4
$\begingroup$

I was wondering if there was a intuitive/heuristic argument to understand why generalizing the QCD gauge group $SU(3)$ to $SU(N)$ and taking $N\rightarrow \infty$ simplifies the analysis of the theory. Since in this limit only planar diagrams survive, the others being suppressed.

At first I would expect things to get a lot nastier by introducing such large number $N$ of colors...

$\endgroup$
  • 1
    $\begingroup$ It's an expansion parameter, like any other. In the large-N limit, many diagrams become sub-dominant. So it's enough to deal with a dominant diagrams, which are easier to deal with. $\endgroup$ – Siva Mar 11 '15 at 16:17
3
$\begingroup$

The intuitive idea is based on the Central Limit Theorem. Because suppose (as is usually the case) that your matter fields are in the fundamental representation of $SU(N)$, then the multiplet contains $N$ independent fields. Now the central limit tells us that the arithmetic average of $N$ independent random variables self-averages to a normal distribution, i.e. have small fluctuations. So in QCD for example, hadrons are always color singlets; a pion is $\pi = \sum_{c = 1}^N q_c\bar{q}_c$, so they have to be an average over all quark colors, so for $N\rightarrow \infty$ the fluctuations of $\pi$ (and hadrons in general) are much smaller than those of $q$ (quarks), because they should self-average according to the CLT. And in particular $$\left\langle \pi(x) \pi(y) \right\rangle \xrightarrow{N\rightarrow\infty}\left\langle \pi(x)\right\rangle\left\langle \pi(y)\right\rangle$$

$\endgroup$

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.