量子物理笔记 - 1：数学基础

统计概念

期望：
$$
E(x)=\langle x \rangle=\sum_ {x} xP(x)
$$
方差：
$$
\sigma^2=\langle (x-\langle x \rangle )^2 \rangle =\sum_ {x}(x^2+\langle x \rangle ^2-2x\langle x\rangle)P(x)=\langle x^2\rangle-\langle x \rangle ^2
$$
标准差：
$$
\sigma=\sqrt{\langle x^2\rangle-\langle x \rangle ^2}
$$

波函数

归一条件：
$$
\int_ {-\infty}^{+\infty} |\Psi(x,t)|^2 dx=1
$$

普朗克常量

光子的能量：
$$
E=h\nu=h \frac{\omega}{2\pi}=\hbar \omega, \quad \hbar=\frac{h}{2\pi}
$$
光子动量：

薛定谔方程

哈密顿算符

哈密顿算符：
$$
\hat{H}=-\frac{\hbar^2}{2m}\nabla^2+V
$$
薛定谔方程：
$$
\begin{aligned}
i\hbar \dot{\Psi}=\hat{H}\Psi=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2}+V\Psi
, \quad \hbar=\frac{h}{2\pi}
\end{aligned}
$$
于是有：
$$
\begin{aligned}
& \begin{cases}
i\hbar \dot{\Psi}=-\frac{\hbar^2}{2m}\nabla^2\Psi+V\Psi \\
-i\hbar \dot{\Psi}^* =-\frac{\hbar^2}{2m}\nabla^2\Psi^* +V\Psi^*
\end{cases} \\
\Rightarrow
& i\hbar(\Psi^* \dot{\Psi}+\Psi \dot{\Psi}^* )=-\frac{\hbar^2}{2m}(\Psi^* \nabla^2\Psi-\Psi\nabla^2\Psi^* ) \\
\Rightarrow
& \Psi^* \dot{\Psi}+\Psi \dot{\Psi}^* =\frac{i\hbar}{2m}(\Psi^* \nabla^2\Psi-\Psi\nabla^2\Psi^* )
\end{aligned}
$$

概率守恒

考虑 $\Psi$ 关于时间的微分：
$$
\begin{aligned}
\frac{d}{dt} \int_ {-\infty}^{+\infty} |\Psi|^2 dx
&=\int_ {-\infty}^{+\infty} \frac{\partial (\Psi^* \Psi)}{\partial t}dx \\
&=\int_ {-\infty}^{+\infty} \left( \Psi^* \dot{\Psi}+\Psi \dot{\Psi}^* \right) dx \\
&=\int_ {-\infty}^{+\infty} \frac{i\hbar}{2m}\left(\Psi^* \frac{\partial^2 \Psi}{\partial x^2}-\Psi \frac{\partial^2 \Psi^* }{\partial x^2} \right)dx \\
&=\int_ {-\infty}^{+\infty} \frac{i\hbar}{2m}\frac{\partial}{\partial x}\left(\Psi^* \frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^* }{\partial x} \right)dx \\
&=\frac{i\hbar}{2m}\underbrace{\left(\Psi^* \frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^* }{\partial x} \right) \bigg|_ {-\infty}^{+\infty}}_ {0} \\
&=0
\end{aligned}
$$
因此 $\Psi$ 的积分为常数：
$$
\int_ {-\infty}^{+\infty} |\Psi(x,t)|^2dx=\int_ {-\infty}^{+\infty} |\Psi(x,0)|^2dx
$$
同理，可以得到粒子出现在 $[a,b]$ 的概率 $P_ {ab}(t)$ 关于时间的导数：
$$
\frac{d}{dt}P_ {ab}(t)
=\int_ {a}^{b}\frac{\partial |\Psi^2|}{\partial t}dx
=\frac{i\hbar}{2m}\left(\Psi^* \frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^* }{\partial x} \right) \bigg|_ {a}^{b}=J(a,t)-J(b,t)
$$
可以定义概率流 $J(x,t)$ 为：
$$
J(x,t)=\frac{i\hbar}{2m}\left(\Psi \frac{\partial \Psi^* }{\partial x}-\Psi^* \frac{\partial \Psi}{\partial x} \right)
$$
若 $\Psi(x,t)=f(x)e^{-iat}$，则：
$$
J(x,t)=\frac{i\hbar}{2m}\left(fe^{-iat}e^{iat}\nabla f-fe^{iat}e^{-iat}\nabla f\right)=0
$$
若 $\Psi_ 1,\Psi_ 2$ 是归一化后的薛定谔方程的解，则：
$$
\begin{aligned}
\frac{d}{dt} \int_ {-\infty}^{\infty}\Psi_ 1^* \Psi_ 2dt
&=\int_ {-\infty}^{\infty}\left(\frac{\partial}{\partial t}(\Psi_ 1^* \Psi_ 2)\right)dx \\
&=\int_ {-\infty}^{\infty}\left(\dot{\Psi}_ 1^* \Psi_ 2+\Psi^* _ 1\dot{\Psi}_ 2\right)dx \\
&=\int_ {-\infty}^{\infty}\left(\frac{\hbar}{2im}\nabla^2\Psi_ 1^* \Psi_ 2+\frac{iV}{\hbar}\Psi_ 1^* \Psi_ 2
-\frac{\hbar}{2im}\nabla^2\Psi_ 2\Psi_ 1^* -\frac{iV_ 0}{\hbar}\Psi_ 2\Psi^* _ 1\right)dx \\
&=0
\end{aligned}
$$

