测度原理下自由粒子的路径积分研究

Path Integral Study of Free Particles under the Principle of Measurement

摘要: 本文研究的是量子力学中关于自由粒子运动的路径积分问题，通过将测度理论中的方法引入对自由粒子运动路径的描述，从而构建不同测度意义下路径积分量之间的关系。

Abstract: In this paper, the path integral of free particle motion in quantum mechanics is studied. By introducing the method of measure theory into the description of free particle motion path, therefore, the relationship between the path product components in different measurement sense is constructed.

1. 引言

物理为数学的研究提供背景，数学为物理提供研究工具。本文正是基于这样的基本思想，通过建立测度论与路径积分的联系，得到路径积分的物理量间的新的关系。

2. 考虑有限的平面路径积分情形

考虑x-t的平面路径积分图(如下)，将其看作为平面矩形图( [1] )。于是坐标 $\left(a,b\right)$ 满足：

I区域： $\{\begin{array}{c}{t}_1\le a\le t_2\\ t_1\le a<t_2\\ t_1<a\le t_2\\ t_1<a<t_2\end{array}$， $\{\begin{array}{c}{x}_1\le b\le x_2\\ x_1\le b<x_2\\ x_1<b\le x_2\\ x_1<b<x_2\end{array}$

II区域： $\{\begin{array}{c}{t}_2\le a\le t_3\\ t_2\le a<t_3\\ t_2<a\le t_3\\ t_2<a<t_3\end{array}$， $\{\begin{array}{c}{x}_2\le b\le x_3\\ x_2\le b<x_3\\ x_2<b\le x_3\\ x_2<b<x_3\end{array}$

Figure 1. Simple plane path integral diagram

图1. 简单平面路径积分图

对于图1，由于区域 $I+II$ 是矩形平面初等集，其测度为：

$m\left(I\right)=\left(x_2-x_1\right)\left(t_2-t_1\right)$

$m\left(II\right)=\left(x_3-x_2\right)\left(t_3-t_2\right)$

这里m是满足 $m:R\times R\to R$ 的Lebesgue测度，所以该平面是坐标 $\left(a,b\right)\in {\vartheta }_m$ 的集族 ${\vartheta }_m$。由于该平面内的点属于 ${\vartheta }_m$，并且满足条件：

1) $I+II=I\cup II$，且 $I\cap II=\varnothing$ ；

2) $I+II$ 的测度为 $m\left(I+II\right)=\left(x_3-x_1\right)\left(t_3-t_1\right)=m\left(I\right)+m\left(II\right)=\left(x_2-x_1\right)\left(t_2-t_1\right)+\left(x_3-x_2\right)\left(t_3-t_2\right)$

因此该平面矩形的测度m满足 $\sigma$ 可加性。

现在考虑一种简单的自由粒子对 $U\left(t\right)$ 的近似情况：

由路径积分( [2] )中的

$U\left(t\right)U\left(x,t;x^{\prime },t^{\prime }\right)=A{\displaystyle \underset{\text{allpath}}{\sum }\exp \left\{is\left[x\left(t\right)\right]/\hslash \right\}}$

计算

$U\left(x,t;x^{\prime },t^{\prime }\right)=A^{\prime }\exp \left\{i{s}_{cl}/\hslash \right\}$

这里假设路径为贯穿两个区域的一条直线，所以有

$\frac{x_{cl}\left(t_3\right)-x_2}{t_3-t_2}=\frac{x_1-x_2}{t_1-t_2}$

由前面的测度理论可得

$m\left(I\right)=\left(x_1-x_2\right)\left(t_1-t_2\right)$

$m\left(II\right)=\left(x_{cl}\left(t_3\right)-x_2\right)\left(t_3-t_2\right)$

$\Rightarrow$ $\frac{m\left(I\right)}{{\left(t_1-t_2\right)}^2}=\frac{m\left(II\right)}{{\left(t_3-t_2\right)}^2}$

$\Rightarrow$ ${\left(t_3-t_2\right)}^{2}m\left(I\right)={\left(t_1-t_2\right)}^{2}m(\; I\; I\; )$

