本系列将推导量子力学课程中常见与常用的公式及定理的推导, 并且将以十个为一组的形式进行更新, 旨在帮助学习本课程的学生以及考研复习的考生.

归一化不随时间改变

归一化的表达式为

$$\int _{-\infty}^{\infty}|\Psi (x,t)|^2 dx =1$$

由于该式为对 $x$ 的积分, 则归一化随时间的变化可写为

$$\frac{d}{dt}\int _{-\infty}^{\infty}|\Psi (x,t)|^2 dx=\int _{-\infty}^{\infty}\frac{\partial}{\partial t}|\Psi (x,t)|^2 dx$$

其中

$$\frac{\partial}{\partial t}|\Psi |^2=\frac{\partial}{\partial t}(\Psi ^{*}\Psi)=\Psi ^{*}\frac{\partial \Psi}{\partial t}+ \frac{\partial \Psi^{*}}{\partial t} \Psi$$

由薛定谔方程及其共轭式有

$$\frac{\partial \Psi}{\partial t}=\frac{i\hbar}{2m}\frac{\partial ^2 \Psi}{\partial x^2}-\frac{i}{\hbar}V\Psi$$

$$\frac{\partial \Psi ^{*}}{\partial t}=-\frac{i\hbar}{2m}\frac{\partial ^2 \Psi ^{*}}{\partial x^2}+\frac{i}{\hbar}V\Psi ^{*}$$

于是

$$\frac{\partial}{\partial t}|\Psi |^2=\frac{i\hbar}{2m}(\Psi ^{*}\frac{\partial ^2 \Psi}{\partial x^2}-\frac{\partial ^2 \Psi ^{*}}{\partial x^2}\Psi) =\frac{\partial}{\partial x}[\frac{i\hbar}{2m}(\Psi ^* \frac{\partial \Psi}{\partial x}-\frac{\partial \Psi ^*}{\partial x}\Psi)]$$

则随时间变化的积分可写作

$$\frac{d}{dt} \int _{-\infty}^{\infty}|\Psi (x,t)|^2 dx=\frac{i\hbar}{2m}(\Psi ^* \frac{\partial \Psi}{\partial x}-\frac{\partial \Psi ^*}{\partial x}\Psi)|_{-\infty}^{\infty}$$

又由于波函数要可归一化, 即 $x$ 趋于无穷时 $\Psi$ 要趋于零, 即:

$$\frac{d}{dt} \int _{-\infty}^{\infty}|\Psi (x,t)|^2 dx=0$$

即归一化不随时间改变.

证毕.

概率流密度

在 “归一化不随时间改变” 中已得到过下式

$$\frac{\partial}{\partial t}|\Psi |^2=\frac{\partial}{\partial x}[\frac{i\hbar}{2m}(\Psi ^* \frac{\partial \Psi}{\partial x}-\frac{\partial \Psi ^*}{\partial x}\Psi)]$$

若换成三维情况, 且令概率 $P=|\Psi |^2$ ,则有

$$\frac{\partial P}{\partial t}=\frac{i\hbar}{2m}\nabla\cdot (\Psi ^* \nabla \Psi - \Psi \nabla \Psi ^* )$$

若令

$$J=\frac{i\hbar}{2m}\nabla\cdot ( \Psi \nabla \Psi ^* - \Psi ^* \nabla \Psi )$$

则有

$$\frac{\partial P}{\partial t}+\nabla \cdot J =0$$

显然这是一个连续性方程的微分形式, 若将其作体积分则有

$$\iiint \frac{\partial P}{\partial t} dV=-\iiint \nabla \cdot J dV$$

由高斯公式(见本博客的矢量分析那一篇)可将散度的体积分换成矢量的曲面积分,即

$$\iiint \frac{\partial P}{\partial t} dV=-\oiint J dS$$

即单位时间内曲面所包围的体积内的概率增加等于单位时间概率流密度通过曲面进入的面积分, 也就是概率. 其中 $J$ 就定义为概率流密度

证毕.

