传播子

演化算符 $\ket{\psi(t)}=\hat{U}(t,t_0)\ket{\psi(t_0)}$ ，薛定谔方程 $i\hbar \partial_t \ket{\psi(t)}=\hat{H}\ket{\psi(t)}$

$\hat{H}\hat{U}=i\hbar\partial_t\hat{U}$，若哈密顿量不显含时间，得$\hat{U}(t,t_0)=\displaystyle \theta(t-t_0)e^{-i\hat{H}(t-t_0)/\hbar}$

$\psi(x,t)=\braket{x}{\psi(t)}=\bra{x}\hat{U}(t,t_0)\ket{\psi(t_0)}=\displaystyle\int_{x_0}\bra{x}\hat{U}(t,t_0)\ket{x_0}\bra{x_0}\ket{\psi(t_0)}\dd x_0$

传播子 $\psi(x,t)=\displaystyle\int_{x_0}U(x,t;x_0,t_0)\psi(x_0,t_0)\dd x_0$

$U(x,t;x_0,t_0)=\bra{x}\hat{U}(t,t_0)\ket{x_0}=\displaystyle\sum_n \bra{x}\hat{U}(t,t_0)\ket{n}\bra{n}\ket{x_0}=\sum_n \bra{x}e^{-i\hat{H}(t-t_0)/\hbar}\ket{n}\bra{n}\ket{x_0}$

$U(x,t;x_0,t_0)=\displaystyle\sum_n e^{-iE_n(t-t_0)/\hbar}\psi_n(x)\psi_n^*(x_0)$

路径积分

$\hat{U}(t,t_0)=\hat{U}(t,t_{n-1})\hat{U}(t_{n-1},t_{n-2})\cdots \hat{U}(t_{1},t_{0})$

$$\begin{aligned}
U(x_{i+1},t_{i+1};x_i,t_i)&=\bra{x_{i+1}}e^{-i\hat{H}{\Delta t}/\hbar}\ket{x_i}\\
\qquad\\
&=\int \dd p_i \bra{x_{i+1}}\ket{p_i}\bra{p_i}e^{-i\hat{H}{\Delta t}/\hbar}\ket{x_i}\\
\qquad\\
&=\int \frac{\dd p_i}{2\pi \hbar}e^{ip_i\Delta x/\hbar}e^{-iH(x_i,p_i){\Delta t}/\hbar}\\
\qquad\\
&\approx\int \frac{\dd p_i}{2\pi \hbar}e^{iL(x_i,p_i)\Delta t/\hbar}
\end{aligned}$$

$$\begin{aligned}U(x,t;x_0,t_0)&\approx \int \prod _{i=0}^{n-1}\frac{\dd p_i}{2\pi \hbar}\prod _{i=0}^{n-1} \dd x_i\exp\left[\sum_{\Delta t}\frac{iL(x_i,p_i)\Delta t}{\hbar}\right]\\
\qquad\\
&\approx \int \prod _{i=0}^{n-1}\frac{\dd p_i}{2\pi \hbar}\prod _{i=0}^{n-1} \dd x_i\exp\left[\frac{i}{\hbar}\sum_{\Delta t} \left(p_i\dot{x}_i-\frac{p_i^2}{2m}-V(x_i)\right)\Delta t\right]\\
\qquad\\
& = \left(\frac{m}{2\pi\hbar i\Delta t}\right)^{n/2}\int \prod _{i=0}^{n-1} \dd x_i\exp\left[\frac{i}{\hbar}\int_{t_0}^t L(x(t’),p(t’))\dd t’\right]\\
\qquad\\
&= \int \mathcal{D}x(t’)\exp\left[\frac{i}{\hbar}S(x,t;x_0,t_0)\right]
\end{aligned}
$$

含时哈密顿量

哈密顿量显含时间时，对 $\hat{H}\hat{U}=i\hbar\partial_t\hat{U}$ 积分得

$U(t,t_0)=1-\displaystyle\frac{i}{\hbar}\int_{t_0}^t H(t’)U(t’,t_0)\dd t’$，迭代逐段积分

$$\begin{aligned}
U(t,t_0)&=1-\displaystyle\frac{i}{\hbar}\int_{t_0}^t H(t_1)\left[1-\displaystyle\frac{i}{\hbar}\int_{t_0}^{t_1} H(t_2)U(t_2,t_0)\dd t_2\right]\dd t_1=\cdots\\
\qquad\\
&=1+\sum_{n=1}^\infty\left(-\frac{i}{\hbar}\right)^n\int_{t_0}^{t}\dd t_1\int_{t_0}^{t_1}\dd t_2\cdots\int_{t_0}^{t_{n-1}}\dd t_n H(t_1)H(t_2)\cdots H(t_n)
\end{aligned}$$