可观测量

位置：
$$
\langle x \rangle =\int_ {-\infty}^{\infty} x |\Psi(x,t)|^2dx=\langle \hat{x}\rangle
$$
速度：
$$
\begin{aligned}
\langle v \rangle
&=\frac{d \langle x \rangle}{dt} \\
&=\int_ {-\infty}^{\infty}x(\Psi^* \dot{\Psi}+\Psi\dot{\Psi}^* )dx \\
&=\int_ {-\infty}^{\infty}x\frac{i\hbar}{2m}\frac{\partial^2}{\partial x^2}(\Psi-\Psi^* )dx \\
&=\frac{i\hbar}{2m}\int_ {-\infty}^{\infty} x\frac{\partial}{\partial x}\left(\Psi^* \frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^* }{\partial x} \right)dx \\
&=\frac{i\hbar}{2m} \left( \left( x\left(\Psi^* \frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^* }{\partial x}\right) \right)\bigg|_ {-\infty}^{\infty}-\int_ {-\infty}^{\infty} \left(\Psi^* \frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^* }{\partial x}\right)dx \right) \\
&=-\frac{i\hbar}{2m}\int_ {-\infty}^{\infty}\Psi^* \frac{\partial \Psi}{\partial x}dx+\frac{i\hbar}{2m}\left((\Psi\Psi^* )|_ {-\infty}^{\infty}-\int_ {-\infty}^{\infty}\Psi^* \frac{\partial \Psi}{\partial x}dx\right) \\
&=-\frac{i\hbar}{m}\int_ {-\infty}^{\infty} \Psi^* \frac{\partial}{\partial x}\Psi dx \\
&=\left\langle \frac{\hbar}{im} \frac{\partial}{\partial x}\right\rangle
\end{aligned}
$$
动量：
$$
\langle p \rangle=m\langle v \rangle=\left\langle \frac{\hbar}{i} \frac{\partial}{\partial x}\right\rangle
$$
动能：
$$
\langle T \rangle=\frac{1}{2m}\langle \hat{p}^2\rangle=-\frac{\hbar^2}{2m^2}\langle \nabla^2 \rangle
$$
合外力：
$$
\begin{aligned}
\langle F \rangle
&=\frac{d \langle p \rangle}{dt} \\
&=\frac{d}{dt}\int_ {-\infty}^{\infty}\Psi^* \frac{\hbar}{i} \frac{\partial \Psi}{\partial x} dx\\
&=-i\hbar \int_ {-\infty}^{\infty} (\dot{\Psi}^* \nabla \Psi+\Psi^* \nabla \dot{\Psi}) dx \\
&=-i\hbar \int_ {-\infty}^{\infty} \left( \frac{\nabla\Psi}{-i\hbar}\left(\frac{-\hbar^2}{2m}\nabla^2\Psi^* +V\Psi^* \right)+\frac{\Psi^* }{i\hbar}\nabla\left(\frac{-\hbar^2}{2m}\nabla^2\Psi+V\Psi\right) \right)dx \\
&=\int_ {-\infty}^{\infty} \left( \left(-\frac{\hbar^2}{2m}\nabla^2\Psi^* +V\Psi^* \right)\nabla \Psi+\Psi^* \nabla\left(\frac{\hbar^2}{2m}\nabla^2\Psi-V\Psi\right) \right)dx \\
