角动量量子化
矩阵力学方法¶
aha moment
1.量子化条件是 \([\hat{x},\hat{P}]=i\hbar\)
2.上升 下降算子
矩阵力学解简谐振子能量量子化¶
研究重点：掌握矩阵力学处理的思维方式
\[
V(x)=\frac 1 2kx^2=\frac 1 2m\omega ^2x^2
\]
Hint1
无单位化
Hint2
\(\hat{a},{\hat{a}}^+\)
Hint3
建立量子化条件
Hint4
\(\hat{H},\hat{a},{\hat{a}}^+\)之间的关系
Hint5
\(\hat{a},{\hat{a}}^+\)的意义
Hint6
基态方程式
Hint7
能量量子化
推导¶
\[
\hat{H}=\frac{\hat{P}^2}{2m}+\frac 1 2m\omega^2\hat{X}^2
\]
\[
\hat{q}=\sqrt{\frac{m\omega}{\hbar}}\hat{X}\,,\,\hat{p}=\frac 1 {\sqrt{m\hbar\omega}}\hat{P}
\]
\[
\Rightarrow \hat{H}=\frac 1 2\hbar\omega(\hat{p}^2+\hat{q}^2)
\tag{1}
\]
\[
\,
\]
\[
量子化条件\text{：}[\hat{X},\hat{P}]=i\hbar
\]
\[
\Rightarrow [\hat{q},\hat{p}]=\sqrt{\frac{m\omega}{\hbar}}\cdot\frac 1{\sqrt{m\hbar\omega}}[\hat{X},\hat{P}]=i
\tag{2}
\]
\[
灵感\text{：}\hat{a}=\frac{1}{\sqrt2}(\hat{q}+i\hat{p})\,\,\,\,\hat{a}^+=\frac{1}{\sqrt2}(\hat{q}-i\hat{p})
\]
\[
\,
\]
\[
\begin{align*}
[\hat{a},\hat{a}^+]&=[\frac{1}{\sqrt 2}(\hat{q}+i\hat{p}),\frac{1}{\sqrt 2}(\hat{q}-i\hat{p})]\\
&=i[\hat{p},\hat{q}]\\
&\overset{(2)}=i\cdot(-i)\\
&=1
\tag{3}
\end{align*}
\]
\[
\,
\]
\[
\begin{align*}
\hat{a}^+\hat{a}&=\frac 1 2(\hat{q}-\hat{p})(\hat{q}+i\hat{p})\\
&=\frac 1 2({\hat q}^2+{\hat p}^2)+\frac 1 2i\,[\hat{q}, \hat{p}]\\
&=\frac 1 2(\hat p^2+\hat q^2)-\frac 1 2
\end{align*}
\]
\[
\Rightarrow \hat H\overset{(1)}=\hbar \omega \cdot\frac 1 2(\hat p^2+\hat q^2)=\hbar \omega(\hat{a}^+\hat{a}+\frac 1 2)
\]
\([AB,C]=[A,C]B+A[B,C]\)
\[
\begin{align*}
[\hat{H},\hat{a}]&=\hbar \omega[\hat a^+\hat a,\hat a]\\
&=\hbar \omega([\hat a^+,\hat a]\hat a\,,\,\hat a^+[\hat a,\hat a])\\
&\overset{(3)}=-\hbar \omega \hat a
\tag{4}\\
[\hat H,\hat a^+]&=\hbar \omega[\hat a^+\hat a\,,\,\hat a^+]\\
&=\hbar \omega([\hat a^+\,,\,\hat a^+]\hat a+\hat a^+[\hat a\,,\,\hat a^+])\\
&\overset{(3)}=\hbar \omega \hat a^+
\end{align*}
\]
\[
\,
\]
\[
\hat H|E\rangle=E|E\rangle
\]
\[
\begin{align*}
\Rightarrow \hat H\hat a|E\rangle&\overset{(4)}=(\hat a\hat H-\hbar\omega\hat a)|E\rangle\\
&=\hat a\hat H|E\rangle-\hbar \omega \hat a|E\rangle\\
&=\hat aE|E\rangle-\hbar \omega \hat a|E\rangle\\
&=E\hat a|E\rangle-\hbar \omega \hat a|E\rangle\\
&=(E-\hbar\omega)\hat a|E\rangle
\end{align*}
\]
\[
\hat a|E\rangle对应的能量为E-\hbar\omega
\]
\[
故\hat a为下降算子，同理\hat a^+为上升算子
\]
\[
\,
\]
\[
最低能量状态为|u_0\rangle
\]
\[
满足\text{：}\hat a|u_0\rangle=0
\]
\[
则\text{：}\hat H|u_0\rangle=\hbar \omega(\hat a^+\hat a+\frac 1 2)|u_0\rangle=\frac 1 2\hbar \omega|u_0\rangle
\]
\[
E_{min}=\frac 1 2\hbar\omega
\]
\[
\,
\]
\[
从基态出发，根据\hat a^+的性质（每次增加\hbar\omega），可推出能量量子化
\]
角动量量子化¶
思路
1.找出角动量对应的运算子（两个）
2. 找出具有上升下降性质的共轭算子
3. 将这两个算子与z轴角动量对应的算子联系起来
4. 确定范围 并代入最小和最大的方程式
\[
\begin{align*}
[J_x,J_y]&=i\hbar J_z\\
