# Two questions on Feynman diagram and Green's function

• A
First, is it suitable to solve a Green's function by one-order self-energy, since it only consider partial high order perturbation, so it's unclear that this calculation corresponding to which order perturbation. In other word, if one wants to use self-energy to get Green's function, he should conclude all self-energies, isn't it?
Second, the effective interaction potential is considered as correction of interaction line of Feynman graph, but some graphs' interaction are not corrected. For example, some graphs only change in particle lines. So for these graph, effective interaction potential make no sense, right?

king vitamin
Gold Member
I'm having a little bit of difficulty understanding your post due to some grammar issues. I'll answer the best I can, but I might need clarification.

First, is it suitable to solve a Green's function by one-order self-energy, since it only consider partial high order perturbation, so it's unclear that this calculation corresponding to which order perturbation. In other word, if one wants to use self-energy to get Green's function, he should conclude all self-energies, isn't it?
Certainly if you want the exact Green's function, you'll need to compute the self-energy to all orders. However, this is almost always impossible, so you usually need to truncate at a low order. Depending on the problem you're looking at, you might need to go to two-loop or higher to get important physical effects.

I think you might also be asking about how computing the self-energy effectively sums a subset of diagrams to infinite order, whereas you are ignoring several lower-orde diagrams. This is true if you consider the perturbation expansion of $G(\omega,k)$, but if you prefer you could say that you are really interested in the perturbation expansion for $G(\omega,k)^{-1}$, in which case the self-energy diagrams are all that appear, and you are doing a normal expansion in the coupling. This perspective is rather justified, since it is the location of the poles of $G(\omega,k)$ which are usually physically important, and one can never get a shift in the pole structure of $G(\omega,k)$ at finite order in perturbation theory.