&=\int_ {-\infty}^{\infty} \left( \left(-\frac{\hbar^2}{2m}\nabla^2\Psi^* +V\Psi^* \right)\nabla \Psi+\Psi^* \left(\frac{\hbar^2}{2m}\nabla^3\Psi-\nabla(V\Psi)\right) \right)dx \\
&=\int_ {-\infty}^{\infty} \left( \left(-\frac{\hbar^2}{2m}\nabla^2\Psi^* +V\Psi^* \right)\nabla \Psi+\Psi^* \left(\frac{\hbar^2}{2m}\nabla^3\Psi-\Psi\nabla V-V\nabla \Psi\right) \right)dx \\
&=\int_ {-\infty}^{\infty} \left( -\frac{\hbar^2}{2m}\nabla^2\Psi^* \nabla \Psi+\Psi^* \frac{\hbar^2}{2m}\nabla^3\Psi-\Psi^* \Psi\nabla V \right)dx \\
&=\int_ {-\infty}^{\infty} \left( -\frac{\hbar^2}{2m}\nabla^2\Psi^* \nabla \Psi+\Psi^* \frac{\hbar^2}{2m}\nabla^3\Psi-|\Psi|^2\nabla V \right)dx \\
&=\langle -\nabla V \rangle+\frac{\hbar^2}{2m}\langle \nabla^3 \rangle-\frac{\hbar^2}{2m}\int_ {-\infty}^{\infty}\nabla^2\Psi^* \nabla\Psi dx \\
&=\langle -\nabla V \rangle+\frac{\hbar^2}{2m}\int_ {-\infty}^{\infty} (\Psi^* \nabla^3\Psi-\nabla^2\Psi^* \nabla\Psi)dx \\
&=\langle -\nabla V \rangle+\frac{\hbar^2}{2m}\left(\int_ {-\infty}^{\infty}\Psi^* \nabla^3\Psi dx-\int_ {-\infty}^{\infty}\nabla^2\Psi^* \nabla\Psi dx\right) \\
&=\langle -\nabla V \rangle+\frac{\hbar^2}{2m}\left((\Psi^* \nabla^2\Psi)|_ {-\infty}^{\infty}-\int_ {-\infty}^{\infty}\nabla^2\Psi\nabla\Psi^* dx-(\nabla\Psi^* \nabla\Psi)|_ {-\infty}^{\infty}+\int_ {-\infty}^{\infty}\nabla^2\Psi\nabla\Psi^* dx\right) \\
&=\langle -\nabla V \rangle
\end{aligned}
$$
常数势 $V_ 0$：
$$
\begin{aligned}
& \begin{cases}
i\hbar \dot{\Psi}=\hat{H} \Psi \\
\Psi_ 0=\Psi \exp\left(-\frac{iV_ 0t}{\hbar}\right)
\end{cases} \\
\Rightarrow
& i\hbar\dot{\Psi}_ 0
=i\hbar \dot{\Psi}\exp\left(-\frac{iV_ 0t}{\hbar}\right)+i\hbar\Psi \frac{-iV_ 0}{\hbar}\exp\left(-\frac{iV_ 0t}{\hbar}\right) \\
&=\hat{H}\Psi\exp\left(-\frac{iV_ 0t}{\hbar}\right)+V_ 0\Psi\exp\left(-\frac{iV_ 0t}{\hbar}\right)\Psi \\
&=-\frac{\hbar^2}{2m}\nabla^2\left(\Psi\exp\left(-\frac{iV_ 0t}{\hbar}\right)\right)+V\Psi\exp\left(-\frac{iV_ 0t}{\hbar}\right)+V_ 0\Psi\exp\left(-\frac{iV_ 0t}{\hbar}\right) \\
&=\left(-\frac{\hbar^2}{2m}\nabla^2+V+V_ 0\right)\Psi_ 0 \\
&=\hat{H}_ 0\Psi_ 0
\end{aligned}
$$

