《手解量子化学》练习题 1-2

判断下列算子是否可交换？

[1] [\hat{x}, {\hat{p}}_{x}] [2] [{\hat{l}}_{x}, \hat{l_{y}}] [3] [{\hat{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} ψ (r) = \hat{g} \hat{f} ψ (r) 或 (\hat{f} \hat{g} - \hat{g} \hat{f}) ψ (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	$x (r)$	$\hat{x} (\hat{x})$
动量	$p_{x} (p)$	${\hat{p}}_{x} = - i ℏ \frac{d}{d x} (\hat{p})$
角动量	$l^{2} = l_{x}^{2} + l_{y}^{2} + l_{z}^{2}$	${\hat{l}}^{2} = {\hat{l}}_{x}^{2} + {\hat{l}}_{y}^{2} + {\hat{l}}_{z}^{2}$
		${\hat{l}}_{x} = - i ℏ (y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y})$
		${\hat{l}}_{y} = - i ℏ (z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z})$
		${\hat{l}}_{z} = - i ℏ (x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x})$

算子组一

将 ${\hat{p}}_{x} = - i ℏ \frac{d}{d x}$ 代入可得

\begin{aligned} [\hat{x}, {\hat{p}}_{x}] ψ (x) & = (\hat{x} {\hat{p}}_{x} - {\hat{p}}_{x} \hat{x}) ψ (x) \\ = - i ℏ x \frac{d}{d x} ψ (x) - (- i ℏ) \frac{d}{d x} [x ψ (x)] \end{aligned}

这里需要注意算子 $\frac{d}{d x}$ 是求导算子，根据链式法则应该对 $[x ψ (x)]$ 分别对 $x$ 求导，于是：

\begin{aligned} 原 式 & = - i ℏ x \frac{d}{d x} ψ (x) + i ℏ [x \frac{d}{d x} ψ (x) + ψ (x)] \\ = i ℏ ψ (x) \neq 0 \end{aligned}

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 ${(- i)}^{2} = - 1$ )

\begin{aligned} {\hat{l}}_{x} {\hat{l}}_{y} & = (- i ℏ)^{2} (y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y}) (z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z}) \\ = - ℏ^{2} (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}) \end{aligned}

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{aligned} {\hat{l}}_{x} {\hat{l}}_{y} & = - ℏ^{2} [y (z \frac{\partial^{2}}{\partial z \partial x} + \frac{\partial}{\partial x}) - z^{2} \frac{\partial^{2}}{\partial y \partial x} - x y \frac{\partial^{2}}{\partial z^{2}} + x z \frac{\partial^{2}}{\partial y \partial z}] \\ = - ℏ^{2} [y z \frac{\partial^{2}}{\partial z \partial x} + y \frac{\partial}{\partial x} - z^{2} \frac{\partial^{2}}{\partial y \partial x} - x y \frac{\partial^{2}}{\partial z^{2}} + x z \frac{\partial^{2}}{\partial y \partial z}] \end{aligned}

同理

\begin{aligned} {\hat{l}}_{y} {\hat{l}}_{x} & = (- i ℏ)^{2} (z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z}) (y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y}) \\ = - ℏ^{2} (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}) \\ = - ℏ^{2} [y z \frac{\partial^{2}}{\partial x \partial z} - x y \frac{\partial^{2}}{\partial z^{2}} - z^{2} \frac{\partial^{2}}{\partial x \partial y} + x (\frac{\partial}{\partial y} + z \frac{\partial^{2}}{\partial z \partial y})] \\ = - ℏ^{2} (y z \frac{\partial^{2}}{\partial x \partial z} - x y \frac{\partial^{2}}{\partial z^{2}} - z^{2} \frac{\partial^{2}}{\partial x \partial y} + x \frac{\partial}{\partial y} + x z \frac{\partial^{2}}{\partial z \partial y}) \end{aligned}

因此交换子为（减法抵消相同项）

\begin{aligned} [{\hat{l}}_{x}, {\hat{l}}_{y}] & = {\hat{l}}_{x} {\hat{l}}_{y} - {\hat{l}}_{y} {\hat{l}}_{x} \\ = - ℏ^{2} (y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y}) \\ = i^{2} ℏ^{2} (y \frac{\partial}{\partial x} - x \frac{\partial}{\partial y}) \\ = i ℏ [- i ℏ (x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x})] \\ = i ℏ {\hat{l}}_{z} \neq 0 \end{aligned}

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}] = i ℏ {\hat{l}}_{x}

[{\hat{l}}_{z}, {\hat{l}}_{x}] = i ℏ {\hat{l}}_{y}

将其代入可得：

\begin{aligned} [{\hat{l}}^{2}, {\hat{l}}_{z}] & = {\hat{l}}^{2} {\hat{l}}_{z} - {\hat{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} + {\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}}_{x} {\hat{l}}_{z} {\hat{l}}_{x} - {\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} - {\hat{l}}_{y} {\hat{l}}_{z} {\hat{l}}_{y} - {\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} \\ = - i ℏ {\hat{l}}_{x} {\hat{l}}_{y} - i ℏ {\hat{l}}_{y} {\hat{l}}_{x} + i ℏ {\hat{l}}_{y} {\hat{l}}_{x} + i ℏ {\hat{l}}_{x} {\hat{l}}_{y} = 0 \end{aligned}

因此，这两个算子可交换。

（采用 CC BY-NC-SA 4.0 许可协议进行授权）

本文标题：《《手解量子化学》练习题 1-2 》

本文链接：https://lisz.me/ac/qc/solve-quantum-chemistry-by-hand-1-2.html

本文最后一次更新为天前，文章中的某些内容可能已过时！

《手解量子化学》练习题 1-2

从练习题推导中加深对量子化学理论的理解

算子组一

算子组二

算子组三

目录

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《手解量子化学》练习题 1-2

从练习题推导中加深对量子化学理论的理解

算子组一

算子组二

算子组三

目录

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Preview: