Free particle propagator in energy space

Question

I want to calculate the Fourier transform of the Feynman propagator for a free particle $$G(q_f,q_i;E) = \left ( -\frac{i}{\hbar} \right ) \int_{0}^{\infty} K(q_f t\mid q_i 0)e^{\frac{iEt}{\hbar}} dt \tag{1} $$

to do that I`m recall that the propagator of a free particle looks like:

$$K(q_f t\mid q_i 0) = \left ( \frac{m}{2\pi i\hbar t} \right )^{1/2}\exp \left ( \frac{im(q_f- q_i)^2}{2\hbar t} \right ) .\tag{2} $$

This expression confused me a lot, because here I have $1/t$ in the exponent. Any tips on how I could do this?

Answer 1

If we insert OP's formula (2) into OP's integral (1), and Wick-rotate $\tau=it$, then OP's integral (1) becomes (up to a sign)

$$\begin{align} &\frac{1}{\hbar}\int_{\mathbb{R}_+}\!d\tau \sqrt{\frac{m}{2\pi\hbar\tau}} \exp\left(-\frac{m(\Delta q)^2}{2\hbar\tau}+\frac{E\tau}{\hbar} \right) \cr ~=~& \frac{1}{\hbar}\sqrt{-\frac{m}{2E}} \exp\left(-\sqrt{-2mE}\frac{|\Delta q|}{\hbar} \right), \qquad E~<~0.\end{align}\tag{A}$$

Eq. (A) matches the Greens function

$$G(\Delta q;E)~=~\frac{m}{\hbar^2\kappa} \exp\left(- \kappa|\Delta q|\right), \qquad \kappa~:=~\frac{\sqrt{-2mE}}{\hbar},\tag{B}$$

for

$$ \left(-\frac{\hbar^2}{2m}\frac{\partial}{\partial q_f^2}-E\right)G(\Delta q;E)~=~\delta(\Delta q), \qquad \Delta q~:=~q_f-q_i.\tag{C}$$