- #1
gonadas91
- 75
- 5
Imagine I want to calculate the INTERACTING single particle Green function in a model:
\begin{eqnarray}
G(t-t')=\langle Td(t)d^{\dagger}(t')\rangle
\end{eqnarray}
We know from Dyson equation that this is:\\
\begin{eqnarray}
G=G_{0} + G_{0}\Sigma G
\end{eqnarray}
Where $\Sigma$ is the self-energy. My question is, if we want to calculate the correlator :\\
\begin{eqnarray}
g(t-t')=\langle Td(t)^{\dagger}d(t')\rangle
\end{eqnarray}
Does the self-energy $\Sigma$ change for this correlator? In my intuition it shouldnt, but maybe I am wrong... Thanks!
\begin{eqnarray}
G(t-t')=\langle Td(t)d^{\dagger}(t')\rangle
\end{eqnarray}
We know from Dyson equation that this is:\\
\begin{eqnarray}
G=G_{0} + G_{0}\Sigma G
\end{eqnarray}
Where $\Sigma$ is the self-energy. My question is, if we want to calculate the correlator :\\
\begin{eqnarray}
g(t-t')=\langle Td(t)^{\dagger}d(t')\rangle
\end{eqnarray}
Does the self-energy $\Sigma$ change for this correlator? In my intuition it shouldnt, but maybe I am wrong... Thanks!