量子力学中的动量为何说动量是平移的生成元

原文: https://zhuanlan.zhihu.com/p/84531862
发布日期: 2019-10-02
分类: 量子力学基础与量子信息

以前看 Sakurai 的量子力学的时候就在这里卡过一下, 现在把自己的理解写出来供个参考.

其实如果你知道平移算符 ${Q}^{\dagger }\left( x \right)\equiv {\rm{e}^{-{\rm{i}\frac{P}{\hbar }\cdot x}$ 的话, 从群论的角度不难看出 $P$ 事单参子群 ${Q^{†} (x) x \in R}$ 的生成元. 但我们这里将要讲的 Sakurai 指的那个生成元, 是经典力学里的概念. 这个生成元的概念与群论中的不是一回事, 但有许多相似之处.

I、II、III 部分引出量子力学中的动量并简洁地给出一些证明. IV、V、VI 部分提供感性的物理图像, 通过建立与经典力学的联系来直观理解算符的形式. VII 综合应用了前面的结论, 一定程度地缓解了上部分的空降感与下部分的强行类比感.

I. 定义平移算符:

构造算符 ${Q}^{\dagger }={\rm{e}^{-{\rm{i}\frac{P}_{x}{\hbar }{x}_{0}.$

利用关系 $\left[ {X}_{i},F\left( {\vec{P} \right) \right]={\rm{i}\hbar \frac{\partial }{\partial {P}_{i}F\left( {\vec{P} \right)$ 可得:

$\left[ X,{Q}^{\dagger } \right]={\rm{i}\hbar \frac{\partial }{\partial {P}_{x}{Q}^{\dagger }={\rm{i}\hbar \frac{\partial }{\partial {P}_{x}{\rm{e}^{-{\rm{i}\frac{P}_{x}{\hbar }{x}_{0}={x}_{0}{Q}^{\dagger }\Rightarrow X{Q}^{\dagger }={x}_{0}{Q}^{\dagger }+{Q}^{\dagger }X.$

关系来源见附录 **[附 1]**.

右乘 $∣ x ⟩$ 有 $X Q^{†} ∣ x ⟩ = x_{0} Q^{†} ∣ x ⟩ + Q^{†} X ∣ x ⟩ = (x + x_{0}) Q^{†} ∣ x ⟩ .$

这意味着 ${Q}^{\dagger }={\rm{e}^{-{\rm{i}\frac{P}_{x}{\hbar }{x}_{0}$ 作用在 $x$ 对应的本征态 $∣ x ⟩$ 上后,
会使 $∣ x ⟩$ 变成 $x + x_{0}$ 对应的本征态 $∣ x + x_{0} ⟩ .$

基于这个特性我们将 ${Q}^{\dagger }={\rm{e}^{-{\rm{i}\frac{P}_{x}{\hbar }{x}_{0}$ 称为位置平移算符.

有人会将 ${Q}^{\dagger }={\rm{e}^{-{\rm{i}\frac{P}_{x}{\hbar }{x}_{0}$ 写作 ${Q}^{\dagger }={\rm{e}^{-{\rm{i}\frac{P}_{x}{\hbar }{x}.$ 但即使是这样也要记住 $x$ 是个常数, 表示平移的量度. 即在坐标表象下是 $\langle x|{Q}^{\dagger }=\langle x|{\rm{e}^{-{\rm{i}\frac{P}_{x}{\hbar }x}={\rm{e}^{-x\frac{\partial }{\partial x}\langle x|.$ 总之注意即使在坐标表象下 ${P}_{x}$ 也是不会作用在这个平移量 $x$ 上的[[1]](#ref_1).

同理可用 $\left[ {P}_{i},G\left( {\vec{R} \right) \right]=\frac{\hbar }{\rm{i}\frac{\partial }{\partial {X}_{i}G\left( {\vec{R} \right)$ 构造动量平移算符 ${T}^{\dagger }={\rm{e}^{\rm{i}\frac{p}_{x}_{0}{\hbar }X}.$

II. 动量的本征矢:

(1). 若动量存在本征值为 0 的本征矢 $∣ p_{x} = 0 ⟩$ , 记作 $\left| {0}_{p}_{x} \right\rangle$ , 就可以推导出其它的动量本征矢:

只需要通过动量平移算符作用在 $∣ 0_{p} ⟩$ 即有: $\left| {p}_{x0} \right\rangle ={T}^{+}\left| {0}_{p}_{x} \right\rangle ={\rm{e}^{\rm{i}\frac{p}_{x}_{0}{\hbar }X}\left| {0}_{p}_{x} \right\rangle .$

这就说明了动量的谱或者说本征值可以连续地取值, 当然这里说的是自由粒子.

显然并非一定要先知道 $\left| {0}_{p}_{x} \right\rangle$ , 实际上给出任何一个本征矢就可以推导出其他的本征矢.

(2). 动量本征矢在坐标表象下的波函数 ${\psi }_{p}_{x}\left( x \right)=\left\langle x | {p}_{x} \right\rangle$ :

$\begin{align} & {\psi }_{p}_{x}\left( x \right)=\left\langle x|{p}_{x} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ =\langle x|{\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }X}\left| {0}_{p}_{x} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ ={\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }x}\left\langle x|{0}_{p}_{x} \right\rangle ={\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }x}\langle {0}_{x}|{\rm{e}^{\rm{i}\frac{P}_{x}{\hbar }x}\left| {0}_{p}_{x} \right\rangle ={\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }x}\langle {0}_{x}|{\rm{e}^{\rm{i}\frac{0}{\hbar }x}\left| {0}_{p}_{x} \right\rangle \\ \end{align}$

$\Rightarrow {\psi }_{p}_{x}\left( x \right)=\left\langle {0}_{x}|{0}_{p}_{x} \right\rangle {\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }x}$ , 那么现在只需求出 $\left\langle {0}_{x} | {0}_{p}_{x} \right\rangle$ 即可:

已知 $\left\langle {0}_{p}_{x}|p \right\rangle =\delta \left( p-0 \right)=\delta \left( p \right),$

$\begin{align} & \delta \left( p \right)=\left\langle {0}_{p}_{x}|p \right\rangle =\langle {0}_{p}_{x}|{\rm{e}^{\rm{i}\frac{p}{\hbar }X}\left| {0}_{p}_{x} \right\rangle \\ & \ \ \ \ \ \ \ \ =\langle {0}_{p}_{x}|{\rm{e}^{\rm{i}\frac{p}{\hbar }X}\int{dx\left| x \right\rangle \left\langle x|{0}_{px} \right\rangle } \\ & \ \ \ \ \ \ \ \ =\int{\left\langle {0}_{p}_{x}|x \right\rangle {\rm{e}^{\rm{i}\frac{p}{\hbar }x}\left\langle x|{0}_{px} \right\rangle }{\rm{d}x \\ & \ \ \ \ \ \ \ \ =\int{\langle {0}_{p}_{x}|{\rm{e}^{-{\rm{i}\frac{P}{\hbar }x}\left| {0}_{x} \right\rangle {\rm{e}^{\rm{i}\frac{p}{\hbar }x}\langle {0}_{x}|{\rm{e}^{\rm{i}\frac{P}{\hbar }x}\left| {0}_{p}_{x} \right\rangle }{\rm{d}x \\ & \ \ \ \ \ \ \ \ =\int{\langle {0}_{p}_{x}|{\rm{e}^{0}\left| {0}_{x} \right\rangle {\rm{e}^{\rm{i}\frac{p}{\hbar }x}\langle {0}_{x}|{\rm{e}^{0}\left| {0}_{p}_{x} \right\rangle }{\rm{d}x \\ & \ \ \ \ \ \ \ \ ={\left| \left\langle {0}_{x}|{0}_{p}_{x} \right\rangle \right|}^{2}\int{\rm{e}^{\rm{i}\frac{p}{\hbar }x}{\rm{d}x \\ & \ \ \ \ \ \ \ \ ={\left| \left\langle {0}_{x}|{0}_{p}_{x} \right\rangle \right|}^{2}2\pi \hbar \delta \left( p \right)\Rightarrow \left\langle {0}_{x}|{0}_{p}_{x} \right\rangle {\rm{=}\frac{\rm{1}{\sqrt{\rm{2}\pi \hbar }{\rm{e}^{\rm{i}\varphi }. \\ \end{align}$

δ 函数的数学表达式见附录 **[附2]**.

取 $φ = 0$ 有 $\left\langle x|{0}_{px} \right\rangle {\rm{=}\frac{\rm{1}{\sqrt{\rm{2}\pi \hbar }\Rightarrow {\psi }_{p}_{x}\left( x \right)=\frac{\rm{1}{\sqrt{\rm{2}\pi \hbar }{\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }x}$ 即得到了所求波函数.

III. 坐标表象下的动量算符:

(1). 动量算符在坐标表象下的矩阵元:

$\begin{align} & \langle x|{P}_{x}\left| {x}'} \right\rangle =\int \langle x|{P}_{x}\left| {p}_{x} \right\rangle \left\langle {p}_{x}|{x}' \right\rangle {\rm{d}{p}_{x}=\int{p}_{x}\left\langle x|{p}_{x} \right\rangle \left\langle {p}_{x}|{x}' \right\rangle {\rm{d}{p}_{x} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{p}_{x}\frac{1}{\sqrt{2\pi \hbar }{\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }x}\frac{1}{\sqrt{2\pi \hbar }{\rm{e}^{-{\rm{i}\frac{p}_{x}{\hbar }{x}'}{\rm{d}{p}_{x} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2\pi \hbar }\int{p}_{x}{\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }\left( x-{x}' \right)}{\rm{d}{p}_{x} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2\pi \hbar }\int{\frac{\hbar }{\rm{i}\frac{\partial }{\partial x}{\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }\left( x-{x}' \right)}{\rm{d}{p}_{x} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\hbar }{\rm{i}\frac{\partial }{\partial x}\frac{1}{2\pi \hbar }\int{\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }\left( x-{x}' \right)}{\rm{d}{p}_{x} \\ & \Rightarrow {P}_{x{x}'}=\langle x|{P}_{x}\left| {x}'} \right\rangle =\frac{\hbar }{\rm{i}\frac{\partial }{\partial x}\delta \left( x-{x}' \right). \\ \end{align}$

δ 函数的定义见附录 **[附 2]**.

