附录：计算对易子
以下计算要反复用到对易子的乘法性质：
$$
\begin{aligned}
\bigl[ \hat{X}, \hat{Y} \hat{Z} \bigr] = \bigl[ \hat{X}, \hat{Y} \bigr] \hat{Z} + \hat{Y} \bigl[ \hat{X}, \hat{Z} \bigr], \\
\bigl[ \hat{X} \hat{Y}, \hat{Z} \bigr] = \bigl[ \hat{X}, \hat{Z} \bigr] \hat{Y} + \hat{X} \bigl[ \hat{Y}, \hat{Z} \bigr].
\end{aligned}
$$
练习 14.4：$\hat{\boldsymbol{A}}$ 和 $\hat{H}$ 的对易关系
证明对易关系 (14.8)：$\bigl[ \hat{\boldsymbol{A}}, \hat{H} \bigr] = \boldsymbol{0}$。
证明辅助式 (1)：
$$
\displaystyle \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}} = 2 \hat{\boldsymbol{p}}^{2} \boldsymbol{r} - 2 (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \hat{\boldsymbol{p}}.
$$
式 (1) 的证明：
$$
\begin{aligned}
\bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{k} &= \epsilon_{i, j, k} \hat{p}_{i} \hat{L}_{j} \\
&\stackrel{\flat}{=} \epsilon_{i, j, k} \epsilon_{j, r, s} \hat{p}_{i} x_{r} \hat{p}_{s} \\
&= (\Lambda_{k, r} \Lambda_{i, s} - \Lambda_{k, s} \Lambda_{i, r}) \hat{p}_{i} x_{r} \hat{p}_{s} \\
&= \hat{p}_{i} x_{k} \hat{p}_{i} - \hat{p}_{i} x_{i} \hat{p}_{k} \\
&= \hat{p}_{i} (\hat{p}_{i} x_{k} + \mathrm{i} \hbar \Lambda_{k, i}) - \hat{p}_{i} x_{i} \hat{p}_{k} \\
&= \hat{p}_{i} \hat{p}_{i} x_{k} - \hat{p}_{i} x_{i} \hat{p}_{k} + \mathrm{i} \hbar \hat{p}_{k} \\
&= \hat{\boldsymbol{p}}^{2} x_{k} - (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \hat{p}_{k} + \mathrm{i} \hbar \hat{p}_{k},
\end{aligned}
$$
同理，
$$
\begin{aligned}
\bigl( \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}} \bigr)_{k} &= \epsilon_{i, j, k} \hat{L}_{i} \hat{p}_{j} \\
&= \epsilon_{i, j, k} \epsilon_{i, r, s} x_{r} \hat{p}_{s} \hat{p}_{j} \\
&= \epsilon_{i, j, k} \epsilon_{i, r, s} x_{r} \hat{p}_{j} \hat{p}_{s} \\
&= \epsilon_{i, j, k} \epsilon_{i, r, s} (\hat{p}_{j} x_{r} + \mathrm{i} \hbar \Lambda_{r, j}) \hat{p}_{s} \\
&= \epsilon_{i, j, k} \epsilon_{i, r, s} \hat{p}_{j} x_{r} \hat{p}_{s} + \mathrm{i} \hbar \epsilon_{i, j, k} \epsilon_{i, j, s} \hat{p}_{s} \\
&= \epsilon_{j, i, k} \epsilon_{j, r, s} \hat{p}_{i} x_{r} \hat{p}_{s} + 2 \mathrm{i} \hbar \Lambda_{k, s} \hat{p}_{s} \\
&= -\epsilon_{i, j, k} \epsilon_{j, r, s} \hat{p}_{i} x_{r} \hat{p}_{s} + 2 \mathrm{i} \hbar \hat{p}_{k} \\
&\stackrel{\natural}{=} -\hat{\boldsymbol{p}}^{2} x_{k} + (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \hat{p}_{k} + \mathrm{i} \hbar \hat{p}_{k}.
\end{aligned}
$$
等号 $\natural$ 处用到了前面的等号 $\flat$。
证明辅助对易子 (2)：
$$
\displaystyle \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] = \mathrm{i} \hbar \frac{1}{r^{3}} \boldsymbol{r}.
$$
式 (2) 的证明：
$$
\begin{aligned}
\biggl[ \hat{p}_{k}, \frac{1}{r} \biggr] \psi &= -\mathrm{i} \hbar \biggl[ \frac{\partial}{\partial x_{k}}, \frac{1}{r} \biggr] \psi \\
&= -\mathrm{i} \hbar \biggl( \frac{\partial}{\partial x_{k}} \dfrac{\psi}{r} - \frac{1}{r} \frac{\partial \psi}{\partial x_{k}} \biggr) \\
&= -\mathrm{i} \hbar \psi \frac{\partial}{\partial x_{k}} \frac{1}{r} = \mathrm{i} \hbar \frac{1}{r^{3}} x_{k} \psi.
\end{aligned}
$$
利用式 (2) 证明辅助对易子 (3)：
$$
\displaystyle \biggl[ \hat{\boldsymbol{p}} \cdot \boldsymbol{r}, \frac{1}{r} \biggr] = \mathrm{i} \hbar \frac{1}{r}.
$$
式 (3) 的证明：
$$
\begin{aligned}
\biggl[ \hat{\boldsymbol{p}} \cdot \boldsymbol{r}, \frac{1}{r} \biggr] &= \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] \cdot \boldsymbol{r} \\
&= \mathrm{i} \hbar \frac{1}{r}.
\end{aligned}
$$
证明辅助对易子 (4)：
$$
\displaystyle \bigl[ \hat{\boldsymbol{p}}^{2}, \boldsymbol{r} \bigr] = -2 \mathrm{i} \hbar \hat{\boldsymbol{p}}.
$$
式 (4) 的证明：
$$
\begin{aligned}
\bigl[ \hat{\boldsymbol{p}}^{2}, \boldsymbol{x}_{k} \bigr] &= \bigl[ \hat{p}_{i} \hat{p}_{i}, \boldsymbol{x}_{k} \bigr] \\
&= [\hat{p}_{i}, \boldsymbol{x}_{k}] \hat{p}_{i} + \hat{p}_{i} [\hat{p}_{i}, \boldsymbol{x}_{k}] \\
&= -2 \mathrm{i} \hbar \hat{p}_{i} \Lambda_{i, k} \\
&= -2 \mathrm{i} \hbar \hat{p}_{k}.
