What is the meaning of the Fourier transform of Feynman propagator?

Question

I know $K(a,b,t)$ is the probability amplitude of find a particle that starts at point a in b in a time t later. There is also an expression that sometimes is called green function:
$$G(a,b,E)=(i/\hbar)\int_{-\infty}^\infty\;\exp(iEt/\hbar)\;K(a,b,t)\;dt$$

or Fourier transform of Feynman propagator
See: Grosche. Handbook of Feynman path integrals page 149. Keller. the feynman integral page 461.
http://arxiv.org/abs/cond-mat/0304290v1

I want to know if $G(a,b,E)$ could be the amplitude that a particle of energy $E$ at the initial point $a$ will appear at some (arbitrary) time at $b$. It seems that Martin Schaden and Larry Spruch use this interpretation in http://arxiv.org/abs/cond-mat/0304290v1 but I have not found this in any book of quantum mechanics.

Answer 1

Yes, the Schaden and Spruch interpretation is correct. The interpretation is not used much because it's not as well connected into how experiments are run.

In the usual text books, the Fourier transform is taken over position and time $(\vec{x},t)$ to get energy and momentum $(E,\vec{p})$. This is done by four integrations, one each getting rid of one of the four variables $x_1, x_2, x_3, t$, and replacing with the corresponding one of the four Fourier transformed variables $p_1, p_2, p_3, E$. Mathematically, nothing is wrong with only doing one of the four transforms, and the interpretation is clear.