习题

1.3

考虑高斯分布：
$$
\rho(x)=Ae^{-\lambda(x-a)^2}
$$
其中 $A,a,\lambda$ 是正实数。

1. 确定 $A$
2. 求 $\langle x \rangle, \langle x^2 \rangle, \sigma$

由归一化条件得：
$$
\begin{aligned}
\frac{1}{A}
&=\int_ {-\infty}^{\infty} e^{-\lambda(x-a)^2}dx, \quad t=\sqrt{\lambda}(x-a) \\
&=\frac{1}{\sqrt{\lambda}}\int_ {-\infty}^{\infty} e^{-t^2}dt \\
&=\frac{2}{\sqrt{\lambda}}\int_ 0^{\infty} e^{-t^2}dt, \quad u=t^2 \\
&=\frac{1}{\sqrt{\lambda}}\int_ 0^{\infty} e^{-u} y^{-\frac{1}{2}} du \\
&=\frac{1}{\sqrt{\lambda}}\Gamma \left(\frac{1}{2}\right) \\
&=\sqrt{\frac{\pi}{\lambda}}
\end{aligned}
$$
即：
$$
A=\sqrt{\frac{\lambda}{\pi}}
$$
考虑如下积分变换：
$$
\begin{aligned}
& \int_ {0}^{\infty} x^n e^{-\lambda x^2}dx, \quad n \in \mathbb{N}, t=\lambda x^2,dt=2\lambda \sqrt{t/\lambda}dx \\
=& \int_ {0}^{\infty} \lambda^{-\frac{n}{2}}t^{\frac{n}{2}} e^{-t}\frac{dt}{2\lambda\sqrt{t/\lambda}} \\
=& \frac{1}{2}\int_ {0}^{\infty} \lambda^{-\frac{n+1}{2}}t^{\frac{n-1}{2}}e^{-t}dt \\
=& \frac{1}{2}\int_ {0}^{\infty} \lambda^{-\frac{n+1}{2}}t^{\frac{n-1}{2}}e^{-t}dt \\
=& \frac{1}{2}\lambda^{-\frac{n+1}{2}}\Gamma\left(\frac{n+1}{2}\right)
\end{aligned}
$$
于是有：
$$
\int_ {-\infty}^{\infty} x^ne^{-\lambda x^2} dx=
\begin{cases}
0, \quad & n \equiv 1\pmod{2} \\
\lambda^{-\frac{n+1}{2}}\Gamma\left(\frac{n+1}{2}\right), \quad &n \equiv 0 \pmod{2}
\end{cases}
$$
由期望定义：
$$
\begin{aligned}
& \begin{aligned}
\langle x \rangle
&= \int_ {-\infty}^{\infty} x Ae^{-\lambda(x-a)^2}dx, \quad t=x-a \\
&=A\underbrace{\int_ {-\infty}^{\infty}te^{-\lambda t^2} dt}_ {0}+a\underbrace{\int_ {-\infty}^{\infty}Ae^{-\lambda (x-a)^2} dx}_ {1} \\
&=a
\end{aligned} \\
& \begin{aligned}
\langle x^2 \rangle
&= \int_ {-\infty}^{\infty} x^2 Ae^{-\lambda(x-a)^2}dx, \quad t=x-a \\
&= A\int_ {-\infty}^{\infty} (t^2+2at+a^2)e^{-\lambda t^2}dt \\
&= A\left( \lambda^{-3/2}\Gamma(3/2)+a^2 \sqrt{\frac{\pi}{\lambda}} \right) \\
&= A\left( \lambda^{-3/2}\frac{1}{2}\sqrt{\pi}+a^2 \sqrt{\frac{\pi}{\lambda}} \right) \\
&= \frac{1}{2\lambda}+a^2
\end{aligned} \\
& \begin{aligned}
\sigma
&= \sqrt{\langle x^2 \rangle-\langle x \rangle^2} \\
&= \sqrt{\frac{1}{2\lambda}+a^2-a^2} \\
&=\frac{1}{\sqrt{2\lambda}}
\end{aligned}
\end{aligned}
$$

1.9

一个质量为 $m$ 的粒子处于态：
$$
\Psi(x,t)=Ae^{-a((mx^2/\hbar)+it)}
$$
其中 $A,a$ 是正实数。

1. 确定 $A$
2. 若 $\Psi$ 满足薛定谔方程，求势 $V(x)$
3. 求 $\langle x \rangle,\langle x^2 \rangle,\langle p \rangle, \langle p^2 \rangle$
4. 求 $\sigma_ x,\sigma_ p$

