科学科技量子力学概论

1.7 从波函数中得到坐标,速度和动量 Position, ve

2020-05-23  本文已影响0人  莎野椰

前言

我们知道量子力学体系的性质都encapsulated 在波函数里面, 波函数涉及到粒子的位置,速度,动量等信息。我们已经知道如何计算粒子在某一位置出现的概率(期望笔记1.5,或者某一区间的概率。那么其他的动力学条件呢?比如动量速度,这就涉及到算符算符是量子力学的基本概念,他们把波函数与物理量联系起来。

1. 运动的重复测量

\int_a^b |\Psi(x)|^2 dx

\langle x \rangle=\int_{-\infty}^{\infty} x \Psi^*(x) \Psi(x) dx

2.1 什么是运动?

2. 运动期望的“速率”

\frac{d}{dt}\langle x \rangle=\frac{d}{dt} \int_{-\infty}^{\infty} x \Psi^*(x) \Psi(x) dx = \int x \underline{\frac{\partial}{\partial t} \Psi^* \Psi dx}
\because \frac{d}{dt} \int_{-\infty}^{\infty} |\Psi(x,t)|^2 dx =...=\int_{-\infty}^{\infty} \frac{i\hbar}{2m}\left( -\frac{\partial^2 \Psi^*}{\partial x^2} \Psi + \Psi^* \frac{\partial^2 \Psi}{\partial x^2} \right)dx \color{red}{见笔记 1.6,part2}
\therefore \frac{d}{dt}|\Psi(x,t)|^2 = \frac{i\hbar}{2m}\left( -\frac{\partial^2 \Psi^*}{\partial x^2} \Psi + \Psi^* \frac{\partial^2 \Psi}{\partial x^2} \right) = \frac{\partial}{\partial x} \left[ \frac{i\hbar}{2m} \left( \frac{\partial \Psi}{\partial x} \Psi^*- \Psi \frac{\partial \Psi^*}{\partial x}\right)\right]

$ = \frac{i\hbar}{2m} \int x \frac{\partial}{\partial x} \left(\frac{\partial \Psi}{\partial x} \Psi^*- \Psi \frac{\partial \Psi^*}{\partial x}\right)$
- Note:这里$\Psi与\frac{\partial \Psi^*}{\partial x}点乘不分先后$

积分:\because u=x , du=dx ; dv = \frac{\partial}{\partial x} \left(\frac{\partial \Psi}{\partial x} \Psi^*- \Psi \frac{\partial \Psi^*}{\partial x}\right)dx , v= \frac{\partial \Psi}{\partial x} \Psi^*- \Psi \frac{\partial \Psi^*}{\partial x}
\therefore \frac{i\hbar}{2m} \int x \frac{\partial}{\partial x} \left(\frac{\partial \Psi}{\partial x} \Psi^*- \Psi \frac{\partial \Psi^*}{\partial x}\right),且归一化的波函数在无穷处归于0
\require{cancel}= \frac{i\hbar}{2m} \left[ \cancel{x (\frac{\partial \Psi}{\partial x} \Psi^*- \Psi \frac{\partial \Psi^*}{\partial x})|_{-\infty}^{\infty}}-\int ( \frac{\partial}{\partial x} \left(\frac{\partial \Psi}{\partial x} \Psi^*- \Psi \frac{\partial \Psi^*}{\partial x}\right)dx ) \right]
= \frac{i\hbar}{2m} \left[\int ( \frac{\partial}{\partial x} \left(\frac{\partial \Psi}{\partial x} \Psi^*- \Psi \frac{\partial \Psi^*}{\partial x}\right)dx ) \right]
积分:\because u=\Psi, du=\frac{\partial \Psi}{\partial x};dv=\frac{\partial \Psi^*}{\partial x}, v = \Psi^*
\therefore = \frac{i\hbar}{m} \int \Psi^* \frac{\partial \Psi}{\partial x} dx = \langle \hat{v} \rangle 这就是速率算符

2.1 算符的期望

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