# 自旋-轨道作用本文重定向自 自旋-軌道作用

（重定向自LS耦合）

## 电子的自旋-轨道作用

### 磁场

${\displaystyle \mathbf {B} =-\,{\frac {\mathbf {v} \times \mathbf {E} }{c^{2}}}\,\!}$(1)

${\displaystyle \mathbf {E} ={\frac {Ze}{4\pi \epsilon _{0}r^{2}}}{\hat {\mathbf {r} }}={\frac {Ze}{4\pi \epsilon _{0}r^{3}}}\mathbf {r} \,\!}$

${\displaystyle \mathbf {p} =m\mathbf {v} \,\!}$

${\displaystyle \mathbf {B} ={\frac {Ze}{4\pi \epsilon _{0}mc^{2}r^{3}}}\,\mathbf {r} \times \mathbf {p} ={\frac {Ze}{4\pi \epsilon _{0}mc^{2}r^{3}}}\,\mathbf {L} \,\!}$(2)

${\displaystyle \mathbf {B} \,\!}$ 是一个正值因子乘以 ${\displaystyle \mathbf {L} \,\!}$ ，也就是说，磁场与电子的轨道角动量平行。

### 磁矩

${\displaystyle {\boldsymbol {\mu }}=\gamma \,\mathbf {S} \,\!}$

${\displaystyle {\boldsymbol {\mu }}=-{\frac {e}{m}}\mathbf {S} \,\!}$(3)

### 哈密顿量摄动项目

${\displaystyle H'=-{\boldsymbol {\mu }}\cdot \mathbf {B} \,\!}$

${\displaystyle H'={\frac {Ze^{2}}{4\pi \epsilon _{0}m^{2}c^{2}}}\ {\frac {\mathbf {L} \cdot \mathbf {S} }{r^{3}}}\,\!}$

${\displaystyle H'={\frac {Ze^{2}}{8\pi \epsilon _{0}m^{2}c^{2}}}\ {\frac {\mathbf {L} \cdot \mathbf {S} }{r^{3}}}\,\!}$

### 能级位移

${\displaystyle \mathbf {J} =\mathbf {L} +\mathbf {S} \,\!}$

${\displaystyle \mathbf {J} ^{2}=\mathbf {L} ^{2}+\mathbf {S} ^{2}+2\mathbf {L} \cdot \mathbf {S} \,\!}$

${\displaystyle \mathbf {L} \cdot \mathbf {S} ={1 \over 2}(\mathbf {J} ^{2}-\mathbf {L} ^{2}-\mathbf {S} ^{2})\,\!}$

{\displaystyle {\begin{aligned}\langle n,j,l,s\,|\,\mathbf {L} \cdot \mathbf {S} \,|\,n,j,l,s\rangle &={1 \over 2}(\langle \mathbf {J} ^{2}\rangle -\langle \mathbf {L} ^{2}\rangle -\langle \mathbf {S} ^{2}\rangle )\\&={\hbar ^{2} \over 2}[j(j+1)-l(l+1)-s(s+1)]\\&={\hbar ^{2} \over 2}[j(j+1)-l(l+1)-3/4]\\\end{aligned}}\,\!}

${\displaystyle \langle n,j,l,s\,|\,r^{-3}\,|\,n,j,l,s\rangle ={\frac {2Z^{3}}{a_{0}^{3}n^{3}l(l+1)(2l+1)}}\,\!}$

${\displaystyle E_{n}^{(1)}={\frac {Z^{4}e^{2}\hbar ^{2}}{8\pi \epsilon _{0}m^{2}c^{2}a_{0}^{3}}}\ {\frac {[j(j+1)-l(l+1)-3/4]}{n^{3}\,l(l+1)(2l+1)}}\,\!}$

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

${\displaystyle Y_{0}^{0}={\frac {1}{\sqrt {4\pi }}}\,\!}$