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Jul
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accepted What's the definition of the time ordering operator for more than two particles?
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15
asked What's the definition of the time ordering operator for more than two particles?
Jun
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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
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2
asked Questions about the Dyson equation
May
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comment Can the work done between two non-equilibrium states be calculated?
@genneth thanks, I'll look at that.
May
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comment When can the source term of a partition function be put in?
@Qmechanic I'm not sure what you mean ... Do you mean that I have to start at equation (1) then diagonalise the Hamiltonian plus source term? And that I can't start with the Hamiltonian in diagonal form?
May
19
accepted Is there a relativistic (quantum) thermodynamics?