一维束缚定态无简并

先考虑一般情况, 设 $\psi _1$ 与 $\psi _2$ 是薛定谔方程同一本征值的两个解, 即

$$-\frac{\hbar ^2}{2m}\frac{d^2}{dx^2} \psi _1 +V \psi _1 =E \psi _1$$

$$-\frac{\hbar ^2}{2m}\frac{d^2}{dx^2} \psi _2 +V \psi _2 =E \psi _2$$

一式乘 $\psi _2$ , 二式乘 $\psi _1$ , 并相减得

$$\psi _1 \psi _2^{''} - \psi _1^{''} \psi _2 =0$$

即

$$\frac{d}{dx}(\psi _1 \psi _2^{'}-\psi _1^{'}\psi _2 )=0$$

即

$$\psi _1 \psi _2^{'}-\psi _1^{'} \psi _2 =constant$$

引入朗斯基(Wronskian)行列式

$$W=\begin{vmatrix}\psi _1 & \psi _2 \\ \psi _1^{'} & \psi _2^{'} \end{vmatrix}$$

当 $W=0$ 时代表 $\psi _1$ 与 $\psi _2$ 线性相关, 当 $W$ 不等于 $0$ 时代表 $\psi _1$ 与 $\psi _2$ 线性无关. 则由上面可得对于同一本征值的两个解是线性无关的, 其朗斯基行列式为一常数.

现在考虑一维束缚定态存在简并, 即存在两个波函数$\psi _1$ 与 $\psi _2$ 对应同一本征值 $E$ ,因其束缚态, 则波函数在 $x$ 趋于无穷时应趋于零, 那么其朗斯基行列式 $W=0$ , 即 $\psi _1$ 与 $\psi _2$ 线性相关, 与上面所证的两解线性无关相矛盾.因此一维束缚定态无简并.

证毕.

Baker-Hausdorff 公式

Baker-Hausdorff 公式为:

$$e^{\hat{A}}\hat{B}e^{-\hat{A}}=\hat{B}+[\hat{A},\hat{B}]+\frac{1}{2!}[\hat{A},[\hat{A},\hat{B}]]+\frac{1}{3!}[\hat{A},[\hat{A},[\hat{A},\hat{B}]]]+...$$

证明过程:

令

$$\hat{F}(\lambda)= e^{\lambda \hat{A}}\hat{B}e^{- \lambda \hat{A}}$$

为将其在 $\lambda =0$ 处进行泰勒展开, 现求其各阶导数

$$\frac{d}{d\lambda} \hat{F}(\lambda)=e^{\lambda \hat{A}}\hat{A}\hat{B}e^{- \lambda \hat{A}}- e^{\lambda \hat{A}}\hat{B}\hat{A}e^{- \lambda \hat{A}}=e^{\lambda \hat{A}}[\hat{A},\hat{B}]e^{- \lambda \hat{A}}$$

$$\frac{d^2}{d\lambda ^2} \hat{F}(\lambda)= e^{\lambda \hat{A}}[\hat{A},[\hat{A},\hat{B}]]e^{- \lambda \hat{A}}$$

$$\frac{d^3}{d\lambda ^3} \hat{F}(\lambda)= e^{\lambda \hat{A}}[\hat{A},[\hat{A},[\hat{A},\hat{B}]]]e^{- \lambda \hat{A}}$$

$$...$$

于是可以将 $\hat{F}(\lambda)$ 在 $\lambda =0$ 处进行展开

$$\hat{F}(\lambda)= \hat{F}(0)+\lambda \frac{d}{d\lambda}\hat{F}(\lambda)|_{\lambda =0}+\frac{\lambda ^2}{2!} \frac{d^2}{d\lambda ^2}\hat{F}(\lambda)|_{\lambda =0} +\frac{\lambda ^3}{3!} \frac{d^3}{d\lambda ^3}\hat{F}(\lambda)|_{\lambda =0} + ...$$

其中

$$\hat{F}(0)=\hat{B}$$

$$\frac{d}{d\lambda}\hat{F}(\lambda)|_{\lambda =0}=[\hat{A},\hat{B}]$$

$$\frac{d^2}{d\lambda ^2}\hat{F}(\lambda)|_{\lambda =0}=[\hat{A},[\hat{A},\hat{B}]]$$

$$\frac{d^3}{d\lambda ^3}\hat{F}(\lambda)|_{\lambda =0}=[\hat{A},[\hat{A},[\hat{A},\hat{B}]]]$$

$$...$$

则

$$\hat{F}(\lambda)=\hat{B}+\lambda [\hat{A},\hat{B}]+\frac{\lambda ^2}{2!}[\hat{A},[\hat{A},\hat{B}]]+\frac{\lambda ^3}{3!} [\hat{A},[\hat{A},[\hat{A},\hat{B}]]]+...$$

