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Ehrenfest定理(狭义)的简单证明

作者:Erewh0n碎月发布时间:2024-10-05

    引入

    经典力学中,物体的运动遵循牛顿第二定律:

    %5Cvec%7BF%7D%20%3D%5Cfrac%7Bd%5Cvec%7Bp%7D%7D%7Bdt%7D

    而对于一个保守系统,力同样可以写成:

    %5Cvec%7BF%7D%3D-%5Cnabla%5Ccdot%20V

    而在量子力学中,粒子的运动遵循Schrodinger方程:

    i%5Chbar%5Cfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial%20t%7D%3D%5Chat%7BH%7D%5CPsi

    力学量可以通过算符来计算其期望值。而Ehrenfest定理(狭义)表明,系统动量的期望和势能梯度的期望仍然可以满足类似牛顿第二定律的形式,这里我们仅考虑一维的情况:

    %5Cfrac%7Bd%5Clangle%20p%5Crangle%7D%7Bdt%7D%3D%5Clangle-%5Cfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20x%7D%5Crangle

    接下来我将仅根据Schrodinger方程以及波函数的一些条件给出简单的证明。

    证明

    %5Chat%7Bp%7D%3D-i%5Chbar%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D,由此可以计算其期望为:

    %5Clangle%20p%5Crangle%3D-i%5Chbar%5Cint%7B%5CPsi%5E*%5Cfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial%20x%7D%7Ddx

    (-%5Cinfty%2C%2B%5Cinfty)书写时均省略。上式对时间求导并用Leibniz法则乘开可以得到:

    %5Cfrac%7Bd%5Clangle%20p%5Crangle%7D%7Bdt%7D%3D%5Cint%7B-i%5Chbar(%5Cdot%7B%5CPsi%5E*%7D%5Cfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial%20x%7D%2B%5CPsi%5E*%5Cfrac%7B%5Cpartial%5Cdot%7B%5CPsi%7D%7D%7B%5Cpartial%20x%7D)%7Ddx

    而势能梯度的期望:

    %5Clangle-%5Cfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20t%7D%5Crangle%3D-%5Cint%7B%5CPsi%5E*%5Cfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20x%7D%5CPsi%7Ddx%3D-%5Cint%7B%5Cvert%20%5CPsi%20%5Cvert%5E2%5Cfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20x%7D%20%7Ddx

    这里使用分部积分可以得到:

    %5Clangle%20-%5Cfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20x%7D%5Crangle%20%3D-(V%5Cvert%20%5CPsi%20%5Cvert%5E2%20)%5E%7B%2B%5Cinfty%7D_%7B-%5Cinfty%7D%2B%5Cint%7BV%7Dd%5Cvert%20%5CPsi%20%5Cvert%20%5E2

    x%5Crightarrow%5Cinfty时,%5CPsi%3D%5CPsi%5E*%3D0,且%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D%5CPsi%3D%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D%5CPsi%5E*%3D0(这一点我们会在后面用到)。因此这里我们可以得到:

    %5Clangle-%5Cfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20x%7D%5Crangle%3D%5Cint%7BV%7Dd(%5CPsi%5E*%5CPsi)%3D%5Cint%7BV(%5CPsi%5E*%5Cfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial%20x%7D%2B%5CPsi%5Cfrac%7B%5Cpartial%5CPsi%5E*%7D%7B%5Cpartial%20x%7D)%7Ddx

    %5Chat%7BH%7D%3D-%5Cfrac%7B%5Chbar%5E2%7D%7B2m%7D%5Cfrac%7B%5Cpartial%5E2%7D%7B%5Cpartial%20x%5E2%7D%2BV)可以得到:

    V%5CPsi%3Di%5Chbar%5Cfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial%20t%7D%2B%5Cfrac%7B%5Chbar%5E2%7D%7B2m%7D%5Cfrac%7B%5Cpartial%5E2%5CPsi%7D%7B%5Cpartial%20x%5E2%7D

    V%5CPsi%5E*%3D-i%5Chbar%5Cfrac%7B%5Cpartial%5CPsi%5E*%7D%7B%5Cpartial%20t%7D%2B%5Cfrac%7B%5Chbar%5E2%7D%7B2m%7D%5Cfrac%7B%5Cpartial%5E2%5CPsi%5E*%7D%7B%5Cpartial%20x%5E2%7D

