Question: The exact fine-structure formula for hydrogen (obtained from the Dirac equation without recourse to perturbation theory) is 16

${\mathbf{E}}_{\mathbf{nj}}\mathbf{=}{\mathbf{mc}}^{\mathbf{2}}\left\{{\left[1+{\left(\frac{a}{n-\left(j+{\displaystyle \frac{1}{2}}\right)+\sqrt{{\left(j+\frac{1}{2}\right)}^2-a^2}}\right)}^2\right]}^{-\frac{1}{2}}-1\right\}$

Expand to order ${\mathbf{\alpha }}^{\mathbf{4}}$(noting that $\mathbf{\alpha }\mathbf{\ll }\mathbf{1}$), and show that you recoverEquation .

The exact fine structure formula for Hydrogen is:

${\mathrm{E}}_{\mathrm{nj}}={\mathrm{mc}}^2\{{\left[1+{\left(\frac{\mathrm{a}}{\mathrm{n}-\left(\mathrm{j}+{\displaystyle \frac{1}{2}}\right)+\sqrt{{\left(\mathrm{j}+\frac{1}{2}\right)}^2-{\mathrm{a}}^2}}\right)}^2\right]}^{\frac{1}{2}}-1\}$

To expand w.r.t. , use Taylor expansion, first arrange the function, the term:

$\sqrt{{\left(\mathrm{j}+\frac{1}{2}\right)}^2-{\mathrm{a}}^2}=\left(\mathrm{j}+\frac{1}{2}\right)\sqrt{1-{\left(\frac{a}{\left(\mathrm{j}+\frac{1}{2}\right)}\right)}^2}$

Now, expand this term by Taylor expansion:

Choose,$x_0=0,x=\frac{a}{\left(\mathrm{j}+\frac{1}{2}\right)}\&f\left(x\right)\sqrt{1-{\left(\frac{a}{\left(\mathrm{j}+\frac{1}{2}\right)}\right)}^2}$

$$\begin{gathered} f\left(x\right)=f\left(x_0\right)+(x-x_0)f^{\text{'}}{\overline{)\left(x\right)}}_{x-x_0}+{\overline{)\frac{\left(x-x_0\right)}{2!}f\left(x\right)}}_{x-x_0}+... \\ \frac{d}{dx}\left(\sqrt{1-x^2}\right)=\frac{-2x}{2\sqrt{1-x^2}} \\ f\left(x\right){\overline{)}}_{x-x_0}=zero \end{gathered}$$

And,

$$\begin{gathered} \frac{d^2}{d{x}^2}\left(\sqrt{1-x^2}\right)=-{\left(1-x^2\right)}^{\frac{-1}{2}}+\frac{x}{2}-{\left(1-x^2\right)}^{\frac{-3}{2}}.(-2x) \\ =\frac{-1}{\sqrt{1-x^2}}-\frac{x^2}{{\left(1-x^2\right)}^{\frac{3}{2}}} \\ \Rightarrow f^{"}\left(x\right){|}_{x-x_0}=-1-zero \\ =-1 \end{gathered}$$

Use Taylor expansion

$$\begin{gathered} \sqrt{1-x^2}=1+\frac{x}{1!}{f}^{\text{'}}\left(x\right){|}_{x-x_0}+\frac{x^2}{2!}{f}^{"}\left(x\right){|}_{x-x_0}+... \\ \approx 1+zero+\frac{x^2}{2!}(-1)+... \\ \Rightarrow \sqrt{1-x^2} \\ =1-\frac{x^2}{2} \end{gathered}$$

