Two questions on Feynman diagram and Green's function

  • #1
16
1
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?
Could anybody answer these please?
 

Answers and Replies

  • #2
king vitamin
Science Advisor
Gold Member
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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 [itex]G(\omega,k)[/itex], but if you prefer you could say that you are really interested in the perturbation expansion for [itex]G(\omega,k)^{-1}[/itex], 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 [itex]G(\omega,k)[/itex] which are usually physically important, and one can never get a shift in the pole structure of [itex]G(\omega,k)[/itex] at finite order in perturbation theory.

I don't understand your question about the interaction potential.
 
  • #3
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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.
Thanks a lot for your reply!
I try to clear my two question once time, hope it helpful.
First question is when one do perturbation, we could only consider low order perturbation, since high order would be smaller. But if one calculate Green's function by one order irreducible self-energy and Dyson equation, it conclude all order perturbation indeed. Since irreducible diagrams could built higher order reducible diagrams. So, could we just compute one order irreducible self-energy?
Second one is that, comparing to low order diagram, some high diagrams could be seen as correction in particle line, while some as in interaction line. For later, it looks like that interacting potential change to effective potential. So when one says about effective potential in many-body system, it only means for some diagrams but not all diagrams, is my understanding right?
For your answer, I think you basically understand my first question. But I am still not fully clear that, do you mean when one adds higher order irreducible self-energy, it can't change the pole of Green's function, but imaginary part? So after one consider higher order self-energy, eigenvalues is unchanged while density of states not. Is it true?
 

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