# How many loops are there in a connected Feynman diagram with a fixed number of vertices and external lines?

For the $\lambda \phi^4$ theory, I checked the connected Feynman diagrams arising in correlation functions for a number of cases; and the following summarises my observation. Let us denote by $E, V$ and $L$ the number of sources (external lines), vertices and loops respectively in a Feynman diagram.

\begin{array}{|c|ccc|ccc|ccc|ccc|} \hline E & & 0 & & & 2 & & & 4 & & & 6 & \\ \hline V & 0 & 1 & \cdots & 1 & 2 & \cdots & 2 & 3 & \cdots & 3 & 4 & \cdots\\ \hline L & 1 & 2 & \cdots & 1 & 2 & \cdots & 1 & 2 & \cdots & 1 & 2 & \cdots\\ \hline \end{array}

This clearly shows that the formula to obtain the number of loops in a connected Feynman diagram with a given number of vertices and external lines is the following.

$$L = 1 + V - E/2$$

So, here are my questions.

1. How do I prove this formula?
2. How do I generalise to any arbitrary theory? For example, what is the corresponding formula for $$\mathcal L = \frac1{2}(\partial \phi)^2 - \frac1{2}\mu_0\phi^2 -\frac{\lambda}{n!} \phi^n \quad ?$$

MOTIVATION: I am looking into this because if the formula stands true, then there is a simple relation between the number of external lines and the superficial degree of divergence $D$ of a Feynman diagram, namely:

$$D = 4 - E \,.$$

This is so because we know, by definition, that $D = 4L - 2I$ where we denote the number of internal lines by $I,$ which is now given by: $I=L+V-1.$ Combining all the information together, we get the above result.

• Related/possible duplicate: physics.stackexchange.com/q/176453/50583, physics.stackexchange.com/q/208553/50583 Jul 9 '17 at 20:11
• @ACuriousMind I am not asking for a proof of what the first link asks (that formula appears briefly in my motivation comment: just ignore it, for the sake of the main question!). The second link is related and interesting, but certainly not a duplicate. :) Jul 9 '17 at 20:16

As you have shown here, for a $\lambda \phi^N$ theory the number of internal lines $I$ is given by

$$I = \frac1{2}(NV-E) \,.$$

Since you also accept that

$$I = L+V-1 \,,$$

you just have to eliminate $I$ from the above equations to get

$$L = 1 + \Big(\frac N{2} -1\Big)V - \frac E{2}\,.$$

For N=4, you have your desired result.