Question: Derive the fine structure formula (Equation 6.66) from the relativistic correction (Equation 6.57) and the spin-orbit coupling (Equation 6.65). Hint: Note tha $\mathrm{j}=\mathrm{l}\pm \frac{1}{2}$t; treat the plus sign and the minus sign separately, and you'll find that you get the same final answer either way.

The relativistic correction in the energy levels is:

$E_r^{\text{'}}=\frac{-(E_n{)}^2}{\left(2m{c}^2\right)}\left[\frac{4n}{\mathrm{l}+{\displaystyle \frac{1}{2}}}-3\right]$

And, the spin-orbit coupling: ${\mathbf{E}}_{50}^{\mathbf{\text{'}}}\mathbf{=}\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mc}}^{\mathbf{2}}}\mathbf{\left[}\frac{\mathbf{n}\mathbf{\left[}\mathbf{j}\mathbf{(}\mathbf{j}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{-}\mathbf{l}\mathbf{(}\mathbf{l}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{-}{\displaystyle \frac{3}{4}}\mathbf{\right]}}{\mathbf{l}\mathbf{}\mathbf{(}\mathbf{l}\mathbf{+}{\displaystyle \frac{1}{2}}\mathbf{)}\mathbf{(}\mathbf{l}\mathbf{+}\mathbf{1}\mathbf{)}}\mathbf{\right]}$

Where,

$$\begin{gathered} \mathrm{j}=\mathrm{l}\pm \frac{1}{2} \\ \Rightarrow \mathrm{l}=\mathrm{j}\pm \frac{1}{2} \end{gathered}$$

The equations 6.65, 6.66, 6.67 are

role="math" localid="1658296542883" $$\begin{gathered} {\mathrm{E}}_{\mathrm{s}0}^1=\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\mathbf{\left[}\frac{\mathbf{n}\mathbf{\left[}\mathbf{j}\mathbf{(}\mathbf{j}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{-}\mathbf{l}\mathbf{(}\mathbf{l}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{-}{\displaystyle \frac{3}{4}}\mathbf{\right]}}{\mathbf{l}\mathbf{}\mathbf{(}\mathbf{l}\mathbf{+}{\displaystyle \frac{1}{2}}\mathbf{)}\mathbf{(}\mathbf{l}\mathbf{+}\mathbf{1}\mathbf{)}}\mathbf{\right]}......(6.65) \\ {\mathrm{E}}_{\mathrm{fs}}^1=\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mc}}^{\mathbf{2}}}\mathbf{[}\mathbf{3}\mathbf{-}\frac{\mathbf{4}\mathbf{n}}{\mathbf{j}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}}\mathbf{]}....(6.66) \\ \mathrm{E}=\frac{13.6\mathrm{eV}}{{\mathrm{n}}^{\mathbf{2}}}\mathbf{[}\mathbf{a}\mathbf{+}\frac{{\mathbf{\alpha }}^{\mathbf{2}}}{{\mathbf{n}}^{\mathbf{2}}}\left(\frac{4n}{j+{\displaystyle \frac{1}{2}}}-\frac{3}{4}\right)\mathbf{]}....(6.67) \end{gathered}$$

Take$l=j-\frac{1}{2}$ and substitute into the relativistic correction equation:

${\mathrm{E}}_{\mathrm{r}}^{\text{'}}=\frac{{\mathbf{-}\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mc}}^{\mathbf{2}}}\mathbf{[}\frac{\mathbf{4}\mathbf{n}}{\mathbf{l}\mathbf{-}{\displaystyle \frac{1}{2}}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}}\mathbf{-}\mathbf{3}\mathbf{]}$

And the spin-orbit coupling has the form:

$$\begin{gathered} {\mathrm{E}}_{so}^{\text{'}}=\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\mathbf{\left[}\frac{\mathbf{n}\left[j(j+1)-\left(j-{\displaystyle \frac{1}{2}}\right)(j-\frac{1}{2}+1)\right]\mathbf{-}{\displaystyle \frac{\mathbf{3}}{\mathbf{4}}}}{\left(j-{\displaystyle \frac{1}{2}}\right)\mathbf{(}\mathbf{j}\mathbf{-}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{)}\mathbf{(}\mathbf{j}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}}\mathbf{\right]} \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\mathbf{\left[}\frac{\mathbf{n}\mathbf{\left[}\mathbf{j}\mathbf{(}\mathbf{j}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{-}\left(j-{\displaystyle \frac{1}{2}}\right)\mathbf{(}\mathbf{j}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathbf{\right]}\mathbf{-}{\displaystyle \frac{\mathbf{3}}{\mathbf{4}}}}{\mathbf{j}\mathbf{(}\mathbf{j}\mathbf{-}{\displaystyle \frac{1}{2}}\mathbf{)}\mathbf{(}\mathbf{j}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{)}}\mathbf{\right]} \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\mathbf{\left[}\frac{\mathbf{n}\mathbf{\left[}{\mathbf{j}}^{\mathbf{2}}\mathbf{+}\mathbf{j}\mathbf{-}\mathbf{(}{\mathbf{j}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{4}\mathbf{)}\mathbf{\right]}\mathbf{-}{\displaystyle \frac{\mathbf{3}}{\mathbf{4}}}}{\mathbf{j}\mathbf{(}{\mathbf{j}}^{\mathbf{2}}\mathbf{+}{\displaystyle \frac{1}{4}}\mathbf{)}}\mathbf{\right]} \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\mathbf{\left[}\frac{\mathbf{n}\left[j-{\displaystyle \frac{1}{2}}\right]}{\mathbf{j}\mathbf{(}\mathbf{j}\mathbf{-}{\displaystyle \frac{1}{2}}\mathbf{)}\mathbf{(}\mathbf{j}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{)}}\mathbf{\right]} \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\mathbf{\left[}\frac{\mathbf{n}}{\mathbf{j}\mathbf{(}\mathbf{j}\mathbf{-}{\displaystyle \frac{1}{2}}\mathbf{)}}\mathbf{\right]} \end{gathered}$$