速度为

$v=\frac{x_1-x_2}{t_1-t_2}=\frac{m\left(I\right)}{{\left(t_1-t_2\right)}^2}=\frac{m\left(II\right)}{{\left(t_3-t_2\right)}^2}$

$\Rightarrow$ $S_{cl}={\displaystyle {\int }_{t_2}^{t_1}L\text{d}{t}_3}=\frac{1}{2}{\displaystyle {\int }_{t_2}^{t_1}m{v}^2\text{d}{t}_3}={m{t}_3\frac{m{\left(I\right)}^2}{2{\left(t_1-t_2\right)}^4}|}_{t_2}^{t_1}=\frac{1}{2}m\frac{m{\left(I\right)}^2}{{\left(t_1-t_2\right)}^3}$

其中 $L=\frac{1}{2}m{v}^2$。将 $x^{\prime },t^{\prime }$ 替换为 $x_2,t_2$ ； $x,t$ 替换为 $x_1,t_1$，所以

$U\left(x_1,t_1;x_2,t_2\right)=A^{\prime }\exp \left[\frac{imm{\left(I\right)}^2}{2{\left(t_1-t_2\right)}^3}/\hslash \right]$

根据delta函数的性质，可得

$U\left(x_1,t_1;x_2,0\right)=U\left(x_1,t_1;x_2\right)={\left(\frac{m}{2\pi \hslash it}\right)}^{1/2}\exp \left[\frac{imm{\left(I\right)}^2}{2t_1\hslash }\right]$.

3. 趋近无穷时的平面路径积分情形

下面研究自由粒子的路径趋近 $\infty$ 时的情形(如下图2)：

根据前面的已知条件， ${\vartheta }_m$ 是初等集集族，测度m具有 $\sigma$ 可加性。考虑图2的情形，不妨令路径间的间距相等，得到的路径积分( [3]、 [4] )为

Figure 2. Multi-planar path integral diagram

图2. 多重平面路径积分图

$U\left(x_N,t_N;x_0,t_0\right)={\displaystyle {\int }_{x_0}^{x_N}\exp \left\{is\left[x\left(t\right)\right]/\hslash \right\}D\left[x\left(t\right)\right]}$

这里 $S={\displaystyle {\int }_{t_0}^{t_N}\frac{1}{2}m{v}^2\text{d}t}$， $v=\frac{x_1-x_0}{t_1-t_0}+\frac{x_2-x_1}{t_2-t_1}+\cdots +\frac{x_N-x_{N-1}}{t_N-t_{N-1}}={\displaystyle \underset{i=0}{\overset{N-1}{\sum }}\frac{x_{i+1}-x_i}{\frac{t_N-t_0}{N}}}$。令 $t_N=t_0+N\theta$，因为

$m\left(S\right)={\displaystyle \underset{i=0}{\overset{N-1}{\sum }}\frac{\left(x_{i+1}-x_i\right)\left(t_N-t_0\right)}{N}}$

将 $v=\frac{m\left(S\right)}{{\theta }^2}$ 代入，就有

$S={\frac{1}{2}m\frac{m{\left(S\right)}^2}{{\theta }^4}t|}_{t_0}^{t_N}\Rightarrow S=\frac{mm{\left(S\right)}^{2}N}{2{\theta }^3}$

代入，得到

$U\left(x_N,t_N;x_0,t_0\right)={\displaystyle {\int }_{x_0}^{x_N}\exp \left\{\frac{imm{\left(S\right)}^{2}N}{2{\theta }^3\hslash }\right\}D\left[x\left(t\right)\right]}$

可以写成

$U\left(x_N,t_N;x_0,t_0\right)=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }A{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\cdots {\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\exp \left[\frac{im{\left(S\right)}^{2}Nm}{2{\theta }^3\hslash }\right]}}}\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}$

若只考虑坐标点之间的关系，可得

$S={\displaystyle \underset{i=0}{\overset{N-1}{\sum }}\frac{m}{2}{\left(\frac{x_{i+1}-x_i}{\theta }\right)}^2\theta }$

$\Rightarrow$ $U\left(x_N,t_N;x_0,t_0\right)=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }A\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{im}{2\hslash }{\displaystyle \underset{i=0}{\overset{N-1}{\sum }}\frac{{\left(x_{i+1}-x_i\right)}^2}{\theta }}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}$

