QM - 路径积分 (Path Integral) PT. 1 - 基本构架

• 原文: https://zhuanlan.zhihu.com/p/275827978
• 发布日期: 2020-11-06
• 分类: 量子力学基础与量子信息

内容很干, 所以有空的时候或许还偶尔会回来这篇文章里指点一下江山···

序言:

量子力学中有三大等价表述:

1. 量子哈密顿-雅可比方程(Hamilton-Jacobi equation), ---人们也称之薛定谔方程(Schrödinger equation).
2. 量子刘维尔方程(Liouville equation), ---人们也称之海森堡方程(Heisenberg equation).
3. 费曼路径积分又提供了一个新的经典 - 量子对应, 上面俩都是哈密顿力学的量子对应, ---等价的拉格朗日力学也该上场了吧, 所以这次我们的主角是作用量.

前面俩都玩腻了, 第三个却始终没提过, 忘却的费曼快要降临了罢, 我正有写一点东西的必要了.

可以尽管放心的是, 本文一定是你能找到的路径积分相关里面最简单易懂的 note 了, 我是说我甚至每一步推导过程都写上去了, 阅读本文甚至不需要动笔, 纯拿眼睛瞪都能瞪懂. 会追求尽可能的简洁, 这样看起来会很舒服, 但如果简洁与易懂发生冲突, 我一定会选择易懂, 当然这也是我一贯的写作风格了, 因为我自己复习的时候也不想动脑子, 当然更不想动手.

越是玄乎复杂的理论就越容易伴随着许多深刻而又玄学的迷思, 这些东西在路径积分中可谓是体现的淋漓尽致, 以至于光是路径积分就可以写本书了. 但这些对初学者是没有好处的, 整的跟听故事一样觉得这里酷那里酷背后思想多么多么深刻, 张口就成名词党, 计算是一问三不知.

所以本文也与二次量子化系列一样, 开局只在乎如何给出最简单明了地从原知识体系中构建出新理论的方法. 重点在于如何最简单地上手用上这个工具. 这么做虽然没学到啥深刻思想, 但至少别人跟你提到相关知识的时候你能觉得清晰明了.

比如说别人提到传播子, 你即刻就应该觉得很清晰: "就是那个时间演化算符的不同时刻坐标表象下的矩阵元嘛, 无限分割很容易写出表达式." 而不是: "噢, 就是那个, 放到积分里面算, 对每一条啊, 路径加权求和的那个系数, 好像还是啥格林函数, 是不是跟电动力学还有点儿关系啊. 表达式? 好像和作用量有关嘛, 对! 最小作用量原理! 其它路径相干相消了. "(然而他根本不知道自己在说什么.)

不过每本书都喜欢给你整这些玄学概念, 这些概念就真的就只是害人吗? 当然不是, 等你学完怎么计算以后再回过头来看看比如费曼亲自写的那本路径积分或者 Sakurai 的 QM 里面的相关讨论都会感到受益匪浅, 物理思想十分深刻.

但这都是后话了, 总之就是和二次量子化系列一样, 本文重点在于如何绕过弯路快速上手[1].

[正樹：二次量子化(Second Quantization)PT.1 - 对称与反对称基矢](https://zhuanlan.zhihu.com/p/95670627)

路径积分二连:

[1~3] QM - 路径积分 (Path Integral) PT. 1 - 基本构架

[4~5] QM - 路径积分 (Path Integral) PT. 2 - 求解实例

当前这篇仅包含目录中的前三章.

目録

1. 路径积分的基本构架

1.1. 幺正时间演化算符 1.2. 传播子及其性质 1.3. 相空间的路径积分 1.4. 位形空间的路径积分 1.5. 传播子的两类表达式

2. 自由粒子 - 相空间的路径积分

2.1. 自由粒子的传播子与其无穷小情况具有相同形式 2.2. 得到自由粒子的传播子

3. 自由粒子 - 位形空间的路径积分

3.1. 自由粒子的传播子与其无穷小情况具有相同形式 3.2. 得到自由粒子的传播子

4. 自由粒子 - 利用对角化求解路径积分

4.1. 对作用量进行变量代换 4.2. 经典路径的分离 4.3. 用对角化的方法计算传播子积分 4.4. 特征多项式与连乘积

5. 谐振子 - 利用对角化求解路径积分

5.1. 对作用量进行变量代换 5.2. 经典路径的分离 5.3. 用对角化的方法计算传播子积分 5.4. 特征多项式与连乘积 5.5. さあ、止めを刺すがいい *5.6. 相位的确定

附录

[附录A] 根号下的虚数单位[附录B] 根号下连乘积[附录C] 为何 $-1={e}^{i\pi }$ 而不是 ${e}^{-i\pi }$ 或 ${e}^{i3\pi }$$\cdots$ To be continued かな?

1. 路径积分的基本构架

所谓路径积分就是研究量子系统态随时间演化的一个手段, 很耳熟对吧.

不错, 它的作用和薛定谔方程是一摸一样的.

研究态的演化无非就是研究如何从给定的态 $|\psi \left(t_0\right)⟩$ 写出任意时刻态 $|\psi \left(t\right)⟩$ 的表达式.

虽然也可以反推, 但是那样子符号会乱七八糟, 所以本文始终默认只研究后续态, 即 $t>t_0$.

1.1. 幺正时间演化算符:

我们先定义一个算符 $U$ , 它的作用是给出演化结果 $U\left(t-t_0\right)|\psi \left(t_0\right)⟩=|\psi \left(t\right)⟩$.