(2). 动量算符在坐标表象下的表现形式:

$\begin{align} & \langle x|{P}_{x}=\langle x|{P}_{x}\int{\rm{d}{p}_{x}\left| {p}_{x} \right\rangle \langle {p}_{x}| \\ & \ \ \ \ \ \ \ \ \ =\int{p}_{x}\left\langle x|{p}_{x} \right\rangle \langle {p}_{x}|{\rm{d}{p}_{x} \\ & \ \ \ \ \ \ \ \ \ =\int{p}_{x}\frac{\rm{1}{\sqrt{\rm{2}\pi \hbar }{\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }x}\langle {p}_{x}|{\rm{d}{p}_{x} \\ & \ \ \ \ \ \ \ \ \ =\frac{\rm{1}{\sqrt{\rm{2}\pi \hbar }\int{\frac{\hbar }{\rm{i}\frac{\partial }{\partial x}{\rm{e}^{\rm{i}\frac{p}_{x}{\hbar }x}\langle {p}_{x}|{\rm{d}{p}_{x} \\ & \ \ \ \ \ \ \ \ \ =\frac{\hbar }{\rm{i}\frac{\partial }{\partial x}\int{\left\langle x|{p}_{x} \right\rangle }\langle {p}_{x}|{\rm{d}{p}_{x}=\frac{\hbar }{\rm{i}\frac{\partial }{\partial x}\langle x|\int{\rm{d}{p}_{x}\left| {p}_{x} \right\rangle \left\langle {p}_{x} \right| \\ \end{align}$

$\Rightarrow \langle x|{P}_{x}=\frac{\hbar }{\rm{i}\frac{\partial }{\partial x}\langle x|.$

下面这四个式子要当作常识:

$\left\{ \begin{align} & \langle \vec{r}|\vec{P}=\frac{\hbar }{\rm{i}\nabla \langle \vec{r}|, \\ & \langle \vec{p}|\vec{P}=\vec{p}\langle \vec{p}|. \\ \end{align} \right.$ $\left\{ \begin{align} & \langle \vec{r}|\vec{R}=\vec{r}\langle \vec{r}|, \\ & \langle \vec{p}|\vec{R}={\rm{i}\hbar {\nabla }_{\vec{p}\langle \vec{p}|. \\ \end{align} \right.$

IV. 如何定义平移算符:

平移算符 ${Q}^{\dagger }\left( {\vec{r}_{0} \right)$ 的作用是使 $\left| {\vec{r} \right\rangle$ 变成 $\left| {\vec{r}_{0}+\vec{r} \right\rangle$ , 不难想象它应该满足以下性质:

(1). 要满足幺正性, 因为变换前后都是坐标算符归一的本征态.

(2). 平移当然要满足 ${Q}^{\dagger }\left( {\vec{r}_{2} \right){Q}^{\dagger }\left( {\vec{r}_{1} \right)={Q}^{\dagger }\left( {\vec{r}_{1} \right){Q}^{\dagger }\left( {\vec{r}_{2} \right)={Q}^{\dagger }\left( {\vec{r}_{1}+{\vec{r}_{2} \right).$

(3). 移过去当然还要能移回来: ${Q}^{\dagger }\left( {\vec{r}_{0} \right)\left| {\vec{r} \right\rangle =\left| \vec{r}+{\vec{r}_{0} \right\rangle \Rightarrow {\left[ {Q}^{\dagger }\left( {\vec{r}_{0} \right) \right]}^{-1}={Q}^{\dagger }\left( -{\vec{r}_{0} \right).$

(4). 不发生平移就什么都别变: $Q^{†} (0) = 1.$

很难一步猜出 ${Q}^{\dagger }\left( {\vec{r}_{0} \right)$ 的表达式, 但退一步猜无穷小算符 ${Q}^{\dagger }\left( {\rm{d}{\vec{r}_{0} \right)$ 还是比较可行的:[2]

根据上面四个性质我们认为 ${Q}^{\dagger }\left( {\rm{d}{\vec{r}_{0} \right)=1-{\rm{i}\vec{K}\cdot {\rm{d}{\vec{r}_{0}$ , 其中 ${\vec{K}$ 是个厄米算符.

不难证明它满足上面四个性质, 证明见附录 **[附 3]**.

然后根据 ${Q}^{\dagger }\left( {\rm{d}{\vec{r}_{0} \right)=1-{\rm{i}\vec{K}\cdot {\rm{d}{\vec{r}_{0}$ 与性质 ${Q}^{\dagger }\left( {\vec{r}_{2} \right){Q}^{\dagger }\left( {\vec{r}_{1} \right)={Q}^{\dagger }\left( {\vec{r}_{1}+{\vec{r}_{2} \right),$

我们就能得到 ${Q}^{\dagger }\left( {\vec{r}_{0} \right)$ :

$\left. \begin{align} & {Q}^{\dagger }\left( {\vec{r}_{0} \right)=\underset{n\to \infty }{\mathop{\lim }\,{Q}^{\dagger }\left( \frac{\vec{r}_{0}{n}+\frac{\vec{r}_{0}{n}+\cdot \cdot \cdot +\frac{\vec{r}_{0}{n} \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ =\underset{n\to \infty }{\mathop{\lim }\,{\left[ {Q}^{\dagger }\left( \frac{\vec{r}_{0}{n} \right) \right]}^{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ =\underset{n\to \infty }{\mathop{\lim }\,{\left( 1-{\rm{i}\vec{K}\cdot \frac{\vec{r}_{0}{n} \right)}^{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ =\underset{n\to \infty }{\mathop{\lim }\,{\left[ {\left( 1-{\rm{i}\vec{K}\cdot \frac{\vec{r}_{0}{n} \right)}^{-\frac{n}{\rm{i}\vec{K}\cdot {\vec{r}_{0} \right]}^{-{\rm{i}\vec{K}\cdot {\vec{r}_{0} \\ \end{align} \right\}\Rightarrow {Q}^{\dagger }\left( {\vec{r}_{0} \right)={\rm{e}^{-{\rm{i}\vec{K}\cdot {\vec{r}_{0}.$

到这里其实就已经剧透了, 显然 ${\vec{K}$ 应该是波矢算符对吧?

