# Understanding Feynman Diagrams for the 2-point correlation function and $\phi^3$

This is a follow-up question (see first one here, second one here). You do not necessarily need to have read those two to follow this post.

I am aimed to understand $$V(\phi) = -\lambda \frac{\phi^3}{3!}$$ theory for the $$2-$$point correlation function in detail. To make it simpler, we will focus on connected diagrams only.

In the previous posts I drew my attention to computations via the partition function $$Z[J] = \int d[\phi] e^{iS[\phi] + i\int d^d x J(x) \phi(x)}$$. Here I want to get more of a conceptual understanding.

Given that we are studying the $$2-$$point correlation function, we can only have two external legs. The number of vertices depends on the order in perturbation theory (i.e. $$\mathcal{O}(\lambda)$$). I have got several conceptual questions that I'll be presenting in due course. Let us get started

• $$\mathcal{O}(\lambda^0)$$ order.

In this case we deal with no vertex (and two external legs of course. I will not be mentioning the latter explicitly, given that we will be dealing all the time with two external legs i.e. the $$2-$$point correlation function $$\langle \phi(x_1) \phi(x_2) \rangle$$). Hence the Feynman diagram is simply the propagator i.e. • $$\mathcal{O}(\lambda^1)$$ order.

I have read that $$\langle \phi(x_1) \phi(x_2) \rangle$$, for $$\phi^3$$, has no diagram for $$\mathcal{O}(\lambda^1)$$ order because it is impossible to have a diagram with two external legs and one vertex. I thought that the reason was that, by definition, an external leg is represented by a vertex with one outgoing line. Hence, if we were to add another external leg we would necessarily need to add another vertex. However this seems to be wrong. Why is such statement true then?

• $$\mathcal{O}(\lambda^2)$$ order.

There is only one diagram at second order: two external legs, a loop in the middle, and each external leg going into its own vertex on the loop, so two vertices Why this is the unique diagram for $$\mathcal{O}(\lambda^2)$$ though? I think the answer relies on Euler's formula $$L=I-V + 1$$ (where $$L$$ is the number of loops, $$I$$ is the number of internal lines and $$V$$ is the number of vertices). We have $$V=2$$, so $$L=I-2 + 1$$. A negative number of loops (I think) does not make sense so we need $$I \geq 1$$. If $$I = 1$$ we simply recover $$\mathcal{O}(\lambda^0)$$ order, so we discard this. If $$I = 2$$, we get the desired diagram. If $$I = 3$$ we would need 2 loops and I think that its is not possible to have 2 loops with only two vertices (I have to ask again: why?)

If the above's argument based on Euler's formula is wrong please let me know.

• $$\mathcal{O}(\lambda^3)$$ order.

Just as for $$\mathcal{O}(\lambda^1)$$, there is no diagram for $$\mathcal{O}(\lambda^3)$$ because it is impossible to have a diagram with two external legs and three vertices. Once I understand why this is the case for $$\mathcal{O}(\lambda^1)$$, I should be able to understand why $$\mathcal{O}(\lambda^3)$$ has no diagram.

• $$\mathcal{O}(\lambda^4)$$ order.

Again, I can only find one contribution, based on Euler's formula: we have $$4$$ vertices and $$5$$ internal lines, hence $$2$$ loops. The question is: why? Again, I should be able to answer once I understand why there is a unique diagram for $$\mathcal{O}(\lambda^2)$$ order.

The orders in perturbation of course continue. However, let us stop at $$\mathcal{O}(\lambda^4)$$ for now :)

PS: Please note this is not a homework question. I am studying Osborn notes, section 2.2. Interacting Scalar Field Theories, and I want to understand how he constructed the Feynman rules via working out the simplest example I could find: $$\phi^3$$ theory and the $$2-$$point correlation function

EDIT 0 Thanks to the provided answer I understand things better! Just let me ask you a couple of quick questions

1. So for $$\mathcal{O}(\lambda^6)$$, the diagram is I.e. $$8$$ internal lines and $$6$$ vertices so, via Euler's formula $$L=8-6+1=3$$ loops.

Mmm I start to see a pattern. I would say that for the $$\mathcal{O}(\lambda^n)$$ we would get

$$\frac{3N}{2}-1$$ internal lines, $$N$$ vertices and $$\frac{N}{2}$$ loops. Do you agree?

1. I only get one diagram for each order in perturbation theory. If I am indeed correct, why we only get one diagram for each order?
• What about the diagram with two loops for $O(\lambda^4)$ (that is, two of your $O(\lambda^2)$ diagrams stuck together? And this is just one example—there should be other diagrams.
– d_b
Mar 13, 2021 at 18:04
• @d_b could you please extend your comment? I do not see what you mean Mar 13, 2021 at 18:08
• Take your diagram for $\lambda^2$. Make a copy of it and label the external lines of the copy $x'$ and $y'$. Then glue the line $y$ from your original diagram to the line $x'$ from the copy. You'll get a new diagram with four vertices, so a contribution at $O(\lambda^4)$.
– d_b
Mar 13, 2021 at 18:12
• @d_b thank you very much, I got it! :) I will probably make another post related to this topic. Mar 13, 2021 at 22:15

The form of the interacting Lagrangian contains a product of three fields $$\mathcal{L}_i \sim \lambda\phi^3$$. Interacting terms contain information about vertices. Here, it means that in every vertex there must be three legs (representing the fields). Thus, it is not possible to have a diagram with two external legs and one vertex, there would not be any Feynman rule for that.
• As the other comment suggests, there are also other diagrams. For example, at order $\lambda^4$ you have also the diagram with two bubbles . Mar 13, 2021 at 21:25