[J^2,J_z]&=0
\end{align*}
\]
详细过程
\[
\boldsymbol{L}=\boldsymbol{r} \times \boldsymbol{P}=
\left[
\begin{matrix}
\hat{i}&\hat{j}&\hat{k}\\
r_x&r_y&r_z\\
P_x&P_y&P_z
\end{matrix}
\right]
\]
\[
\Rightarrow \hat{L_x}=\hat{Y}\hat{P_z}-\hat{Z}\hat{P_y}\,\,\hat{L_y}=\hat{Z}\hat{P_x}-\hat{X}\hat{P_z}\,\,
\hat{L_z}=\hat{X}\hat{P_y}-\hat{Y}\hat{P_x}
\]
\[
由\text{:}\,[\hat{x},\hat{P_x}]=i\hbar\,\,[\hat{y},\hat{P_y}]=i\hbar\,\,[\hat{z},\hat{P_z}]=i\hbar
\]
\[
\begin{align*}
\Rightarrow[\hat{L_x},\hat{L_y}]&=\hat L_x\hat L_y-\hat L_y\hat L_x \\
&=(\hat Y\hat P_z-\hat Z\hat P_y)(\hat Z\hat P_x-\hat X\hat P_z)-(\hat Z\hat P_x-\hat X\hat P_z)(\hat Y\hat P_z-\hat Z\hat P_y)\\
&=\hat P_z\hat Z\hat Y\hat P_x+\hat Z\hat P_z\hat P_y\hat X-\hat Z\hat P_z\hat P_x\hat Y-\hat P_z\hat Z\hat X\hat P_y+0\\
&=\hat P_z\hat Z(\hat Y\hat P_x-\hat X\hat P_y)-\hat Z\hat P_z(\hat Y\hat P_x-\hat X\hat P_y)\\
&=(\hat P_z\hat Z-\hat Z\hat P_z)(\hat Y\hat P_x-\hat X\hat P_y)\\
&=-i\hbar (-\hat L_z)\\
&=i\hbar \hat L_z
\end{align*}
\]
\[
\boldsymbol{{\hat J}^2={\hat J_x}^2+{\hat J_y}^2+{\hat J_z}^2}
\]
\[
\begin{align*}
\Rightarrow [{\hat J}^2,\hat J_x]&=[{\hat J_y}^2,\hat J_x]+[{\hat J_z}^2,\hat J_x]\\
&=[\hat J_y,\hat J_x]\hat J_y+\hat J_y[\hat J_y,\hat J_x]+[\hat J_z,\hat J_x]\hat J_z+\hat J_z[\hat J_z,\hat J_x]\\
&=-i\hbar \hat J_z\hat J_y-i\hbar \hat J_y\hat J_z+i\hbar \hat J_y\hat J_z+i\hbar \hat J_z\hat J_y\\
&=0
\end{align*}
\]
\[
同理\text{:}\,[{\hat J}^2,\hat J_z]=0
\]
\[
选定可同时测量的物理量\,{\hat J}^2和\hat J_z
\]
\[
J_+=J_x+iJ_y\,\,\,\,\,\,\,\,J_-=J_x-iJ_y
\]
\[
关系\text{：}[J_z,J_+]=\hbar J_+\,,\,[J_z,J_-]=-\hbar J_-
\]
\[
J_+J_-=J^2-J_z^2+\hbar J_z\,,\,J_-J_+=J^2-J_z^2-\hbar J_z
\]
\[
\begin{align*}
J|\lambda\,\,m\rangle&=\lambda|\lambda\,\,m\rangle\\
J_z|\lambda\,\,m\rangle&=m\hbar |\lambda \,\,m\rangle
\end{align*}\\
\]
下证\(\text{：}J_+\)为上升算子
\[
\begin{align*}
J_zJ_+|\lambda\,\,m\rangle&=J_+(J_z+\hbar)|\lambda \,\,m\rangle\\
&=(m+1)\hbar J_+|\lambda\,\,m\rangle
\end{align*}
\]
证毕.
\[
J^2=J_x^2+J_y^2+J_z^2
\]
\[
\Rightarrow \lambda-{(m\hbar)}^2\geq0
\]
\[
\begin{align*}
J_-|\lambda\,\,m_{min}\rangle&=0\\
J_+J_-|\lambda\,\,m_{min}\rangle&=0\\
\end{align*}
\]
\[
\begin{align*}
\lambda-m_{min}(m_{min}-1)\hbar ^2&=0\\
同理\text{:}\,\lambda-m_{max}(m_{max}+1)\hbar ^2&=0
\end{align*}
\]
\[
\Rightarrow m_{max}=-m_{min}=j
\]
\[
\Rightarrow \lambda=j(j+1)\hbar ^2
\]
含义
\(j(=0,1,2,\ldots,n-1)\)确定角量子数，在此基础上确定磁量子数的范围\(，-j\leq m\leq j\)
结论¶
\[
\boldsymbol{J^2 |j\,\,m\rangle=j(j+1)\hbar ^2|j\,\,m\rangle}
\]
$$
\boldsymbol{J_z|j\,\,m\rangle=m\hbar|j\,\,m\rangle}
$$
说明：角动量量子化
1.总角动量量子化 只能取离散的值\(\,\sqrt{j(j+1)}\,\hbar.\)
2.\(z\)角动量方向投影量子化，取值为\(\,m\hbar.\)