V. 为什么说动量是平移的生成元:

这种说法是从分析力学的哈密顿-雅可比理论中类比而来的.

我们将无穷小平移看作是这样的一个正则变换: $\left\{ \begin{align} & {\vec{r}'=\vec{r}+{\rm{d}{\vec{r}_{0}, \\ & {\vec{p}'=\vec{p}. \\ \end{align} \right.$

正则变换满足关系 $\left\{ \begin{align} & {\rm{d}{U}_{2}=\sum\limits_{\alpha =1}^{s}{\left( {p}_{\alpha }{\rm{d}{q}_{\alpha }+{q}'}_{\alpha }{\rm{d}{p}'}_{\alpha } \right)}{\rm{+}\left( {H}'-H \right){\rm{d}t, \\ & \frac{\partial {U}_{2}{\partial {q}_{\alpha }={p}_{\alpha },\ \frac{\partial {U}_{2}{\partial {p}'}_{\alpha }={q}'}_{\alpha },\ \frac{\partial {U}_{2}{\partial t}={H}'-H. \\ \end{align} \right.$

前情提要: [東雲正樹: 正则变换与哈密顿-雅可比理论 / 作用量竟然是··· 真相令人潸然泪下](https://zhuanlan.zhihu.com/p/85001265).

不难看出来生成函数是 ${U}_{2}=\vec{r}\cdot {\vec{p}'+{\vec{p}'\cdot {\rm{d}{\vec{r}_{0}=\vec{r}\cdot \vec{p}+\vec{p}\cdot {\rm{d}{\vec{r}_{0}.$

假如说没有 $\vec{p}\cdot {\rm{d}\vec{r}$ 这部分, $U_{2} = r \cdot p$ 就是一个恒等变换 $\left\{ \begin{align} & {\vec{r}'=\vec{r}, \\ & {\vec{p}'=\vec{p}. \\ \end{align} \right.$

再看回前面我们提到过的 ${Q}^{\dagger }\left( {\rm{d}\vec{r} \right)=1-{\rm{i}\vec{K}\cdot {\rm{d}{\vec{r}_{0}$ 和 ${U}_{2}=\vec{r}\cdot \vec{p}+\vec{p}\cdot {\rm{d}{\vec{r}_{0},$

这二者是不是有着很高的相似度?

那么我们就会考虑 ${\vec{K}$ 与动量 ${\vec{P}$ 有何联系, 从量纲看它们显然不相等, 但可以差一个常系数 $ℏ$ .

这也就暗示了德布罗意关系 $\vec{p}=\hbar \vec{k}=\frac{h}{\lambda }{\vec{e}_{k}$ .

然后嗯猜可以推测出 ${Q}^{\dagger }\left( {\vec{r}_{0} \right)={\rm{e}^{-{\rm{i}\vec{K}\cdot {\vec{r}_{0}={\rm{e}^{-{\rm{i}\frac{\vec{P}{\hbar }\cdot {\vec{r}_{0}.$

VI. 坐标动量对易关系:

前面猜出了表达式 ${Q}^{\dagger }\left( {\rm{d}{\vec{r}_{0} \right)=1-{\rm{i}\vec{K}\cdot {\rm{d}{\vec{r}_{0}=1-{\rm{i}\frac{\vec{P}{\hbar }\cdot {\rm{d}{\vec{r}_{0}.$

不难计算如下对易关系:

$\begin{align} & \left[ \vec{R},{Q}^{\dagger }\left( d{\vec{r}_{0} \right) \right]\left| {\vec{r} \right\rangle =\vec{R}{Q}^{\dagger }\left( {\rm{d}{\vec{r}_{0} \right)\left| {\vec{r} \right\rangle -{Q}^{\dagger }\left( {\rm{d}{\vec{r}_{0} \right)\vec{R}\left| {\vec{r} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( \vec{r}+{\rm{d}{\vec{r}_{0} \right)\left| \vec{r}+{\rm{d}{\vec{r}_{0} \right\rangle -\vec{r}\left| \vec{r}+{\rm{d}{\vec{r}_{0} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\rm{d}{\vec{r}_{0}\left| \vec{r}+{\rm{d}{\vec{r}_{0} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\rm{d}{\vec{r}_{0}\left[ \left| {\vec{r} \right\rangle +O\left( {\rm{d}{\vec{r}_{0} \right) \right] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\rm{d}{\vec{r}_{0}\left| {\vec{r} \right\rangle +{\rm{d}{\vec{r}_{0}O\left( {\rm{d}{\vec{r}_{0} \right)={\rm{d}{\vec{r}_{0}\left| {\vec{r} \right\rangle +o\left( {\rm{d}{\vec{r}_{0} \right)={\rm{d}{\vec{r}_{0}\left| {\vec{r} \right\rangle \\ \end{align}$

