1
$\begingroup$

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.

$\endgroup$
2
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.