# Two questions on Feynman diagram and Green's function

• A
• howl
In summary, when one does perturbation, they only consider low order perturbation, so they would not be able to get the exact Green's function by using one-order self-energy. However, if they calculate Green's function by using self-energy to all orders, they will get the effects of all the perturbation. Some high diagrams are seen as corrections in particle line, while some are seen as in interaction line. The effective interaction potential is only meant for some diagrams, but not all.
howl
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?

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.

howl said:
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.

king vitamin said:
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.
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?

## What is a Feynman diagram?

A Feynman diagram is a graphical representation of the mathematical expressions used to calculate the probability amplitudes of particle interactions in quantum field theory. It was developed by physicist Richard Feynman in the 1940s as a way to visualize complex mathematical equations.

## How do you interpret a Feynman diagram?

In a Feynman diagram, time flows from left to right and particles are represented by lines connecting interaction vertices. The direction of the arrow on the line indicates the direction of the particle's momentum, while the type and color of the line represent the type of particle. Interaction vertices represent the exchange of energy and momentum between particles.

## What is the purpose of using Green's function in Feynman diagrams?

Green's functions are mathematical tools used to solve differential equations in quantum field theory. In Feynman diagrams, they represent the propagation of particles between interaction vertices and help calculate the probability amplitudes of particle interactions.

## How do Feynman diagrams relate to quantum mechanics?

Feynman diagrams are a visual representation of the mathematical equations used in quantum field theory, which combines the principles of quantum mechanics with special relativity. They are used to calculate the probabilities of particle interactions and are essential in understanding the behavior of subatomic particles.

## What are the limitations of using Feynman diagrams?

Although Feynman diagrams are a useful tool for visualizing particle interactions, they have limitations in their ability to accurately represent all aspects of quantum field theory, such as higher-order corrections and interactions involving more than three particles. Additionally, they do not take into account the effects of gravity and are primarily used for calculations in particle physics rather than cosmology.

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