判断下列算子是否可交换？
\[[1]\ [\hat{x},\hat{p}_x] \quad [2]\ [\hat{l}_x, \hat{l_y}] \quad [3]\ [\hat{\boldsymbol{l}}^2, \hat{l}_z]\]
解决本题首先要了解可交换的定义，对于任意两个算子有：
\[\hat{f}\hat{g}\psi(\boldsymbol{r})=\hat{g}\hat{f}\psi(\boldsymbol{r})\ 或\ (\hat{f}\hat{g}-\hat{g}\hat{f})\psi(\boldsymbol{r})=0\]那么这两个算子可交换，否则不可交换。其中下列式子称为交换子：
\[[\hat{f}, \hat{g}] \equiv \hat{f}\hat{g}-\hat{g}\hat{f}\]除此之外，还需要了解以下观测量在古典力学中的变量和量子力学中的算子对应：
| 观测量 | 变量 | 算子 |
|---|---|---|
| 位置 | \(x\ (\boldsymbol{r})\) | \(\hat{x}\ (\hat{\boldsymbol{x}})\) |
| 动量 | \(p_x\ (\boldsymbol{p})\) | \(\hat{p}_x=-\mathrm{i}\hbar{d \over dx}\ (\hat{\boldsymbol{p}})\) |
| 角动量 | \(\boldsymbol{l}^2=l_x^2+l_y^2+l_z^2\) | \(\hat{\boldsymbol{l}}^2=\hat{l}_x^2+\hat{l}_y^2+\hat{l}_z^2\) |
| \(\hat{l}_x=-\mathrm{i}\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\) | ||
| \(\hat{l}_y=-\mathrm{i}\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\) | ||
| \(\hat{l}_z=-\mathrm{i}\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)\) |
算子组一
将 \(\hat{p}_x=-\mathrm{i}\hbar{d \over dx}\) 代入可得
\[\begin{align} [\hat{x}, \hat{p}_x]\psi(x)&=(\hat{x}\hat{p}_x-\hat{p}_x\hat{x})\psi(x) \\\\ &= -\mathrm{i}\hbar x \frac{d}{dx}\psi(x)-(-\mathrm{i}\hbar)\frac{d}{dx}[x\psi(x)] \end{align}\]这里需要注意算子 \(\frac{d}{dx}\) 是求导算子，根据链式法则应该对 \([x\psi(x)]\) 分别对 \(x\) 求导，于是：
\[\begin{align} 原式&=-\mathrm{i}\hbar x \frac{d}{dx}\psi(x) + \mathrm{i}\hbar\left[x\frac{d}{dx}\psi(x)+\psi(x)\right] \\\\ &= \mathrm{i}\hbar\psi(x) ≠ 0 \end{align}\]因此，这两个算子不可交换。
算子组二
将 \(\hat{l}_x\) 和 \(\hat{l}_y\) 代入可得 (注意 \(\mathrm{(-i)}^2=-1\))
\[\begin{align} \hat{l}_x\hat{l}_y&=(-\mathrm{i}\hbar)^2\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right) \\\\ &=-\hbar^2\left(y\frac{\partial}{\partial z}z\frac{\partial}{\partial x}-z\frac{\partial}{\partial y}z\frac{\partial}{\partial x}-y\frac{\partial}{\partial z}x\frac{\partial}{\partial z}+z\frac{\partial}{\partial y}x\frac{\partial}{\partial z}\right) \end{align}\]这里需要注意“求偏导的函数中是否包含了偏导的对象”，如果不包含则可以直接将变量左移，如果包含则需要根据链式法则分别求导。
接着有
\[\begin{align} \hat{l}_x\hat{l}_y&=-\hbar^2\left[y\left(z\frac{\partial^2}{\partial z \partial x}+\frac{\partial}{\partial x}\right)-z^2\frac{\partial^2}{\partial y \partial x}-xy\frac{\partial^2}{\partial z^2}+xz\frac{\partial^2}{\partial y \partial z}\right] \\\\ &=-\hbar^2\left[yz\frac{\partial^2}{\partial z \partial x}+y\frac{\partial}{\partial x}-z^2\frac{\partial^2}{\partial y \partial x}-xy\frac{\partial^2}{\partial z^2}+xz\frac{\partial^2}{\partial y \partial z}\right] \end{align}\]同理
