Question

Consider the (eight) $n=2$states, $|2lj{m}_j⟩$. Find the energy of each state, under weak-field Zeeman splitting, and construct a diagram like Figure 6.11 to show how the energies evolve as$B_{\text{ext}}$ increases. Label each line clearly, and indicate its slope.

Short answer

The energy of each state is,

$\begin{array}{l}{E}_1=-3.4\text{eV}(1+5{\alpha }^2/16)+{\mu }_B{B}_{\text{ext}}\\ E_2=-3.4\text{eV}(1+5{\alpha }^2/16)-{\mu }_B{B}_{\text{ext}}\\ E_3=-3.4\text{eV}(1+5{\alpha }^2/16)+{\mu }_B{B}_{\text{ext}}/3\\ E_4=-3.4\text{eV}(1+5{\alpha }^2/16)-{\mu }_B{B}_{\text{ext}}/3\\ E_5=-3.4\text{eV}(1+{\alpha }^2/16)+2{\mu }_B{B}_{\text{ext}}\\ E_6=-3.4\text{eV}(1+{\alpha }^2/16)+2{\mu }_B{B}_{\text{ext}}/3\\ E_7=-3.4\text{eV}(1+{\alpha }^2/16)-2{\mu }_B{B}_{\text{ext}}/3\\ E_8=-3.4\text{eV}(1+{\alpha }^2/16)-2{\mu }_B{B}_{\text{ext}}\end{array}$

Step-by-step solution

Identification of given data

The given data is shown below,

The number of possible states is 8 which are possible for $n=2$.

Definition of weak field Zeeman splitting

Fine structural splitting is shown on the left side. Because of spin-orbit coupling, this splitting happens even in the absence of a magnetic field. The additional Zeeman splitting that happens in the presence of magnetic fields is depicted on the right side.

Determination of all eight states

It is required to determine the states for $n=2$, $(j=\frac{1}{2})$, $l=1$$(j=\frac{1}{2}\text{\hspace{0.17em}or}\frac{3}{2})$. Determine all eight states.

$|1⟩=|20\frac{1}{2}\frac{1}{2}⟩$

$|2⟩=|20\frac{1}{2}-\frac{1}{2}⟩$

$|3⟩=|21\frac{1}{2}\frac{1}{2}⟩$

$|4⟩=|21\frac{1}{2}-\frac{1}{2}⟩$

$|5⟩=|21\frac{3}{2}\frac{3}{2}⟩$

$|6⟩=|21\frac{3}{2}\frac{1}{2}⟩$

$|7⟩=|21\frac{3}{2}-\frac{1}{2}⟩$

$|8⟩=|21\frac{3}{2}-\frac{3}{2}⟩$

Determination of the equation of Bohr magneton and the Lande g-factor

$E_n$The sum of the fine-structure part $E_{nj}$ and the Zeeman part $E_z$gives the weak-field Zeeman energy.So, it can be represented as follows,

$E=E_{nj}+E_z$

Here,the value of $E_{nj}=\frac{-13.6}{n^2}(1+\frac{{\alpha }^2}{n^2}(\frac{n}{j+\frac{1}{2}}-\frac{3}{4}))$

Write the value of $E_n$.

$E_n={\mu }_n{g}_j{B}_{ext}{m}_j$

Here, ${\mu }_n$ is the Bohr magneton, $g_j$isthe Lande g-factor, $B_{ext}$ is the external magnetic field and$n$ isthe principle quantum number.

Write the equation of the Lande$g$-factor.

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

Write the value of Bohr magneton.

${\mu }_g=\frac{e\hslash }{2m}$

Step 5:Determination of the Lande g-factor for eight states

For each of the eight states, the Lande g-factors are determined as follows:

For first two states,

$\begin{array}{c}{g}_j=1+\frac{\frac{1}{2}(\frac{1}{2}+1)-(0+1)+\frac{3}{4}}{2\left(\frac{1}{2}\right)(\frac{1}{2}+1)}\\ =1+\frac{\frac{3}{4}+\frac{3}{4}}{\frac{3}{2}}\\ =2\end{array}$

For next two states,

$\begin{array}{c}{g}_j=1+\frac{\frac{1}{2}(\frac{1}{2}+1)-1(1+1)+\frac{3}{4}}{2\left(\frac{1}{2}\right)(\frac{1}{2}+1)}\\ =1+\frac{\frac{3}{4}-2+\frac{3}{4}}{\frac{3}{2}}\\ =\frac{2}{3}\end{array}$

For next four steps,

$\begin{array}{c}{g}_j=1+\frac{\left(\frac{3}{2}\right)(\frac{3}{2}+1)-1(1+1)+\frac{3}{4}}{2\left(\frac{3}{2}\right)(\frac{3}{2}+1)}\\ =1+\frac{\left(\frac{3}{2}\right)\left(\frac{5}{2}\right)-2+\frac{3}{4}}{3\left(\frac{5}{2}\right)}\\ =\frac{4}{3}\end{array}$

Determination of the Zeeman energy for eight states

The Zeeman part will be calculated to get the value of$E_z$.

Write the general equation for Zeeman energy.