$\Rightarrow$ $U=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }A\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left\{i{\displaystyle \underset{i=0}{\overset{N-1}{\sum }}\left[{\left(\frac{m}{2\hslash \theta }\right)}^{1/2}{x}_{i+1}-{\left(\frac{m}{2\hslash \theta }\right)}^{1/2}{x}_i\right]}\right\}$ (1)

其中令 $y_i={\left(\frac{m}{2\hslash \theta }\right)}^{1/2}{x}_i$

$\Rightarrow$ $U^{\prime }=\underset{N\to \infty }{\lim }{A}^{\prime }\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[i{\displaystyle \underset{i=0}{\overset{N-1}{\sum }}{\left(y_{i+1}-y_i\right)}^2}\right]\text{d}{y}_1\cdots \text{d}{y}_{N-1}$ (2)

将(1)式与(2)式进行比较，得到

$A^{\prime }=A{\left(\frac{2\hslash \theta }{m}\right)}^{\left(N-1\right)/2}$

当 $t=2$ 时，得到

${\displaystyle {\int }_{-\infty }^{\infty }\exp \left\{i\left[{\left(y_2-y_1\right)}^2+{\left(y_1-y_0\right)}^2\right]\right\}\text{d}{y}_1}={\left(\frac{i\pi }{2}\right)}^{1/2}{\text{e}}^{-{\left(y_2-y_0\right)}^2/2i}$

当 $t=3$ 时，得到

${\left(\frac{i\pi }{2}\right)}^{1/2}{\displaystyle {\int }_{-\infty }^{\infty }{\text{e}}^{-{\left(y_3-y_2\right)}^2/i}{\text{e}}^{-{\left(y_2-y_0\right)}^2/2i}\text{d}{y}_2}=\frac{i\pi }{3}{\text{e}}^{-{\left(y_3-y_0\right)}^2/3i}$

当 $t=N$ 时，得到

${\left(\frac{i\pi }{2}\right)}^{\left(N-2\right)/2}{\displaystyle {\int }_{-\infty }^{\infty }{\text{e}}^{-{\left(y_N-y_{N-1}\right)}^2/i}{\text{e}}^{-{\left(y_{N-1}-y_0\right)}^2/\left(N-1\right)i}\text{d}{y}_{N-1}=\frac{{\left(i\pi \right)}^{\left(N-1\right)/2}}{N^{1/2}}{\text{e}}^{-{\left(y_N-y_0\right)}^2/Ni}}$

所以由

$\{\begin{array}{l}\frac{{\left(i\pi \right)}^{\left(N-1\right)/2}}{N^{1/2}}{\text{e}}^{-{\left(y_N-y_0\right)}^2/Ni}\\ y_N={\left(\frac{m}{2\hslash \theta }\right)}^{1/2}{x}_N,y_0={\left(\frac{m}{2\hslash \theta }\right)}^{1/2}{x}_0\end{array}$

$\Rightarrow$ $\frac{{\left(i\pi \right)}^{\left(N-1\right)/2}}{N^{1/2}}{\text{e}}^{-\left[{\left(\frac{m}{2\hslash \theta }\right)}^{1/2}{x}_N-{\left(\frac{m}{2\hslash \theta }\right)}^{1/2}{x}_0\right]/Ni}=\frac{{\left(i\pi \right)}^{\left(N-1\right)/2}}{N^{1/2}}{\text{e}}^{-m{\left(x_N-x_0\right)}^2/2\hslash \theta Ni}$

所以

$U=A{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{N/2}{\left(\frac{m}{2\pi \hslash \theta iN}\right)}^{1/2}\exp \left[\frac{im{\left(x_N-x_0\right)}^2}{2\hslash N\theta }\right]$

前面已令 $N\theta =t_N-t_0$，所以知 $A={\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}$，所以

$U\left(x_N,t_N;x_0,t_0\right)=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{imm{\left(s\right)}^{2}N}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}$