\[\begin{align} \hat{l}_y\hat{l}_x&=(-\mathrm{i}\hbar)^2\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right) \\\\ &=-\hbar^2\left(z\frac{\partial}{\partial x}y\frac{\partial}{\partial z}-x\frac{\partial}{\partial z}y\frac{\partial}{\partial z}-z\frac{\partial}{\partial x}z\frac{\partial}{\partial y}+x\frac{\partial}{\partial z}z\frac{\partial}{\partial y}\right) \\\\ &=-\hbar^2\left[yz\frac{\partial^2}{\partial x \partial z}-xy\frac{\partial^2}{\partial z^2}-z^2\frac{\partial^2}{\partial x \partial y}+x\left(\frac{\partial}{\partial y}+z\frac{\partial^2}{\partial z \partial y}\right)\right] \\\\ &=-\hbar^2\left(yz\frac{\partial^2}{\partial x \partial z}-xy\frac{\partial^2}{\partial z^2}-z^2\frac{\partial^2}{\partial x \partial y}+x\frac{\partial}{\partial y}+xz\frac{\partial^2}{\partial z \partial y}\right) \end{align}\]因此交换子为（减法抵消相同项）
\[\begin{align} [\hat{l}_x, \hat{l}_y]&=\hat{l}_x\hat{l}_y-\hat{l}_y\hat{l}_x \\\\ &=-\hbar^2\left(y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}\right) \\\\ &=\mathrm{i}^2\hbar^2\left(y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}\right) \\\\ &=\mathrm{i}\hbar\left[-\mathrm{i}\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)\right] \\\\ &=\mathrm{i}\hbar\hat{l}_z≠0 \end{align}\]因此，这两个算子不可交换。
算子组三
通过算子组二可以类似推理得到：
\[[\hat{l}_y, \hat{l}_z]=\mathrm{i}\hbar\hat{l}_x\] \[[\hat{l}_z, \hat{l}_x]=\mathrm{i}\hbar\hat{l}_y\]将其代入可得：
\[\begin{align} [\hat{\boldsymbol{l}}^2, \hat{l}_z]&=\hat{\boldsymbol{l}}^2\hat{l}_z-\hat{\boldsymbol{l}}^2\hat{l}_z \\\\ &=(\hat{l}_x^2+\hat{l}_y^2+\hat{l}_z^2)\hat{l}_z-\hat{l}_z(\hat{l}_x^2+\hat{l}_y^2+\hat{l}_z^2) \\\\ &=\hat{l}_x\hat{l}_x\hat{l}_z+\hat{l}_y\hat{l}_y\hat{l}_z+\hat{l}_z\hat{l}_z\hat{l}_z-\hat{l}_z\hat{l}_x\hat{l}_x-\hat{l}_z\hat{l}_y\hat{l}_y-\hat{l}_z\hat{l}_z\hat{l}_z \\\\ &=\hat{l}_x\hat{l}_x\hat{l}_z+\hat{l}_y\hat{l}_y\hat{l}_z+\bcancel{\hat{l}_z\hat{l}_z\hat{l}_z}-\hat{l}_z\hat{l}_x\hat{l}_x-\hat{l}_z\hat{l}_y\hat{l}_y-\bcancel{\hat{l}_z\hat{l}_z\hat{l}_z} \\\\ &=\hat{l}_x\hat{l}_x\hat{l}_z+\color{red}{\hat{l}_x\hat{l}_z\hat{l}_x-\hat{l}_x\hat{l}_z\hat{l}_x}\color{black}{-\hat{l}_z\hat{l}_x\hat{l}_x}+\hat{l}_y\hat{l}_y\hat{l}_z+\color{red}{\hat{l}_y\hat{l}_z\hat{l}_y-\hat{l}_y\hat{l}_z\hat{l}_y}\color{black}{-\hat{l}_z\hat{l}_y\hat{l}_y} \\\\ &=\hat{l}_x(\hat{l}_x\hat{l}_z-\hat{l}_z\hat{l}_x)+(\hat{l}_x\hat{l}_z-\hat{l}_z\hat{l}_x)\hat{l}_x+\hat{l}_y(\hat{l}_y\hat{l}_z-\hat{l}_z\hat{l}_y)+(\hat{l}_y\hat{l}_z-\hat{l}_z\hat{l}_y)\hat{l}_y \\\\ &=\hat{l}_x(-[\hat{l}_z, \hat{l}_x])+(-[\hat{l}_z, \hat{l}_x])\hat{l}_x+\hat{l}_y[\hat{l}_y, \hat{l}_z]+[\hat{l}_y, \hat{l}_z]\hat{l}_y \\\\ &=-\mathrm{i}\hbar\hat{l}_x\hat{l}_y-\mathrm{i}\hbar\hat{l}_y\hat{l}_x+\mathrm{i}\hbar\hat{l}_y\hat{l}_x+\mathrm{i}\hbar\hat{l}_x\hat{l}_y = 0 \end{align}\]因此，这两个算子可交换。
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本文标题：《 《手解量子化学》练习题 1-2 》
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