令 $\lambda =1$ ,可得

$$\hat{F}(1)=e^{\hat{A}}\hat{B}e^{-\hat{A}}=\hat{B}+[\hat{A},\hat{B}]+\frac{1}{2!}[\hat{A},[\hat{A},\hat{B}]]+\frac{1}{3!}[\hat{A},[\hat{A},[\hat{A},\hat{B}]]]+...$$

证毕.

Glauber 公式

Glauber 公式为如果 $\hat{A}$ 与 $\hat{B}$ 都与 $[\hat{A},\hat{B}]$ 对易, 那么:

$$e^{\hat{A}+\hat{B}}=e^{\hat{A}}e^{\hat{B}}e^{-\frac{1}{2}[\hat{A},\hat{B}]} \qquad or \qquad e^{\hat{A}}e^{\hat{B}}=e^{\hat{A}+\hat{B}+\frac{1}{2}[\hat{A},\hat{B}]}$$

证明过程:

令

$$\hat{F}(\lambda)=e^{\lambda \hat{A}}e^{\lambda \hat{B}}$$

将其对 $\lambda$ 求导, 得

$$\begin{aligned} \frac{d}{d\lambda}\hat{F}(\lambda) &= \hat{A}e^{\lambda \hat{A}}e^{\lambda \hat{B}}+e^{\lambda \hat{A}}\hat{B}e^{\lambda \hat{B}} \\ &= (\hat{A}e^{\lambda \hat{A}}e^{\lambda \hat{B}}e^{-\lambda \hat{B}}e^{-\lambda \hat{A}}+e^{\lambda \hat{A}}\hat{B}e^{\lambda \hat{B}}e^{-\lambda \hat{B}}e^{-\lambda \hat{A}})e^{\lambda \hat{A}}e^{\lambda \hat{B}} \\ &= (\hat{A}+ e^{\lambda \hat{A}}\hat{B}e^{-\lambda \hat{A}})\hat{F}(\lambda) \end{aligned}$$

由 Baker-Hausdorff 公式和 $\hat{A}$ 与 $[\hat{A},\hat{B}]$ 的对易关系得

$$\begin{aligned} e^{\lambda \hat{A}}\hat{B}e^{- \lambda \hat{A}} &= \hat{B}+\lambda [\hat{A},\hat{B}] +\lambda ^2 [\hat{A},[\hat{A},\hat{B}]] +... \\ &= \hat{B}+\lambda [\hat{A},\hat{B}] \end{aligned}$$

即

$$\frac{d}{d\lambda}\hat{F}(\lambda) =(\hat{A}+\hat{B}+\lambda [\hat{A},\hat{B}] ) \hat{F}(\lambda)$$

解微分方程可得

$$\hat{F}(\lambda)=Ce^{\lambda (\hat{A}+\hat{B})+\frac{1}{2}\lambda ^2 [\hat{A},\hat{B}] }$$

由 $\hat{F}(0)=1$ , 得系数 $C=1$ , 则

$$\hat{F}(\lambda)=e^{\lambda (\hat{A}+\hat{B})+\frac{1}{2}\lambda ^2 [\hat{A},\hat{B}] }$$

令 $\lambda =1$ ,则

$$\hat{F}(1)=e^{\hat{A}+\hat{B}+\frac{1}{2} [\hat{A},\hat{B}] }= e^{\hat{A}}e^{\hat{B}}$$

证毕.

