1
$\begingroup$

Consider $\varphi^p$ theory in dimension $D$. For a Feyman diagram $\Gamma$ one can introduce the superficial degree of divergence $deg(\Gamma)$. It is defined as $DL-2I$ where $I$ is the number of internal edges of $\Gamma$ and $L$ is the loop number. This value can be easily understood by examining the power of the momenta under the integral corresponding to $\Gamma$. By a simple combinatorial argument one can show that $\deg(\Gamma)=(D-\frac{2p}{p-2})L+\frac{2}{p-2}(p-N)$ where $N$ is the number of external edges of $\Gamma$. For example for $p=4$ one gets $deg(\Gamma)=(D-4)L+4-N$. In particular this degree decreases when $N$ increases and increases when $D$ increases: but the most interesting phenomenon is related to the behaviour wih respect to $L$: in small dimension many loops implies smaller degree of divergence while in higher dimension increasing the number of loops causes problems with convergence. There is also this critical value of dimension $\frac{2p}{p-2}$ in which the loop number does not make a difference. The only values $p$ for which this critical dimension appears to be an integer are $p=3,4,6$ with corresponding $D=6,4,3$.

Is there any physical reason (not just mere power counting+simple combinatorics) for which one should expect such behaviour of this degree of divergence with respect to $N,D$ and in particular $L$? By a physical reason I mean some heuristic (not necessarilly rigorous) argument for finiteness/divergence, not involving Feynman rules and power counting.

I am aware that my question is rather vaque nevertheless I hope that my point is clear and I will be grateful for any explanation.

$\endgroup$
3
  • $\begingroup$ Those are the values for which the theory is scale-invariant (cf. this PSE post). $\endgroup$ Aug 20, 2019 at 12:09
  • $\begingroup$ Related: physics.stackexchange.com/q/481249/2451 $\endgroup$
    – Qmechanic
    Aug 21, 2019 at 16:15
  • $\begingroup$ Thank you for the links: however the explanations given in these discussions still heavily relies on power counting and combinatorics and is not exactly what I had in mind $\endgroup$
    – truebaran
    Aug 23, 2019 at 2:32

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.