$
\phantom{V=(\langle 2p,m)\quad}2s\phantom{|V|2s\rangle\quad\langle 2s|\quad}2p\quad m=0\phantom{0\rangle\qquad}2p\quad m=0\phantom{\quad}2p\qquad m=-1 \\
V=\left(\begin{array}{cccc}
0 & \langle 2s|V|2p,m=0\rangle & 0 & 0 \\
\langle 2p,m=0|V|2s\rangle & 0 & 0 & 0 \\
0 & 0 & \phantom{\langle 2s|V|2}0\phantom{,m=0} & 0 \\
0 & 0 & 0 & \phantom{\langle 2s|V|2}0\phantom{,m=0}
\end{array}\right)
$

$
\phantom{V=(\langle 2p,m)\quad}2s\phantom{|V|2s\rangle\quad\langle 2s|\quad}2p\quad m=0\phantom{0\rangle\qquad}2p\quad m=0\phantom{a\qquad}2p\quad m=-1 \\
V=\left(\begin{array}{cccc}
0 & \langle 2s|V|2p,m=0\rangle & 0 & 0 \\
\langle 2p,m=0|V|2s\rangle & 0 & 0 & 0 \\
0 & 0 & \phantom{\langle 2s|V|2}0\phantom{,m=0} & 0 \\
0 & 0 & 0 & \phantom{\langle 2s|V|2}0\phantom{,m=0}
\end{array}\right)
$

$
\phantom{V=(\langle 2p,m)\quad}2s\phantom{|V|2s\rangle\quad\langle 2s|\quad}2p\quad m=0\phantom{0\rangle\qquad}2p\quad m=0\phantom{a\qquad}2p\quad m=-1 \\
V=\left(\begin{array}{cccc}
0 & \langle 2s|V|2p,m=0\rangle & 0 & 0 \\
\langle 2p,m=0|V|2s\rangle & 0 & 0 & 0 \\
0 & 0 & \phantom{\langle 2s|V|2|}0\phantom{,m=0} & 0 \\
0 & 0 & 0 & \phantom{\langle 2s|V|2|}0\phantom{,m=0}
\end{array}\right)
$

$
\phantom{V=(\langle 2p,m)\quad}2s\phantom{|V|2s\rangle\quad\langle 2s|\quad}2p\quad m=0\phantom{0\rangle\qquad}2p\quad m=0\phantom{a\qquad}2p\quad m=-1 \\
V=\left(\begin{array}{cccc}
0 & \langle 2s|V|2p,m=0\rangle & 0 & 0 \\
\langle 2p,m=0|V|2s\rangle & 0 & 0 & 0 \\
0 & 0 & \phantom{\langle 2s|V|2-}0\phantom{,m=0} & 0 \\
0 & 0 & 0 & \phantom{\langle 2s|V|2-}0\phantom{,m=0}
\end{array}\right)
$

$\phantom{V=(\langle 2p,m)\quad}2s\phantom{|V|2s\rangle\quad\langle 2s|\quad}2p\quad m=0\phantom{0\rangle\qquad}2p\quad m=0\phantom{a\qquad}2p\quad m=-1 \\
V=\left(\begin{array}{cccc}
0 & \langle 2s|V|2p,m=0\rangle & 0 & 0 \\
\langle 2p,m=0|V|2s\rangle & 0 & 0 & 0 \\
0 & 0 & \phantom{\langle 2s|V|2}0\phantom{,m=0} & 0 \\
0 & 0 & 0 & \phantom{\langle 2s|V|2}0\phantom{,m=0}
\end{array}\right)$