由归一化条件得：
$$
\begin{aligned}
\frac{1}{|A|^2}
&=\int_ {-\infty}^{\infty}e^{-a((mx^2/\hbar)-it)}e^{-a((mx^2/\hbar)+it)}dx \\
&=\int_ {-\infty}^{\infty} e^{-2amx^2/\hbar}dx \\
&=2\int_ {0}^{\infty}e^{-2amx^2/\hbar}dx, \quad t=\frac{2amx^2}{\hbar},dt=\frac{4am}{\hbar}xdx \\
&=2\int_ {0}^{\infty} e^{-t}\frac{\hbar}{4am} \sqrt{\frac{2am}{\hbar t}}dt \\
&=\sqrt{\frac{\hbar}{2am}}\int_ {0}^{\infty} t^{-1/2}e^{-t}dt \\
&=\sqrt{\frac{\hbar}{2am}} \Gamma(1/2) \\
&=\sqrt{\frac{\pi \hbar}{2am}}
\end{aligned}
$$
即：
$$
A=\left(\frac{2am}{\pi \hbar}\right)^{1/4}
$$
将 $\Psi$ 对时间求导：
$$
\begin{aligned}
& \dot{\Psi}=Ae^{-amx^2/\hbar}(-ia)e^{-ait}=-ia\Psi \\
\Rightarrow
& i\hbar\dot{\Psi}=a\hbar \Psi \\
\Rightarrow
& -\frac{\hbar^2}{2m}\nabla^2\Psi+V\Psi=a\hbar\Psi \\
\end{aligned}
$$
即：
$$
\begin{aligned}
\frac{2m(V-a\hbar)}{\hbar^2}\Psi
&= \nabla^2\Psi \\
&=\nabla \left( \frac{-2amx}{\hbar} e^{-iamx^2/\hbar}e^{-iat} \right) \\
&=\frac{-2am}{\hbar}e^{-iat} \left( e^{-iamx^2/\hbar}+\frac{-2amx^2}{\hbar}e^{-iamx^2/\hbar} \right) \\
&=\left(-\frac{2am}{\hbar}+\frac{4a^2m^2}{\hbar^2}x^2\right)\Psi
\end{aligned}
$$
即：
$$
\begin{aligned}
V(x)
&=\frac{\hbar^2}{2m}\left(-\frac{2am}{\hbar}+\frac{4a^2m^2}{\hbar^2}x^2\right)+a\hbar \\
&=-a\hbar+2a^2mx^2+a\hbar \\
&=2ma^2x^2
\end{aligned}
$$
由期望定义得：
$$
\begin{aligned}
\langle x \rangle
&=\int_ {-\infty}^{\infty} x |\Psi|^2 dx\\
&=|A|^2\int_ {-\infty}^{\infty} x e^{-amx^2/\hbar^2}dx \\
&=0 \\
\langle x^2 \rangle
&=\int_ {-\infty}^{\infty} x^2 |\Psi|^2 dx \\
&=|A|^2\int_ {-\infty}^{\infty} x^2 e^{-2amx^2/\hbar}dx, t=\frac{2amx^2}{\hbar} \\
&=|A|^2\int_ {-\infty}^{\infty} \frac{\hbar t}{2am}e^{-t} \frac{\hbar}{4am} \sqrt{\frac{2am}{\hbar t}}dt \\
&=|A|^2 \sqrt{\frac{\hbar}{2am}}\frac{\hbar}{4am} \int_ {-\infty}^{\infty} t^{1/2}e^{-t}dt \\
&=\sqrt{\frac{2am}{\pi \hbar}\frac{\hbar}{2am}}\frac{\hbar}{2am}\Gamma(3/2) \\
&=\frac{1}{\sqrt{\pi}}\frac{\hbar}{2am}\frac{1}{2}\sqrt{\pi} \\
&=\frac{\hbar}{4am} \\
\langle p \rangle
&=m\frac{d \langle x \rangle}{dt} \\
&=0 \\
\langle p^2 \rangle
&=\int_ {-\infty}^{\infty}\Psi^* (-i\hbar)^2 \frac{\partial^2 \Psi}{\partial x^2} dx \\
&=-\hbar^2|A|^2\int_ {-\infty}^{\infty}e^{-amx^2/\hbar}e^{iat}e^{-iat} \nabla \left( \frac{-2amx}{\hbar}e^{-amx^2/\hbar} \right)dx \\
&=2am\hbar|A|^2\int_ {-\infty}^{\infty}e^{-amx^2/\hbar}\left( e^{-amx^2/\hbar}+\frac{-2amx^2}{\hbar}e^{-amx^2/\hbar} \right)dx \\
&=2am\hbar|A|^2\int_ {-\infty}^{\infty}\left(1-\frac{2amx^2}{\hbar}\right)e^{-2amx^2/\hbar}dx \\
&=2am\hbar\left( \langle 1 \rangle-\frac{2am}{\hbar}\langle x^2 \rangle \right) \\
&=2am\hbar\left( 1-\frac{2am}{\hbar}\frac{\hbar}{4am} \right) \\
&=am\hbar
\end{aligned}
$$
由方差定义得：
$$
\begin{aligned}
& \sigma_ x=\sqrt{\langle x^2\rangle-\langle x \rangle ^2}=\sqrt{\frac{\hbar}{4am}} \\
& \sigma_ p=\sqrt{\langle p^2\rangle-\langle p \rangle ^2}=\sqrt{am\hbar}
\end{aligned}
$$
因此：
$$
\sigma_ x\sigma_ p=\frac{\hbar}{2} \ge \frac{\hbar}{2}
$$

