# What's the radius of convergence for 1/N expansion in QCD?

1. Mar 14, 2010

### petergreat

I'm asking this question as someone who has not studied this topic technically. By radius of convergence, I mean exact results, not just approximations, can be obtained by summing sufficiently many terms. (does this ever happen?) I don't mind if you need 1000 terms, as long as the series is convergent mathematically. Is N=3 QCD within the radius of convergence? And how about N=2, N=1?

2. Mar 15, 2010

### Physics Monkey

Though I know of no precise statement to this effect, I think the answer is that the 1/N series does not converge. I am aware of statements that some perturbative results in the 't Hooft coupling can be convergent exactly at large N, but this is different expansion. There definitely can be non-perturbative in 1/N corrections; one has to worry about such things in AdS/CFT, for example.

There are lots of subtle issues relating to large N. For example, the limits $$T \rightarrow 0$$ and $$N \rightarrow \infty$$ may not commute. More generally, the limits of infinite volume and large N may not commute as in Eguchi-Kawai reduction. The limits $$\omega \rightarrow 0$$ and $$N \rightarrow \infty$$ relevant for response functions do not commute. There is plenty more.

Hope this helps.

3. Mar 15, 2010

### tom.stoer

Interesting question; I studied large-N 1+1 dim. QCD up to first loop order (= with mesonic fluctuations) w/o ever wondering about convergence ...