\end{aligned}
$$
证明辅助式 (5)：
$$
\displaystyle \boldsymbol{r} \cdot \hat{\boldsymbol{p}} - \hat{\boldsymbol{p}} \cdot \boldsymbol{r} = 3 \mathrm{i} \hbar.
$$
证明辅助对易子 (6)：
$$
\displaystyle \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r^{3}} \biggr] = 3 \mathrm{i} \hbar \frac{1}{r^{5}} \boldsymbol{r}.
$$
式 (6) 的证明：
$$
\begin{aligned}
\biggl[ \hat{\boldsymbol{p}}, \frac{1}{r^{3}} \biggr] &= \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] \frac{1}{r^{2}} + \frac{1}{r} \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] \frac{1}{r} + \frac{1}{r^{2}} \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] \\
&= 3 \mathrm{i} \hbar \frac{1}{r^{5}} \boldsymbol{r}.
\end{aligned}
$$
证明辅助对易子 (7)：
$$
\displaystyle \biggl[ \hat{\boldsymbol{p}} \cdot \boldsymbol{r}, \frac{1}{r^{3}} \biggr] = 3 \mathrm{i} \hbar \frac{1}{r^{3}}.
$$
式 (7) 的证明：
$$
\begin{aligned}
\biggl[ \hat{\boldsymbol{p}} \cdot \boldsymbol{r}, \frac{1}{r^{3}} \biggr] &= \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r^{3}} \biggr] \cdot \boldsymbol{r} \\
&= 3 \mathrm{i} \hbar \frac{1}{r^{3}}.
\end{aligned}
$$
利用式 (5)–(7) 证明辅助式 (8)：
$$
\displaystyle \frac{1}{r^{3}} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) \boldsymbol{r} = (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \frac{1}{r^{3}} \boldsymbol{r}.
$$
式 (8) 的证明：
$$
\begin{aligned}
\frac{1}{r^{3}} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) \boldsymbol{r} &= \frac{1}{r^{3}} (\hat{\boldsymbol{p}} \cdot \boldsymbol{r} + 3 \mathrm{i} \hbar) \boldsymbol{r} \\
&= \biggl( \frac{1}{r^{3}} (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) + \frac{3 \mathrm{i} \hbar}{r^{3}} \biggr) \boldsymbol{r} \\
&= (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \frac{1}{r^{3}} \boldsymbol{r}.
\end{aligned}
$$
式 (14.8) 的证明：
$$
\begin{aligned}
\bigl[ \hat{\boldsymbol{A}}, \hat{H} \bigr] &= \biggl[ \frac{1}{2 \mu} \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}} \bigr) - \frac{\kappa}{r} \boldsymbol{r}, \frac{\hat{\boldsymbol{p}}^{2}}{2 \mu} - \frac{\kappa}{r} \biggr] \\
&= \frac{1}{4 \mu^{2}} \bigl[ \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}}, \hat{p}^{2} \bigr] \\
&\phantom{{} = {}} - \frac{\kappa}{2 \mu} \biggl( \biggl[ \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] + \biggl[ \frac{1}{r} \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \biggr] \biggr),
\end{aligned}
$$
由 $\kappa$ 的任意性，只需分别证明
$$
\begin{aligned}
& \bigl[ \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}}, \hat{\boldsymbol{p}}^{2} \bigr] = 0, \\
& \biggl[ \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] + \biggl[ \frac{1}{r} \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \biggr] = 0.
\end{aligned}
$$
由式 (4)，
$$
\begin{aligned}
&\phantom{{} = {}} \bigl[ \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}}, \hat{\boldsymbol{p}}^{2} \bigr] \\
&= 2 \bigl[ \hat{\boldsymbol{p}}^{2} \boldsymbol{r} - (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \hat{\boldsymbol{p}}, \hat{\boldsymbol{p}}^{2} \bigr] \\
&= 2 \bigl( \bigl[ \hat{\boldsymbol{p}}^{2} \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \bigr] - \bigl[ (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \hat{\boldsymbol{p}}, \hat{\boldsymbol{p}}^{2} \bigr] \bigr) \\
&= 2 \bigl( \hat{\boldsymbol{p}}^{2} \bigl[ \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \bigr] - \bigl[ \hat{\boldsymbol{p}} \cdot \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \bigr] \hat{\boldsymbol{p}} \bigr) \\
&= 2 \bigl( \hat{\boldsymbol{p}}^{2} \bigl[ \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \bigr] - \hat{\boldsymbol{p}} \cdot \bigl[ \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \bigr] \hat{\boldsymbol{p}} \bigr) \\
&\stackrel{(4)}{=} 4 \mathrm{i} \hbar \bigl( \hat{\boldsymbol{p}}^{2} \hat{\boldsymbol{p}} - (\hat{\boldsymbol{p}} \cdot \hat{\boldsymbol{p}}) \hat{\boldsymbol{p}} \bigr) \\
&= 0.
\end{aligned}
$$
最后，
$$
\begin{aligned}
&\phantom{{} = {}} \biggl[ \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] + \biggl[ \frac{1}{r} \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \biggr] \\
&\stackrel{(1)}{=} 2 \biggl[ \hat{\boldsymbol{p}}^{2} \boldsymbol{r} - (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] + \biggl[ \frac{1}{r} \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \biggr] \\
&= 2 \biggl[ \hat{\boldsymbol{p}}^{2} \boldsymbol{r}, \frac{1}{r} \biggr] - 2 \biggl[ (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] + \biggl[ \frac{1}{r} \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \biggr] \\
&= 2 \biggl[ \hat{\boldsymbol{p}}^{2}, \frac{1}{r} \biggr] \boldsymbol{r} - 2 \biggl[ \hat{\boldsymbol{p}} \cdot \boldsymbol{r}, \frac{1}{r} \biggr] \hat{\boldsymbol{p}} - 2 (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] + \biggl[ \frac{1}{r}, \hat{\boldsymbol{p}}^{2} \biggr] \boldsymbol{r} + \frac{1}{r} \biggl[ \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \biggr] \\
&= \biggl[ \hat{\boldsymbol{p}}^{2}, \frac{1}{r} \biggr] \boldsymbol{r} - 2 \biggl( \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] \cdot \boldsymbol{r} \biggr) \hat{\boldsymbol{p}} - 2 (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] + \frac{1}{r} \biggl[ \boldsymbol{r}, \hat{\boldsymbol{p}}^{2} \biggr] \\
&\stackrel{\sharp}{=} \biggl[ \hat{\boldsymbol{p}}^{2}, \frac{1}{r} \biggr] \boldsymbol{r} - \frac{2 \mathrm{i} \hbar}{r^{3}} (\boldsymbol{r} \cdot \boldsymbol{r}) \hat{\boldsymbol{p}} - 2 (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] + \frac{2 \mathrm{i} \hbar}{r} \hat{\boldsymbol{p}} \\
&= \biggl[ \hat{\boldsymbol{p}}^{2}, \frac{1}{r} \biggr] \boldsymbol{r} - 2 (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] \\
&= \biggl( \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] \cdot \boldsymbol{p} \biggr) \boldsymbol{r} + \biggl( \boldsymbol{p} \cdot \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] \biggr) \boldsymbol{r} - 2 (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \biggl[ \hat{\boldsymbol{p}}, \frac{1}{r} \biggr] \\
&\stackrel{(2)}{=} \mathrm{i} \hbar \biggl( \frac{1}{r^{3}} (\boldsymbol{r} \cdot \boldsymbol{p}) \boldsymbol{r} + (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \frac{1}{r^{3}} \boldsymbol{r} - 2 (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \frac{1}{r^{3}} \boldsymbol{r} \biggr) \\
&= \mathrm{i} \hbar \biggl( \frac{1}{r^{3}} (\boldsymbol{r} \cdot \boldsymbol{p}) \boldsymbol{r} - (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \frac{1}{r^{3}} \boldsymbol{r} \biggr) \\
&\stackrel{(8)}{=} \boldsymbol{0}.
