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The quantum numbers for the three fermions are $j_{1}=j_{2}=j_{3}=\frac{1}{2}$.

Asked

It is asked of us to find the allowed values for the total quantum number $j_{123}$.

For two fermions with quantum numbers $j_1$and $j_2$ , the allowed values for the total quantum number j are between$\left|j_1-j_2\right|$ and $j_1+j_2$in integral steps.

鈥︹��. (2)

The approximate radius $r_n$ of an atom is given as, $r_n=n^2{a}_0$. 鈥︹��. (4)

With n being the principal quantum number, and being the Bohr radius, given as, . $a_0=\frac{\left(4蟺{蔚}_0\right){\mathrm{鈩�}}^2}{m_e{e}^2}$ 鈥︹��. (5)

With the variables having the same meanings as in the other expressions.

The speed of light c in vacuum can be written as, $c^2=\frac{1}{{渭}_0{蔚}_0}$. 鈥︹��. (6)

With ${铀�}_0$ and${铀�}_0$ being the permeability and permittivity of free space, respectively.

The magnitudes of the total and orbital angular momentums J and L are given as follows:

$\begin{array}{rcl}J& =& \sqrt{j(j+1)}\mathrm{鈩�}L\\ & =& \sqrt{\mathrm{鈩�}(\mathrm{鈩�}+1)}\mathrm{鈩�}\\ & & \end{array}$ 鈥︹��. (7)

With $\mathrm{鈩�}$being Planck's reduced constant, and j and I being the quantum numbers for total and orbital angular momentums, respectively.

If the spin-orbit interaction energy is proportional to the hydrogen energy, equations (1) and (3) would be related by a proportionality constant k .

$\begin{array}{rcl}U& =& kE\frac{{渭}_0{e}^2}{4蟺m_e^2{r}^3}(S.L)\\ & =& k\left(-\frac{m_e{e}^4}{2{\left(4蟺{蔚}_0\right)}^2{\mathrm{鈩�}}^2{n}^2}\right)\\ & =& -\frac{k{m}_e{e}^4}{2{\left(4蟺{蔚}_0\right)}^2{\mathrm{鈩�}}^2{n}^2}\\ & & \end{array}$

Here k supposedly involving the fine structure constant squared.

It is assumed that the radius of interaction r will be approximately the atomic radius, so equation (4) for r .

$$\begin{gathered} \frac{{渭}_0{e}^2}{4蟺m_e^2{r}^3}(S.L)=-\frac{k{m}_e{e}^4}{2{\left(4蟺{蔚}_0\right)}^2{\mathrm{鈩�}}^2{n}^2} \\ \frac{{渭}_0{e}^2}{4蟺m_e^2{\left(n^2{a}_0\right)}^3}(S.L)=-\frac{k{m}_e{e}^4}{2{\left(4蟺{蔚}_0\right)}^2{\mathrm{鈩�}}^2{n}^2} \\ \frac{{渭}_0{e}^2}{4蟺m_e^2{n}^6{a}_0^3}(S.L=-\frac{k{m}_e{e}^4}{2{\left(4蟺{蔚}_0\right)}^2{\mathrm{鈩�}}^2{n}^2} \end{gathered}$$

Equation (5)is men used to fill in the Bohr radius.

$\begin{array}{rcl}\frac{{渭}_0{e}^2}{4蟺m_e^2{n}^6{a}_0^3}(S.L)& =& -\frac{k{m}_e{e}^4}{2{\left(4蟺{蔚}_0\right)}^2{\mathrm{鈩�}}^2{n}^2}\\ \frac{{渭}_0{e}^2}{4蟺m_e^2{n}^6{\left(\frac{\left(4蟺c_0\right){\mathrm{鈩�}}^2}{m_e{e}^2}\right)}^3}(S.L)& =& -\frac{k{m}_e{e}^4}{2{\left(4蟺{蔚}_0\right)}^2{\mathrm{鈩�}}^2{n}^2}\\ \frac{{渭}_0{e}^2}{4蟺m_e^2{n}^6{\left(\frac{{\left(4蟺{蔚}_0\right)}^3{\mathrm{鈩�}}^2}{{m^3}_e{e}^2}\right)}^3}(S.L)& =& -\frac{k{m}_e{e}^4}{2{\left(4蟺{蔚}_0\right)}^2{\mathrm{鈩�}}^2{n}^2}\\ \frac{{渭}_0{m}_e{e}^8}{4蟺{\left(4蟺{蔚}_0\right)}^3{\mathrm{鈩�}}^6{n}^6}(S.L)& =& -\frac{k{m}_e{e}^4}{2{\left(4蟺{蔚}_0\right)}^2{\mathrm{鈩�}}^2{n}^2}\\ & & \end{array}$

Then the like terms are cancelled, and then rewritten as:

$$\begin{gathered} \frac{{渭}_0{m}_e{e}^8}{4蟺{\left(4{蟺}_{{蔚}_0}\right)}^3{\mathrm{鈩�}}^6{n}^6}(S.L)=-\frac{k{m}_e{e}^4}{2{\left(4{蟺}_{{蔚}_0}\right)}^2{\mathrm{鈩�}}^2{n}^2} \\ \frac{{渭}_0{e}^4}{4蟺\left(4蟺{蔚}_0\right){\mathrm{鈩�}}^4{n}^4}(S.L)=-\frac{k}{2} \\ \frac{k}{2}=-\frac{{渭}_0{e}^4}{4蟺\left(4蟺{蔚}_0\right){\mathrm{鈩�}}^4{n}^4}(S.L) \end{gathered}$$

The 2 with the k isn't cancelled so as to make a future part a little easier.

