The hydrogen spin-orbit interaction energy given in equation (8-25) is $\left({\mu }_0{e}^2/4\pi m_r^2{r}^3\right)\mathit{S}$. L. Using a reasonable value for in terms of $a_0$and the relationships $\mathit{S}\mathbf{=}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}$and $\mathit{L}\mathbf{=}\sqrt{\mathbf{\epsilon }\mathbf{(}\mathbf{\ell }\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{h}}$, show that this energy is proportional to a typical hydrogen atom energy by the factor${\mathit{\alpha }}^{\mathbf{2}}$ . where $\mathit{\alpha }$is the fine structure constant.

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.

……. (2)

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

With n being the principal quantum number, and being the Bohr radius, given as, . $a_0=\frac{\left(4\pi {\epsilon }_0\right){\hslash }^2}{m_e{e}^2}$ ……. (5)

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

The speed of light c in vacuum can be written as, $c^2=\frac{1}{{\mu }_0{\epsilon }_0}$. ……. (6)

With ${♢}_0$ and${♢}_0$ 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:

$\begin{array}{rcl}J& =& \sqrt{j(j+1)}\hslash L\\ & =& \sqrt{\ell (\ell +1)}\hslash \\ & & \end{array}$ ……. (7)

With $\hslash$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 .

$\begin{array}{rcl}U& =& kE\frac{{\mu }_0{e}^2}{4\pi m_e^2{r}^3}(S.L)\\ & =& k\left(-\frac{m_e{e}^4}{2{\left(4\pi {\epsilon }_0\right)}^2{\hslash }^2{n}^2}\right)\\ & =& -\frac{k{m}_e{e}^4}{2{\left(4\pi {\epsilon }_0\right)}^2{\hslash }^2{n}^2}\\ & & \end{array}$

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 .

$$\begin{gathered} \frac{{\mu }_0{e}^2}{4\pi m_e^2{r}^3}(S.L)=-\frac{k{m}_e{e}^4}{2{\left(4\pi {\epsilon }_0\right)}^2{\hslash }^2{n}^2} \\ \frac{{\mu }_0{e}^2}{4\pi m_e^2{\left(n^2{a}_0\right)}^3}(S.L)=-\frac{k{m}_e{e}^4}{2{\left(4\pi {\epsilon }_0\right)}^2{\hslash }^2{n}^2} \\ \frac{{\mu }_0{e}^2}{4\pi m_e^2{n}^6{a}_0^3}(S.L=-\frac{k{m}_e{e}^4}{2{\left(4\pi {\epsilon }_0\right)}^2{\hslash }^2{n}^2} \end{gathered}$$

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

$\begin{array}{rcl}\frac{{\mu }_0{e}^2}{4\pi m_e^2{n}^6{a}_0^3}(S.L)& =& -\frac{k{m}_e{e}^4}{2{\left(4\pi {\epsilon }_0\right)}^2{\hslash }^2{n}^2}\\ \frac{{\mu }_0{e}^2}{4\pi m_e^2{n}^6{\left(\frac{\left(4\pi c_0\right){\hslash }^2}{m_e{e}^2}\right)}^3}(S.L)& =& -\frac{k{m}_e{e}^4}{2{\left(4\pi {\epsilon }_0\right)}^2{\hslash }^2{n}^2}\\ \frac{{\mu }_0{e}^2}{4\pi m_e^2{n}^6{\left(\frac{{\left(4\pi {\epsilon }_0\right)}^3{\hslash }^2}{{m^3}_e{e}^2}\right)}^3}(S.L)& =& -\frac{k{m}_e{e}^4}{2{\left(4\pi {\epsilon }_0\right)}^2{\hslash }^2{n}^2}\\ \frac{{\mu }_0{m}_e{e}^8}{4\pi {\left(4\pi {\epsilon }_0\right)}^3{\hslash }^6{n}^6}(S.L)& =& -\frac{k{m}_e{e}^4}{2{\left(4\pi {\epsilon }_0\right)}^2{\hslash }^2{n}^2}\\ & & \end{array}$