其中 $S=S_1\cup S_2\cup \cdots \cup S_N$。

取矩形 $A\subset S$，已知A的外测度是 ${\mu }^{\ast }\left(A\right)=\underset{A\subset S}{\inf }{\displaystyle \underset{N}{\sum }m\left(S_N\right)}=\underset{A\subset S}{\inf }m\left(S\right)$。则任何与A相邻的矩形B，使得 $A\cap B$ 有图3所示的几种情形：

Figure 3. Several adjacent rectangular graphs

图3. 几种相邻矩形图

所以 ${\mu }^{\ast }\left(A△B\right)={\mu }^{\ast }\left(A\cup B-\left(t_i,x\left(t_i\right)\right)\right)=\underset{A\cup B-\left(t_i,x\left(t_i\right)\right)\subset A\cup B}{\inf }m\left(A\cup B\right)<m\left(A\cup B\right)$，这里

${\mu }^{\ast }\left(A△B\right)={\mu }^{\ast }\left(A\cup B-\left\{\left(x,y\right)|x=t_i,x\left(t_{i-1}\right)\le y\le x\left(t_i\right)\right\}\right)$

因为

$m\left(A\cup B\right)=m\left(A\right)+m\left(B\right)={\displaystyle \sum \frac{\left(x\left(t_i\right)-x\left(t_{i-p}\right)\right)\left(t_i-t_{i-p}\right)}{p}}+{\displaystyle \sum \frac{\left(x\left(t_{i+q}\right)-x\left(t_i\right)\right)\left(t_{i+q}-t_i\right)}{q}<\epsilon }$

所以 ${\mu }^{\ast }\left(A△B\right)<\epsilon$，同时 ${\mu }^{\ast }\left(A△B\right)\le {\mu }^{\ast }\left(A\right)+{\mu }^{\ast }\left(B\right)=2\underset{A,B\subset S}{\inf }m\left(S\right)<\epsilon$。所以有路径积分

$U^{\ast }\left(x_N,t_N;x_0,t_0\right)=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\exp \left[\frac{im{\mu }^{\ast }{\left(A△B\right)}^{2}N}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}$

$\begin{array}{l}{U}^{\ast }\left(x_N,t_N;x_0,t_0\right)\\ \le \underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{im\left({\mu }^{\ast }{\left(A\right)}^2+2{\mu }^{\ast }\left(A\right){\mu }^{\ast }\left(B\right)+{\mu }^{\ast }{\left(B\right)}^2\right)N}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}\\ \le \underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{im\left(2{\mu }^{\ast }{\left(A\right)}^2+2{\mu }^{\ast }{\left(B\right)}^2\right)N}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}\\ =\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{2im{\left(\underset{A,B\subset S}{\inf }m\left(S\right)\right)}^{2}N}{\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}\\ <\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{imN{\epsilon }^2}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}\end{array}$

令

$U_L\left(x_N,t_N;x_0,t_0\right)=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{im\mu {\left(A\right)}^{2}N}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}$

为Lebesgue意义( [1]、 [5] )下的路径积分。再令

$U_{\epsilon }\left(x_N,t_N;x_0,t_0\right)=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{im{\epsilon }^{2}N}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}$

$U_A\left(x_N,t_N;x_0,t_0\right)=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/4}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{im{\mu }^{\ast }{\left(A\right)}^{2}N}{4\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}$

$U_B\left(x_N,t_N;x_0,t_0\right)=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/4}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{im{\mu }^{\ast }{\left(B\right)}^{2}N}{4\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}$

就有 $U_A^2{U}_B^2<U_{\epsilon }$。

考虑矩形 $A,B,C$ 形成具有m测度的环 ${\Re }_m$，且 $C\subset B\subset A$，可知

$A-C=S_1\cup S_2-\left(t_a,x\left(t_a\right)\right)-\left(t_{a+d},x\left(t_{a+d}\right)\right)\Rightarrow A-C\subset S_1\cup S_2$

所以

$m\left(A-C\right)\le {\displaystyle \underset{i=1,2}{\sum }m\left(S_i\right)}={\displaystyle \underset{p=0}{\overset{j-1}{\sum }}\frac{\left(x\left(t_{a-p}\right)-x\left(t_{a-p-1}\right)\right)\left(t_a-t_{a-j}\right)}{j}+{\displaystyle \underset{q=0}{\overset{f-d-1}{\sum }}\frac{\left(x\left(t_{a+d+q+1}\right)-x\left(t_{a+d+q}\right)\right)\left(t_{a+d+q+1}-t_{a+d+q}\right)}{f-d}}}$