.$E_z=g_j{m}_j{\mu }_B{B}_{\text{ext}}$

For $|1⟩$,

$\begin{array}{c}{E}_z=\left(2\right)\left(\frac{1}{2}\right){\mu }_B{B}_{ext}\\ ={\mu }_B{B}_{ext}\end{array}$

For$|2⟩$,

$\begin{array}{c}{E}_z=\left(2\right)\left(\frac{-1}{2}\right){\mu }_B{B}_{ext}\\ =-{\mu }_B{B}_{ext}\end{array}$

For$|3⟩$,

$\begin{array}{c}{E}_n=\left(\frac{2}{3}\right)\left(\frac{1}{2}\right){\mu }_B{B}_{ext}\\ =\frac{1}{3}{\mu }_B{B}_{ext}\end{array}$

For$|4⟩$,

$\begin{array}{c}{E}_z=\left(\frac{2}{3}\right)\left(\frac{-1}{2}\right){\mu }_B{B}_{ext}\\ =-\frac{1}{3}{\mu }_B{B}_{ext}\end{array}$

For$|5⟩$,

$\begin{array}{c}{E}_z=\left(\frac{4}{3}\right)\left(\frac{3}{2}\right){\mu }_B{B}_{ext}\\ =2{\mu }_B{B}_{ext}\end{array}$

For$|6⟩$,

$\begin{array}{c}{E}_z=\left(\frac{4}{3}\right)\left(\frac{1}{2}\right){\mu }_B{B}_{ext}\\ =\frac{2}{3}{\mu }_B{B}_{ext}\end{array}$

For $|7⟩$ ,

$\begin{array}{c}{E}_z=\left(\frac{4}{3}\right)\left(\frac{-1}{2}\right){\mu }_B{B}_{ext}\\ =-\frac{2}{3}{\mu }_B{B}_{ext}\end{array}$

For$|8⟩$,

$\begin{array}{c}{E}_z=\left(\frac{4}{3}\right)\left(\frac{-3}{2}\right){\mu }_B{B}_{ext}\\ =-2{\mu }_B{B}_{ext}\end{array}$

Step 7:Determination of the fine structure for eight states

Write the general expression for the fine structure.

$E_{nf}=\frac{-13.6}{n^2}(1+\frac{{\alpha }^2}{n^2}(\frac{n}{j+\frac{1}{2}}-\frac{3}{4}))$

For the first four states, the value of$E_{nf}$is the same that can be represented as follows,

$\begin{array}{c}{E}_{nf}=\frac{-13.6}{2^2}(1+\frac{{\alpha }^2}{4}(\frac{2}{\frac{1}{2}+\frac{1}{2}}-\frac{3}{4}))\\ =\frac{-13.6}{4}(1+\frac{{\alpha }^2}{4}(2-\frac{3}{4}))\\ =\frac{-13.6}{4}(1+\frac{{\alpha }^2}{4}\left(\frac{5}{4}\right))\\ =\frac{-13.6}{4}(1+\frac{5}{16}{\alpha }^2)\\ =-3.4(1+\frac{5{\alpha }^2}{16})\end{array}$

For the next four states, the value of $E_{nf}$ is the samethat can be represented as follows,

$\begin{array}{c}{E}_{nf}=\frac{-13.6}{2^2}(1+\frac{{\alpha }^2}{4}(\frac{2}{\frac{3}{2}+\frac{1}{2}}-\frac{3}{4}))\\ =-3.4(1+\frac{{\alpha }^2}{4}(1-\frac{3}{4}))\\ =-3.4(1+\frac{{\alpha }^2}{4}\left(\frac{1}{4}\right))\\ =-3.4(1+\frac{{\alpha }^2}{16})\end{array}$

Step 8:Determination of the value of total Zeeman energy

Write the expression for the total Zeeman energy.

$E=E_{nj}+E_z$

So, the total Zeeman energy for first four states can be represented as follows,

$\begin{array}{l}{E}_1=-3.4(1+\frac{5{\alpha }^2}{16})+{\mu }_B{B}_{ext}\\ E_2=-3.4(1+\frac{5{\alpha }^2}{16})-{\mu }_B{B}_{ext}\\ E_3=-3.4(1+\frac{5{\alpha }^2}{16})+\frac{1}{3}{\mu }_B{B}_{ext}\\ E_4=-3.4(1+\frac{5{\alpha }^2}{16})-\frac{1}{3}{\mu }_B{B}_{ext}\end{array}$

So, the total Zeeman energy for next four states can be represented as follows,

$\begin{array}{l}{E}_5=-3.4(1+\frac{1}{16}{\alpha }^2)+2{\mu }_B{B}_{ext}\\ E_6=-3.4(1+\frac{1}{16}{\alpha }^2)+\frac{2}{3}{\mu }_B{B}_{ext}\\ E_7=-3.4(1+\frac{1}{16}{\alpha }^2)-\frac{2}{3}{\mu }_B{B}_{ext}\\ E_8=-3.4(1+\frac{1}{16}{\alpha }^2)-2{\mu }_B{B}_{ext}\end{array}$

Thus, only two energy values, $E_{nj}$are used to indicate the total energy values on the graph, and splitting occurs with an increase of $B_{ext}$.

The representation of the total energies on graph is as follows,