    将其带入后重新整理可以得到:

    %5Clangle-%5Cfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20x%7D%5Crangle%3D%5Cint%7B-i%5Chbar(%5Cdot%7B%5CPsi%5E*%7D%5Cfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial%20x%7D-%5Cdot%7B%5CPsi%7D%5Cfrac%7B%5Cpartial%5CPsi%5E*%7D%7B%5Cpartial%20x%7D)%7Ddx%2B%5Cint%7B%5Cfrac%7B%5Chbar%5E2%7D%7B2m%7D(%5Cfrac%7B%5Cpartial%5E2%5CPsi%5E*%7D%7B%5Cpartial%20x%5E2%7D%5Cfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial%20x%7D%2B%5Cfrac%7B%5Cpartial%5E2%5CPsi%7D%7B%5Cpartial%20x%5E2%7D%5Cfrac%7B%5Cpartial%5CPsi%5E*%7D%7B%5Cpartial%20x%7D)%7Ddx

    先考虑其中的第一项,由于:

    %5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D(%5Cdot%7B%5CPsi%7D%5CPsi%5E*)%3D%5Cdot%7B%5CPsi%7D%5Cfrac%7B%5Cpartial%5CPsi%5E*%7D%7B%5Cpartial%20x%7D%2B%5CPsi%5E*%5Cfrac%7B%5Cpartial%5Cdot%7B%5CPsi%7D%7D%7B%5Cpartial%20x%7D

    移项后两边积分就可以得到:

    %5Cint%7B%5CPsi%5E*%5Cfrac%7B%5Cpartial%20%5Cdot%7B%5CPsi%7D%7D%7B%5Cpartial%20x%7D%7Ddx%3D(%5Cdot%7B%5CPsi%7D%5CPsi%5E*)%5E%7B%2B%5Cinfty%7D_%7B-%5Cinfty%7D-%5Cint%7B%5Cdot%7B%5CPsi%7D%5Cfrac%7B%5Cpartial%5CPsi%5E*%7D%7B%5Cpartial%20x%7D%7Ddx%3D-%5Cint%7B%5Cdot%7B%5CPsi%7D%5Cfrac%7B%5Cpartial%5CPsi%5E*%7D%7B%5Cpartial%20x%7D%7Ddx

    %5CPsi%5E*%3D0。将其带入后我们发现其实积分的第一项的值就是%5Cfrac%7Bd%7D%7Bdt%7D%5Clangle%20p%5Crangle,接下来我们只需证明第二项积分的值为0。

    %5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D(%5Cfrac%7B%5Cpartial%5CPsi%5E*%7D%7B%5Cpartial%20x%7D%5Cfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial%20x%7D)%3D%5Cfrac%7B%5Cpartial%5E2%5CPsi%5E*%7D%7B%5Cpartial%20x%5E2%7D%5Cfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial%20x%7D%2B%5Cfrac%7B%5Cpartial%5E2%5CPsi%7D%7B%5Cpartial%20x%5E2%7D%5Cfrac%7B%5Cpartial%5CPsi%5E*%7D%7B%5Cpartial%20x%7D,因此第二项可以写成:

    %5Cint%7B%5Cfrac%7B%5Chbar%5E2%7D%7B2m%7D%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D(%5Cfrac%7B%5Cpartial%5CPsi%5E*%7D%7B%5Cpartial%20x%7D%5Cfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial%20x%7D)%7Ddx

    即(这里用到了无穷远处波函数导数的条件):

    %5Cfrac%7B%5Chbar%5E2%7D%7B2m%7D(%5Cfrac%7B%5Cpartial%5CPsi%5E*%7D%7B%5Cpartial%20x%7D%5Cfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial%20x%7D)%5E%7B%2B%5Cinfty%7D_%7B-%5Cinfty%7D%3D0

    %5Cfrac%7Bd%5Clangle%20p%5Crangle%7D%7Bdt%7D%3D%5Clangle-%5Cfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20x%7D%5Crangle

    结语

    Ehrefest定理初步地表明了量子力学和经典力学存在的某种联系。以上证明过程实际是Griffiths量子力学教材上的一道作业,写着玩的,没有看参考答案。广义的Ehrenfest定理的证明还涉及对易关系之类的知识,还没学到。具体证得对不对我也不清楚,如果有错误的话恳请大佬指正。




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