Convert time operator from momentum space to position space I'm trying to transform the time evolution operator from momentum space to position space. I know that 
$$ U(t) = e^{-iHt/h} = \int_{-\infty}^\infty e^{-ip^2t/2uh} 
| p \rangle \langle p | dp
$$
and I'm trying to find the form of 
$$ \langle x | U(t) | x' \rangle $$
I'm given the hint (paraphrased):

To evaluate this explicitly, use the Fourier transform of a Gaussian
  function, with imaginary $a$

I'm trying to apply the time operator to a momentum space wave function:
$$ U(t)|\psi (p,0)  \rangle  = \int_{-\infty}^\infty e^{-ip^2t/2uh} \psi (p,0) 
| p \rangle  dp
$$
But I'm not sure how to simplify to a form where a fourier transform would be straightforward 
 A: This appears in propagators, so you can google for any documents or look up any book on propagators.
$$\begin{align}
\langle x_2|e^{-\frac{ip^2t}{2m}}|x_1\rangle &=\int dp \langle x_2|p\rangle\langle p|e^{-\frac{ip^2t}{2m}}|x_1\rangle\\
&=\int dp \left( \frac{e^{ipx_2/\hbar}}{\sqrt[2]{2\pi \hbar}}\right)\langle p|e^{-\frac{ip^2t}{2m}}|x_1\rangle\\
&=\int dp \left( \frac{e^{ipx_2/\hbar}}{\sqrt[2]{2\pi \hbar}}\right)\left(e^{-\frac{ip^2t}{2m}} \right)\langle p|x_1\rangle\\
&=\int dp \left( \frac{e^{ipx_2/\hbar}}{\sqrt[2]{2\pi \hbar}}\right)\left(e^{-\frac{ip^2t}{2m}} \right)\left( \frac{e^{-ipx_1/\hbar}}{\sqrt[2]{2\pi \hbar}} \right)\\
&=\int dp \left( \frac{e^{ipx/\hbar}}{2\pi \hbar}\right)\left(e^{-\frac{ip^2t}{2m}} \right) \hspace{1.0cm}{(x=x_2-x_1)}\\
&=\int dp \left( \frac{1}{2\pi \hbar}\right)\left(e^{-\frac{it}{2m\hbar}\left( p-\frac{mx}{t}\right)^2+\frac{imx^2}{2\hbar t}} \right) \hspace{1.0cm}{(x=x_2-x_1)}\\
&=\sqrt[2]{\frac{m}{2\pi i \hbar t}}e^{\frac{imx^2}{2\hbar t}} \hspace{1.0cm}{(x=x_2-x_1)}\\
\end{align}
$$
