I've heard about people talking about the 1/N expansion as a mothod of nonperturbative QCD. This seems strange. In the 1/N expansion all you have is Feynman diagrams, which are by definition perturbative. Just resumming them to infinite orders doesn't really count as nonperturbative physics. Is it really jusified to think resummation has anything to do with noperturbative physics? Is there some subtle connection betweeen the two that I missed?


1) On one hand, perturbative QCD means an expansion in the Yang-Mills coupling constant $g_{YM}$ for a fixed number of colors $N_c$.

2) On the other hand, the $1/N_c$ expansion means an expansion in $1/N_c$ for fixed 't Hooft coupling constant


(Here it is implicitly assumed that all dependence of $g_{YM}$ have been replaced, so that there are only dependences of $\lambda$ and $N_c$.)

3) The two expansions (1) and (2) are different. There are perturbative terms in the latter expansion that correspond to non-perturbative terms from the former point of view.

4) To convey the idea without going into the complications of actual QCD, consider for simplicity the function


On one hand, the expansion (1) of $f$ yields an expansion that is zero to all orders perturbatively in $g_{YM}$. On the other hand, the expansion (2) of $f$ is exact already after the zeroth-order term in $1/N_c$.

| cite | improve this answer | |
  • $\begingroup$ Summation of all perturbative orders into a finite formula means finding an exact (non perturbative, not approximate) solution. $\endgroup$ – Vladimir Kalitvianski Nov 18 '11 at 21:13
  • 2
    $\begingroup$ @Vladimir Kalitvianski: 1/N expansion, just like ordinary $g_{YM}$ expansion, is an asymptotic series rather than a converging series. It's not finite in any sense. $\endgroup$ – felix Nov 18 '11 at 21:49
  • 1
    $\begingroup$ @Qmechanic: My objection is that there are effects which are zero to all orders in perturbation theory, but only non-zero nonperturbatively. Baryon number violation for example. I'd like to correct your last sentence as "There are low-order perturbative terms in the latter expansion that correspond to all-order perturbative terms in the latter expansion, and vice versa, but there are no nonperturbative terms whatsoever." Am I wrong? $\endgroup$ – felix Nov 18 '11 at 21:53
  • 2
    $\begingroup$ I updated the answer with a toy-example that is non-perturbative in the sense (1) but tree-level exact in the sense (2). $\endgroup$ – Qmechanic Dec 19 '11 at 12:36

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.