比如一个对谐振子的微扰H^=H0^+H′^\hat{H}=\hat{H_0}+\hat{H'}，其中 \\
H0^=−ℏ22μd2dx2+12kx2\hat{H_0}=-\dfrac{\hbar^2}{2\mu}\dfrac{d^2}{dx^2}+\dfrac{1}{2}kx^2，H′^=kx\hat{H'}=kx。
用 \\
⟨v′|x|v⟩=(ℏ2μω0)1/2[v+1−−−−√δv′v+1+v√δv′v−1]\langle v'|x|v\rangle=\left(\dfrac{\hbar}{2\mu \omega_0}\right)^{1/2} [\sqrt{v+1} \delta_{v'v+1}+\sqrt{v} \delta_{v'v-1}]
就可以得到 \\
E(1)v=⟨ψ(0)v|H′^|ψ(0)0⟩=b⟨ψ(0)v|x|ψ(0)v⟩=b⟨v|x|v⟩=0E_v^{(1)}=\langle \psi_v^{(0)} | \hat{H'} | \psi_0^{(0)} \rangle = b\langle \psi_v^{(0)} |x| \psi_v^{(0)} \rangle = b\langle v|x|v \rangle = 0\\
E(2)v=∑v′≠v|H′v′v|2E(0)v−E(0)v′=|H′v+1v|2E(0)v−E(0)v+1+|H′v−1v|2E(0)v−E(0)v−1=ℏb22μω0(|v+1−−−−√|2ℏω0(v+1/2)−ℏω0(v+1+1/2)+|v√|2ℏω0(v+1/2)−ℏω0(v−1+1/2))=ℏb22μω0(v+1−ℏω0+vℏω0)=−b22μω0=−b22kE_v^{(2)}=\sum_{v' \neq v} \dfrac{|H'_{v'v}|^2}{E_v^{(0)}-E_{v'}^{(0)}} =\dfrac{|H'_{v+1v}|^2}{E_v^{(0)}-E_{v+1}^{(0)}}+\dfrac{|H'_{v-1v}|^2}{E_v^{(0)}-E_{v-1}^{(0)}} = \dfrac{\hbar b^2}{2\mu \omega_0}\left(\dfrac{|\sqrt{v+1}|^2}{\hbar \omega_0(v+1/2)-\hbar \omega_0(v+1+1/2)}+\dfrac{|\sqrt{v}|^2}{\hbar \omega_0(v+1/2)-\hbar \omega_0(v-1+1/2)}\right) = \dfrac{\hbar b^2}{2\mu \omega_0}\left(\dfrac{v+1}{-\hbar \omega_0}+\dfrac{v}{\hbar \omega_0}\right)=-\dfrac{b^2}{2\mu \omega_0} = -\dfrac{b^2}{2k}\\
这里面用到了ω0=kμ−−√\omega_0 = \sqrt{\dfrac{k}{\mu}}\\
所以有 \\
Ev=E(0)v+E(1)v+E(2)v=ℏω0(v+12)−b22kE_v = E_v^{(0)} + E_v^{(1)} + E_v^{(2)} = \hbar \omega_0(v+\frac{1}{2})-\dfrac{b^2}{2k}

或者尝试求严格解： \\
H^=H0^+H′^=−ℏ22μd2dx2+12kx2+bx=−ℏ22μd2dx2+12k(x2+2bkx)=−ℏ22μd2dx2+12k(x+bk)2−b22k\hat{H}=\hat{H_0}+\hat{H'}=-\dfrac{\hbar^2}{2\mu}\dfrac{d^2}{dx^2}+\dfrac{1}{2}kx^2+bx=-\dfrac{\hbar^2}{2\mu}\dfrac{d^2}{dx^2}+\dfrac{1}{2}k(x^2+\dfrac{2b}{k}x)=-\dfrac{\hbar^2}{2\mu}\dfrac{d^2}{dx^2}+\dfrac{1}{2}k(x+\dfrac{b}{k})^2-\dfrac{b^2}{2k} \\
令y=x+bky=x+\dfrac{b}{k} \\
有H^=−ℏ22μd2dy2+12ky2−b22k\hat{H}=-\dfrac{\hbar^2}{2\mu}\dfrac{d^2}{dy^2}+\dfrac{1}{2}ky^2-\dfrac{b^2}{2k} \\
于是有: \\
Eexactv=ℏω0(v+12)−b22kE_v^{exact}=\hbar\omega_0(v+\dfrac{1}{2})-\dfrac{b^2}{2k}

$\displaystyle\int \mathcal{Y}^{\dag} \boldsymbol{S\cdot L}\mathcal{Y}d\Omega= \dfrac{1}{2}\left[j(j+1)-l(l+1)-\dfrac{3}{4}\right]\hbar^2 =\dfrac{\hbar^2}{2}\left\{\begin{array}{c}
l \\
-(l+1)\end{array}\right\}\begin{array}{c}
j=l+\frac{1}{2} \\
j=l-\frac{1}{2} \end{array}$