# 精細結構

$H=H^{(0)}+H_{kinetic}+H_{so}\,\!$

## 相對論性修正

$T=\frac{p^{2}}{2m}\,\!$

$T=\sqrt{p^{2}c^{2}+m^{2}c^{4}} - mc^{2}\,\!$

$T=\frac{p^{2}}{2m} - \frac{p^{4}}{8m^{3}c^{2}}+\dots\,\!$

$H_{kinetic}= - \frac{p^{4}}{8m^{3}c^{2}}\,\!$

$E_{n}^{(1)}=\langle\psi_n^{(0)}\vert H_{kinetic}\vert\psi_n^{(0)}\rangle= - \frac{1}{8m^{3}c^{2}}\langle\psi_n^{(0)}\vert p^{4}\vert\psi_n^{(0)}\rangle= - \frac{1}{8m^{3}c^{2}}\langle\psi_n^{(0)}\vert p^{2}p^{2}\vert\psi_n^{(0)}\rangle\,\!$

$H^{(0)}\vert\psi_n^{(0)}\rangle=E_{n}^{(0)}\vert\psi_n^{(0)}\rangle \,\!$

$\left(\frac{p^{2}}{2m}+V\right)\vert\psi_n^{(0)}\rangle =E_{n}^{(0)}\vert\psi_n^{(0)}\rangle \,\!$

$p^{2}\vert\psi_n^{(0)}\rangle=2m(E_{n}^{(0)} - V) \vert\psi_n^{(0)}\rangle\,\!$

\begin{align} E_{n}^{(1)} & = - \frac{1}{8m^{3}c^{2}}\langle\psi_n^{(0)}\vert p^{2}p^{2}\vert\psi_n^{(0)}\rangle \\ & = - \frac{1}{8m^{3}c^{2}}\langle\psi_n^{(0)}\vert (2m)^{2}(E_{n}^{(0)} - V)^{2}\vert\psi_n^{(0)}\rangle \\ & = - \frac{1}{2mc^{2}}[(E_{n}^{(0)})^{2} - 2E_{n}^{(0)}\langle V\rangle +\langle V^{2}\rangle] \\ \end{align}\,\!

$\langle V\rangle=\frac{Z^2 e^{2}}{4\pi\epsilon_0 a_{0}n^{2}}\,\!$
$\langle V^{2}\rangle=\frac{Z^4 e^{4}}{(l+1/2)(4\pi\epsilon_0 a_{0})^{2}n^{3}}\,\!$

\begin{align}E_{n}^{(1)} & = - \frac{1}{2mc^{2}}\left[(E_{n}^{(0)})^{2} - 2E_{n}^{(0)}\frac{Z^2 e^{2}}{4\pi\epsilon_0 a_{0}n^{2}} +\frac{Z^4 e^{4}}{(l+1/2)(4\pi\epsilon_0 a_{0})^{2}n^{3}}\right] \\ & = - \frac{(E_{n}^{(0)})^{2}}{2mc^{2}}\left(\frac{4n}{l+1/2} - 3\right) \\ \end{align} \,\!

## 自旋-軌道修正

$H_{so}=\frac{Ze^2}{8\pi\epsilon_0 m^2 c^2 r^3}\, (\mathbf{L}\cdot\mathbf{S}) \,\!$

$E_n^{(1)} =\frac{(E_n^{(0)})^2}{mc^2}\ \frac{2n[j(j+1) - l(l+1) - 3/4]}{l(l+1)(2l+1)}\,\!$

## 總合

$E_n^{(1)} = - \frac{(E_{n}^{(0)})^{2}}{2mc^{2}}\left(\frac{4n}{l+1/2} - 3\right) +\frac{(E_n^{(0)})^2}{mc^2}\ \frac{2n[j(j+1) - l(l+1) - 3/4]}{l(l+1)(2l+1)}\,\!$

$j\,\!$ 的這兩個數值分別代入總合方程式裏，經過一番運算，可以得到同樣的結果：

$E_n^{(1)} =\frac{(E_{n}^{(0)})^2}{mc^2}\left(\frac{3}{2} - \frac{4n}{2j+1}\right)\,\!$

$E_n =\frac{E_{1}^{(0)}}{n^2}\left(1 +\left(\frac{Z\alpha}{n}\right)^2 \left( \frac{2n}{2j+1} - \frac{3}{4}\right)\right)\,\!$ ;

## 更精确的结果

$E_n = -mc^2\left[1-\left(1+\left[\dfrac{Z\alpha}{n-j-\frac{1}{2}+\sqrt{\left(j+\frac{1}{2}\right)^2-Z^2\alpha^2}}\right]^2\right)^{-1/2}\right]$

## 注

1. ^ 参考英文版氢原子条目相关小节

## 參考文獻

1. ^ Griffiths, David J. Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. 2004: pp. 266–276. ISBN 0-13-111892-7.
2. ^
• Liboff, Richard L. Introductory Quantum Mechanics. Addison-Wesley. 2002. ISBN 0-8053-8714-5.