# How do we understand the results of $1/N$ or $\epsilon$ expansion beyond leading orders?

When we do $$1/N$$ expansions in, say, 2+1$$D$$ $$O(N)$$ models and try to extract all kinds of critical exponents from it, we get the following results for the scaling dimensions of various operators up to order $$1/N^2$$ (Here $$\phi$$ is the vector field, $$S$$ is the scalar operator and $$T$$ is the symmetric traceless operator) $$\Delta_\phi = 0.5+\frac{0.135}{N}-\frac{0.097}{N^2}$$ $$\Delta_S = 2-\frac{1.081}{N}-\frac{3.048}{N^2}$$ $$\Delta_T = 1+\frac{1.081}{N}-\frac{0.195}{N^2}$$ We see that especially for $$S$$ the expansion is not very good since the coefficient for $$1/N^2$$ is very big and it will not give us a sensible scaling dimensions. Similar problems will be encountered in epsilon expansions. How do we make sense of these results beyond leading orders?

• One can use suitable resummation methods. E.g. the simple fact that for small $N$ an expansion in positive powers of $N$ also exists (even if these expansion coefficients are not known), means that you can re-expand the series by putting $\frac{1}{N} = \lambda \frac{u}{1-u}$ where $\lambda$ is an arbitrary parameter that can be optimized by choosing it such that the last known term of the expansion becomes zero. – Count Iblis Feb 6 at 20:25
• As shown by Zinn-Justin, this method yields rapidly converging approximations in the sense that with more and more terms and summing till the last term set to zero using $\lambda$, makes the error in the approximation smaller and smaller. – Count Iblis Feb 6 at 20:26
• Thank you very much! Do you have a reference for that? – Weicheng Ye Feb 6 at 20:28
• You can check out arxiv.org/abs/1001.0675 and references in there. – Count Iblis Feb 6 at 20:31