# Why is the self-energy for quarks in $d=2$ Large $N$ QCD only order $g^2$?

In an interesting article by 't Hooft , he is able to find the exact quark propagator, in the large $$N$$ limit of QCD. He finds that the full 1PI self-energy is given by: $$\Gamma(p)=-\frac{g^2}{2\pi} \int dk_{-} \frac{1}{k_{-}^2}sgn(k_{-}+p_{-})$$ Where $$x_{\pm}=\frac{x_{1}\pm x_{0}}{\sqrt{2}}.$$ Although he claims, using a convincing diagrammatic argument that he has indeed obtained the full, non-perturbative self-energy contribution in the large $$N$$ limit, his answer only appears to be $$O(g^2)$$. Is there any explanation for this suspicious answer?

• Can you explain what you think is around with the expression? Are you saying that something proportional to $g^2$ can't be the exact answer to any question, and if so, why not? Jan 17 '20 at 5:02
• An order $g^2$ answer suggests that instead of doing a full non-perturbative calculation, one could merely do perturbation theory to two orders and get the same answer. Maybe this reasoning is too naive? Also, one can explicitly draw diagrams that contribute at $O(g^4)$. What happened to these diagrams? Jan 17 '20 at 5:11
• That definitely sounds too naive. If perturbation theory breaks down, you can't think of the answer as being made up of contributions at each order in perturbation theory, by definition. The order $g^2$ perturbative contribution is not necessarily the exact answer, even if the exact answer happens to be proportional to $g^2$. Jan 17 '20 at 5:40
• @knzhou thanks. That makes sense. I think I may have fooled my self into thinking there is a deep conspiracy here, when there really isn't. Non perturbative results, generically don't have anything to with perturbation theory and nor should they. Jan 17 '20 at 6:14