$$\begin{align} U(t) =& \langle \vec{x} | e^{-\mathrm{i} H t} | \vec{x}_0\rangle \\ =& \int \frac{\mathrm{d}^3\vec{p}}{(2\pi)^3} \langle \vec{x}| e^{-\mathrm{i} \frac{p^2}{2m}t}|\vec{p}\rangle \langle \vec{p} | \vec{x}_0\rangle \\ =& \int \frac{\mathrm{d}^3\vec{p}}{(2\pi)^3} e^{-\mathrm{i} \frac{p^2}{2m}t} e^{\mathrm{i}\vec{p}\cdot(\vec{x} - \vec{x}_0)} \\ =& \left( \frac{m}{2\pi \mathrm{i}t} \right)^{3/2} e^{\mathrm{i}m\frac{(\vec{x} - \vec{x}_0)^2}{2t}} \end{align}$$

最后一个等号的过程如下. 记 $|\vec{x} - \vec{x}_0| = \Delta x$

$$\begin{align} U(t) =& \frac{1}{(2\pi)^2} \int_0^{\infty}\mathrm{d}p\cdot p^2\int_{-1}^1\mathrm{d}\cos\theta\cdot e^{-\mathrm{i} \frac{p^2}{2m}t}e^{\mathrm{i} p\Delta x \cos\theta} \\ =& \frac{1}{(2\pi)^2} \int_0^{\infty}\mathrm{d}p\cdot p e^{-\mathrm{i} \frac{p^2}{2m}t} \frac{1}{\mathrm{i}\Delta x} [e^{\mathrm{i} p\Delta x} - e^{-\mathrm{i} p\Delta x}]\\ =& \frac{1}{(2\pi)^2} \int_{-\infty}^{\infty}\mathrm{d}p\cdot p e^{-\mathrm{i} \frac{p^2}{2m}t} \frac{1}{\mathrm{i}\Delta x} e^{\mathrm{i} p\Delta x} \\ =& \frac{1}{(2\pi)^2} \int_{-\infty}^{\infty}\mathrm{d}p e^{-\mathrm{i} \frac{p^2}{2m}t} \frac{p}{\mathrm{i}\Delta x}\cdot \frac{1}{\mathrm{i}p} \frac{\partial}{\partial\Delta x} e^{\mathrm{i} p\Delta x} \\ =& \frac{1}{(2\pi)^2} \frac{-1}{\mathrm{i}\Delta x}\frac{\partial}{\partial\Delta x} \int_{-\infty}^{\infty}\mathrm{d}p e^{-\mathrm{i} \frac{p^2}{2m}t} e^{\mathrm{i} p\Delta x} \\ =& \frac{1}{(2\pi)^2} \frac{-1}{\mathrm{i}\Delta x}\frac{\partial}{\partial\Delta x} \sqrt{\frac{2\pi m}{\mathrm{i}t}} e^{-\frac{\Delta x^{2} m}{2it}}\\ =& \left( \frac{m}{2\pi \mathrm{i}t} \right)^{3/2} e^{\mathrm{i}m\frac{\Delta x^2}{2t}} \end{align}$$

注

记住这个还不是错的, 不用老配方了...

$$\begin{align} \int_{-\infty}^{\infty} e^{-a x^2 + bx} = \sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}} \end{align}$$

Reference