$\Rightarrow \left[ \vec{R},{Q}^{\dagger }\left( {\rm{d}{\vec{r}_{0} \right) \right]={\rm{d}{\vec{r}_{0}.$

代入 ${Q}^{\dagger }\left( {\rm{d}{\vec{r}_{0} \right)=1-{\rm{i}\frac{\vec{P}{\hbar }\cdot {\rm{d}{\vec{r}_{0}\Rightarrow \vec{R}\vec{P}\cdot {\rm{d}{\vec{r}_{0}-\vec{P}\cdot {\rm{d}{\vec{r}_{0}\vec{R}={\rm{i}\hbar {\rm{d}{\vec{r}_{0}.$

假定上式 ${\rm{d}{\vec{r}_{0}={\rm{d}{r}_{0j}{\vec{e}_{j},$

则有 $\vec{R}{P}_{j}{\rm{d}{r}_{0j}-{\vec{P}_{j}{\rm{d}{r}_{0j}\vec{R}={\rm{i}\hbar {\rm{d}{r}_{0j}{\vec{e}_{j}\Rightarrow \vec{R}{P}_{j}-{P}_{j}\vec{R}={\rm{i}\hbar {\vec{e}_{j}.$

上式再取一个与 ${\vec{e}_{j}$ 正交的方向 ${\vec{e}_{i}$ , 即点乘 ${\vec{e}_{i},$

就得到 ${R}_{i}{P}_{j}-{P}_{j}{R}_{i}=\left[ {R}_{i},{P}_{j} \right]={\rm{i}\hbar {\delta }_{ij}$ , 即经典对易关系.

VII. 坐标表象下动量算符与平移算符的另一个理解角度:

这一段是 [@Jeon](https://www.zhihu.com/people/d0fb95c9f968bbefc0045b5782481e00) 提供的一个角度, 我认为对初学者有益就融入一些自己观点地誊写上来了.

已知无穷小平移算符的定义式 $Q\left( {\rm{d}{\vec{r}_{0} \right)\psi \left( {\vec{r} \right)=\psi \left( \vec{r}+{\rm{d}{\vec{r}_{0} \right).$

展开后忽略高阶项: $Q\left( {\rm{d}{\vec{r}_{0} \right)\psi \left( {\vec{r} \right)=\psi \left( \vec{r}+{\rm{d}{\vec{r}_{0} \right)=\psi \left( {\vec{r} \right)+\frac{\rm{d}{\rm{d}\vec{r}\psi \left( {\vec{r} \right)\cdot {\rm{d}{\vec{r}_{0}.$

引入记号 $\frac{\rm{d}{\rm{d}\vec{r}\equiv \nabla$ 则有 $Q\left( {\rm{d}{\vec{r}_{0} \right)\psi \left( {\vec{r} \right)=\left( 1+{\rm{d}{\vec{r}_{0}\cdot \nabla \right)\psi \left( {\vec{r} \right).$

令 ${\rm{d}{\vec{r}_{0}=\frac{\vec{r}_{0}{n}$ 可计算出 $Q\left( {\vec{r}_{0} \right)$ 的表达式:

$\begin{align} & \psi \left( \vec{r}+{\vec{r}_{0} \right)=Q\left( {\vec{r}_{0} \right)\psi \left( {\vec{r} \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{n\to \infty }{\mathop{\lim }\,{Q}^{n}\left( {\rm{d}{\vec{r}_{0} \right)\psi \left( {\vec{r} \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{n\to \infty }{\mathop{\lim }\,{\left( 1+\frac{\vec{r}_{0}{n}\cdot \nabla \right)}^{n}\psi \left( {\vec{r} \right)={\rm{e}^{\vec{r}_{0}\cdot \nabla }\psi \left( {\vec{r} \right). \\ \end{align}$

其中用到了一个显然成立的性质 $Q\left( {\vec{r}_{2} \right)Q\left( {\vec{r}_{1} \right)=Q\left( {\vec{r}_{1}+{\vec{r}_{2} \right).$

$\begin{align} & \Rightarrow \psi \left( \vec{r}+{\vec{r}_{0} \right)=Q\left( {\vec{r}_{0} \right)\psi \left( {\vec{r} \right)={\rm{e}^{\vec{r}_{0}\cdot \nabla }\psi \left( {\vec{r} \right) \\ & \Rightarrow \left\langle \vec{r}+{\vec{r}_{0}|\psi \right\rangle =Q\left( {\vec{r}_{0} \right)\left\langle \vec{r}|\psi \right\rangle ={\rm{e}^{\vec{r}_{0}\cdot \nabla }\left\langle \vec{r}|\psi \right\rangle ={\rm{e}^{\frac{\rm{i}{\hbar }{\vec{r}_{0}\cdot \frac{\hbar }{\rm{i}\nabla }\left\langle \vec{r}|\psi \right\rangle \\ & \Rightarrow \left\langle \vec{r}+{\vec{r}_{0}|\psi \right\rangle =\langle \vec{r}|Q\left( {\vec{r}_{0} \right)\left| \psi \right\rangle =\langle \vec{r}|{\rm{e}^{\frac{\rm{i}{\hbar }{\vec{r}_{0}\cdot \vec{P}\left| \psi \right\rangle \\ & \Rightarrow \langle \vec{r}+{\vec{r}_{0}|=\langle \vec{r}|Q\left( {\vec{r}_{0} \right)=\langle \vec{r}|{\rm{e}^{\frac{\rm{i}{\hbar }{\vec{r}_{0}\cdot \vec{P}. \\ \end{align}$

