# 路徑積分表述

## 數學方法

### 哈密頓算符在量子力學里的意義

$U(t_b,t_a)=e^{-\frac{i}{\hbar}(t_b-t_a)H}$

$iG(x_b,t_b;x_a,t_a)\equiv \left\langle x_b \right| U(t_b,t_a) \left| x_a \right\rangle$

$U(t_b,t_a)=U(t_b,t)U(t,t_a)$

$iG(x_b,t_b;x_a,t_a) = \int dx i G(x_b,t_b; x, t) iG(x, t; x_a,t_a)$

### 時間切片

\begin{align} \left\langle x_{j} \left| e^{-i\frac{\Delta}{\hbar} H(\hat{p},\hat{x})} \right| x_{j-1} \right\rangle &= \int d p_{j} \langle x_{j} | p_{j} \rangle \left\langle p_{j} \left| e^{-i\frac{\Delta}{\hbar} H(\hat{p},\hat{x})} \right| x_{j-1} \right\rangle \end{align}

$e^{-i\frac{\Delta}{\hbar} H(\hat{p},\hat{x})} = :e^{-i\frac{\Delta}{\hbar} H(\hat{p},\hat{x})}: + O(\Delta^2)$

\begin{align}\left\langle x_{j} \left| e^{-i\frac{\Delta}{\hbar} H(\hat{p},\hat{x})} \right| x_{j-1} \right\rangle &= \int \frac{d p_{j}}{2\pi\hbar} e^{i \frac{p_j}{\hbar} (x_j-x_{j-1})} \, e^{-i\frac{\Delta}{\hbar} H(p_j,x_{j-1})}\\ &= \int \frac{d p_{j}}{2\pi\hbar} e^{i \frac{\Delta}{\hbar} \left( p_j \frac{x_j-x_{j-1}}{\Delta} - H(p_j,x_{j-1}) \right)} \\ \end{align}

\begin{align} i G(x_b,t_b; x_a, t_a) &= \int dx_1\cdots dx_{n-1} \prod_{i=1}^{n-1} dp_i \exp\left[\frac{i}{\hbar} \sum_{j=1}^{n-1} \Delta\, L \left(t_j, \frac{x_{j}+x_{j-1}}{2},\frac{x_{j}-x_{j-1}}{\Delta} \right) \right] \\ &= \int \mathcal{D}\left[ x(t) \right] e^{\frac{i}{\hbar} S[x(t)]} \end{align}

$S$是路徑$x(t)$的作用量，拉格朗日量$L(t,x,\dot{x})$的時間積分：

$S=\int L(t, x, \dot{x}) dt$

## 简單例子

### 自由粒子

$S = \int \frac{\dot{x}^2}{2} dt$

$G(x-y;T) = \int_{x(0)=x}^{x(T)=y} e^{-\int_0^T \frac{\dot{x}^2}{2} dt} \mathcal{D}x = \int_{x(0)=x}^{x(T)=y} \prod_{t} e^{-\frac{1}{2} \left( \frac{(x(t+\epsilon)-x(t)}{\epsilon} \right)^2 \epsilon} \mathcal{D}x$

$\mathcal{D}x$是以上時間切成有限片的積分。連乘裡每一項都是平均值為$x(t)$方差為c的高斯函數。多重積分是相鄰時間高斯函數$G_\epsilon$的卷積：

$G(x-y;T) = G_\epsilon * G_\epsilon * G_\epsilon \cdots G_\epsilon$

$\tilde{G}(p; T) = \tilde{G}_\epsilon(p)^{T/\epsilon}$

$\tilde{G}_\epsilon(p)=e^{-\epsilon \frac{p^2}{2}}$

$\tilde{G}(p; T) = e^{-T \frac{p^2}{2}}$

$G(x-y; T) \propto e^{-\frac{(x-y)^2}{2T}}$

$\int G(x-y; T) dy = 1$

$\frac{d}{dt} G(x;t) = \frac{\nabla^2}{2} G$

$G(x-y; T) \propto e^{\frac{i(x-y)^2}{2T}}$

$\frac{d}{dt} G(x;t) = \frac{i\nabla^2}{2} G$

$\varphi_t(x) = \int \varphi_0(y) G(x-y; t) dy$

$G$一樣服從薛定諤方程式：

$i\frac{d}{dt}\varphi_t = -\frac{\nabla^2}{2} \varphi_t (x)$