\end{aligned}
$$
其中等号 $\sharp$ 处用到了式 (2) 和 (4)。
练习 14.5：标量积 $\hat{\boldsymbol{L}} \cdot \hat{\boldsymbol{M}}$
证明标量积 (14.9)：$\hat{\boldsymbol{L}} \cdot \hat{\boldsymbol{A}} = 0$。
证明辅助式 (9)：
$$
\displaystyle \hat{\boldsymbol{L}} \cdot \boldsymbol{r} = 0.
$$
式 (9) 的证明：
$$
\begin{aligned}
\hat{\boldsymbol{L}} \cdot \boldsymbol{r} &= \hat{L}_{k} x_{k} \\
&= \epsilon_{i, j, k} x_{i} \hat{p}_{j} x_{k} \\
&= \epsilon_{i, j, k} (\hat{p}_{j} x_{i} + \mathrm{i} \hbar \Lambda_{i, j}) x_{k} \\
&= \epsilon_{i, j, k} \hat{p}_{j} x_{i} x_{k} + \mathrm{i} \hbar \epsilon_{i, j, k} \Lambda_{i, j} x_{k} \\
&= 0.
\end{aligned}
$$
证明辅助式 (10)：
$$
\displaystyle \hat{\boldsymbol{p}} \cdot \hat{\boldsymbol{L}} = 0.
$$
式 (10) 的证明：
$$
\begin{aligned}
\hat{\boldsymbol{p}} \cdot \hat{\boldsymbol{L}} &= \hat{p}_{k} \hat{L}_{k} \\
&= \epsilon_{i, j, k} \hat{p}_{k} x_{i} \hat{p}_{j} \\
&= 0.
\end{aligned}
$$
证明辅助式 (11)：
$$
\displaystyle \bigl[ \hat{L}_{i}, \hat{p}_{j} \bigr] = \mathrm{i} \hbar \epsilon_{i, j, k} \hat{p}_{k}.
$$
式 (11) 的证明：
$$
\begin{aligned}
\bigl[ \hat{L}_{i}, \hat{p}_{j} \bigr] &= \epsilon_{m, n, i} \bigl[ x_{m} \hat{p}_{n}, \hat{p}_{j} \bigr] \\
&= \epsilon_{m, n, i} \bigl[ x_{m}, \hat{p}_{j} \bigr] \hat{p}_{n} \\
&= \mathrm{i} \hbar \epsilon_{m, n, i} \Lambda_{m, j} \hat{p}_{n} \\
&= \mathrm{i} \hbar \epsilon_{i, j, n} \hat{p}_{n}.
\end{aligned}
$$
利用式 (10) 和 (11) 证明辅助式 (12)：
$$
\displaystyle \hat{\boldsymbol{L}} \cdot \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) = 0.
$$
式 (12) 的证明：
$$
\begin{aligned}
\hat{\boldsymbol{L}} \cdot \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) &= \epsilon_{i, j, k} \hat{L}_{k} \hat{p}_{i} \hat{L}_{j} \\
&= \epsilon_{i, j, k} \bigl( \hat{p}_{i} \hat{L}_{k} + \mathrm{i} \hbar \epsilon_{k, i, n} \hat{p}_{n} \bigr) \hat{L}_{j} \\
&= \epsilon_{i, j, k} \hat{p}_{i} \hat{L}_{k} \hat{L}_{j} + \mathrm{i} \hbar \epsilon_{i, j, k} \epsilon_{k, i, n} \hat{p}_{n} \hat{L}_{j} \\
&= 2 \mathrm{i} \hbar \Lambda_{j, n} \hat{p}_{n} \hat{L}_{j} \\
&= 2 \mathrm{i} \hbar \hat{p}_{j} \hat{L}_{j} \\
&= 2 \mathrm{i} \hbar \bigl( \hat{\boldsymbol{p}} \cdot \hat{\boldsymbol{L}} \bigr) \\
&= 0.
\end{aligned}
$$
证明辅助式 (13)：
$$
\displaystyle \hat{\boldsymbol{L}} \cdot \bigl( \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}} \bigr) = 0.
$$
式 (13) 的证明：
$$
\begin{aligned}
\hat{\boldsymbol{L}} \cdot \bigl( \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}} \bigr) &= \epsilon_{i, j, k} \hat{L}_{k} \hat{L}_{i} \hat{p}_{j} \\
&= 0.
\end{aligned}
$$
于是
$$
\begin{aligned}
\hat{\boldsymbol{L}} \cdot \hat{\boldsymbol{A}} &= \hat{\boldsymbol{L}} \cdot \biggl( \frac{1}{2 \mu} \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}} \bigr) - \frac{\kappa}{r} \boldsymbol{r} \biggr) \\
&= \frac{1}{2 \mu} \hat{\boldsymbol{L}} \cdot \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) - \frac{1}{2 \mu} \hat{\boldsymbol{L}} \cdot \bigl( \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}} \bigr) - \kappa \bigl( \hat{\boldsymbol{L}} \cdot \boldsymbol{r} \bigr) \frac{1}{r} \\
&= 0.