Now, calculate the fine structure formula

$$\begin{gathered} E_{fs}^{\text{'}}=E_r^{\text{'}}+E_{so}^{\text{'}} \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{4n}{j}-3\right]+\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{2n}{j\left(j+{\displaystyle \frac{1}{2}}\right)}\right] \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{2n}{j\left(j+{\displaystyle \frac{1}{2}}\right)}-\frac{4n}{j}+3\right] \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{2nj-4n\left(j+\frac{1}{2}\right)}{j^2\left(j+\frac{1}{2}\right)}+3\right] \end{gathered}$$

Solve further the equation

$$\begin{gathered} E_{fs}^{\text{'}}=\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{2nj-4n^2-2nj}{j^2\left(j+\frac{1}{2}\right)}+3\right] \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\left[3-\frac{4n}{\left(j+\frac{1}{2}\right)}\right] \end{gathered}$$

Now take $l=j+\frac{1}{2}$ and substitute into the relativistic correction equation:

And the spin-orbit coupling has the form,

$$\begin{gathered} E_{so}^{\text{'}}=\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{n\left[j\left(j+1\right)-\left(j+\frac{1}{2}\right)\left(j+\frac{1}{2}+1\right)-{\displaystyle \frac{3}{4}}\right]}{\left(j+\frac{1}{2}\right)\left(j+\frac{1}{2}+\frac{1}{2}\right)\left(j+\frac{1}{2}+1\right)}\right] \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{n\left[j^2+j-j^2-{\displaystyle \frac{3}{2}j-\frac{1}{2}j-\frac{3}{4}-\frac{3}{4}}\right]}{\left(j+\frac{1}{2}\right)\left(j+1\right)\left(j+\frac{3}{2}\right)}\right] \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{n\left[-j-{\displaystyle \frac{3}{2}}\right]}{\left(j+\frac{1}{2}\right)\left(j+1\right)\left(j+\frac{3}{2}\right)}\right] \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{-n\left[j+{\displaystyle \frac{3}{2}}\right]}{\left(j+\frac{1}{2}\right)\left(j+1\right)\left(j+\frac{3}{2}\right)}\right] \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{-n}{\left(j+\frac{1}{2}\right)\left(j+1\right)}\right] \end{gathered}$$

Now, calculate the fine structure formula:

$$\begin{gathered} E_{fs}^{\text{'}}=E_r^{\text{'}}+E_{so}^{\text{'}} \\ =\frac{{\mathbf{-}\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{4n}{\left(j+1\right)}-3\right]-\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mc}}^{\mathbf{2}}}\left[\frac{-2n}{\left(j+1\right)\left(j+{\displaystyle \frac{1}{2}}\right)}\right] \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mc}}^{\mathbf{2}}}\left[3-\frac{4n}{\left(j+1\right)}+\frac{2n}{\left(j+1\right)\left(j+{\displaystyle \frac{1}{2}}\right)}\right] \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mc}}^{\mathbf{2}}}\left[3-\frac{4n\left(j+1\right)\left(j+\frac{1}{2}\right)+2n\left(j+1\right)}{\left(j+1\right)\left(j+1\right)\left(j+\frac{1}{2}\right)}\right] \end{gathered}$$

Solve the equation further

$$\begin{gathered} =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mc}}^{\mathbf{2}}}\left[3-\frac{4n+2n+2n}{\left(j+1\right)\left(j+\frac{1}{2}\right)}\right] \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mc}}^{\mathbf{2}}}\left[3-\frac{4n\left(j+1\right)}{\left(j+1\right)\left(j+\frac{1}{2}\right)}\right] \\ =\frac{{\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{mc}}^{\mathbf{2}}}\left[3-\frac{4n}{\left(j+\frac{1}{2}\right)}\right] \end{gathered}$$

This is the same answer for $l=j-\frac{1}{2}$.