Then,

$$\begin{gathered} \sqrt{{\left(\mathrm{j}+\frac{1}{2}\right)}^2-{\mathrm{a}}^2}=\left(\mathrm{j}+\frac{1}{2}\right)\sqrt{1-{\left(\frac{a}{\left(\mathrm{j}+\frac{1}{2}\right)}\right)}^2} \\ \approx \left(\mathrm{j}+\frac{1}{2}\right)\left[1-\frac{1}{2}{\left(\frac{a}{\left(\mathrm{j}+\frac{1}{2}\right)}\right)}^2\right] \\ \approx \left(\mathrm{j}+\frac{1}{2}\right)-\frac{1}{2}\frac{a^2}{\mathrm{j}+\frac{1}{2}} \end{gathered}$$

Now, write the full terms as:

$$\begin{gathered} {\left(\frac{a}{n-\left(\mathrm{j}+\frac{1}{2}\right)\sqrt{{\left(\mathrm{j}+\frac{1}{2}\right)}^2-{\mathrm{a}}^2}}\right)}^2\approx {\left(\frac{a}{n-\left(\mathrm{j}+\frac{1}{2}\right)+\left(\mathrm{j}+\frac{1}{2}\right)-{\displaystyle \frac{a^2}{2\left(\mathrm{j}+\frac{1}{2}\right)}}}\right)}^2 \\ \approx {\left(\frac{a}{n-{\displaystyle \frac{a^2}{2\left(\mathrm{j}+\frac{1}{2}\right)}}}\right)}^2 \end{gathered}$$

Expand the term:

$\frac{a}{n-{\displaystyle \frac{a^2}{2\left(\mathrm{j}+\frac{1}{2}\right)}}}$

First, re-arrange that term:

$\frac{\alpha }{n}\left[\frac{1}{1-\left(n-{\displaystyle \frac{a^2}{2n\left(\mathrm{j}+\frac{1}{2}\right)}}\right)}\right]=\frac{\alpha }{n}{\left[1-\left(\frac{a^2}{2n\left(\mathrm{j}+\frac{1}{2}\right)}\right)\right]}^{-1}$

Take,

$$\begin{gathered} x=\frac{a^2}{2n\left(\mathrm{j}+\frac{1}{2}\right)}\Rightarrow f\left(x\right)=\frac{1}{{\left(1-x\right)}^{\text{'}}},x_0=0 \\ {f}^{\text{'}}\left(x\right)=\frac{1}{{\left(1-x\right)}^2} \\ \Rightarrow \mathbf{}{f}^{\text{'}}\left(x\right){|}_{x-x_0}=1 \end{gathered}$$

And,

$$\begin{gathered} \mathbf{}{f}^{\text{'}}\left(x\right)=\frac{-2\left(1-x\right)\left(-1\right)}{{\left(1-x\right)}^4} \\ =\frac{2-2x}{{\left(1-x\right)}^4} \\ \mathbf{}{f}^{"}\left(x\right){|}_{x-x_0}=2 \end{gathered}$$

Solve further

$$\begin{gathered} \frac{1}{1-x}=f\left(x_0\right)+\frac{x}{1!}{f}^{\text{'}}\left(x\right){|}_{x-x_0}+\frac{x^2}{2!}{f}^{"}\left(x\right)\mathbf{}{|}_{x-x_0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\approx 1+x+x^2 \\ \Rightarrow \frac{1}{1-x} \\ \approx 1+x \end{gathered}$$

Then,

$\frac{\alpha }{n}{\left[1-\left(\frac{a^2}{2n\left(\mathrm{j}+\frac{1}{2}\right)}\right)\right]}^{-1}\approx \frac{\alpha }{n}\left[1+\left(\frac{a^2}{2n\left(\mathrm{j}+\frac{1}{2}\right)}\right)\right]$

Now, write:

${\left(\frac{a}{n-{\displaystyle \frac{a^2}{2n\left(\mathrm{j}+\frac{1}{2}\right)}}}\right)}^2\approx {\left(\frac{\alpha }{n}\left[1+\frac{a^2}{2n\left(\mathrm{j}+\frac{1}{2}\right)}\right]\right)}^2$

Now, the exact fine structure can take the form:

$\left[{\left(1+{\left(\frac{\alpha }{n}\left[1+\frac{a^2}{2n\left(\mathrm{j}+\frac{1}{2}\right)}\right]\right)}^2\right)}^{\frac{1}{2}}-1\right]$

Preserve the order ${\alpha }^4$, so

width="359">αn1+a22nj+122=α2n21+a22nj+12+0(a4)≈α2n21+a2nj+12

Now, expand the term,

${\left[1+\frac{{\alpha }^2}{n^2}\left(1+\frac{a^2}{n\left(\mathrm{j}+\frac{1}{2}\right)}\right)\right]}^{\frac{-1}{2}}$

Take:

$$\begin{gathered} x=\frac{{\alpha }^2}{n^2}\left(1+\frac{a^2}{n\left(\mathrm{j}+\frac{1}{2}\right)}\right),x_0=0 \\ f\left(x\right)=\frac{1}{\sqrt{1+x}}={\left(1+x\right)}^{\frac{-1}{2}} \\ {f}^{\text{'}}\left(x\right)=\frac{-1}{2}{(1+x)}^{{}^{\frac{-3}{2}}}\Rightarrow f^{\text{'}}\left(x\right){|}_{x-x_0}=\frac{-1}{2} \\ {f}^{"}\left(x\right)=\frac{3}{4}{(1+x)}^{{}^{\frac{-5}{2}}} \\ \Rightarrow f^{\text{'}}\left(x\right){|}_{x-x_0}=\frac{3}{4} \\ f\left(x\right)=1+\frac{x}{1!}\left(\frac{-1}{2}\right)+\frac{x^2}{2!}\left(\frac{3}{4}\right)+... \\ f\left(x\right)\approx 1-\frac{1}{2}x+\frac{3}{8}{x}^2 \\ \Rightarrow {\left[1+\frac{{\alpha }^2}{n^2}\left(1+\frac{a^2}{n\left(\mathrm{j}+\frac{1}{2}\right)}\right)\right]}^{\frac{-1}{2}} \end{gathered}$$

Solve further,

$$\begin{gathered} \approx 1-\frac{1}{2}\frac{{\alpha }^2}{n^2}\left(1+\frac{a^2}{n\left(\mathrm{j}+\frac{1}{2}\right)}\right)+\frac{3}{8}\frac{{\alpha }^4}{n^4}\left(1+\frac{a^2}{n\left(\mathrm{j}+\frac{1}{2}\right)}\right) \\ \approx 1-\frac{1}{2}\frac{{\alpha }^2}{n^2}\left(1+\frac{a^2}{n\left(\mathrm{j}+\frac{1}{2}\right)}\right)+\frac{3}{8}\frac{{\alpha }^4}{n^4}+0\left(a^6\right) \end{gathered}$$

Then,

$$\begin{gathered} E_{mj}\approx m{c}^2\left[1-\frac{1}{2}\frac{{\alpha }^2}{n^2}\left(1+\frac{a^2}{n\left(\mathrm{j}+\frac{1}{2}\right)}\right)+\frac{3}{8}\frac{{\alpha }^4}{n^4}-1\right] \\ \approx \frac{-m{c}^2{a}^2}{2n^2}\left[1+\frac{a^2}{n\left(\mathrm{j}+\frac{1}{2}\right)}+\frac{3}{8}\frac{{\alpha }^2}{n^2}\right] \\ \approx \frac{-m{c}^2{a}^2}{2n^2}\left[1+\frac{a^2}{n^2}\left(\frac{n}{\mathrm{j}+\frac{1}{2}}-\frac{3}{4}\right)\right] \end{gathered}$$

$$\begin{gathered} \alpha =\frac{1}{137.036} \\ m{c}^2=0.511MeV \\ {E}_{nj}=\frac{-13.6}{n^2}1+\frac{a^2}{n^2}\left(\frac{n}{\mathrm{j}+\frac{1}{2}}-\frac{3}{4}\right) \end{gathered}$$

As the same as equation (6.67).