那接下来就求解 $U$ 的表达式, 研究时间演化我们最熟悉的还是薛定谔方程, 老朋友了:

\begin{align} & \ \ \ \ \ \ i\hbar \frac{\text{d}{\text{d}t}\left| \psi \left( t \right) \right\rangle =H\left| \psi \left( t \right) \right\rangle \\ & \Rightarrow \frac{\text{d}{\text{d}t}U\left( t-{t}_{0} \right)\left| \psi \left( {t}_{0} \right) \right\rangle =-i\frac{H}{\hbar }U\left( t-{t}_{0} \right)\left| \psi \left( {t}_{0} \right) \right\rangle \\ & \Rightarrow \frac{\text{d}{\text{d}t}U\left( t-{t}_{0} \right)=-i\frac{H}{\hbar }U\left( t-{t}_{0} \right) \\ & \Rightarrow \text{d}\ln U\left( t-{t}_{0} \right)=-i\frac{H}{\hbar }\text{d}t\Rightarrow U\left( t-{t}_{0} \right)={e}^{-i\frac{H}{\hbar }t}\cdot const \\ \end{align} 如果时间不变自然不发生演化, 以此作为初始条件, 即 $t=t_0\Rightarrow U=I$.

很轻松地确定了系数之后得到了幺正时间演化算符的表达式:

$U\left(t-t_0\right)=e^{-i\frac{H}{\hslash }\left(t-t_0\right)}$

于是我们就终于可以摆脱玩腻了的老套薛定谔方程.

而开始形式上地使用 $|\psi \left(t\right)⟩=U\left(t-t_0\right)|\psi \left(t_0\right)⟩$ 这个方便的式子了.

1.2. 传播子及其性质:

接下来对上式取坐标表象, 即左乘 \left\langle {\vec{r} \right| , 再插入完备性关系式 I=\int{\left| {\vec{r}_{0} \right\rangle \left\langle {\vec{r}_{0} \right|\text{d}{\vec{r}_{0}, 可得:

\psi \left( \vec{r},t \right)=\int \langle \vec{r}|U\left( t,{t}_{0} \right)\left| {\vec{r}_{0} \right\rangle \left\langle {\vec{r}_{0}|\psi \left( {t}_{0} \right) \right\rangle \text{d}{\vec{r}_{0}=\int{K\left( \vec{r},t;{\vec{r}_{0},{t}_{\text{0} \right)\psi \left( {\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{0}

其中K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right) 称作传播子, 作用如上所示, 路径积分的最终目的就是求解它.

先设置一个全文适用的时间节点 {t}_{0}<{t}_{n}\le t={t}_{N},\ n\in \left\{ 1,2,3,\cdot \cdot \cdot ,N \right\} .

我们不平均分割时间, 这样做的好处很快就会体现出来.

由传播子的定义, 我们不难发现它具有如下特性:

\begin{align} & K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\langle \vec{r}|U\left( t,{t}_{0} \right)\left| {\vec{r}_{0} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\langle \vec{r}|U\left( t,{t}_{1} \right)U\left( {t}_{1},{t}_{0} \right)\left| {\vec{r}_{0} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{\langle \vec{r}|U\left( t,{t}_{1} \right)\left| {\vec{r}_{1} \right\rangle \langle {\vec{r}_{1}|U\left( {t}_{1},{t}_{0} \right)\left| {\vec{r}_{0} \right\rangle \text{d}{\vec{r}_{1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{\langle \vec{r}|U\left( t,{t}_{2} \right)\left| {\vec{r}_{2} \right\rangle \langle {\vec{r}_{2}|U\left( {t}_{2},{t}_{1} \right)\left| {\vec{r}_{1} \right\rangle \langle {\vec{r}_{1}|U\left( {t}_{1},{t}_{0} \right)\left| {\vec{r}_{0} \right\rangle \text{d}{\vec{r}_{1}\text{d}{\vec{r}_{2} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{K\left( \vec{r},t;{\vec{r}_{2},{t}_{2} \right)K\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)K\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1}\text{d}{\vec{r}_{2} \\ \end{align}

注意我们要求上式中恒满足关系 {\vec{r}_{N}=\vec{r},{t}_{N}=t,\ a>b\Leftrightarrow {t}_{a}>{t}_{b} .

物理学永远是极端情况最好计算, 让我们来走一个极端, 利用上述性质进行无穷分割可得:

K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\int{\left[ \prod\limits_{n=1}^{N}{K\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)} \right]\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1} .

注意本文所有的运算全部都取了极限$\max \left\{t_n-t_{n-1}\right\}\to 0$, 主要是想省写极限符号.

但我们却偏不这么写, 我们硬要将积分号里边儿的那个无穷小过程传播子换一个符号, 记为:

K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\int{\left[ \prod\limits_{n=1}^{N}{\mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)} \right]\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1} [2].

1.3. 相空间的路径积分:

记 $\Delta t_n=t_n-t_{n-1}$, 且全文沿用.

显然想求得 K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right) 就得先对 \mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right) 下手:

\mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)=\langle {\vec{r}_{n}|U\left( {t}_{n},{t}_{n-1} \right)\left| {\vec{r}_{n-1} \right\rangle =\langle {\vec{r}_{n}|{e}^{-\frac{iH}{\hbar }\Delta {t}_{n}\left| {\vec{r}_{n-1} \right\rangle

考虑到

{e}^{-\frac{i}{\hbar }H\Delta {t}_{n}={e}^{-\frac{i}{\hbar }\left[ \frac{P}^{2}{2m}+V\left( \vec{R},{t}_{n} \right) \right]\Delta {t}_{n}\overset{\Delta {t}_{n}\to 0}{\mathop{=}\,{e}^{-\frac{i}{\hbar }V\left( \vec{R},{t}_{n} \right)\Delta {t}_{n}{e}^{-\frac{i}{\hbar }\frac{P}^{2}{2m}\Delta {t}_{n}

就是从这一刻开始恒有 $K\ne \mathcal{K}$ , 除非你是自由粒子, 这一点后面会讲到.

