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 Aug 21 awarded Popular Question Jun 8 awarded Popular Question Feb 3 awarded Popular Question Jan 15 awarded Favorite Question Jul 2 awarded Curious May 30 awarded Nice Question May 16 awarded Popular Question Feb 18 awarded Notable Question Jan 9 awarded Popular Question Dec 17 awarded Yearling Nov 19 awarded Nice Question Sep 2 awarded Popular Question Dec 17 awarded Yearling Jul 19 accepted What's the definition of the time ordering operator for more than two particles? Jul 15 asked What's the definition of the time ordering operator for more than two particles? Jun 5 comment Questions about the Dyson equation @marek The $\tau$ is imaginary time within the finite temperature/equilibrium framework. Which Fourier transformation do you mean to perform first? Can you in general not do partial summations to infinite order then? (Eg in the Hartree-Fock approximation, you're left with two diagrams in the self energy: is it usually only possible to calculate $G$ using the integral equation to as many orders as you can calculate? Or is there something I'm missing which means they can be summed to infinite order?) Jun 4 comment Questions about the Dyson equation One more comment.. If the integral equation can't be written like my equation 1, then don't the iterations only give the usual perturbatively expansion and not an actual summation? I.e. Without the integrals, to first order perturbation, $G=G_0 + G_0 \Sigma G_0$ and to second order $G=G_0 + G_0 \Sigma G_0 + G_0 \Sigma G_0 \Sigma G_0$. Jun 4 comment Questions about the Dyson equation I do mean in thermal equilibrium. For the last question, the sign of each diagram appears to be different in each book. In Mattuck you only need to multiply by $-1$ for loops, but in Negele, the sign is also related to the order of perturbation and the sign of permutation (though I'm not 100% sure what that means). Jun 4 comment Questions about the Dyson equation @Marek thanks, from the book, it sounds like you can only get $(G_^{-1} - \Sigma)^{-1}$ under certain special circumstances, otherwise you get an integral equation rather than an algebraic one which can't be written like that.. Is that not the case? Jun 3 revised Questions about the Dyson equation edited tags