1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

This appears in propagators, so you can google for any documents or look up any book on propagators. $$\begin{align} \langle x_2|e^{-\frac{ip^2t}{2m}}|x_1\rangle &=\int dp \langle x_2|p\rangle\langle p|e^{-\frac{ip^2t}{2m}}|x_1\rangle\\ &=\int dp \left( \frac{e^{ipx_2/\hbar}}{\sqrt[2]{2\pi \hbar}}\right)\langle p|e^{-\frac{ip^2t}{2m}}|x_1\rangle\\ &=\int dp \left( \frac{e^{ipx_2/\hbar}}{\sqrt[2]{2\pi \hbar}}\right)\left(e^{-\frac{ip^2t}{2m}} \right)\langle p|x_1\rangle\\ &=\int dp \left( \frac{e^{ipx_2/\hbar}}{\sqrt[2]{2\pi \hbar}}\right)\left(e^{-\frac{ip^2t}{2m}} \right)\left( \frac{e^{-ipx_1/\hbar}}{\sqrt[2]{2\pi \hbar}} \right)\\ &=\int dp \left( \frac{e^{ipx/\hbar}}{2\pi \hbar}\right)\left(e^{-\frac{ip^2t}{2m}} \right) \hspace{1.0cm}{(x=x_2-x_1)}\\ &=\int dp \left( \frac{1}{2\pi \hbar}\right)\left(e^{-\frac{it}{2m\hbar}\left( p-\frac{mx}{t}\right)^2+\frac{imx^2}{2\hbar t}} \right) \hspace{1.0cm}{(x=x_2-x_1)}\\ &=\sqrt[2]{\frac{m}{2\pi i \hbar t}}e^{\frac{imx^2}{2\hbar t}} \hspace{1.0cm}{(x=x_2-x_1)}\\ \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.