就可以推知

\begin{align} & \mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)=\langle {\vec{r}_{n}|{e}^{-\frac{i}{\hbar }V\left( \vec{R},{t}_{n} \right)\Delta {t}_{n}{e}^{-\frac{i}{\hbar }\frac{P}^{2}{2m}\Delta {t}_{n}\left| {\vec{r}_{n-1} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={e}^{-\frac{i}{\hbar }V\left( {\vec{r}_{n},{t}_{n} \right)\Delta {t}_{n}\langle {\vec{r}_{n}|{e}^{-\frac{i}{\hbar }\frac{P}^{2}{2m}\Delta {t}_{n}\left| {\vec{r}_{n-1} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int {e}^{-\frac{i}{\hbar }V\left( {\vec{r}_{n},{t}_{n} \right)\Delta {t}_{n}\left\langle {\vec{r}_{n}|{\vec{p}_{n} \right\rangle \langle {\vec{p}_{n}|{e}^{-\frac{i}{\hbar }\frac{P}^{2}{2m}\Delta {t}_{n}\left| {\vec{r}_{n-1} \right\rangle \text{d}{\vec{p}_{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{e}^{-\frac{i}{\hbar }V\left( {\vec{r}_{n},{t}_{n} \right)\Delta {t}_{n}\left\langle {\vec{r}_{n}|{\vec{p}_{n} \right\rangle {e}^{-\frac{i}{\hbar }\frac{p_{n}^{2}{2m}\Delta {t}_{n}\left\langle {\vec{p}_{n}|{\vec{r}_{n-1} \right\rangle \text{d}{\vec{p}_{n} \\ \end{align}

记 \Delta {\vec{r}_{n}={\vec{r}_{n}-{\vec{r}_{n-1}, 且全文沿用. 记 \ {\dot{\vec{r}_{n}={\Delta {\vec{r}_{n}/{\Delta {t}_{n}\;, 但这只是个记号, 其实路径基本都是不可微的所以这里不表征速度. 同理下面的 Lagrangian 也只是一个记号, 并不是真正的 Lagrangian[[3]](#ref_3).

\begin{align} & \Rightarrow \mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)=\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{-\frac{i}{\hbar }V\left( {\vec{r}_{n},{t}_{n} \right)\Delta {t}_{n}{e}^{i\frac{\vec{p}_{n}{\hbar }\cdot \Delta {\vec{r}_{n}{e}^{-\frac{i}{\hbar }\frac{p_{n}^{2}{2m}\Delta {t}_{n}\text{d}{\vec{p}_{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{\frac{i}{\hbar }\left[ {\vec{p}_{n}\cdot \frac{\Delta {\vec{r}_{n}{\Delta {t}_{n}-H\left( {\vec{r}_{n},{\vec{p}_{n},{t}_{n} \right) \right]\Delta {t}_{n}\text{d}{\vec{p}_{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{\frac{i}{\hbar }\left[ {\vec{p}_{n}\cdot {\dot{\vec{r}_{n}-H\left( {\vec{r}_{n},{\vec{p}_{n},{t}_{n} \right) \right]\Delta {t}_{n}\text{d}{\vec{p}_{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{i\frac{L\left( {\vec{r}_{n},{\vec{p}_{n},{t}_{n} \right)\Delta {t}_{n}{\hbar }\text{d}{\vec{p}_{n} \\ \end{align}

将上式代回传播子的表达式就得到相空间的路径积分:

\begin{align} & K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\int{\left[ \prod\limits_{n=1}^{N}{\mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)} \right]\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{\prod\limits_{n=1}^{N}{\left[ \frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{\frac{i}{\hbar }L\left( {\vec{r}_{n},{\vec{p}_{n},{t}_{n} \right)\Delta {t}_{n}\text{d}{\vec{p}_{n} \right]}\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\left( 2\pi \hbar \right)}^{3N}\int{\left[ \int{e}^{\frac{i}{\hbar }\sum\limits_{n=1}^{N}{L\left( {\vec{r}_{n},{\vec{p}_{n},{t}_{n} \right)\Delta {t}_{n}\text{d}{\vec{p}_{1}\cdot \cdot \cdot \text{d}{\vec{p}_{N} \right]\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overset{\text{limit}{\mathop{=}\,\,\,\,\,\frac{1}{\left( 2\pi \hbar \right)}^{3N}\int{e}^{\frac{i}{\hbar }\int{L\left( \vec{r},\vec{p},\tau \right)\text{d}\tau }\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1}\text{d}{\vec{p}_{1}\cdot \cdot \cdot \text{d}{\vec{p}_{N} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\left( 2\pi \hbar \right)}^{3N}\int{\mathcal{D}\vec{r}\mathcal{D}\vec{p}{e}^{i\frac{S\left( \vec{r},\vec{p},t \right)}{\hbar } \\ \end{align}

其中的 $\mathcal{D}$ 就是一个十分感性的记号, 表征我们对包括各种闪现在内的所有怪异路径进行积分. 注意位置积分重数比动量少一重, 从 $1$ 积到 $N-1$ 是理所当然的, 因为端点是固定的嘛.

1.4. 位形空间的路径积分:

我们更常用的其实不是相空间的路径积分, 而是位形空间的, 其实就是把动量先积掉罢了:

\begin{align} & \mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)=\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{-\frac{i}{\hbar }V\left( {\vec{r}_{n},{t}_{n} \right)\Delta {t}_{n}{e}^{i\frac{\vec{p}_{n}{\hbar }\cdot \Delta {\vec{r}_{n}{e}^{-\frac{i}{\hbar }\frac{p_{n}^{2}{2m}\Delta {t}_{n}\text{d}{\vec{p}_{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\left( 2\pi \hbar \right)}^{3}{e}^{-\frac{i}{\hbar }V\left( {\vec{r}_{n},{t}_{n} \right)\Delta {t}_{n}\int{e}^{i\frac{\vec{p}_{n}{\hbar }\cdot \Delta {\vec{r}_{n}{e}^{-\frac{i}{\hbar }\frac{p_{n}^{2}{2m}\Delta {t}_{n}\text{d}{\vec{p}_{n} \\ \end{align}

显然后面的积分在自由粒子的格林函数中求解过, 这里直接而代入结论:

