第四章 力学量随时间的演化与对称性

求证 $\frac{\mathrm{d}\hat{A}}{\mathrm{d}t}=0$ ，

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\braket{\psi(t)\vert\hat{A}\vert\psi(t)}&=\braket{\dot\psi(t)\vert\hat{A}\vert\psi(t)}+\braket{\psi(t)\vert\frac{\mathrm{d}\hat{A}}{\mathrm{d}t}\vert\psi(t)}+\braket{\psi(t)\vert\hat{A}\vert\dot\psi(t)} \\ &=\braket{\frac{\mathrm{d}\psi}{\mathrm{d}t}\vert\hat{A}\vert\psi(t)}+\braket{\psi(t)\vert\hat{A}\vert\frac{\mathrm{d}\psi}{\mathrm{d}t}} \\ &=\braket{\frac{1}{i\hbar}\hat{H}\psi(t)\vert\hat{A}\vert\psi(t)}+\braket{\psi(t)\vert\hat{A}\vert\frac{1}{i\hbar}\hat{H}\psi(t)} \\ &=\left(\frac{1}{i\hbar}\hat{H}\psi,\hat{A}\psi\right)+\left(\psi,\hat{A}\left(\frac{1}{i\hbar}\hat{H}\psi\right)\right)\\ &=\left(\frac{1}{i\hbar}\right)^{*}\left(\hat{H}\psi,\hat{A}\psi\right)+\frac{1}{i\hbar}\left(\psi,\hat{A}\hat{H}\psi\right)\\ &=-\frac{1}{i\hbar}\left(\hat{H}\psi,\hat{A}\psi\right)+\frac{1}{i\hbar}\left(\psi,\hat{A}\hat{H}\psi\right)\\ &=-\frac{1}{i\hbar}\left(\hat{H}^{\dagger}\psi,\hat{A}\psi\right)+\frac{1}{i\hbar}\left(\psi,\hat{A}\hat{H}\psi\right)\\ &=-\frac{1}{i\hbar}\left(\psi,\hat{H}\hat{A}\psi\right)+\frac{1}{i\hbar}\left(\psi,\hat{A}\hat{H}\psi\right)\\ &=\frac{1}{i\hbar}\left(\psi,\left(\hat{A}\hat{H}-\hat{H}\hat{A}\right)\psi\right)\\ &=\frac{1}{i\hbar}\left(\psi,\left[\hat{A},\hat{H}\right]\psi\right)\\ &=0 \end{aligned}$$

对易子在任何态下的平均值都是 $0$ 。

例：在氢原子中，电子的哈密顿量为 $\hat{H}=\frac{\hat{P}^2}{2m}-\frac{e^2}{r}$ ，考虑角动量算符 $\hat{\vec{L}}=\hat{L}_x\vec{i}+\hat{L}_y\vec{j}+\hat{L}_z\vec{k}$，$\left[\hat{L}_x,\hat{H}\right]=\left[\hat{L}_y,\hat{H}\right]=\left[\hat{L}_z,\hat{H}\right]=0$ 。验证 $\left[\hat{L}_x,\hat{P}^2\right]=0$ ，$\left[\hat{L}_x,\frac{1}{r}\right]=0$ 。

$\left[\hat{u}_1\hat{u}_2,\hat{v}\right]=\left[\hat{u}_1,\hat{v}\right]\hat{u}_2+\hat{u}_1\left[\hat{u}_2,\hat{v}\right]\,$
$\left[\hat{u},\hat{v}_1\hat{v}_2\right]=\left[\hat{u},\hat{v}_1\right]\hat{v}_2+\hat{v}_1\left[\hat{u},\hat{v}_2\right]\,$

$$\begin{aligned} \left[\hat{L}_x,\hat{P}^2\right]&=\left[\hat{y}\hat{P}_z-\hat{z}\hat{P}_y,\hat{P}_x^2+\hat{P}_y^2+\hat{P}_z^2\right]\\ &=\left[\hat{y}\hat{P}_z-\hat{z}\hat{P}_y,\hat{P}_y^2+\hat{P}_z^2\right]\\ &=\left[\hat{y}\hat{P}_z,\hat{P}_y^2\right]+\underbrace{\left[\hat{y}\hat{P}_z,\hat{P}_z^2\right]}_{=0}+\underbrace{\left[-\hat{z}\hat{P}_y,\hat{P}_y^2\right]}_{=0}+\left[-\hat{z}\hat{P}_y,\hat{P}_z^2\right] \\ &=\left[\hat{y},\hat{P}_y^2\right]\hat{P}_z+\hat{y}\underbrace{\left[\hat{P}_z,\hat{P}_y^2\right]}_{=0}-\left[\hat{z},\hat{P}_z^2\right]\hat{P}_y-\hat{z}\underbrace{\left[\hat{P}_y,\hat{P}_z^2\right]}_{=0} \\ &=\left\{\underbrace{\left[\hat{y},\hat{P}_y\right]}_{=i\hbar\hat{I}}\hat{P}_y+\hat{P}_y\underbrace{\left[\hat{y},\hat{P}_y\right]}_{=i\hbar\hat{I}}\right\}\hat{P}_z-\left\{\underbrace{\left[\hat{z},\hat{P}_z\right]}_{=i\hbar\hat{I}}\hat{P}_z+\hat{P}_z\underbrace{\left[\hat{z},\hat{P}_z\right]}_{=i\hbar\hat{I}}\right\}\hat{P}_y \\ &=\underbrace{\left\{i\hbar\hat{I}\hat{P}_y+i\hbar\hat{P}_y\hat{I}\right\}}_{=2i\hbar\hat{P}_y}\hat{P}_z-\underbrace{\left\{i\hbar\hat{I}\hat{P}_z+i\hbar\hat{P}_z\hat{I}\right\}}_{=2i\hbar\hat{P}_z}\hat{P}_y\\ &=2i\hbar\left(\hat{P}_y\hat{P}_z-\hat{P}_z\hat{P}_y\right)\\ &=2i\hbar\left[\hat{P}_y,\hat{P}_z\right]\\ &=0 \end{aligned}$$

$\left[\hat{L}_x,\frac{1}{r}\right]=\left[\hat{y}\hat{P}_z-\hat{P}_z\hat{y},\frac{1}{\sqrt{x^2+y^2+z^2}}\right]$ ，考虑 $\frac{1}{r}$ 中，$y$ 和 $z$ 是对称的，则左式为 $0$，另一方面 $\left[\hat{P},\frac{1}{r}\right]=\frac{\hbar}{i}\nabla\frac{1}{r}\neq 0$ 。

在中心力场中，角动量是一个守恒量，动量却不是一个守恒量。

Noether 定理（经典力学）：
一个给定的力学体系的守恒量是由该体系的对称性决定的。

1. 空间平移不变性 $\implies$ 总动量守恒
2. 时间平移不变性 $\implies$ 总能量守恒
3. 转动对称性（各向同性） $\implies$ 总角动量守恒
4. 镜像对称 $\implies$ 宇称守恒

{}