ =\frac{\beta }{\hslash }\{\sqrt{\frac{4}{5}}2-1|\frac{1}{2}-\frac{1}{2}|+\sqrt{\frac{1}{5}}2-2|⟨\frac{1}{2}\frac{1}{2}|\}(L_Z+2S_Z)\left\{\sqrt{\frac{1}{5}}|2-1⟩|\frac{1}{2}-\frac{1}{2}⟩-\sqrt{\frac{4}{5}}|2-2⟩|\frac{1}{2}\frac{1}{2}⟩\right\}⟩⟩⟩\\ =\beta \left\{\sqrt{\frac{4}{5}}2-1|\frac{1}{2}-\frac{1}{2}|+\sqrt{\frac{1}{5}}2-2|⟨\frac{1}{2}\frac{1}{2}|\left\}\right\{(-1-1)\sqrt{\frac{1}{5}}|2-1⟩|\frac{1}{2}-\frac{1}{2}⟩-(-2+1\left)\sqrt{\frac{4}{5}}\right|2-2⟩|\frac{1}{2}\frac{1}{2}⟩⟩⟩⟩⟩\right\}\\ =\beta \{\sqrt{\frac{4}{5}}2-1|\frac{1}{2}-\frac{1}{2}|+\sqrt{\frac{1}{5}}2-2|⟨\frac{1}{2}\frac{1}{2}|\}\left\{(-2)\sqrt{\frac{1}{5}}|2-1⟩|\frac{1}{2}-\frac{1}{2}⟩+\sqrt{\frac{4}{5}}|2-2⟩|\frac{1}{2}\frac{1}{2}⟩\right\}⟩⟩⟩\\ =\beta (-\frac{4}{5}+\frac{2}{5})\\ =-\frac{2}{5}\beta \end{array}$

For $j=-\frac{1}{2}$,

$\begin{array}{c}5/2,-1/2{⟨H_Z^{\text{'}}⟩}_{3/2,-1/2}{=}_{3/2,-1/2}{⟨H_Z^{\text{'}}⟩}_{5/2,-1/2}\\ =-\frac{\sqrt{6}}{5}\beta \end{array}$

$\begin{array}{c}5/2,1/2{⟨H_Z^{\text{'}}⟩}_{3/2,1/2}{=}_{3/2,1/2}{⟨H_Z^{\text{'}}⟩}_{5/2,1/2}\\ =-\frac{\sqrt{6}}{5}\beta \end{array}$

$\begin{array}{c}5/2,3/2{⟨H_Z^{\text{'}}⟩}_{3/2,3/2}{=}_{3/2,3/2}{⟨H_Z^{\text{'}}⟩}_{5/2,3/2}\\ =-\frac{2}{5}\beta \end{array}$

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

$\frac{2}{5}\beta ,\frac{\sqrt{6}}{5}\beta$

Step 11:Determination of Matrix  −W

Matrix$-W$(which has$18\times 18$elements) has six$1\times 1$ blocksand six$2\times 2$ blocks:

$9\gamma -\beta ,9\gamma +\beta ,3\gamma -2\beta ,3\gamma +2\beta ,\gamma +3\beta ,\gamma -3\beta$

$\left(\begin{array}{cc}3\gamma -\frac{2}{3}\beta & \frac{\sqrt{2}}{3}\beta \\ \frac{\sqrt{2}}{3}\beta & 9\gamma -\frac{1}{3}\beta \end{array}\right),\left(\begin{array}{cc}3\gamma +\frac{2}{3}\beta & \frac{\sqrt{2}}{3}\beta \\ \frac{\sqrt{2}}{3}\beta & 9\gamma +\frac{1}{3}\beta \end{array}\right)$

$\begin{array}{l}\left(\begin{array}{cc}\gamma +\frac{9}{5}\beta & \frac{2}{5}\beta \\ \frac{2}{5}\beta & 3\gamma +\frac{6}{5}\beta \end{array}\right),\left(\begin{array}{cc}\gamma -\frac{9}{5}\beta & \frac{2}{5}\beta \\ \frac{2}{5}\beta & 3\gamma -\frac{6}{5}\beta \end{array}\right)\\ \left(\begin{array}{cc}\gamma -\frac{3}{5}\beta & \frac{\sqrt{6}}{5}\beta \\ \frac{\sqrt{6}}{5}\beta & 3\gamma -\frac{2}{5}\beta \end{array}\right),\left(\begin{array}{cc}\gamma +\frac{3}{5}\beta & \frac{\sqrt{6}}{5}\beta \\ \frac{\sqrt{6}}{5}\beta & 3\gamma +\frac{2}{5}\beta \end{array}\right)\end{array}$

Step 12:Determination of Eigen values

Eigen values of$2\times 2$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.

$\begin{array}{c}\left|\begin{array}{cc}3\gamma -\frac{2}{3}\beta -\lambda & \frac{\sqrt{2}}{3}\beta \\ \frac{\sqrt{2}}{3}\beta & 9\gamma -\frac{1}{3}\beta -\lambda \end{array}\right|=0\\ {\lambda }^2+\lambda (\beta -12\gamma )+\gamma (27\gamma -7\beta )=0\\ \lambda =6\gamma -\frac{\beta }{2}\pm \sqrt{9{\gamma }^2+\beta \gamma +\frac{{\beta }^2}{4}}\end{array}$

$\begin{array}{c}\left|\begin{array}{cc}\gamma -\frac{9}{5}\beta -\lambda & \frac{2}{5}\beta \\ \frac{2}{5}\beta & 3\gamma -\frac{6}{5}\beta -\lambda \end{array}\right|=0\\ {\lambda }^2+\lambda (3\beta -4\gamma )+3{\gamma }^2-\frac{33}{5}\beta \gamma +2{\beta }^2=0\\ \lambda =2\gamma -\frac{3}{2}\beta \pm \sqrt{{\gamma }^2+\frac{3}{5}+\frac{{\beta }^2}{4}}\end{array}$

$\begin{array}{c}\left|\begin{array}{cc}\gamma -\frac{3}{5}\beta -\lambda & \frac{\sqrt{6}}{5}\beta \\ \frac{\sqrt{6}}{5}\beta & 3\gamma -\frac{2}{5}\beta -\lambda \end{array}\right|=0\\ {\lambda }^2+\lambda (\beta -4\gamma )+\gamma (3\gamma -\frac{11}{5}\beta )=0\\ \lambda =2\gamma -\frac{\beta }{2}\pm \sqrt{{\gamma }^2+\frac{1}{5}\beta \gamma +\frac{{\beta }^2}{4}}\end{array}$

Step 13:Determination of energies in intermediate field

Finally, energies in intermediate field are as follows,

${ϵ}_1=E_3-9\gamma +\beta$

${ϵ}_2=E_3-3\gamma +2\beta$

${ϵ}_3=E_3-\gamma +3\beta$

${ϵ}_4=E_3-6\gamma +\frac{\beta }{2}+\sqrt{9{\gamma }^2+\beta \gamma +\frac{{\beta }^2}{4}}$

${ϵ}_5=E_3-6\gamma +\frac{\beta }{2}-\sqrt{9{\gamma }^2+\beta \gamma +\frac{{\beta }^2}{4}}$

${ϵ}_6=E_3-2\gamma +\frac{3}{2}\beta +\sqrt{{\gamma }^2+\frac{3}{5}+\frac{{\beta }^2}{4}}$

${ϵ}_7=E_3-2\gamma +\frac{3}{2}\beta -\sqrt{{\gamma }^2+\frac{3}{5}+\frac{{\beta }^2}{4}}$

${ϵ}_8=E_3-2\gamma +\frac{\beta }{2}+\sqrt{{\gamma }^2+\frac{1}{5}\beta \gamma +\frac{{\beta }^2}{4}}$

${ϵ}_9=E_3-2\gamma +\frac{\beta }{2}-\sqrt{{\gamma }^2+\frac{1}{5}\beta \gamma +\frac{{\beta }^2}{4}}$

Other nine energies can be obtained by changing the sign of$\beta <<\gamma$ .For weak field limit is obtained and for strong $\beta ⩾>\gamma$field limit is obtained.