\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{i\frac{\vec{p}_{n}{\hbar }\cdot \Delta {\vec{r}_{n}{e}^{-\frac{i}{\hbar }\frac{p_{n}^{2}{2m}\Delta {t}_{n}\text{d}{\vec{p}_{n}={\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}\exp \left[ i\frac{m{\left( \Delta {\vec{r}_{n} \right)}^{2}{2\hbar \Delta {t}_{n} \right]

上面是凭印象写出来的, 系数不太可能出错, 但过两天仍会写篇文验算一下, 并给出解法. 先提醒一下, 配方后使用高斯积分公式可是不合法的噢. 如果你用高斯积分公式去算, 那我问你, 根号下的 $i$ 是什么? {e}^{i\frac{\pi }{2}? 为什么不是 $e^{i\frac{5}{2}\pi }$ ? 为什么不是 $e^{-i\frac{3}{2}\pi }$ ? 什么你觉得都一样? 笑了. 好奇这个问题可以翻到第二篇最后看看**[附录A]**.

将上式代回无穷小间隔传播子表达式得:

\begin{align} & \mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)=\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{-\frac{i}{\hbar }V\left( {\vec{r}_{n},{t}_{n} \right)\Delta {t}_{n}{e}^{i\frac{\vec{p}_{n}{\hbar }\cdot \Delta {\vec{r}_{n}{e}^{-\frac{i}{\hbar }\frac{p_{n}^{2}{2m}\Delta {t}_{n}\text{d}{\vec{p}_{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}{e}^{-\frac{i}{\hbar }V\left( {\vec{r}_{n},{t}_{n} \right)\Delta {t}_{n}{e}^{i\frac{m{\left( \Delta {\vec{r}_{n} \right)}^{2}{2\hbar \Delta {t}_{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}{e}^{\frac{i}{\hbar }\left[ \frac{1}{2}m\frac{\left( \Delta {\vec{r}_{n} \right)}^{2}{\left( \Delta {t}_{n} \right)}^{2}-V\left( {\vec{r}_{n},{t}_{n} \right) \right]\Delta {t}_{n} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}{e}^{\frac{i}{\hbar }\left[ \frac{1}{2}m\dot{\vec{r}_{n}^{2}-V\left( {\vec{r}_{n},{t}_{n} \right) \right]\Delta {t}_{n} \\ \end{align}

将上述结果代回传播子的表达式得到位形空间的路径积分:

\begin{align} & K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\int{\left[ \prod\limits_{n=1}^{N}{\mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)} \right]\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int{\left[ \prod\limits_{n=1}^{N}{\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}{e}^{\frac{i}{\hbar }\left[ \frac{1}{2}m\dot{\vec{r}_{n}^{2}-V\left( {\vec{r}_{n},{t}_{n} \right) \right]\Delta {t}_{n} \right]\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\prod\limits_{n=1}^{N}{\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}\int{e}^{\frac{i}{\hbar }\sum\limits_{n=1}^{N}{\left[ \frac{1}{2}m\dot{\vec{r}_{n}^{2}-V\left( {\vec{r}_{n},{t}_{n} \right) \right]\Delta {t}_{n}\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overset{\text{limit}{\mathop{=}\,\,\,\,\,\prod\limits_{n=1}^{N}{\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}\int{e}^{\frac{i}{\hbar }\int_{t}_{0}^{t}{\frac{1}{2}m\dot{\vec{r}_{n}^{2}-V\left( \vec{r},\tau \right)\text{d}\tau }\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\prod\limits_{n=1}^{N}{\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}\int{\mathcal{D}\vec{r}{e}^{i\frac{S\left( \vec{r},\dot{\vec{r},t \right)}{\hbar } \\ \end{align}

1.5. 传播子的两类表达式:

综上所述, 路径积分中的传播子就表达为:

\left\{ \begin{align} & K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\frac{1}{\left( 2\pi \hbar \right)}^{3N}\int{\mathcal{D}\vec{r}\mathcal{D}\vec{p}{e}^{i\frac{S\left( \vec{r},\vec{p},t \right)}{\hbar } \\ & K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\prod\limits_{n=1}^{N}{\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}\int{\mathcal{D}\vec{r}{e}^{i\frac{S\left( \vec{r},\dot{\vec{r},t \right)}{\hbar } \\ \end{align} \right.

但说是这么说啦, 实际上你真要去算那还得回到:

\left\{ \begin{align} & K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\frac{1}{\left( 2\pi \hbar \right)}^{3N}\int{e}^{\frac{i}{\hbar }\sum\limits_{n=1}^{N}{\left[ {\vec{p}_{n}\cdot {\dot{\vec{r}_{n}-H\left( {\vec{r}_{n},{\vec{p}_{n},{t}_{n} \right) \right]\Delta {t}_{n}\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1}\text{d}{\vec{p}_{1}\cdot \cdot \cdot \text{d}{\vec{p}_{N} \\ & K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\prod\limits_{n=1}^{N}{\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}\int{e}^{\frac{i}{\hbar }\sum\limits_{n=1}^{N}{\left[ \frac{1}{2}m\dot{\vec{r}_{n}^{2}-V\left( {\vec{r}_{n},{t}_{n} \right) \right]\Delta {t}_{n}\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1} \\ \end{align} \right.

其中 \left\{ \begin{align} & \Delta {t}_{n}={t}_{n}-{t}_{n-1} \\ & \Delta {\vec{r}_{n}={\vec{r}_{n}-{\vec{r}_{n-1} \\ & {\dot{\vec{r}_{n}=\Delta {\vec{r}_{n}/\Delta {t}_{n} \\ & n\in \left\{ 1,2,3,\cdot \cdot \cdot ,N \right\} \\ \end{align} \right.且规定 \left\{ \begin{align} & {\vec{r}_{N}=\vec{r} \\ & {t}_{N}=t \\ & {t}_{0}<{t}_{n}\le t \\ & a>b\Leftrightarrow {t}_{a}>{t}_{b} \\ & \min \left\{ {t}_{n}-{t}_{n-1} \right\}\to 0 \\ \end{align} \right.根号下的虚数单位 $i$ 定义为 $\exp \left[ {i\pi }/{2}\; \right]$ 详情参考 **附录A**.

