正则变换与哈密顿-雅可比理论 作用量竟做出这种事⋯ 背后原因令人暖心

• 原文: https://zhuanlan.zhihu.com/p/85001265
• 发布日期: 2019-10-02
• 分类: 经典力学 / 分析力学

经典力学感觉好像唯一的作用就是用来引出量子力学的(暴论). (或者用来巩固微分几何?) 所以这篇文章实际上只是其它文章的附录.

保守系哈密顿量可表达为 H=\sum\limits_{\alpha =1}^{s}{\frac{\partial L}{\partial {\dot{q}_{\alpha }{\dot{q}_{\alpha }-L 且记 \frac{\partial L}{\partial {\dot{q}_{\alpha }={p}_{\alpha }

正则方程则被表达为 \left\{ \begin{align} & \frac{\partial H}{\partial {p}_{\alpha }={\dot{q}_{\alpha } \\ & \frac{\partial H}{\partial {q}_{\alpha }=-{\dot{p}_{\alpha } \\ \end{align} \right.

若哈密顿量 $H\left(q_{\alpha },p_{\alpha }\right)$ 不显含 $q_{\alpha }$ 则有 \frac{\partial H}{\partial {q}_{\alpha }=-{\dot{p}_{\alpha }=0 , 此时 {\dot{p}_{\alpha }是体系的守恒量, 我们将相应的广义坐标 $q_{\alpha }$ 称作循环坐标. 显然循环坐标越多体系就越容易被求解. 而循环坐标的数量是与我们选择的坐标系相关的, 不同的坐标系下哈密顿量的表现形式不同. 而这些不同形式的哈密顿量之间的变换就被称作正则变换. 正则变换可以保证正则方程的简洁形式不被改变, 要不然就达不到构造守恒量的目的了.

我们将新的广义坐标、广义动量、哈密顿量分别记作 {q}'}_{\alpha },{p}'}_{\alpha },{H}'\left( {q}'}_{\alpha },{p}'}_{\alpha } \right).

要保证新的哈密顿量仍然满足正则方程则要求新的哈密顿量必须遵循最小作用量原理:

\delta {S}'=\delta \left[ \int{L}'\left( {q}'}_{\alpha },{p}'}_{\alpha } \right)dt} \right]=\delta \left[ \int{\left( \sum\limits_{\alpha =1}^{s}{p}'}_{\alpha }{\dot{q}'}_{\alpha }-{H}' \right)dt} \right]=\delta \left( \int{\sum\limits_{\alpha =1}^{s}{p}'}_{\alpha }d{q}'}_{\alpha }-{H}'dt} \right)=0

最小作用量原理即哈密顿原理: [正樹：由哈密顿原理得到哈密顿正则方程](https://zhuanlan.zhihu.com/p/52990974)

而原来的哈密顿量肯定是满足了最小作用量原理的, 即 $\delta \left(\int \sum _{\alpha =1}^s{p}_{\alpha }d{q}_{\alpha }-Hdt\right)=0$

结合上面两行式子可以得到 \delta \left[ \int{\sum\limits_{\alpha =1}^{s}{\left( {p}_{\alpha }d{q}_{\alpha }-{p}'}_{\alpha }d{q}'}_{\alpha } \right)}+\left( {H}'-H \right)dt} \right]=0

若被积函数为某函数 $U_1$ 的全微分, 即: d{U}_{1}=\sum\limits_{\alpha =1}^{s}{\left( {p}_{\alpha }d{q}_{\alpha }-{p}'}_{\alpha }d{q}'}_{\alpha } \right)}+\left( {H}'-H \right)dt

观察上式不难得到 \frac{\partial {U}_{1}{\partial {q}_{\alpha }={p}_{\alpha }\ \ ,\ \ \frac{\partial {U}_{1}{\partial {q}'}_{\alpha }=-{p}'}_{\alpha }\ \ ,\ \ \frac{\partial {U}_{1}{\partial t}={H}'-H

由上一行前面俩式子可解出 \left\{ \begin{align} & {p}'}_{\alpha }={p}'}_{\alpha }\left( {p}_{1},\cdot \cdot \cdot ,{p}_{s};{q}_{1},\cdot \cdot \cdot ,{q}_{s},t \right) \\ & {q}'}_{\alpha }={q}'}_{\alpha }\left( {p}_{1},\cdot \cdot \cdot ,{p}_{s};{q}_{1},\cdot \cdot \cdot ,{q}_{s},t \right) \\ \end{align} \right. 而 {H}'=H+\frac{\partial {U}_{1}{\partial t}

即一个正则变换可以被一个函数 {U}_{\text{1}\left( {q}_{\alpha },{q}'}_{\alpha },t \right) 完全描述, 我们称其为生成函数或母函数.