最后一行取厄米共轭就得到平移算符的表达式: ${Q}^{\dagger }\left( {\vec{r}_{0} \right)\left| {\vec{r} \right\rangle ={\rm{e}^{-{\rm{i}\frac{\vec{P}{\hbar }\cdot {\vec{r}_{0}\left| {\vec{r} \right\rangle =\left| \vec{r}+{\vec{r}_{0} \right\rangle .$

值得注意的是幂级数展开只能是对函数进行的操作, 而不能是对右矢. 你不能写出这样的式子: $Q\left( {\rm{d}{\vec{r}_{0} \right)\left| {\vec{r} \right\rangle =\left| \vec{r}+{\rm{d}{\vec{r}_{0} \right\rangle =\left| {\vec{r} \right\rangle +\frac{\partial }{\partial \vec{r}\left| {\vec{r} \right\rangle \cdot {\rm{d}{\vec{r}_{0}=\left( 1+{\rm{d}{\vec{r}_{0}\cdot \nabla \right)\left| {\vec{r} \right\rangle .$ 但选择了表象后, 右矢就可以表达为波函数, 所以可以这样表达: $\begin{align} & \ \ \ \ \ \ Q\left( {\rm{d}{\vec{r}_{0} \right)\langle \vec{r}|=\langle \vec{r}+{\rm{d}{\vec{r}_{0}|=\langle \vec{r}|+\frac{\partial }{\partial \vec{r}\langle \vec{r}|\cdot {\rm{d}{\vec{r}_{0}=\left( 1+{\rm{d}{\vec{r}_{0}\cdot \nabla \right)\langle \vec{r}| \\ & \Rightarrow \langle \vec{r}+{\vec{r}_{0}|=Q\left( {\vec{r}_{0} \right)\langle \vec{r}|={Q}^{n}\left( {\rm{d}{\vec{r}_{0} \right)\langle \vec{r}|={\left( 1+\frac{\vec{r}_{0}{n}\cdot \nabla \right)}^{n}\langle \vec{r}|={\rm{e}^{\vec{r}_{0}\cdot \nabla }\langle \vec{r}| \\ & \Rightarrow \langle \vec{r}+{\vec{r}_{0}|=Q\left( {\vec{r}_{0} \right)\langle \vec{r}|={\rm{e}^{\vec{r}_{0}\cdot \nabla }\langle \vec{r}|={\rm{e}^{\frac{\rm{i}{\hbar }{\vec{r}_{0}\cdot \frac{\hbar }{\rm{i}\nabla }\langle \vec{r}| \\ & \Rightarrow \langle \vec{r}+{\vec{r}_{0}|=\langle \vec{r}|Q\left( {\vec{r}_{0} \right)=\langle \vec{r}|{\rm{e}^{\rm{i}{\vec{r}_{0}\cdot \frac{\vec{P}{\hbar }. \\ \end{align}$ 因为给出了基左矢就等同于选择了表象, 左矢作为右矢的线性泛函只要作用到右矢就能得到一个复数, 这里的级数展开也不是对左矢进行展开, 而只是一个展开的操作, 具体展开的是谁取决于作用的右矢. 右矢与左矢合成的波函数就是被展开的对象. 这里也用到了 III - (2) 提到的一个重要结论 $\langle \vec{r}|\vec{P}=\frac{\hbar }{\rm{i}\nabla \langle \vec{r}|.$ 当然最后还是得到这个结果: ${Q}^{\dagger }\left( {\vec{r}_{0} \right)\left| {\vec{r} \right\rangle ={\rm{e}^{-{\rm{i}{\vec{r}_{0}\cdot \frac{\vec{P}{\hbar }\left| {\vec{r} \right\rangle =\left| \vec{r}+{\vec{r}_{0} \right\rangle .$

VIII. 附录:

[附 1] 仅从 $\left[ {X}_{i},{P}_{j} \right]={\rm{i}\hbar {\delta }_{ij}$ 出发证明 $\left[ {X}_{i},F\left( {\vec{P} \right) \right]={\rm{i}\hbar \frac{\partial }{\partial {P}_{i}F\left( {\vec{P} \right):$

类似的, 我们同样可以证明 $\left[ {P}_{i},G\left( {\vec{R} \right) \right]=-{\rm{i}\hbar \frac{\partial }{\partial {X}_{i}G\left( {\vec{R} \right).$

(1). 已知 $\left[ {X}_{i},{P}_{i} \right]={\rm{i}\hbar \Rightarrow {X}_{i}{P}_{i}={\rm{i}\hbar +{P}_{i}{X}_{i},$