\end{aligned}
$$
练习 14.6：确定 $\hat{\boldsymbol{A}}^{2}$
证明式 (14.10)：$\hat{\boldsymbol{A}}^{2} = \dfrac{2}{\mu} \hat{H} \bigl( \hat{\boldsymbol{L}}^{2} + \hbar^{2} \bigr) + \kappa^{2}$。
利用式 (11) 证明辅助式 (14)：
$$
\displaystyle \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}} = -\hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} + 2 \mathrm{i} \hbar \hat{\boldsymbol{p}}.
$$
式 (14) 的证明：
$$
\begin{aligned}
\bigl( \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}} \bigr)_{k} &= \epsilon_{i, j, k} \hat{L}_{i} \hat{p}_{j} \\
&= \epsilon_{i, j, k} \bigl( \hat{p}_{j} \hat{L}_{i} + \mathrm{i} \hbar \epsilon_{i, j, r} p_{r} \bigr) \\
&= \epsilon_{i, j, k} \hat{p}_{j} \hat{L}_{i} + \mathrm{i} \hbar \epsilon_{i, j, k} \epsilon_{i, j, r} \hat{p}_{r} \\
&= -\epsilon_{j, i, k} \hat{p}_{j} \hat{L}_{i} + 2 \mathrm{i} \hbar \Lambda_{k, r} \hat{p}_{r} \\
&= -\bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{k} + 2 \mathrm{i} \hbar \hat{p}_{k}.
\end{aligned}
$$
利用式 (10) 证明辅助式 (15)：
$$
\displaystyle \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)^{2} = \hat{\boldsymbol{p}}^{2} \hat{\boldsymbol{L}}^{2}.
$$
式 (15) 的证明：
$$
\begin{aligned}
\bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)^{2} &= \bigl( \epsilon_{i, j, k} \hat{p}_{i} \hat{L}_{j} \bigr) \bigl( \epsilon_{r, s, k} \hat{p}_{r} \hat{L}_{s} \bigr) \\
&= \epsilon_{i, j, k} \epsilon_{r, s, k} \hat{p}_{i} \hat{L}_{j} \hat{p}_{r} \hat{L}_{s} \\
&= (\Lambda_{i, r} \Lambda_{j, s} - \Lambda_{i, s} \Lambda_{j, r}) \hat{p}_{i} \hat{L}_{j} \hat{p}_{r} \hat{L}_{s} \\
&= \hat{p}_{i} \hat{L}_{j} \hat{p}_{i} \hat{L}_{j} - \hat{p}_{i} \underbrace{\hat{L}_{j} \hat{p}_{j}}_{0} \hat{L}_{i} \\
&= \hat{p}_{i} \bigl( \hat{p}_{i} \hat{L}_{j} + \mathrm{i} \hbar \epsilon_{j, i, k} \hat{p}_{k} \bigr) \hat{L}_{j} \\
&= \hat{p}_{i} \hat{p}_{i} \hat{L}_{j} \hat{L}_{j} + \mathrm{i} \hbar \epsilon_{j, i, k} \hat{p}_{i} \hat{p}_{k} \hat{L}_{j} \\
&= \hat{\boldsymbol{p}}^{2} \hat{\boldsymbol{L}}^{2}.
\end{aligned}
$$
利用式 (11) 证明辅助式 (16)：
$$
\displaystyle \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) \cdot \boldsymbol{\hat{p}} = 2 \mathrm{i} \hbar \hat{\boldsymbol{p}}^{2}.
$$
式 (16) 的证明：
$$
\begin{aligned}
\bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) \cdot \boldsymbol{\hat{p}} &= \epsilon_{i, j, k} \hat{p}_{i} \hat{L}_{j} \hat{p}_{k} \\
&= \epsilon_{i, j, k} \hat{p}_{i} \bigl( \hat{p}_{k} \hat{L}_{j} + \mathrm{i} \hbar \epsilon_{j, k, n} \hat{p}_{n} \bigr) \\
&= \epsilon_{i, j, k} \hat{p}_{i} \hat{p}_{k} \hat{L}_{j} + \mathrm{i} \hbar \epsilon_{i, j, k} \epsilon_{j, k, n} \hat{p}_{i} \hat{p}_{n} \\
&= 2 \mathrm{i} \hbar \Lambda_{i, n} \hat{p}_{i} \hat{p}_{n} \\
&= 2 \mathrm{i} \hbar \hat{\boldsymbol{p}}^{2}.
\end{aligned}
$$
证明辅助式 (17)：
$$
\displaystyle \hat{\boldsymbol{V}}_{1} \cdot \bigl( \hat{\boldsymbol{V}}_{1} \times \hat{\boldsymbol{V}}_{2} \bigr) = \bigl( \hat{\boldsymbol{V}}_{2} \times \hat{\boldsymbol{V}}_{1} \bigr) \cdot \hat{\boldsymbol{V}}_{1} = 0.
$$
利用式 (11) 证明辅助式 (18)：
$$
\displaystyle \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) \cdot \boldsymbol{r} = \hat{\boldsymbol{L}}^{2} + 2 \mathrm{i} \hbar (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}).
$$
式 (18) 的证明：
$$
\begin{aligned}
\bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) \cdot \boldsymbol{r} &= \epsilon_{i, j, k} \hat{p}_{i} \hat{L}_{j} x_{k} \\
&= \epsilon_{i, j, k} \bigl( \hat{L}_{j} \hat{p}_{i} - \mathrm{i} \hbar \epsilon_{j, i, n} \hat{p}_{n} \bigr) x_{k} \\
&= \epsilon_{i, j, k} \hat{L}_{j} \hat{p}_{i} x_{k} + \mathrm{i} \hbar \epsilon_{i, j, k} \epsilon_{i, j, n} \hat{p}_{n} x_{k} \\
&= \epsilon_{i, j, k} \hat{L}_{j} (x_{k} \hat{p}_{i} - \mathrm{i} \hbar \Lambda_{k, i}) + 2 \mathrm{i} \hbar \Lambda_{k, n} \hat{p}_{n} x_{k} \\
&= \epsilon_{i, j, k} \hat{L}_{j} x_{k} \hat{p}_{i} + 2 \mathrm{i} \hbar \hat{p}_{k} x_{k} \\
&= \hat{L}_{j} \hat{L}_{j} + 2 \mathrm{i} \hbar \hat{p}_{k} x_{k} \\
&= \hat{\boldsymbol{L}}^{2} + 2 \mathrm{i} \hbar (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}).
\end{aligned}
$$
证明辅助式 (19)：
$$
\displaystyle \boldsymbol{r} \cdot \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) = \hat{\boldsymbol{L}}^{2}.
$$
式 (19) 的证明：
$$
\begin{aligned}
\boldsymbol{r} \cdot \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) &= \epsilon_{i, j, k} x_{k} \hat{p}_{i} \hat{L}_{j} \\
&= \epsilon_{k, i, j} x_{k} \hat{p}_{i} \hat{L}_{j} \\
&= \hat{L}_{j} \hat{L}_{j} \\
&= \hat{\boldsymbol{L}}^{2}.