[x,f(p)] [f(x),p] 对易关系

对于 $[\hat{x},f(p)]$ 有:

$$\begin{aligned} [\hat{x},f(p)]\phi (p) &= \hat{x} f(p) \phi (p) -f(p)\hat{x}\phi (p) \\ &= i\hbar \frac{d}{dp}[f(p) \phi (p)] -f(p)i\hbar \frac{d}{dp}\phi (p) \\ &= [i\hbar \frac{d}{dp}f(p)]\phi (p) \end{aligned}$$

即

$$[\hat{x},f(p)]= i\hbar \frac{d}{dp}f(p)$$

反之则

$$[f(p),\hat{x}]= -i\hbar \frac{d}{dp}f(p)$$

对于 $[f(x),\hat{p}]$ 同理有:

$$\begin{aligned} [f(x),\hat{p}]\psi (x) &= f(x)\hat{p}\psi (x)-\hat{p}f(x)\psi (x) \\ &=-f(x)i\hbar \frac{d}{dx}\psi (x) + i\hbar \frac{d}{dx}[f(x)\psi (x)] \\ &= [i\hbar \frac{d}{dx}f(x)] \psi (x) \end{aligned}$$

即

$$[f(x),\hat{p}]= i\hbar \frac{d}{dx}f(x)$$

反之则

$$[\hat{p},f(x)]= -i\hbar \frac{d}{dx}f(x)$$

厄米算符本征函数性质

厄米算符的本征值为实数, 证明过程:

$$\langle \alpha | \hat{Q} \alpha \rangle =a \langle \alpha |\alpha \rangle$$

$$\langle \hat{Q} \alpha | \alpha \rangle =a \langle \alpha |\alpha \rangle$$

又由厄米算符性质, 有

$$\langle \alpha | \hat{Q} \alpha \rangle = \langle \hat{Q} \alpha | \alpha \rangle$$

则

$$a \langle \alpha |\alpha \rangle = a^{*} \langle \alpha |\alpha \rangle$$

由于本征函数与自身的内积不能为0, 则

$$a=a^{*}$$

即厄米算符的本征值为实数, 证毕.

厄米算符属于不同本征值的本征函数相互正交, 证明过程:

$$\langle \alpha |\hat{Q} \beta \rangle =b \langle \alpha | \beta \rangle$$

$$\langle \hat{Q} \alpha | \beta \rangle =a^{*} \langle \alpha | \beta \rangle$$

由于 $a^{*}=a$ ,且 $a$ 不等于 $b$ ,于是

$$\langle \alpha | \beta \rangle =0$$

即厄米算符属于不同本征值的本征函数相互正交, 证毕.

Hellmann–Feynman 定理

Hellmann–Feynman 定理为( 其中 $\lambda$ 为与 $\hat{H}$ , $E$ 相关的任意参数 ):

$$\frac{\partial E}{\partial \lambda}=\langle \psi |\frac{\partial \hat{H}}{\partial \lambda}|\psi \rangle$$

证明过程:

$$\begin{aligned} \frac{\partial E}{\partial \lambda} &= \frac{\partial}{\partial \lambda}(\langle \psi |\hat{H}| \psi \rangle) \\ &= \langle \frac{\partial}{\partial \lambda} \psi |\hat{H}| \psi \rangle + \langle \psi |\frac{\partial \hat{H}}{\partial \lambda} | \psi \rangle +\langle \psi |\hat{H}| \frac{\partial}{\partial \lambda} \psi \rangle \\ &= E \langle \frac{\partial}{\partial \lambda} \psi | \psi \rangle + \langle \psi |\frac{\partial \hat{H}}{\partial \lambda} | \psi \rangle + E \langle \psi | \frac{\partial}{\partial \lambda} \psi \rangle \\ &= E\frac{\partial}{\partial \lambda} (\langle \psi | \psi \rangle) + \langle \psi |\frac{\partial \hat{H}}{\partial \lambda} | \psi \rangle \\ &= \langle \psi |\frac{\partial \hat{H}}{\partial \lambda} | \psi \rangle \end{aligned}$$

证毕.

Ehrenfest 定理

Ehrenfest 定理为

$$\frac{d}{dt} \langle \hat{Q} \rangle =\frac{1}{i \hbar} \langle [\hat{Q},\hat{H}] \rangle +\langle \frac{\partial \hat{Q}}{\partial t}\rangle$$