2. 自由粒子 - 相空间的路径积分

相空间传播子表达式即

K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\int{\left[ \prod\limits_{n=1}^{N}{\mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)} \right]\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1}

其中$\mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)=\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{\frac{i}{\hbar }\left[ {\vec{p}_{n}\cdot \Delta {\vec{r}_{n}-H\left( {\vec{r}_{n},{\vec{p}_{n},{t}_{n} \right)\Delta {t}_{n} \right]}\text{d}{\vec{p}_{n}$

自由粒子 Hamiltonian 很爽, 无势能项: H=\frac{P}^{2}{2m} \Rightarrow \mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)=\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{\frac{i}{\hbar }\left( {\vec{p}_{n}\cdot \Delta {\vec{r}_{n}-\frac{p_{n}^{2}{2m}\Delta {t}_{n} \right)}\text{d}{\vec{p}_{n}

2.1. 自由粒子的传播子与其无穷小情况具有相同形式:

接下来我将证明 \mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1}[4].

要知道 K\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1} 是当然的, 而 \mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1} 则是要证明的.

证明:

\begin{align} & \ \ \ \ \ \int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1} \\ & =\frac{1}{\left( 2\pi \hbar \right)}^{6}\int{\left[ \int{e}^{\frac{i}{\hbar }\left( {\vec{p}_{2}\cdot \Delta {\vec{r}_{2}-\frac{p_{2}^{2}{2m}\Delta {t}_{2} \right)}\text{d}{\vec{p}_{2} \right]\left[ \int{e}^{\frac{i}{\hbar }\left( {\vec{p}_{1}\cdot \Delta {\vec{r}_{1}-\frac{p_{1}^{2}{2m}\Delta {t}_{1} \right)}\text{d}{\vec{p}_{1} \right]\text{d}{\vec{r}_{1} \\ & =\frac{1}{\left( 2\pi \hbar \right)}^{6}\int{\text{d}{\vec{p}_{1}\text{d}{\vec{p}_{2}\text{d}{\vec{r}_{1}{e}^{\frac{i}{\hbar }\left[ {\vec{p}_{1}\cdot \left( {\vec{r}_{1}-{\vec{r}_{0} \right)+{\vec{p}_{2}\cdot \left( {\vec{r}_{2}-{\vec{r}_{1} \right)-\frac{p_{2}^{2}{2m}\left( {t}_{2}-{t}_{1} \right)-\frac{p_{1}^{2}{2m}\left( {t}_{1}-{t}_{0} \right) \right]} \\ & =\frac{1}{\left( 2\pi \hbar \right)}^{6}\int{\text{d}{\vec{p}_{1}\text{d}{\vec{p}_{2}\text{d}{\vec{r}_{1}{e}^{\frac{i}{\hbar }\left[ {\vec{p}_{1}\cdot \left( {\vec{r}_{1}-{\vec{r}_{0} \right)+{\vec{p}_{2}\cdot \left( {\vec{r}_{2}-{\vec{r}_{0}+{\vec{r}_{0}-{\vec{r}_{1} \right)-\frac{p_{2}^{2}{2m}\left( {t}_{2}-{t}_{0}+{t}_{0}-{t}_{1} \right)-\frac{p_{1}^{2}{2m}\left( {t}_{1}-{t}_{0} \right) \right]} \\ \end{align} \ =\frac{1}{\left( 2\pi \hbar \right)}^{6}\int{\text{d}{\vec{p}_{1}\text{d}{\vec{p}_{2}{e}^{\frac{i}{\hbar }\left[ {\vec{p}_{2}\cdot \left( {\vec{r}_{2}-{\vec{r}_{0} \right)-\frac{p_{2}^{2}{2m}\left( {t}_{2}-{t}_{0} \right) \right]}{e}^{i\frac{p_{2}^{2}-p_{1}^{2}{2m\hbar }\left( {t}_{1}-{t}_{0} \right)}\left[ \int{e}^{-i\frac{\left( {\vec{p}_{2}-{\vec{p}_{1} \right)}{\hbar }\cdot \left( {\vec{r}_{1}-{\vec{r}_{0} \right)}\text{d}{\vec{r}_{1} \right]}

接下来我们要利用最右边那重积分构造一个三维 delta function. delta function 的定义: \delta \left( t-{t}' \right)=\frac{1}{2\pi }\int_{-\infty }^{\infty }{e}^{i\omega \left( t-{t}' \right)}\text{d}\omega }, 这不用我多说了吧. $\Rightarrow \delta \left( {\vec{p}_{2}-{\vec{p}_{1} \right)=\frac{1}{\left( 2\pi \right)}^{3}\int{e}^{-i\frac{\left( {\vec{r}_{1}-{\vec{r}_{0} \right)}{\hbar }\cdot \left( {\vec{p}_{2}-{\vec{p}_{1} \right)}\frac{\text{d}\left( {\vec{r}_{1}-{\vec{r}_{0} \right)}{\hbar }^{3}$ [[5]](#ref_5)

\begin{align} & \Rightarrow \int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1} \\ & =\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{\text{d}{\vec{p}_{1}\text{d}{\vec{p}_{2}{e}^{\frac{i}{\hbar }\left[ {\vec{p}_{2}\cdot \left( {\vec{r}_{2}-{\vec{r}_{0} \right)-\frac{p_{2}^{2}{2m}\left( {t}_{2}-{t}_{0} \right) \right]}{e}^{i\frac{p_{2}^{2}-p_{1}^{2}{2m\hbar }\left( {t}_{1}-{t}_{0} \right)}\delta \left( {\vec{p}_{2}-{\vec{p}_{1} \right)} \\ & =\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{\frac{i}{\hbar }\left[ {\vec{p}_{2}\cdot \left( {\vec{r}_{2}-{\vec{r}_{0} \right)-\frac{p_{2}^{2}{2m}\left( {t}_{2}-{t}_{0} \right) \right]}\text{d}{\vec{p}_{2}=\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{0},{t}_{0} \right) \\ \end{align}