\end{aligned}
$$
利用式 (4) 证明辅助对易子 (20)：
$$
\displaystyle \bigl[ \hat{\boldsymbol{p}}^{2}, \hat{\boldsymbol{L}} \bigr] = \boldsymbol{0}.
$$
式 (20) 的证明：
$$
\begin{aligned}
\bigl[ \hat{\boldsymbol{p}}^{2}, \hat{L}_{k} \bigr] &= \epsilon_{m, n, k} \bigl[ \hat{\boldsymbol{p}}^{2}, x_{m} \hat{p}_{n} \bigr] \\
&= \epsilon_{m, n, k} \bigl[ \hat{\boldsymbol{p}}^{2}, x_{m} \bigr] \hat{p}_{n} \\
&= -2 \mathrm{i} \hbar \epsilon_{m, n, k} \hat{p}_{k} \hat{p}_{n} \\
&= 0.
\end{aligned}
$$
利用式 (20) 证明辅助对易子 (21)：
$$
\displaystyle \biggl[ \frac{1}{r}, \hat{\boldsymbol{L}} \biggr] = \boldsymbol{0}.
$$
式 (21) 的证明：
$$
\begin{aligned}
\biggl[ \frac{1}{r}, \hat{\boldsymbol{L}} \biggr] = \frac{1}{\kappa} \biggl[ \frac{\hat{\boldsymbol{p}}^{2}}{2 \mu} - \hat{H}, \hat{\boldsymbol{L}} \biggr] = \boldsymbol{0}.
\end{aligned}
$$
利用式 (3)、(5) 证明辅助式 (22)：
$$
\displaystyle (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \frac{1}{r} - \frac{1}{r} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) = -2 \mathrm{i} \hbar \frac{1}{r}.
$$
式 (22) 的证明：
$$
\begin{aligned}
&\phantom{{} = {}} (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \frac{1}{r} - \frac{1}{r} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) \\
&= (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \frac{1}{r} - \frac{1}{r} \bigl( \boldsymbol{p} \cdot \hat{\boldsymbol{r}} + 3 \mathrm{i} \hbar \bigr) \\
&= \biggl[ \hat{\boldsymbol{p}} \cdot \boldsymbol{r}, \frac{1}{r} \biggr] - 3 \mathrm{i} \hbar \frac{1}{r} \\
&= -2 \mathrm{i} \hbar \frac{1}{r}.
\end{aligned}
$$
于是，
$$
\begin{aligned}
\hat{\boldsymbol{A}}^{2} &= \biggl( \frac{1}{2 \mu} \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \hat{\boldsymbol{L}} \times \hat{\boldsymbol{p}} \bigr) - \kappa \frac{\boldsymbol{r}}{r} \biggr)^{2} \\
&\stackrel{(14)}{=} \biggl( \frac{1}{\mu} \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \mathrm{i} \hbar \hat{\boldsymbol{p}} \bigr) - \kappa \frac{\boldsymbol{r}}{r} \biggr)^{2} \\
&= \frac{1}{\mu^{2}} \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \mathrm{i} \hbar \hat{\boldsymbol{p}} \bigr)^{2} + \Bigl( \kappa \frac{\boldsymbol{r}}{r} \Bigr)^{2} \\
&\phantom{{} = {}} - \frac{\kappa}{\mu} \biggl( \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \mathrm{i} \hbar \hat{\boldsymbol{p}} \bigr) \cdot \frac{\boldsymbol{r}}{r} + \frac{\boldsymbol{r}}{r} \cdot \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \mathrm{i} \hbar \hat{\boldsymbol{p}} \bigr) \biggr) \\
&= \frac{1}{\mu^{2}} \biggl( \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)^{2} - \mathrm{i} \hbar \Bigl( \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) \cdot \hat{\boldsymbol{p}} + \hat{\boldsymbol{p}} \cdot \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) \Bigr) - \hbar^{2} \hat{\boldsymbol{p}}^{2} \biggr) + \kappa^{2} \\
&\phantom{{} = {}} - \frac{\kappa}{\mu} \biggl( \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) \cdot \frac{\boldsymbol{r}}{r} + \frac{\boldsymbol{r}}{r} \cdot \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr) - \mathrm{i} \hbar \biggl( \hat{\boldsymbol{p}} \cdot \frac{\boldsymbol{r}}{r} + \frac{\boldsymbol{r}}{r} \cdot \hat{\boldsymbol{p}} \biggr) \biggr) \\
&\stackrel{\clubsuit}{=} \frac{\hat{\boldsymbol{p}}^{2}}{\mu^{2}} \bigl( \hat{\boldsymbol{L}}^{2} + \hbar^{2} \bigr) + \kappa^{2} - \frac{\kappa}{\mu} \biggl( 2 \frac{\boldsymbol{L}^{2}}{r} + \mathrm{i} \hbar \biggl( (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \frac{1}{r} - \frac{1}{r} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) \biggr) \biggr) \\
&\stackrel{(22)}{=} \frac{2}{\mu} \biggl( \frac{\hat{\boldsymbol{p}}^{2}}{2 \mu} - \frac{\kappa}{r} \biggr) \bigl( \hat{\boldsymbol{L}}^{2} + \hbar^{2} \bigr) + \kappa^{2} \\
&= \frac{2}{\mu} \hat{H} \bigl( \hat{\boldsymbol{L}}^{2} + \hbar^{2} \bigr) + \kappa^{2}.
\end{aligned}
$$
其中等号 $\clubsuit$ 处用到了辅助对易式 (15)–(19)。辅助对易子 (21)、(20) 使得 $\dfrac{\boldsymbol{L}^{2}}{r}$ 不产生歧义。
练习 14.7：对易子 $\bigl[ \hat{A}_{i}, \hat{L}_{j} \bigr]$
证明对易子 (14.12)：$\bigl[ \hat{A}_{i}, \hat{L}_{j} \bigr] = \mathrm{i} \hbar \epsilon_{i, j, k} \hat{A}_{k}$。
证明辅助对易子 (23)：
$$
\displaystyle \bigl[ x_{i}, \hat{L}_{j} \bigr] = \mathrm{i} \hbar \epsilon_{i, j, k} x_{k}.
$$
式 (23) 的证明：
$$
\begin{aligned}
\bigl[ x_{i}, \hat{L}_{j} \bigr] &= \epsilon_{m, n, j} [x_{i}, x_{m} \hat{p}_{n}] \\
&= \epsilon_{m, n, j} x_{m} [x_{i}, \hat{p}_{n}] \\
&= \mathrm{i} \hbar \epsilon_{m, n, j} \Lambda_{i, n} x_{m} \\
&= \mathrm{i} \hbar \epsilon_{i, j, m} x_{m}.