由于

$A-C=\{\begin{array}{l}{S}_1\cup S_2-\left(t_a,x\left(t_a\right)\right)-\left(t_{a+d},x\left(t_{a+d}\right)\right)\\ S_1\cup S_2-\left\{\left(x,y\right)|x=t_a,x\left(t_{a-1}\right)\le y\le x\left(t_a\right)\right\}-\left\{\left(x,y\right)|x=t_{a+d},x\left(t_{a+d}\right)\le y\le x\left(t_{a+d+1}\right)\right\}\end{array}$,

在此条件下令

$\xi ={\displaystyle \underset{p=0}{\overset{j-1}{\sum }}\frac{\left(x\left(t_{a-p}\right)-x\left(t_{a-p-1}\right)\right)\left(t_a-t_{a-j}\right)}{j}+{\displaystyle \underset{q=0}{\overset{f-d-1}{\sum }}\frac{\left(x\left(t_{a+d+q+1}\right)-x\left(t_{a+d+q}\right)\right)\left(t_{a+d+q+1}-t_{a+d+q}\right)}{f-d}}}$

所以 $m\left(A-C\right)<\xi \Rightarrow B$ 是Jordan集。得到Jordan意义( [1]、 [5] )下的路径积分

$U_J\left(x_N,t_N;x_0,t_0\right)=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{im\mu {\left(B\right)}^{2}N}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}$.

下面考虑由图4给出的两种情形：

测度m给定在X里的集半环 ${\vartheta }_m$， $P=P_1\cup P_2$， $P\subset X$，在图(I)情形下， ${\mu }^{\ast }\left(P\right)={\mu }^{\ast }\left(P\cap P_1\right)+{\mu }^{\ast }\left(P-P_1\right)$，由初等集的性质就得到

$m\left(P\right)=m\left(P\cap P_1\right)+m\left(P-P_1\right)\le m\left(P_1\right)+m(\; P\; 2\; )$

因为

Figure 4. Arbitrary path integral diagram

图4. 任意路径积分图

$m\left(P_1\right)={\displaystyle \underset{k=1}{\overset{a}{\sum }}\frac{\left(t_{i+k}-t_{i+k-1}\right)\left(x\left(t_{i+k}\right)-x\left(t_{i+k-1}\right)\right)}{a}}$

$m\left(P_2\right)={\displaystyle \underset{k=1}{\overset{b}{\sum }}\frac{\left(t_{i-k+1}-t_{i-k}\right)\left(x\left(t_{i-k+1}\right)-x\left(t_{i-k}\right)\right)}{b}}$

所以就有

$m\left(P\right)\le {\displaystyle \underset{k=1}{\overset{a,b}{\sum }}\left[\frac{\left(t_{i+k}-t_{i+k-1}\right)\left(x\left(t_{i+k}\right)-x\left(t_{i+k-1}\right)\right)}{a}+\frac{\left(t_{i-k+1}-t_{i-k}\right)\left(x\left(t_{i-k+1}\right)-x\left(t_{i-k}\right)\right)}{b}\right]}$

所以在P区域的自由粒子的路径积分为：

$U_P=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{imm{\left(P\right)}^{2}N}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}$

并且满足

$U_P\le \underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{im{\left(m\left(P_1\right)+m\left(P_2\right)\right)}^{2}N}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}$ (3)

$(3)\ge \underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}{\displaystyle \underset{\begin{array}{l}-\infty \\ \left(N-1\right)\end{array}}{\overset{\infty }{\int }}\exp \left[\frac{2imm\left(P_1\right)m\left(P_2\right)N}{\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}}$

当“=”成立时， $m\left(P_2\right)=m\left(P_1\right)=\frac{1}{2}m\left(P\right)$，所以在此条件下有Caratheodory意义( [1]、 [5] )下可测的路径积分，并且 $A^3{U}_P=U_{P_1}^2{U}_{P_2}^2$。