给出了一个正则变换, 就能找到等价的另外三种形式的正则变换:

若给出 d{U}_{1}=\sum\limits_{\alpha =1}^{s}{\left( {p}_{\alpha }d{q}_{\alpha }-{p}'}_{\alpha }d{q}'}_{\alpha } \right)}+\left( {H}'-H \right)dt

则可以通过勒让德变换得到:

\begin{align} & d{U}_{1}=\sum\limits_{\alpha =1}^{s}{\left( {p}_{\alpha }d{q}_{\alpha }-{p}'}_{\alpha }d{q}'}_{\alpha } \right)}+\left( {H}'-H \right)dt \\ & \ \ \ \ \ \ =\sum\limits_{\alpha =1}^{s}{\left\{ {p}_{\alpha }d{q}_{\alpha }-\left[ d\left( {p}'}_{\alpha }{q}'}_{\alpha } \right)-{q}'}_{\alpha }d{p}'}_{\alpha } \right] \right\}+\left( {H}'-H \right)dt \\ & \ \ \ \ \ \ =\sum\limits_{\alpha =1}^{s}{\left( {p}_{\alpha }d{q}_{\alpha }+{q}'}_{\alpha }d{p}' \right)}-\sum\limits_{\alpha =1}^{s}{d\left( {p}'}_{\alpha }{q}'}_{\alpha } \right)}+\left( {H}'-H \right)dt \\ \end{align} \begin{align} & \Rightarrow d{U}_{2}=d\left( {U}_{1}+\sum\limits_{\alpha =1}^{s}{p}'}_{\alpha }{q}'}_{\alpha } \right)=\sum\limits_{\alpha =1}^{s}{\left( {p}_{\alpha }d{q}_{\alpha }+{q}'}_{\alpha }d{p}'}_{\alpha } \right)}+\left( {H}'-H \right)dt \\ & \Rightarrow {U}_{2}={U}_{1}+\sum\limits_{\alpha =1}^{s}{p}'}_{\alpha }{q}'}_{\alpha } \\ \end{align}

此时有 \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

同理我们还可以有: \left\{ \begin{align} & {U}_{3}={U}_{1}-\sum\limits_{\alpha =1}^{s}{p}_{\alpha }{q}_{\alpha } \\ & {U}_{4}={U}_{1}+\sum\limits_{\alpha =1}^{s}{p}'}_{\alpha }{q}'}_{\alpha }-\sum\limits_{\alpha =1}^{s}{p}_{\alpha }{q}_{\alpha } \\ \end{align} \right.

哈密顿-雅可比理论:

薛定谔方程在 $\hslash \to 0$ 的极限下能过渡到哈密顿-雅可比方程, 所以这里也提一下. 既然评论区有对这个问题感兴趣的, 就把这个过渡写在了文章最后.

前面提到给定一个正则变换或者说给定一个生成函数 $U$ 就有 $H^{\prime }=H+\frac{\partial U}{\partial t}$.

这时很自然地会产生一个比较大胆的想法:

万一整出个 $H^{\prime }=H+\frac{\partial U}{\partial t}=0$ 呢！

那岂不是全部新广义坐标、新广义动量全都是守恒量了？

更奇妙的是, 这个生成函数其实一直就在身边, 它就是:

作用量 $S=\int L\left(q_{\alpha },p_{\alpha }\right)dt=\int \sum _{\alpha =1}^s{p}_{\alpha }d{q}_{\alpha }-Hdt$

显然存在关系 \frac{\partial S}{\partial t}=-H\ \ \ \frac{\partial S}{\partial {q}_{\alpha }={p}_{\alpha } , 这恰好是 $U_2$ 满足的式子.

所以就得到了一个震撼亲妈的方程:

$\frac{\partial S}{\partial t}+H=0$ (Hamilton-Jacobi equation)

对这个方程稍加变形即得到: $i^2\hslash \frac{\partial }{\partial t}\frac{S}{\hslash }=H$ 是不是很眼熟? 其中 $\frac{S}{\hslash }$ 其实是波函数的相位, 这个在樱井的《现代量子力学》上有提到过. 现在, 本文最后也展示了一点儿 - 2019.10.23.