\end{aligned}
$$
利用式 (11)、(23) 证明辅助对易子 (24)：
$$
\displaystyle \bigl[ \hat{\boldsymbol{p}} \cdot \boldsymbol{r}, \hat{\boldsymbol{L}} \bigr] = \boldsymbol{0}.
$$
式 (24) 的证明：
$$
\begin{aligned}
\bigl[ \hat{p}_{k} x_{k}, \hat{L}_{j} \bigr] &= \bigl[ \hat{p}_{k}, \hat{L}_{j} \bigr] x_{k} + \hat{p}_{k} \bigl[ x_{k}, \hat{L}_{j} \bigr] \\
&= \mathrm{i} \hbar (-\epsilon_{j, k, n} \hat{p}_{n} x_{k} + \epsilon_{k, j, n} \hat{p}_{k} x_{n}) \\
&= \mathrm{i} \hbar (-\epsilon_{j, n, k} \hat{p}_{k} x_{n} + \epsilon_{k, j, n} \hat{p}_{k} x_{n}) \\
&= 0.
\end{aligned}
$$
于是，
$$
\begin{aligned}
\bigl[ \hat{A}_{i}, \hat{L}_{j} \bigr] &\stackrel{(1)}{=} \biggl[ \frac{1}{\mu} \bigl( \hat{\boldsymbol{p}}^{2} x_{i} - (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \hat{p}_{i} \bigr) - \kappa \frac{x_{i}}{r}, \hat{L}_{j} \biggr] \\
&= \biggl[ \biggl( \frac{\hat{\boldsymbol{p}}^{2}}{\mu} - \frac{\kappa}{r} \biggr) x_{i}, \hat{L}_{j} \biggr] - \frac{1}{\mu} \bigl[ (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \hat{p}_{i}, \hat{L}_{j} \bigr] \\
&\stackrel{\blacklozenge}{=} \biggl( \frac{\hat{\boldsymbol{p}}^{2}}{\mu} - \frac{\kappa}{r} \biggr) \bigl[ x_{i}, \hat{L}_{j} \bigr] - \frac{1}{\mu} \bigl[ (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}), \hat{L}_{j} \bigr] \hat{p}_{i} - \frac{1}{\mu} (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \bigl[ \hat{p}_{i}, \hat{L}_{j} \bigr] \\
&\stackrel{\spadesuit}{=} \mathrm{i} \hbar \biggl( \frac{\hat{\boldsymbol{p}}^{2}}{\mu} - \frac{\kappa}{r} \biggr) \epsilon_{i, j, k} x_{k} + \frac{\mathrm{i} \hbar}{\mu} (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \epsilon_{j, i, k} \hat{p}_{k} \\
&= \mathrm{i} \hbar \epsilon_{i, j, k} \biggl( \frac{1}{\mu} \bigl( \hat{\boldsymbol{p}}^{2} x_{k} - (\hat{\boldsymbol{p}} \cdot \boldsymbol{r}) \hat{p}_{k} \bigr) - \kappa \frac{x_{k}}{r} \biggr) \\
&\stackrel{(1)}{=} \mathrm{i} \hbar \epsilon_{i, j, k} \hat{A}_{k}.
\end{aligned}
$$
其中等号 $\blacklozenge$ 处用到了辅助对易子 (20)、(21)、等号 $\spadesuit$ 处用到了辅助对易子 (11)、(23) 和 (24)。
练习 14.8：对易子 $\bigl[ \hat{A}_{i}, \hat{A}_{j} \bigr]$
证明对易子 (14.13)：$\bigl[ \hat{A}_{i}, \hat{A}_{j} \bigr] = -\dfrac{2}{\mu} \mathrm{i} \hbar \epsilon_{i, j, k} \hat{H} \hat{L}_{k}$。
利用式 (11) 证明辅助对易子 (25)：
$$
\displaystyle \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \hat{p}_{j} \bigr] = \mathrm{i} \hbar \bigl( \Lambda_{i, j} \hat{\boldsymbol{p}}^{2} - \hat{p}_{i} \hat{p}_{j} \bigr).
$$
式 (25) 的证明：
$$
\begin{aligned}
\bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \hat{p}_{j} \bigr] &= \epsilon_{m, n, i} \bigl[ \hat{p}_{m} \hat{L}_{n}, \hat{p}_{j} \bigr] \\
&= \epsilon_{m, n, i} \hat{p}_{m} \bigl[ \hat{L}_{n}, \hat{p}_{j} \bigr] \\
&= \mathrm{i} \hbar \epsilon_{n, i, m} \epsilon_{n, j, k} \hat{p}_{m} \hat{p}_{k} \\
&= \mathrm{i} \hbar (\Lambda_{i, j} \Lambda_{m, k} - \Lambda_{i, k} \Lambda_{m, j}) \hat{p}_{m} \hat{p}_{k} \\
&= \mathrm{i} \hbar \bigl( \Lambda_{i, j} \hat{\boldsymbol{p}}^{2} - \hat{p}_{i} \hat{p}_{j} \bigr).
\end{aligned}
$$
补充：
$$
\displaystyle \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \hat{p}_{j} \bigr] - \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{j}, \hat{p}_{i} \bigr] = 0.
$$
利用式 (11) 证明辅助对易子 (26)：
$$
\displaystyle \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \hat{L}_{j} \bigr] = \mathrm{i} \hbar \bigl( \hat{p}_{i} \hat{L}_{j} - \hat{p}_{j} \hat{L}_{i} \bigr).
$$
式 (26) 的证明：
$$
\begin{aligned}
\bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \hat{L}_{j} \bigr] &= \epsilon_{m, n, i} \bigl[ \hat{p}_{m} \hat{L}_{n}, \hat{L}_{j} \bigr] \\
&= \epsilon_{m, n, i} \bigl( \bigl[ \hat{p}_{m}, \hat{L}_{j} \bigr] \hat{L}_{n} + \hat{p}_{m} \bigl[ \hat{L}_{n}, \hat{L}_{j} \bigr] \bigr) \\
&= \mathrm{i} \hbar \epsilon_{m, n, i} \bigl( -\epsilon_{j, m, k} \hat{p}_{k} \hat{L}_{n} + \epsilon_{n, j, k} \hat{p}_{m} \hat{L}_{k} \bigr) \\
&= \mathrm{i} \hbar \bigl( -\epsilon_{m, n, i} \epsilon_{j, m, k} \hat{p}_{k} \hat{L}_{n} + \epsilon_{m, n, i} \epsilon_{n, j, k} \hat{p}_{m} \hat{L}_{k} \bigr) \\
&= \mathrm{i} \hbar \bigl( -\epsilon_{n, m, i} \epsilon_{j, n, k} \hat{p}_{k} \hat{L}_{m} + \epsilon_{m, n, i} \epsilon_{n, j, k} \hat{p}_{m} \hat{L}_{k} \bigr) \\
&= \mathrm{i} \hbar \epsilon_{m, n, i} \epsilon_{j, n, k} \bigl( \hat{p}_{k} \hat{L}_{m} - \hat{p}_{m} \hat{L}_{k} \bigr) \\
&= \mathrm{i} \hbar (\Lambda_{m, j} \Lambda_{i, k} - \Lambda_{m, k} \Lambda_{i, j}) \bigl( \hat{p}_{k} \hat{L}_{m} - \hat{p}_{m} \hat{L}_{k} \bigr) \\
&= \mathrm{i} \hbar \bigl( \hat{p}_{i} \hat{L}_{j} - \Lambda_{i, j} \hat{p}_{k} \hat{L}_{k} - \hat{p}_{j} \hat{L}_{i} + \Lambda_{i, j} \hat{p}_{k} \hat{L}_{k} \bigr) \\
&= \mathrm{i} \hbar \bigl( \hat{p}_{i} \hat{L}_{j} - \hat{p}_{j} \hat{L}_{i} \bigr).