1.15

设某自发衰变的不稳定粒子的寿命为 $\tau$，其被发现的概率为：
$$
P(t)=\int_ {-\infty}^{\infty} |\Psi|^2dx=e^{-t/\tau}
$$
当势 $V$ 是实函数时，有概率守恒关系：
$$
\frac{d}{dt}P(t)=0
$$
现在考虑将势 $V$ 添加虚数部分：
$$
V=V_ 0-i\Gamma, \quad \Gamma \in \mathbb{R}
$$

1. 证明 $\frac{d P}{dt}=-\frac{2\Gamma}{\hbar} P$
2. 求 $P(t),\tau$

设粒子处于态 $\Psi$，由薛定谔方程得：
$$
\begin{aligned}
& i\hbar \dot{\Psi}=-\frac{\hbar^2}{2m}\nabla^2\Psi+(V_ 0-i\Gamma)\Psi \\
& -i\hbar \dot{\Psi}^* =-\frac{\hbar^2}{2m}\nabla^2\Psi^* +(V_ 0+i\Gamma)\Psi^*
\end{aligned}
$$
于是有：
$$
\begin{aligned}
\frac{\partial}{\partial t}|\Psi|^2
&=\dot{\Psi}^* \Psi+\Psi^* \dot{\Psi} \\
&=\frac{\hbar}{2im}\nabla^2\Psi^* \Psi+\frac{iV_ 0-\Gamma}{\hbar}\Psi^* \Psi
-\frac{\hbar}{2im}\nabla^2\Psi\Psi^* -\frac{iV_ 0+\Gamma}{\hbar}\Psi\Psi^* \\
&=\frac{\hbar}{2im}\left(\nabla^2\Psi^* \Psi-\nabla^2\Psi\Psi^* \right)-\frac{2\Gamma}{\hbar}|\Psi|^2 \\
\end{aligned}
$$
积分得：
$$
\begin{aligned}
\frac{dP}{dt}
&=\int_ {-\infty}^{\infty} \frac{\partial}{\partial t}|\Psi|^2 dx \\
&=-\frac{2\Gamma}{\hbar}\int_ {-\infty}^{\infty}|\Psi|^2dx+\frac{\hbar}{2im}\int_ {-\infty}^{\infty} \left(\nabla^2\Psi^* \Psi-\nabla^2\Psi\Psi^* \right)dx \\
&=-\frac{2\Gamma}{\hbar}P+\frac{\hbar}{2im}\underbrace{\int_ {-\infty}^{\infty}\nabla(\Psi\nabla \Psi^* -\Psi^* \nabla\Psi)dx}_ {0} \\
&=-\frac{2\Gamma}{\hbar}P
\end{aligned}
$$
由 $\dot{P}=-2\Gamma P/\hbar$ 得：
$$
P(t)=P(0)e^{-\frac{2\Gamma}{\hbar}t}=e^{-\frac{2\Gamma}{\hbar}t}, \quad P(0)=1
$$
即：
$$
\tau=\frac{\hbar}{2\Gamma}
$$