2.2. 得到自由粒子的传播子:

由前面证明过程可知显然可以推广这个关系:

$[1].\ \mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1}$$[2].\ \mathcal{K}\left( {\vec{r}_{3},{t}_{3};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{3}{t}_{3};{\vec{r}_{2},{t}_{2} \right)\mathcal{K}\left( {\vec{r}_{2}{t}_{2};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{2}$$\cdot \cdot \cdot$$[\text{N}].\ \mathcal{K}\left( {\vec{r}_{N},{t}_{N};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{N}{t}_{N};{\vec{r}_{N-1},{t}_{N-1} \right)\mathcal{K}\left( {\vec{r}_{N-1}{t}_{N-1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{N-1}$将 [1] 代入 [2] 后再将 [2] 代入 [3] 直到第 N 式, 我们将得到下述关系:$\mathcal{K}\left( {\vec{r}_{N},{t}_{N};{\vec{r}_{0},{t}_{0} \right)=\int{\left[ \prod\limits_{n=1}^{N}{\mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)} \right]\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1}=K\left( \vec{r},t;{\vec{r}_{0},t \right)$

上述关系说明自由粒子的传播子与其对应的无穷小过程的传播子具有相同的形式, 这就可得:

K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\mathcal{K}\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{\frac{i}{\hbar }\left[ \vec{p}\cdot \left( \vec{r}-{\vec{r}_{0} \right)-\frac{p}^{2}{2m}\left( t-{t}_{0} \right) \right]}\text{d}\vec{p}

废了这么大劲才整出这么个理所当然的结论来, 路径积分在这个角度上确实是多少有点儿捞. 如果我们不用路径积分呢? 那么将如下所示: $\begin{align} & \left| \psi \left( t \right) \right\rangle =U\left( t-{t}_{0} \right)\left| \psi \left( {t}_{0} \right) \right\rangle \Rightarrow \psi \left( \vec{r},t \right)=\int \langle \vec{r}|{e}^{-i\frac{H}{\hbar }\left( t-{t}_{0} \right)}\left| {\vec{r}_{0} \right\rangle \psi \left( {\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{0} \\ & K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\langle \vec{r}|{e}^{-i\frac{H}{\hbar }\left( t-{t}_{0} \right)}\left| {\vec{r}_{0} \right\rangle =\int \left\langle \vec{r}|\vec{p} \right\rangle \langle \vec{p}|{e}^{-i\frac{P}^{2}{2m\hbar }\left( t-{t}_{0} \right)}\left| {\vec{p}'} \right\rangle \left\langle {\vec{p}'|{\vec{r}_{0} \right\rangle \text{d}\vec{p}\text{d}{\vec{p}' \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{i\frac{\vec{p}{\hbar }\cdot \vec{r}{e}^{-i\frac{p}^{2}{2m\hbar }\left( t-{t}_{0} \right)}{e}^{-i\frac{\vec{p}{\hbar }\cdot {\vec{r}_{0}\text{d}p}\int{\delta \left( \vec{p}-{\vec{p}' \right)\text{d}{\vec{p}'} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\left( 2\pi \hbar \right)}^{3}\int{e}^{\frac{i}{\hbar }\left[ \vec{p}\cdot \left( \vec{r}-{\vec{r}_{0} \right)-\frac{\vec{p}^{2}{2m}\left( t-{t}_{0} \right) \right]}\text{d}\vec{p} \\ \end{align}$ 是的, 我写的再细致也就这么点儿运算量.

那么下一步就是把积分积出来, 得到:

K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)={\left[ \frac{m}{2\pi i\hbar \left( t-{t}_{0} \right)} \right]}^{3}/{2}\;}\exp \left[ i\frac{m{\left( \vec{r}-{\vec{r}_{0} \right)}^{2}{2\hbar \left( t-{t}_{0} \right)} \right].

会有人觉得这个积分很难计算吗? 其实这个在前面位形空间路径积分那一段碰到过了. 这里提示一下, 配方后配合欧拉公式使用那个应用范围极广的 Fresnel 积分.

3. 自由粒子 - 位形空间的路径积分

位形空间传播子表达式即

K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\int{\left[ \prod\limits_{n=1}^{N}{\mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)} \right]\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1}

其中$\mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)={\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}{e}^{\frac{i}{\hbar }\left[ \frac{1}{2}m\dot{\vec{r}_{n}^{2}-V\left( {\vec{r}_{n},{t}_{n} \right) \right]\Delta {t}_{n}$

自由粒子 Lagrangian 很爽, 无势能项: L=\frac{P}^{2}{2m} \Rightarrow \mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)={\left( \frac{m}{2\pi i\hbar \Delta {t}_{n} \right)}^{\frac{3}{2}{e}^{i\frac{m}{2\hbar }\frac{\left( \Delta {\vec{r}_{n} \right)}^{2}{\Delta {t}_{n}

3.1. 自由粒子的传播子与其无穷小情况具有相同形式:

接下来我们将证明 \mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1}.

要知道 K\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1} 是当然的, 而 \mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1} 则是要证明的.

