# 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

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\pi \hbar}}\right)\langle p|e^{-\frac{ip^2t}{2m}}|x_1\rangle\\ &=\int dp \left( \frac{e^{ipx_2/\hbar}}{\sqrt{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\pi \hbar}}\right)\left(e^{-\frac{ip^2t}{2m}} \right)\left( \frac{e^{-ipx_1/\hbar}}{\sqrt{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{\frac{m}{2\pi i \hbar t}}e^{\frac{imx^2}{2\hbar t}} \hspace{1.0cm}{(x=x_2-x_1)}\\ \end{align}