# Diagrammatic Representation of non-Gaussian perturbation expansions

I have no experience in graph theory and am a little confused with how Hugh Osborn represents a perturbation expansion with diagrams on page 15 of these notes.

We have a perturbation expansion My questions are: 1. In the diagrammatic interpretation part, 2 points are joined to create one line. So shouldn't a vertex with $$k$$ indices be at the intersection of $$k/2$$ lines? 2. If $$A_{ij}^{-1}$$ is a straight line from $$i$$ to $$j$$, how do we have a vertex with only one line? 3. Where do the points $$i,j,k,l$$, etc. go on the loops?

• Related post by OP: physics.stackexchange.com/q/492098/2451 Jul 19, 2019 at 10:33
• @Qmechanic This one is from the same set of notes. There, I'm asking about how to evaluate a series algebraically by differentiating. Here, I'm asking about how the resulting expansion corresponds to the diagrams shown.
Jul 19, 2019 at 10:38
• I find the questions hard to understand. Have you try to draw the diagrams from 1.133 by yourself? For example, the first term is a vertex with two legs, that you can connect with a line, so you get the first bubble.
Jul 19, 2019 at 11:20
• $A_{ij}^{-1}$ corresponds to points $i$ and $j$ connected by a line. How do I create a vertex V_{ij}out of this straight line?
• It works the other way around. Starting from $V_{ij}A^{-1}_{ij}$, you can draw a picture by drawing $V_{ij}$ as a dot with two legs (one with label $i$, the other with label $j$) connected by a line representing $A^{-1}_{ij}$.
Note that the reason why you have bubbles is that you have many repeated (i.e. dummy) indices. In the first diagram for example, i and j label the two ($$k=2$$) lines joined by $$-V_{ij}$$, but they also label the points joined by $$A^{-1}_{ij}$$, so that is why the two lines close to form a bubble.