\end{aligned}
$$
利用式 (25)、(26) 证明辅助对易子 (27)：
$$
\displaystyle \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{j} \bigr] = -\mathrm{i} \hbar \hat{\boldsymbol{p}}^{2} \epsilon_{i, j, k} \hat{L}_{k}.
$$
式 (27) 的证明：
$$
\begin{aligned}
&\phantom{{} = {}} \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{j} \bigr] \\
&= \epsilon_{m, n, j} \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \hat{p}_{m} \hat{L}_{n} \bigr] \\
&= \epsilon_{m, n, j} \Bigl( \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \hat{p}_{m} \bigr] \hat{L}_{n} + \hat{p}_{m} \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \hat{L}_{n} \bigr] \Bigr) \\
&= \mathrm{i} \hbar \epsilon_{m, n, j} \Bigl( \bigl( \Lambda_{i, m} \hat{\boldsymbol{p}}^{2} - \hat{p}_{i} \hat{p}_{m} \bigr) \hat{L}_{n} + \hat{p}_{m} \bigl( \hat{p}_{i} \hat{L}_{n} - \hat{p}_{n} \hat{L}_{i} \bigr) \Bigr) \\
&= \mathrm{i} \hbar \epsilon_{m, n, j} \bigl( \Lambda_{i, m} \hat{\boldsymbol{p}}^{2} \hat{L}_{n} - \hat{p}_{i} \hat{p}_{m} \hat{L}_{n} + \hat{p}_{m} \hat{p}_{i} \hat{L}_{n} - \hat{p}_{m} \hat{p}_{n} \hat{L}_{i} \bigr) \\
&= -\mathrm{i} \hbar \hat{\boldsymbol{p}}^{2} \epsilon_{i, j, n} \hat{L}_{n}.
\end{aligned}
$$
利用式 (23) 证明辅助对易子 (28)：
$$
\begin{aligned}
\bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, x_{j} \bigr] &= -\mathrm{i} \hbar \bigl( \epsilon_{i, j, k} \hat{L}_{k} + \hat{p}_{j} x_{i} - \Lambda_{i, j} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) + 3 \mathrm{i} \hbar \Lambda_{i, j} \bigr).
\end{aligned}
$$
式 (28) 的证明：
$$
\begin{aligned}
\bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, x_{j} \bigr] &= \epsilon_{m, n, i} \bigl[ \hat{p}_{m} \hat{L}_{n}, x_{j} \bigr] \\
&= \epsilon_{m, n, i} \bigl( \bigl[ \hat{p}_{m}, x_{j} \bigr] \hat{L}_{n} + \hat{p}_{m} \bigl[ \hat{L}_{n}, x_{j} \bigr] \bigr) \\
&= -\mathrm{i} \hbar \bigl( \epsilon_{m, n, i} \Lambda_{m, j} \hat{L}_{n} + \epsilon_{m, n, i} \epsilon_{j, n, k} \hat{p}_{m} x_{k} \bigr) \\
&= -\mathrm{i} \hbar \bigl( \epsilon_{i, j, n} \hat{L}_{n} + (\Lambda_{m, j} \Lambda_{i, k} - \Lambda_{m, k} \Lambda_{i, j}) \hat{p}_{m} x_{k} \bigr) \\
&= -\mathrm{i} \hbar \bigl( \epsilon_{i, j, n} \hat{L}_{n} + \hat{p}_{j} x_{i} - \Lambda_{i, j} \hat{p}_{k} x_{k} \bigr) \\
&= -\mathrm{i} \hbar \bigl( \epsilon_{i, j, n} \hat{L}_{n} + \hat{p}_{j} x_{i} - \Lambda_{i, j} x_{k} \hat{p}_{k} + 3 \mathrm{i} \hbar \Lambda_{i, j} \bigr).
\end{aligned}
$$
利用式 (28) 证明辅助对易子 (29)：
$$
\begin{aligned}
\bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, x_{j} \bigr] &= -\mathrm{i} \hbar \bigl( \epsilon_{i, j, k} \hat{L}_{k} + \hat{p}_{j} x_{i} - \Lambda_{i, j} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) + 2 \mathrm{i} \hbar \Lambda_{i, j} \bigr).
\end{aligned}
$$
式 (29) 的证明：
$$
\begin{aligned}
\bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, x_{j} \bigr] &= \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, x_{j} \bigr] - \mathrm{i} \hbar \bigl[ \hat{p}_{i}, x_{j} \bigr] \\
&= -\mathrm{i} \hbar \bigl( \epsilon_{i, j, n} \hat{L}_{n} + \hat{p}_{j} x_{i} - \Lambda_{i, j} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) + 2 \mathrm{i} \hbar \Lambda_{i, j} \bigr).
\end{aligned}
$$
利用式 (2)、(21) 证明辅助对易子 (30)：
$$
\displaystyle \biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \frac{1}{r} \biggr] = \mathrm{i} \hbar \frac{1}{r^{3}} \bigl( x_{i} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) - r^{2} \hat{p}_{i} \bigr).
$$
式 (30) 的证明：
$$
\begin{aligned}
\biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \frac{1}{r} \biggr] &= \epsilon_{m, n, i} \biggl[ \hat{p}_{m} \hat{L}_{n}, \frac{1}{r} \biggr] \\
&= \epsilon_{m, n, i} \biggl[ \hat{p}_{m}, \frac{1}{r} \biggr] \hat{L}_{n} \\
&= \mathrm{i} \hbar \frac{1}{r^{3}} \epsilon_{m, n, i} x_{m} \hat{L}_{n} \\
&= \mathrm{i} \hbar \frac{1}{r^{3}} \epsilon_{i, m, n} \epsilon_{r, s, n} x_{m} x_{r} \hat{p}_{s} \\
&= \mathrm{i} \hbar \frac{1}{r^{3}} (\Lambda_{i, r} \Lambda_{m, s} - \Lambda_{i, s} \Lambda_{m, r}) x_{m} x_{r} \hat{p}_{s} \\
&= \mathrm{i} \hbar \frac{1}{r^{3}} \bigl( x_{i} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) - r^{2} \hat{p}_{i} \bigr).
