Single-particle Green's function

Question

Define a single-particle Green's function as \begin{equation} i\hbar G(xt;x't') = \langle x| e^{-iH(t-t')/\hbar} | x'\rangle. \end{equation} By inserting the completeness relation, we have \begin{equation} i\hbar G(xt;x't') = \sum_n \langle x|n\rangle \langle n| x'\rangle e^{-iE_n(t-t')/\hbar}, \end{equation} where $|n\rangle$ and $E_n$ are the eigenstates and eigenvalue of the Hamiltonian $H$. By using the Fourier transform, the Green's function in the energy domain can be calculated as \begin{equation} \begin{split} G(x,x';E) &= \int_{-\infty}^{\infty} G(xt;x't') e^{iE(t-t')/\hbar} dt\\ &= \frac{1}{i\hbar}\int_{-\infty}^{\infty} \left\{ \sum_n \langle x|n\rangle \langle n| x'\rangle e^{-iE_n(t-t')/\hbar} \right\} e^{iE(t-t')/\hbar} dt\\ &= \frac{1}{i\hbar}\sum_n \langle x|n\rangle \langle n| x'\rangle \int_{-\infty}^\infty e^{i(E-E_n)(t-t')/\hbar} dt\\ &= -2\pi i\sum_n \langle x|n\rangle \langle n| x'\rangle \delta(E-E_n) \end{split} \end{equation} However, from my knowledge, $G(x,x';E)$ is usually defined as \begin{equation} G(x,x';E) = \sum_n\frac{\langle x|n\rangle\langle n| x'\rangle}{E - E_n}. \end{equation} So my question is how can the above two equations be related? Or is there anything wrong with my derivation?

Answer 1

Both are correct, but are different quantities.

The latter is the (Fourier transform of the) retarded Green's function $G_R$. It is related to your "Green's function" (which really is a kernel) $G$ through

$$ G_R(x,t; x', t') = \frac{1}{i\hbar}\Theta(t-t')G(x,t; x',t') $$

See this wonderful answer about kernels, propagators and Green's functions.