Equation (6) can then be used to rewrite the${铀�}_0$(rearrange the equation to give${渭}_0=\frac{1}{{蔚}_n{c}^2}$).

$\begin{array}{rcl}\frac{k}{2}& =& -\frac{{渭}_0{e}^4}{4蟺\left(4蟺{蔚}_0\right){\mathrm{鈩�}}^4{n}^4}(S.L)\\ & =& -\frac{\left(\frac{1}{c_0{c}^2}\right)e^4}{4蟺\left(4蟺{蔚}_0\right)n^4{n}^4}(S.L)\\ & =& -\frac{e^4}{4蟺{蔚}_0\left(4蟺{蔚}_0\right)c^2{n}^4{n}^4}(S.L)\frac{k}{2}\\ & =& -\frac{e^4}{{\left(4蟺{蔚}_0\right)}^2{c}^2{\mathrm{鈩�}}^4{n}^4}(S.L)\\ & & \end{array}$.....(8)

An expression for the spin-orbit coupling needs to be found, based on how me angular momentum vectors combine. The total angular momentum vector J can be written as $J=L卤S$.

$A卤$Is used to account for the vectors L and S being aligned or anti-aligned.

Then dot each side with itself and simplify further.

$$\begin{gathered} J=L卤S \\ J路J=(L卤S)(L卤S) \\ J路J=L路L卤2S路L+S路S \\ {J}^2=L^2+S^2卤2S路L \end{gathered}$$

$$\begin{gathered} J^2=L^2+S^2卤2S.L \\ {J}^2-L^2-S^2=卤2S.L \\ 卤\frac{1}{2}\left(J^2-L^2-S^2\right)=S.L \end{gathered}$$

That is then used to fill in for the spin-orbit couplingS.L in equation (8).

$\begin{array}{rcl}k& =& -\frac{2e^4}{{\left(4蟺{蔚}_0\right)}^2{c}^2{\mathrm{鈩�}}^4{n}^4}(S路L)\\ & =& -\frac{2e^4}{{\left(4蟺{蔚}_0\right)}^2{c}^2{\mathrm{鈩�}}^4{n}^4}\left[卤\frac{1}{2}\left(J^2-L^2-s^2\right)\right]\\ & =& 鈭�\frac{e^4}{{\left(4蟺{蔚}_0\right)}^2{c}^2{\mathrm{鈩�}}^4{n}^4}\left(J^2-L^2-s^2\right)\\ & & \end{array}$

Then fill in for J and L use equation ,(7) and that the magnitude of the electron spin S is$\frac{\sqrt{3}}{2}\mathrm{鈩�}$ with simplify and cancel the common${\mathrm{鈩�}}^{2\text{'}}s$

$\begin{array}{rcl}k& =& 鈭�\frac{e^4}{{\left(4蟺{蔚}_0\right)}^2{c}^2{\mathrm{鈩�}}^4{n}^4}\left(J^2-L^2-s^2\right)\\ & =& 鈭�\frac{e^4}{{\left(4蟺{蔚}_0\right)}^2{c}^2{\mathrm{鈩�}}^4{n}^4}\left[{\left(\sqrt{j(j+1)\mathrm{鈩�}}\right)}^2-{\left(\sqrt{l(l+1)\mathrm{鈩�}}\right)}^2-{\left(\frac{\sqrt{3}}{2}\mathrm{鈩�}\right)}^2\right]\\ & =& 鈭�\frac{e^4}{{\left(4蟺{蔚}_0\right)}^2{c}^2{\mathrm{鈩�}}^4{n}^4}\left[\left(j(j+1){\mathrm{鈩�}}^2\right)-\left(l(l+1){\mathrm{鈩�}}^2-\frac{3}{4}{\mathrm{鈩�}}^2\right)\right]\\ & =& 鈭�\frac{e^4}{{\left(4蟺{蔚}_0\right)}^2{c}^2{\mathrm{鈩�}}^4{n}^4}\left[\left(j(j+1)\right)-\left(l(l+1)-\frac{3}{4}\right)\right]\\ & & \end{array}$

The coefficient of the brackets can then just be simplified and rewritten in terms of the fine structure constant $铀�铀�$ from equation (2).

$\begin{array}{rcl}k& =& 鈭�\frac{e^4}{{\left(4蟺{蔚}_0\right)}^2{c}^2{\mathrm{鈩�}}^4{n}^4}\left[j(j+1)-\mathrm{鈩�}(\mathrm{鈩�}+1)-\frac{3}{4}\right]\\ & =& 鈭�{\left[\frac{e^2}{\left(4蟺{蔚}_0\right)蟺c}\right]}^2\left[\frac{j(j+1)-\mathrm{鈩�}(\mathrm{鈩�}+1)-\frac{3}{4}}{n^4}\right]\\ & =& 鈭�{伪}^2\left[\frac{j(j+1)-\mathrm{鈩�}(\mathrm{鈩�}+1)-\frac{3}{4}}{n^4}\right]\\ & & \end{array}$

So that shows that the magnetic interaction energy is proportional to the hydrogen electron energy via the fine structure constant squared, along the quantum numbers for the system.