\end{aligned}
$$
利用式 (2)、(30) 证明辅助对易子 (31)：
$$
\displaystyle \biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, \frac{1}{r} \biggr] = \mathrm{i} \hbar \frac{1}{r^{3}} \bigl( x_{i} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) - \mathrm{i} \hbar x_{i} - r^{2} \hat{p}_{i} \bigr).
$$
式 (31) 的证明：
$$
\begin{aligned}
\biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, \frac{1}{r} \biggr] &= \biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \frac{1}{r} \biggr] - \mathrm{i} \hbar \biggl[ \hat{p}_{i}, \frac{1}{r} \biggr] \\
&= \mathrm{i} \hbar \frac{1}{r^{3}} \bigl( x_{i} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) - \mathrm{i} \hbar x_{i} - r^{2} \hat{p}_{i} \bigr).
\end{aligned}
$$
利用式 (29)、(31) 证明辅助对易子 (32)：
$$
\begin{aligned}
\biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, \frac{x_{j}}{r} \biggr] &= \mathrm{i} \hbar \frac{1}{r^{3}} \bigl( x_{i} x_{j} + r^{2} \Lambda_{i, j} \bigr) (\boldsymbol{r} \cdot \hat{\boldsymbol{p}} - \mathrm{i} \hbar) \\
&\phantom{{} = {}} - \mathrm{i} \hbar \frac{1}{r} \bigl( \epsilon_{i, j, k} \hat{L}_{k} + \hat{p}_{i} x_{j} + \hat{p}_{j} x_{i} + \mathrm{i} \hbar \Lambda_{i, j} \bigr).
\end{aligned}
$$
式 (32) 的证明：
$$
\begin{aligned}
&\phantom{{} = {}} \biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, \frac{x_{j}}{r} \biggr] \\
&= \biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, \frac{1}{r} \biggr] x_{j} + \frac{1}{r} \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, x_{j} \bigr] \\
&= \mathrm{i} \hbar \frac{1}{r^{3}} \bigl( x_{i} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) - \mathrm{i} \hbar x_{i} - r^{2} \hat{p}_{i} \bigr) x_{j} \\
&\phantom{{} = {}} - \mathrm{i} \hbar \frac{1}{r} \bigl( \epsilon_{i, j, n} \hat{L}_{n} + \hat{p}_{j} x_{i} - \Lambda_{i, j} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) + 2 \mathrm{i} \hbar \Lambda_{i, j} \bigr) \\
&= \mathrm{i} \hbar \frac{1}{r^{3}} \bigl( x_{i} x_{j} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}}) - \mathrm{i} \hbar x_{i} x_{j} \bigr) - \mathrm{i} \hbar \frac{1}{r} \hat{p}_{i} x_{j} \\
&\phantom{{} = {}} - \mathrm{i} \hbar \frac{1}{r} \bigl( \epsilon_{i, j, n} \hat{L}_{n} + \hat{p}_{j} x_{i} + \mathrm{i} \hbar \Lambda_{i, j} \bigr) + \mathrm{i} \hbar \frac{1}{r} \Lambda_{i, j} (\boldsymbol{r} \cdot \hat{\boldsymbol{p}} - \mathrm{i} \hbar) \\
&= \mathrm{i} \hbar \frac{1}{r^{3}} \bigl( x_{i} x_{j} + r^{2} \Lambda_{i, j} \bigr) (\boldsymbol{r} \cdot \hat{\boldsymbol{p}} - \mathrm{i} \hbar) \\
&\phantom{{} = {}} - \mathrm{i} \hbar \frac{1}{r} \bigl( \epsilon_{i, j, k} \hat{L}_{k} + \hat{p}_{i} x_{j} + \hat{p}_{j} x_{i} + \mathrm{i} \hbar \Lambda_{i, j} \bigr).
\end{aligned}
$$
补充：
$$
\displaystyle \biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, \frac{x_{j}}{r} \biggr] - \biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{j} - \mathrm{i} \hbar \hat{p}_{j}, \frac{x_{i}}{r} \biggr] = -2 \mathrm{i} \hbar \frac{1}{r} \epsilon_{i, j, k} \hat{L}_{k}.
$$
于是，
$$
\begin{aligned}
\bigl[ \hat{A}_{i}, \hat{A}_{j} \bigr] &\stackrel{(14)}{=} \biggl[ \frac{1}{\mu} \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \mathrm{i} \hbar \hat{\boldsymbol{p}} \bigr)_{i} - \kappa \frac{x_{i}}{r}, \frac{1}{\mu} \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} - \mathrm{i} \hbar \hat{\boldsymbol{p}} \bigr)_{j} - \kappa \frac{x_{j}}{r} \biggr] \\
&= \frac{1}{\mu^{2}} \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{j} - \mathrm{i} \hbar \hat{p}_{j} \bigr] \\
&\phantom{{} = {}} - \frac{\kappa}{\mu} \biggl( \biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, \frac{x_{j}}{r} \biggr] - \biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{j} - \mathrm{i} \hbar \hat{p}_{j}, \frac{x_{i}}{r} \biggr] \biggr) \\
&= \frac{1}{\mu^{2}} \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{j} \bigr] \\
&\phantom{{} = {}} - \frac{\mathrm{i} \hbar}{\mu^{2}} \Bigl( \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i}, \hat{p}_{j} \bigr] - \bigl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{j}, \hat{p}_{i} \bigr] \Bigr) \\
&\phantom{{} = {}} - \frac{\kappa}{\mu} \biggl( \biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{i} - \mathrm{i} \hbar \hat{p}_{i}, \frac{x_{j}}{r} \biggr] - \biggl[ \bigl( \hat{\boldsymbol{p}} \times \hat{\boldsymbol{L}} \bigr)_{j} - \mathrm{i} \hbar \hat{p}_{j}, \frac{x_{i}}{r} \biggr] \biggr) \\
&\stackrel{\bigstar}{=} -\frac{1}{\mu} \mathrm{i} \hbar \biggl( \frac{\hat{\boldsymbol{p}}^{2}}{\mu} - 2 \frac{\kappa}{r} \biggr) \epsilon_{i, j, k} \hat{L}_{k} \\
&= -\frac{2}{\mu} \mathrm{i} \hbar \hat{H} \epsilon_{i, j, k} \hat{L}_{k}.
\end{aligned}
$$
其中等号 $\bigstar$ 处用到了辅助对易子 (25)、(27) 和 (32) 的关于 $i, j$ 反对称的部分。