证明过程:

$$\begin{aligned} \frac{d}{dt} \langle \hat{Q} \rangle &= \frac{d}{dt} (\langle \psi |\hat{Q} | \psi \rangle) \\ &= \langle \frac{\partial \psi}{\partial t} |\hat{Q} | \psi \rangle + \langle \psi |\frac{\partial \hat{Q} }{\partial t}| \psi \rangle + \langle \psi |\hat{Q} | \frac{\partial \psi}{\partial t} \rangle \\ &=-\frac{1}{i \hbar} \langle \hat{H}\psi |\hat{Q}|\psi \rangle + \langle \psi |\frac{\partial \hat{Q} }{\partial t}| \psi \rangle +\frac{1}{i \hbar}\langle \psi |\hat{Q}| \hat{H} \psi \rangle \qquad 代入 i\hbar \frac{\partial}{\partial t}\psi =\hat{H} \psi \\ &= -\frac{1}{i \hbar} \langle \psi |\hat{H}\hat{Q}|\psi \rangle + \frac{1}{i \hbar}\langle \psi |\hat{Q}\hat{H}| \psi \rangle + \langle \psi |\frac{\partial \hat{Q} }{\partial t}| \psi \rangle \\ &=\frac{1}{i \hbar}\langle [\hat{Q},\hat{H}]\rangle + \langle \frac{\partial \hat{Q}}{\partial t}\rangle \end{aligned}$$

证毕.

Virial 定理

Virial 定理为:

$$2 \langle T \rangle =\langle r \cdot \nabla V(r) \rangle$$

证明方式(1):

若 $E$ 不显含 $\lambda$ , 而 $\hat{H}(\lambda x)$ , $\psi (\lambda x)$ 与 $\lambda$ 相关,则

$$\frac{\partial E}{\partial \lambda} =0$$

由Hellmann–Feynman 定理得

$$\begin{aligned} 0 &= \langle \psi |\frac{\partial \hat{H}(\lambda x)}{\partial \lambda} | \psi \rangle \\ &= \langle \psi | \frac{\partial}{\partial \lambda}(-\frac{\hbar ^2}{2m}\frac{d^2}{d(\lambda x)^2}+V(\lambda x))|\psi \rangle \\ &= \langle \psi | \frac{\partial}{\partial \lambda}(-\frac{1}{\lambda ^2}\frac{\hbar ^2}{2m}\frac{d^2}{dx^2}+V(\lambda x))|\psi \rangle \\ &=\langle \psi |\frac{2}{\lambda ^3}\frac{\hbar ^2}{2m}\frac{d^2}{dx^2}+\frac{\partial V}{\partial (\lambda x)}\frac{\partial {\lambda x}}{\partial \lambda} |\psi \rangle \\ &= \langle \psi |\frac{2}{\lambda ^3}\frac{\hbar ^2}{2m}\frac{d^2}{dx^2}+x \frac{\partial V}{\partial (\lambda x)} |\psi \rangle \\ &= \langle \psi |2\frac{\hbar ^2}{2m}\frac{d^2}{dx^2}+x \frac{\partial V}{\partial x} |\psi \rangle \qquad 令 \lambda =1 \\ &= \langle \psi |-2 \hat{T} +x\frac{\partial V}{\partial x} |\psi \rangle \\ &=-2 \langle T \rangle + \langle x \frac{\partial V}{\partial x} \rangle \end{aligned}$$

即

$$2 \langle T \rangle = \langle x \frac{\partial V}{\partial x} \rangle$$

证毕.

证明方式(2):

由 Ehrenfest 定理与定态下力学量不随时间改变可得:

$$0=\frac{d}{dt} \langle r \cdot \hat{p} \rangle =\frac{1}{i\hbar}\langle [r \cdot \hat{p},\hat{H}]\rangle$$

其中

$$[r \cdot \hat{p},\hat{H}]=[r \cdot \hat{p},\frac{\hat{p}^2}{2m}]+[r \cdot \hat{p},V(r)]$$

其中

$$[r \cdot \hat{p},\hat{p}^2]=r[\hat{p},\hat{p}^2]+[r,\hat{p}^2]\hat{p} =[r,\hat{p}^2]\hat{p}=2i\hbar \hat{p}^2$$

$$[r \cdot \hat{p},V(r)]= r[\hat{p},V(r)]+[r,V(r)]\hat{p} = r[\hat{p},V(r)]=-i\hbar r \frac{\partial V}{\partial r}$$

即

$$\langle 2i\hbar \frac{\hat{p}^2}{2m} \rangle =\langle i\hbar r \frac{\partial V}{\partial r}\rangle$$

即

$$2\langle T \rangle = \langle r \cdot \nabla \rangle$$

证毕.

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