不难得到 $[X_{i}, P_{i}^{n}] = P_{i} [X_{i}, P_{i}^{n - 1}] + [X_{i}, P_{i}] P_{i}^{n - 1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={P}_{i}\left[ {X}_{i},P_{i}^{n-1} \right]+{\rm{i}\hbar P_{i}^{n-1}$
$\Rightarrow \left[ {X}_{i},P_{i}^{n-1} \right]={P}_{i}\left[ {X}_{i},P_{i}^{n-2} \right]+{\rm{i}\hbar P_{i}^{n-2}$
$\Rightarrow \left[ {X}_{i},P_{i}^{n} \right]={P}_{i}\left( {P}_{i}\left[ {X}_{i},P_{i}^{n-2} \right]+{\rm{i}\hbar P_{i}^{n-2} \right)+{\rm{i}\hbar P_{i}^{n-1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =P_{i}^{2}\left[ {X}_{i},P_{i}^{n-2} \right]+2{\rm{i}\hbar P_{i}^{n-1}$
$\cdot \cdot \cdot$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =P_{i}^{n}\left[ {X}_{i},1 \right]+n{\rm{i}\hbar P_{i}^{n-1}=n{\rm{i}\hbar P_{i}^{n-1}.$

现在直接对算符的函数进行幂级数展开即得证:

$\begin{align} & \left[ {X}_{i},F\left( {\vec{P} \right) \right]=\sum\limits_{n=0}^{\infty }{\frac{1}{n!}{\left. \frac{\partial }^{n}F}{\partial P_{i}^{n} \right|}_{P}_{i}=0}\left[ {X}_{i},P_{i}^{n} \right]} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{n=0}^{\infty }{\frac{1}{n!}{\left. \frac{\partial }^{n}F}{\partial P_{i}^{n} \right|}_{P}_{i}=0}n{\rm{i}\hbar P_{i}^{n-1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\rm{i}\hbar \sum\limits_{n=0}^{\infty }{\frac{1}{n!}{\left. \frac{\partial }^{n}F}{\partial P_{i}^{n} \right|}_{P}_{i}=0}\frac{\partial }{\partial {P}_{i}P_{i}^{n}={\rm{i}\hbar \frac{\partial }{\partial {P}_{i}F\left( {\vec{P} \right). \\ \end{align}$

(2). 若使用结论 $\langle \vec{p}|{X}_{i}={\rm{i}\hbar \frac{\partial }{\partial {p}_{i}\langle \vec{p}|$ , 则可以这样证明:

$\begin{align} & \left[ {X}_{i},F\left( {\vec{P} \right) \right]\left| \psi \right\rangle =\int{\rm{d}\vec{p}\left| {\vec{p} \right\rangle \langle \vec{p}|}\left[ {X}_{i},F\left( {\vec{P} \right) \right]\left| \psi \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{\rm{d}\vec{p}\left| {\vec{p} \right\rangle \langle \vec{p}|}\left[ {X}_{i}F\left( {\vec{P} \right)-F\left( {\vec{P} \right){X}_{i} \right]\left| \psi \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{\rm{d}\vec{p}\left| {\vec{p} \right\rangle \langle \vec{p}|}\left[ {X}_{i}F\left( {\vec{P} \right)\left| \psi \right\rangle -F\left( {\vec{P} \right){X}_{i}\left| \psi \right\rangle \right] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{\rm{d}\vec{p}\left| {\vec{p} \right\rangle }\left\{ {\rm{i}\hbar \frac{\partial }{\partial {p}_{i}\left[ F\left( {\vec{p} \right)\psi \left( {p}_{i} \right) \right]-F\left( {\vec{p} \right){\rm{i}\hbar \frac{\partial }{\partial {p}_{i}\psi \left( {\vec{p} \right) \right\} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{\rm{d}\vec{p}\left| {\vec{p} \right\rangle }{\rm{i}\hbar \frac{\partial }{\partial {p}_{i}F\left( {\vec{p} \right)\left\langle \vec{p}|\psi \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{d\vec{p}\left| {\vec{p} \right\rangle }\langle \vec{p}|{\rm{i}\hbar \frac{\partial }{\partial {P}_{i}F\left( {\vec{P} \right)\left| \psi \right\rangle ={\rm{i}\hbar \frac{\partial }{\partial {P}_{i}F\left( {\vec{P} \right)\left| \psi \right\rangle \\ & \Rightarrow \left[ {X}_{i},F\left( {\vec{P} \right) \right]={\rm{i}\hbar \frac{\partial }{\partial {P}_{i}F\left( {\vec{P} \right). \\ \end{align}$

[附 2] δ 函数的数学表达式:

傅里叶变换: $g\left( \omega \right)=\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{f\left( t \right){\rm{e}^{-{\rm{i}\omega t}{\rm{d}t}.$

傅里叶积分: $f\left( t \right)=\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{g\left( \omega \right){\rm{e}^{\rm{i}\omega t}{\rm{d}\omega }.$

将变换式代入积分式:

$\begin{align} & f\left( t \right)=\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{g\left( \omega \right){\rm{e}^{\rm{i}\omega t}{\rm{d}\omega } \\ & \ \ \ \ \ \ \ =\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{\left[ \frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{f\left( {t}'} \right){\rm{e}^{-{\rm{i}\omega {t}'}{\rm{d}{t}'} \right]{\rm{e}^{\rm{i}\omega t}{\rm{d}\omega } \\ & \ \ \ \ \ \ \ =\int_{-\infty }^{\infty }{f\left( {t}'} \right)\left[ \frac{1}{2\pi }\int_{-\infty }^{\infty }{\rm{e}^{\rm{i}\omega \left( t-{t}' \right)}{\rm{d}\omega } \right]{\rm{d}{t}'}. \\ \end{align}$

这种选择性实际上就是是 δ 函数的定义, 所以我们可以说:

$\delta \left( t-{t}' \right)=\frac{1}{2\pi }\int_{-\infty }^{\infty }{\rm{e}^{\rm{i}\omega \left( t-{t}' \right)}{\rm{d}\omega }.$

或者如果你总觉得不太接受上面那个就是 δ 函数的定义的话, 还可以这样想: 对 δ 函数做傅里叶变换 $\tilde{\delta }\left( \omega \right)=\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{\rm{e}^{-{\rm{i}\omega t}\delta \left( t \right){\rm{d}t}=\frac{1}{\sqrt{2\pi }$ 是个常数. 这样再积分回去就有 $\delta \left( t \right)=\frac{1}{\sqrt{2\pi }\int_{-\infty }^{\infty }{\tilde{\delta }\left( \omega \right){\rm{e}^{\rm{i}\omega t}{\rm{d}\omega }=\frac{1}{2\pi }\int_{-\infty }^{\infty }{\rm{e}^{\rm{i}\omega t}{\rm{d}\omega }.$

[附 3] 四个性质的验证:

$\begin{align} & {\rm{i}.\ {Q}^{\dagger }\left( {\rm{d}\vec{r} \right)Q\left( {\rm{d}\vec{r} \right)=\left( 1-{\rm{i}\vec{K}\cdot {\rm{d}\vec{r} \right)\left( 1+{\rm{i}\vec{K}\cdot {\rm{d}\vec{r} \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+{\rm{i}\vec{K}\cdot {\rm{d}\vec{r}-i\vec{K}\cdot {\rm{d}\vec{r}+o\left( {\rm{d}\vec{r} \right)=1. \\ \end{align}$ $\begin{align} & {\rm{ii}.\ {Q}^{\dagger }\left( {\rm{d}{\vec{r}_{2} \right){Q}^{\dagger }\left( {\rm{d}{\vec{r}_{1} \right)=\left( 1-{\rm{i}\vec{K}\cdot {\rm{d}{\vec{r}_{2} \right)\left( 1-{\rm{i}\vec{K}\cdot {\rm{d}{\vec{r}_{1} \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1-i\vec{K}\cdot \left( {\rm{d}{\vec{r}_{1}+{\rm{d}{\vec{r}_{2} \right)+o\left( {\rm{d}{\vec{r}_{i} \right)={Q}^{\dagger }\left( {\rm{d}{\vec{r}_{1}+{\rm{d}{\vec{r}_{2} \right). \\ \end{align}$ $\begin{align} & {\rm{iii}.\ {Q}^{\dagger }\left( {\rm{d}\vec{r} \right)=1-i\vec{K}\cdot {\rm{d}\vec{r} \\ & \Rightarrow {\left[ {Q}^{\dagger }\left( {\vec{r} \right) \right]}^{-1}={\left[ {Q}^{\dagger }\left( {\vec{r} \right) \right]}^{\dagger }=Q\left( {\rm{d}\vec{r} \right)=1+i\vec{K}\cdot {\rm{d}\vec{r}={Q}^{\dagger }\left( -{\rm{d}\vec{r} \right). \\ \end{align}$ ${\rm{iv}.\ {Q}^{\dagger }\left( {\rm{d}\vec{r} \right)=1-{\rm{i}\vec{K}\cdot {\rm{d}\vec{r}\Rightarrow {Q}^{\dagger }\left( 0 \right)=1.$

[附 4] 对易关系的计算:

$\begin{align} & \left[ \vec{R},{Q}^{\dagger }\left( {\rm{d}{\vec{r}_{0} \right) \right]\left| {\vec{r} \right\rangle =\vec{R}{Q}^{\dagger }\left( {\rm{d}{\vec{r}_{0} \right)\left| {\vec{r} \right\rangle -{Q}^{\dagger }\left( {\rm{d}{\vec{r}_{0} \right)\vec{R}\left| {\vec{r} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( \vec{r}+{\rm{d}{\vec{r}_{0} \right)\left| \vec{r}+{\rm{d}{\vec{r}_{0} \right\rangle -\vec{r}\left| \vec{r}+{\rm{d}{\vec{r}_{0} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\rm{d}{\vec{r}_{0}\left| \vec{r}+{\rm{d}{\vec{r}_{0} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\rm{d}{\vec{r}_{0}\left[ \left| {\vec{r} \right\rangle +O\left( {\rm{d}{\vec{r}_{0} \right) \right] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\rm{d}{\vec{r}_{0}\left| {\vec{r} \right\rangle +o\left( {\rm{d}{\vec{r}_{0} \right)={\rm{d}{\vec{r}_{0}\left| {\vec{r} \right\rangle . \\ \end{align}$

参考

^因为从逻辑上来说, 你左矢〈x| 里的 x 和指数上的参数 x 并不应该是同一个 x, 所以这种写法实际上是有一点微妙的问题的.
^因为可以忽略高阶无穷小量.

量子力学中的动量 为何说动量是平移的生成元 ​

目录: ​

I. 定义平移算符: ​

II. 动量的本征矢: ​

III. 坐标表象下的动量算符: ​

IV. 如何定义平移算符: ​

V. 为什么说动量是平移的生成元: ​

VI. 坐标动量对易关系: ​

VII. 坐标表象下动量算符与平移算符的另一个理解角度: ​

VIII. 附录: ​

参考 ​