证明:

\begin{align} & \ \ \ \ \ \int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1} \\ & =\int{\left( \frac{m}{2\pi i\hbar \Delta {t}_{2} \right)}^{\frac{3}{2}{\left( \frac{m}{2\pi i\hbar \Delta {t}_{1} \right)}^{\frac{3}{2}{e}^{i\frac{m}{2\hbar }\frac{\left( \Delta {\vec{r}_{2} \right)}^{2}{\Delta {t}_{2}{e}^{i\frac{m}{2\hbar }\frac{\left( \Delta {\vec{r}_{1} \right)}^{2}{\Delta {t}_{1}\text{d}{\vec{r}_{1} \\ & ={\left( \frac{m}{2\pi i\hbar \Delta {t}_{2} \right)}^{\frac{3}{2}{\left( \frac{m}{2\pi i\hbar \Delta {t}_{1} \right)}^{\frac{3}{2}\int{e}^{i\frac{m}{2\hbar }\left[ \frac{\left( {\vec{r}_{2}-{\vec{r}_{1} \right)}^{2}{\Delta {t}_{2}+\frac{\left( {\vec{r}_{1}-{\vec{r}_{0} \right)}^{2}{\Delta {t}_{1} \right]}\text{d}{\vec{r}_{1} \\ \end{align}

不难看出下一步就是要想办法先把 {\vec{r}_{1} 配方后积掉再说, 那就先算指数部分:

$\begin{align} & \ \ \ \ \ \ \ \frac{\left( {\vec{r}_{2}-{\vec{r}_{1} \right)}^{2}{\Delta {t}_{2}+\frac{\left( {\vec{r}_{1}-{\vec{r}_{0} \right)}^{2}{\Delta {t}_{1} \\ & =\left( \frac{1}{\Delta {t}_{2}+\frac{1}{\Delta {t}_{1} \right)\vec{r}_{1}^{2}-2\left( \frac{\vec{r}_{2}{\Delta {t}_{2}+\frac{\vec{r}_{0}{\Delta {t}_{1} \right){\vec{r}_{1}+\frac{\vec{r}_{0}^{2}{\Delta {t}_{1}+\frac{\vec{r}_{2}^{2}{\Delta {t}_{2} \\ & =\left( \frac{1}{\Delta {t}_{2}+\frac{1}{\Delta {t}_{1} \right)\left[ \vec{r}_{1}^{2}-2{\left( \frac{1}{\Delta {t}_{2}+\frac{1}{\Delta {t}_{1} \right)}^{-1}\left( \frac{\vec{r}_{2}{\Delta {t}_{2}+\frac{\vec{r}_{0}{\Delta {t}_{1} \right){\vec{r}_{1} \right]+\frac{\vec{r}_{0}^{2}{\Delta {t}_{1}+\frac{\vec{r}_{2}^{2}{\Delta {t}_{2} \\ \end{align}$ $\begin{align} & =\left( \frac{1}{\Delta {t}_{2}+\frac{1}{\Delta {t}_{1} \right){\left[ {\vec{r}_{1}-{\left( \frac{1}{\Delta {t}_{2}+\frac{1}{\Delta {t}_{1} \right)}^{-1}\left( \frac{\vec{r}_{2}{\Delta {t}_{2}+\frac{\vec{r}_{0}{\Delta {t}_{1} \right) \right]}^{2} \\ & \ \ \ \ \ -{\left( \frac{1}{\Delta {t}_{2}+\frac{1}{\Delta {t}_{1} \right)}^{-1}{\left( \frac{\vec{r}_{2}{\Delta {t}_{2}+\frac{\vec{r}_{0}{\Delta {t}_{1} \right)}^{2}+\frac{\vec{r}_{0}^{2}{\Delta {t}_{1}+\frac{\vec{r}_{2}^{2}{\Delta {t}_{2} \\ & =\left( \frac{1}{\Delta {t}_{2}+\frac{1}{\Delta {t}_{1} \right){\left[ {\vec{r}_{1}-{\left( \frac{1}{\Delta {t}_{2}+\frac{1}{\Delta {t}_{1} \right)}^{-1}\left( \frac{\vec{r}_{2}{\Delta {t}_{2}+\frac{\vec{r}_{0}{\Delta {t}_{1} \right) \right]}^{2}+\frac{\vec{r}_{0}^{2}+\vec{r}_{2}^{2}-2{\vec{r}_{2}{\vec{r}_{0}{\Delta {t}_{1}+\Delta {t}_{2} \\ \end{align}$

将上述指数代回原式得到[6]:

\begin{align} & \ \ \ \ \ \int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1} \\ & ={\left( \cdot \right)}^{\frac{3}{2}{\left( \cdot \right)}^{\frac{3}{2}{e}^{i\frac{m}{2\hbar }\frac{\left( {\vec{r}_{2}-{\vec{r}_{0} \right)}^{2}{\Delta {t}_{2}+\Delta {t}_{1}\int{e}^{i\frac{m}{2\hbar }\left( \frac{1}{\Delta {t}_{2}+\frac{1}{\Delta {t}_{1} \right){\left[ {\vec{r}_{1}-\left( \frac{\Delta {t}_{1}\Delta {t}_{2}{\Delta {t}_{1}+\Delta {t}_{2} \right)\left( \frac{\vec{r}_{2}{\Delta {t}_{2}+\frac{\vec{r}_{0}{\Delta {t}_{1} \right) \right]}^{2}\text{d}{\vec{r}_{1} \\ & ={\left( \frac{m}{2\pi i\hbar \Delta {t}_{2} \right)}^{\frac{3}{2}{\left( \frac{m}{2\pi i\hbar \Delta {t}_{1} \right)}^{\frac{3}{2}{e}^{i\frac{m}{2\hbar }\frac{\left( {\vec{r}_{2}-{\vec{r}_{0} \right)}^{2}{t}_{2}-{t}_{0}\int{e}^{i\frac{m}{2\hbar }\left( \frac{1}{\Delta {t}_{2}+\frac{1}{\Delta {t}_{1} \right){\vec{x}^{2}\text{d}\vec{x} \\ & ={\left( \frac{m}{2\pi i\hbar \Delta {t}_{2} \right)}^{\frac{3}{2}{\left( \frac{1}{\pi i\Delta {t}_{1} \right)}^{\frac{3}{2}{e}^{i\frac{m}{2\hbar }\frac{\left( {\vec{r}_{2}-{\vec{r}_{0} \right)}^{2}{t}_{2}-{t}_{0}\int{e}^{-\frac{\Delta {t}_{2}+\Delta {t}_{1}{i\Delta {t}_{2}\Delta {t}_{1}{\vec{y}^{2}\text{d}{\vec{y}^{2} \\ \end{align} \begin{align} & ={\left( \frac{m}{2\pi i\hbar \Delta {t}_{2} \right)}^{\frac{3}{2}{\left( \frac{1}{\pi i\Delta {t}_{1} \right)}^{\frac{3}{2}{e}^{i\frac{m}{2\hbar }\frac{\left( {\vec{r}_{2}-{\vec{r}_{0} \right)}^{2}{t}_{2}-{t}_{0}{\left( \frac{\pi i\Delta {t}_{2}\Delta {t}_{1}{t}_{2}-{t}_{0} \right)}^{\frac{3}{2} \\ & ={\left[ \frac{m}{2\pi i\hbar \left( {t}_{2}-{t}_{0} \right)} \right]}^{\frac{3}{2}{e}^{i\frac{m}{2\hbar }\frac{\left( {\vec{r}_{2}-{\vec{r}_{0} \right)}^{2}{t}_{2}-{t}_{0}=\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{0},{t}_{0} \right) \\ \end{align}