在图(II)的情形下：

$P=P_1\cup P_2\cup P_3,P\subset X,{\mu }^{\ast }\left(P\right)={\mu }^{\ast }\left(P\cap P_1\right)+{\mu }^{\ast }\left(P-P_1\right)$

$\Rightarrow$ $m\left(P\right)=m\left(P\cap P_1\right)+m\left(P-P_1\right)\le m\left(P_1\right)+m\left(P_2\right)+m(\; P\; 3\; )$

因为

$m\left(P_1\right)={\displaystyle \underset{s=1}{\overset{L}{\sum }}\frac{\left(t_{i+s}-t_{i+s-1}\right)\left(x\left(t_{i+s}\right)-x\left(t_{i+s-1}\right)\right)}{L}}$

$m\left(P_2\right)={\displaystyle \underset{s=1}{\overset{M-L}{\sum }}\frac{\left(t_{i+L+s}-t_{i+L+s-1}\right)\left(x\left(t_{i+L+s}\right)-x\left(t_{i+L+s-1}\right)\right)}{M-L}}$

$m\left(P_3\right)={\displaystyle \underset{s=1}{\overset{K}{\sum }}\frac{\left(t_{i-s+1}-t_{i-s}\right)\left(x\left(t_{i-s+1}\right)-x\left(t_{i-s}\right)\right)}{K}}$

所以

$\begin{array}{c}m\left(P\right)\le {\displaystyle \underset{s=1}{\overset{L}{\sum }}\frac{\left(t_{i+s}-t_{i+s-1}\right)\left(x\left(t_{i+s}\right)-x\left(t_{i+s-1}\right)\right)}{L}}\\ +{\displaystyle \underset{s=1}{\overset{M-L}{\sum }}\frac{\left(t_{i+L+s}-t_{i+L+s-1}\right)\left(x\left(t_{i+L+s}\right)-x\left(t_{i+L+s-1}\right)\right)}{M-L}}\\ +{\displaystyle \underset{s=1}{\overset{K}{\sum }}\frac{\left(t_{i-s+1}-t_{i-s}\right)\left(x\left(t_{i-s+1}\right)-x\left(t_{i-s}\right)\right)}{K}}\end{array}$

即

$\begin{array}{c}{U}_P=\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{imm{\left(P\right)}^{2}N}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}\\ \le \underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{im{\left(m\left(P_1\right)+m\left(P_2\right)+m\left(P_3\right)\right)}^{2}N}{2\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}\end{array}$ (4)

因为

$\begin{array}{c}(4)\ge \underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}{\displaystyle \underset{\begin{array}{l}-\infty \\ \left(N-1\right)\end{array}}{\overset{\infty }{\int }}\exp \left[\frac{4imm\left(P_1\right)\sqrt{m\left(P_2\right)m\left(P_3\right)}N}{\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}}\\ =\underset{\begin{array}{l}N\to \infty \\ \theta \to 0\end{array}}{\lim }{\left(\frac{2\pi \hslash \theta i}{m}\right)}^{-N/2}\underset{\left(N-1\right)}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}}}\exp \left[\frac{2imm{\left(P_1\right)}^{2}N}{\hslash {\theta }^3}\right]\times \text{d}{x}_1\cdots \text{d}{x}_{N-1}\end{array}$

当且仅当 $m\left(P_1\right)=2\sqrt{m\left(P_2\right)m\left(P_3\right)}$，即 $m\left(P_2\right)=m\left(P_3\right)$ 时，“=”成立。所以就有

$\{\begin{array}{l}m\left(P_1\right)=2m\left(P_2\right)=2m\left(P_3\right)\\ m\left(P\right)=2m(\; P\; 1\; )\end{array}$

在此条件下有Caratheodory意义( [1]、 [5] )下可测的路径积分，且 $A^7{U}_P=U_{P_1}^3{U}_{P_2}^2{U}_{P_3}^2$。

4. 总结

通过以上的论述和计算得到了在Hilbert紧空间变换条件下的两个从不同方向趋近的能级表达式，所得到的两个能级表达式体现为在谐波振子状态下的距离不对称性。然而相关的深入研究以及应用问题还需要进一步地加深探索。