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

$$\begin{gathered} \frac{{\mu }_0{m}_e{e}^8}{4\pi {\left(4{\pi }_{{\epsilon }_0}\right)}^3{\hslash }^6{n}^6}(S.L)=-\frac{k{m}_e{e}^4}{2{\left(4{\pi }_{{\epsilon }_0}\right)}^2{\hslash }^2{n}^2} \\ \frac{{\mu }_0{e}^4}{4\pi \left(4\pi {\epsilon }_0\right){\hslash }^4{n}^4}(S.L)=-\frac{k}{2} \\ \frac{k}{2}=-\frac{{\mu }_0{e}^4}{4\pi \left(4\pi {\epsilon }_0\right){\hslash }^4{n}^4}(S.L) \end{gathered}$$

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${♢}_0$(rearrange the equation to give${\mu }_0=\frac{1}{{\epsilon }_n{c}^2}$).

$\begin{array}{rcl}\frac{k}{2}& =& -\frac{{\mu }_0{e}^4}{4\pi \left(4\pi {\epsilon }_0\right){\hslash }^4{n}^4}(S.L)\\ & =& -\frac{\left(\frac{1}{c_0{c}^2}\right)e^4}{4\pi \left(4\pi {\epsilon }_0\right)n^4{n}^4}(S.L)\\ & =& -\frac{e^4}{4\pi {\epsilon }_0\left(4\pi {\epsilon }_0\right)c^2{n}^4{n}^4}(S.L)\frac{k}{2}\\ & =& -\frac{e^4}{{\left(4\pi {\epsilon }_0\right)}^2{c}^2{\hslash }^4{n}^4}(S.L)\\ & & \end{array}$.....(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 $J=L\pm S$.

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

Then dot each side with itself and simplify further.

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

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

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

$\begin{array}{rcl}k& =& -\frac{2e^4}{{\left(4\pi {\epsilon }_0\right)}^2{c}^2{\hslash }^4{n}^4}(S·L)\\ & =& -\frac{2e^4}{{\left(4\pi {\epsilon }_0\right)}^2{c}^2{\hslash }^4{n}^4}\left[\pm \frac{1}{2}\left(J^2-L^2-s^2\right)\right]\\ & =& \mp \frac{e^4}{{\left(4\pi {\epsilon }_0\right)}^2{c}^2{\hslash }^4{n}^4}\left(J^2-L^2-s^2\right)\\ & & \end{array}$

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

$\begin{array}{rcl}k& =& \mp \frac{e^4}{{\left(4\pi {\epsilon }_0\right)}^2{c}^2{\hslash }^4{n}^4}\left(J^2-L^2-s^2\right)\\ & =& \mp \frac{e^4}{{\left(4\pi {\epsilon }_0\right)}^2{c}^2{\hslash }^4{n}^4}\left[{\left(\sqrt{j(j+1)\hslash }\right)}^2-{\left(\sqrt{l(l+1)\hslash }\right)}^2-{\left(\frac{\sqrt{3}}{2}\hslash \right)}^2\right]\\ & =& \mp \frac{e^4}{{\left(4\pi {\epsilon }_0\right)}^2{c}^2{\hslash }^4{n}^4}\left[\left(j(j+1){\hslash }^2\right)-\left(l(l+1){\hslash }^2-\frac{3}{4}{\hslash }^2\right)\right]\\ & =& \mp \frac{e^4}{{\left(4\pi {\epsilon }_0\right)}^2{c}^2{\hslash }^4{n}^4}\left[\left(j(j+1)\right)-\left(l(l+1)-\frac{3}{4}\right)\right]\\ & & \end{array}$

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

$\begin{array}{rcl}k& =& \mp \frac{e^4}{{\left(4\pi {\epsilon }_0\right)}^2{c}^2{\hslash }^4{n}^4}\left[j(j+1)-\ell (\ell +1)-\frac{3}{4}\right]\\ & =& \mp {\left[\frac{e^2}{\left(4\pi {\epsilon }_0\right)\pi c}\right]}^2\left[\frac{j(j+1)-\ell (\ell +1)-\frac{3}{4}}{n^4}\right]\\ & =& \mp {\alpha }^2\left[\frac{j(j+1)-\ell (\ell +1)-\frac{3}{4}}{n^4}\right]\\ & & \end{array}$

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.