3.2. 得到自由粒子的传播子:

由证明过程可知显然可以推广这个关系:

$[1].\ \mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{2},{t}_{2};{\vec{r}_{1},{t}_{1} \right)\mathcal{K}\left( {\vec{r}_{1},{t}_{1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{1}$$[2].\ \mathcal{K}\left( {\vec{r}_{3},{t}_{3};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{3}{t}_{3};{\vec{r}_{2},{t}_{2} \right)\mathcal{K}\left( {\vec{r}_{2}{t}_{2};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{2}$$\cdot \cdot \cdot$$[\text{N}].\ \mathcal{K}\left( {\vec{r}_{N},{t}_{N};{\vec{r}_{0},{t}_{0} \right)=\int{\mathcal{K}\left( {\vec{r}_{N}{t}_{N};{\vec{r}_{N-1},{t}_{N-1} \right)\mathcal{K}\left( {\vec{r}_{N-1}{t}_{N-1};{\vec{r}_{0},{t}_{0} \right)\text{d}{\vec{r}_{N-1}$将 [1] 代入 [2] 后再将 [2] 代入 [3] 直到第 N 式, 我们将得到下述关系:$\mathcal{K}\left( {\vec{r}_{N},{t}_{N};{\vec{r}_{0},{t}_{0} \right)=\int{\left[ \prod\limits_{n=1}^{N}{\mathcal{K}\left( {\vec{r}_{n},{t}_{n};{\vec{r}_{n-1},{t}_{n-1} \right)} \right]\text{d}{\vec{r}_{1}\cdot \cdot \cdot \text{d}{\vec{r}_{N-1}=K\left( \vec{r},t;{\vec{r}_{0},t \right)$

上述关系说明自由粒子的传播子与其对应的无穷小过程传播子具有相同的形式, 这就可得[7] :

K\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)=\mathcal{K}\left( \vec{r},t;{\vec{r}_{0},{t}_{0} \right)={\left[ \frac{m}{2\pi i\hbar \left( t-{t}_{0} \right)} \right]}^{3/2\ }\exp \left[ i\frac{m{\left( \vec{r}-{\vec{r}_{0} \right)}^{2}{2\hbar \left( t-{t}_{0} \right)} \right].

整个 PT. 1 其实就只是讲了一下两个基本公式, 你以为求解了俩路径积分? 硬要说的话确实也求解了俩自由粒子的, 不过这没有意义啊, 上面这招只对没有势能项的自由粒子起效果, 为什么要写呢, 主要是想通过这俩例子感受一下这个公式的结构组成.

所以其实我们还没有讲如何具体的使用路径积分公式, 而这就是 PT.2 的内容了, 在那里我们会用一个叫对角化方法的工业化流程先(又一次地)解决自由粒子, 再解决一个谐振子.

参考

• ^我多希望我初学的时有这篇文啊.
• ^为啥定义都一样却不用统一的记号呢? 因为后面(指下面几行紧跟着的推导过程)我们将要利用无穷小过程这个特性将无穷小过程的传播子进一步化简, 而这个化简在有限过程一般是不成立的, だからね.
• ^这些问题其实还是比较棘手的, 它们会导致黎曼定义的积分失效, 我们必须采取其它比如说勒贝格积分的定义. 我们的非法操作时常会导致数学定义失效, 不过定义总可以改良, 方法总可以推广, 我们的思想才是不会失效的东西. 所以你问我你这个积分是什么定义的积分啊, 我愿称之为物理人积分. 它的小名叫做"先这么用着吧, 出了问题再去请数学人帮帮忙想个好用点儿的定义积分."
• ^看完这个简洁而清晰的证明过程, 你就会明白为何我们没有跟绝大多数人一样将时间均分, 而采取 t(b) - t(a) 的形式. 我们这个做法至少为我们省掉了两次积分与一个数学归纳法, 且能让过程推演显得十分 naive. 当然你均分时间也不会带来本质上的区别, 也可以用同样的方法证明, 但这样会有很多很暧昧的地方让证明过程变得令读者摸不着头脑.
• ^这里比较尴尬的是 ℏ 的问题, 如果用波矢 k 来代替动量 p 就可以解决这个问题, 但我觉得那样会更加混乱故作罢. 还望理解一下, 物理人积分符号就是会有很多不严谨的地方, 那为什么不会算错呢? 不告诉你哈哈.
• ^下面那个 dy 部分的积分也因为虚系数而不能套高斯积分公式, 虽然就当下的情况而言套了不会错, 但不严谨, 同时也很容易找到一些会出错的例子.
• ^ 不难看出我们无论怎样都逃脱不了那个长得像高斯积分一样复函数的积分, 应用位形空间的情形等于说还碰到了两次, 第一次是从相空间转移到位形空间的时候, 第二次是证明传播子与其对应的无穷小过程的传播子具有相同的形式的时侯. 所以这个积分公式还是蛮重要的.