#作用量与波函数相位之间的联系:

\left. \begin{align} & i\hbar \frac{d}{dt}\left| \psi \left( t \right) \right\rangle =H\left| \psi \left( t \right) \right\rangle \\ & H=\frac{P}^{2}{2m}+V\left( {\vec{r} \right) \\ \end{align} \right\}\Rightarrow i\hbar \frac{\partial }{\partial t}\psi \left( \vec{r},t \right)=-\frac{\hbar }^{2}{2m}{\nabla }^{2}\psi \left( \vec{r},t \right)+V\left( {\vec{r} \right)\psi \left( \vec{r},t \right)

现在把波函数的相位拉出来观察: \left\langle {\vec{r} | \psi \left( t \right) \right\rangle =\psi \left( \vec{r},t \right)={e}^{i\frac{S\left( \vec{r},t \right)}{\hbar }\left| \psi \left( \vec{r},t \right) \right|={e}^{i\frac{S}{\hbar }\left| \psi \right|

代入薛定谔方程: i\hbar \frac{\partial }{\partial t}\left( {e}^{i\frac{S}{\hbar }\left| \psi \right| \right)=-\frac{\hbar }^{2}{2m}{\nabla }^{2}\left( {e}^{i\frac{S}{\hbar }\left| \psi \right| \right)+V\left( {\vec{r} \right){e}^{i\frac{S}{\hbar }\left| \psi \right|

\Rightarrow i\hbar \left[ \frac{i}{\hbar }{e}^{i\frac{S}{\hbar }\left| \psi \right|\frac{\partial S}{\partial t}+{e}^{i\frac{S}{\hbar }\frac{\partial \left| \psi \right|}{\partial t} \right]=-\frac{\hbar }^{2}{2m}\left( \left| \psi \right|{\nabla }^{2}{e}^{i\frac{S}{\hbar }+2\nabla {e}^{i\frac{S}{\hbar }\nabla \left| \psi \right|+{e}^{i\frac{S}{\hbar }{\nabla }^{2}\left| \psi \right| \right)+V\left( {\vec{r} \right){e}^{i\frac{S}{\hbar }\left| \psi \right|

\Rightarrow i\hbar \left[ \frac{i}{\hbar }{e}^{i\frac{S}{\hbar }\left| \psi \right|\frac{\partial S}{\partial t}+{e}^{i\frac{S}{\hbar }\frac{\partial \left| \psi \right|}{\partial t} \right]=-\frac{\hbar }^{2}{2m}(-\frac{1}{\hbar }^{2}{e}^{i\frac{S}{\hbar }\left| \psi \right|{\left( \nabla S \right)}^{2}+\frac{i}{\hbar }{e}^{i\frac{S}{\hbar }\left| \psi \right|{\nabla }^{2}S+2\frac{i}{\hbar }{e}^{i\frac{S}{\hbar }\nabla S\nabla \left| \psi \right|\cdot \cdot \ \cdot

接下来取极限 $\hslash \to \text{0}$ 得到:

-{e}^{i\frac{S}{\hbar }\left| \psi \right|\frac{\partial S}{\partial t}=\frac{1}{2m}{e}^{i\frac{S}{\hbar }\left| \psi \right|{\left( \nabla S \right)}^{2}+V\left( {\vec{r} \right){e}^{i\frac{S}{\hbar }\left| \psi \right|

\Rightarrow \frac{\partial S}{\partial t}+\frac{\left( \nabla S \right)}^{2}{2m}+V\left( {\vec{r} \right)=0

接下来看看 $\nabla S$ 是什么

概率流密度公式: $\vec{j}=\frac{\hslash }{2mi}\left({\psi }^{\ast }\nabla \psi -\psi \nabla {\psi }^{\ast }\right)=\frac{\hslash }{m}\mathrm{Im}\left({\psi }^{\ast }\nabla \psi \right)$

将 \psi \left( \vec{r},t \right)={e}^{i\frac{S}{\hbar }\left| \psi \right| 代入得到: $\vec{j}=\frac{\hslash }{m}\mathrm{Im}\left(\frac{i}{\hslash }{\left|\psi \right|}^2\nabla S+\left|\psi \right|\nabla \left|\psi \right|\right)={\left|\psi \right|}^2\frac{\nabla S}{m}$

类比电流密度: $\vec{J}=\rho \vec{v}$ 即[ 电流密度=电荷密度*电荷速度 ]

而量子力学中的概率性解释将 ${\left|\psi \right|}^2$ 视作概率密度, 自然地, $\frac{\nabla S}{m}$ 应该被视作某种速度.

由此看来前面得到的式子 \frac{\partial S}{\partial t}+\frac{\left( \nabla S \right)}^{2}{2m}+V\left( {\vec{r} \right)=0

应该是回归了 Hamilton-Jacobi equation \frac{\partial S}{\partial t}+\frac{P}^{2}{2m}+V=0