$\operatorname{O}(N)$ sigma model at large $N$

I would like to better understand the main principles of large-$N$ expansion in quantum field theory. To this end I decided to consider simple toy-model with lagrangian (from Wikipedia)

$\mathcal{L} = \frac{1}{2}(\partial_{\mu} \phi_a)^2-\frac{m^2}{2}\phi_a^2 - \frac{\lambda}{8N}(\phi_a \, \phi_a)^2$

My aim was to renormalize this theory in all orders of perturbation theory in leading order of $\frac{1}{N}$. The calculation of counterterms in two loops in leading order of $\frac{1}{N}$ almost coincides with the corresponding calculation in $\phi^4$ theory. In leading order of $1/N$ the counterterms are (using MS-scheme):

1 Loop:

$\Delta \mathcal{L}_{\phi^4}^{1} = -\lambda^2 \mu^{2\epsilon} \frac{1}{32 \pi^2 \epsilon} \frac{(\phi_a \, \phi_a)^2}{8N}$

$\Delta \mathcal{L}_{\phi^2}^{1} = - \frac{\lambda}{32 \pi^2 \epsilon} \frac{m^2 \phi_a^2}{2}$

2 Loops:

$\Delta \mathcal{L}_{\phi^4}^{1} = -\lambda^3 \mu^{2\epsilon} (\frac{1}{32 \pi^2 \epsilon})^2 \frac{(\phi_a \, \phi_a)^2}{8N}$

$\Delta \mathcal{L}_{\phi^2}^{1} = - (\frac{\lambda}{32 \pi^2 \epsilon})^2 \frac{m^2 \phi_a^2}{2}$

I am quite sure (though I haven't proven it properly yet) that in $n$ loops the leading contribution to counterterms comes from a chain of "fish" diagrams for 4-point Green's function and chain of bubbles for 2-point Green's function (I think, it's quite easy to imagine):

$\Delta \mathcal{L}_{\phi^4}^{1} = -\lambda^{n+1} \mu^{2\epsilon} (\frac{1}{32 \pi^2 \epsilon})^n \frac{(\phi_a \, \phi_a)^2}{8N}+O(\frac{1}{N})$

$\Delta \mathcal{L}_{\phi^2}^{1} = - (\frac{\lambda}{32 \pi^2 \epsilon})^n \frac{m^2 \phi_a^2}{2}+O(\frac{1}{N})$

If this speculation is correct, the summation of perturbation series is quite trivial (we have geometric series). When we do it and then take limit $\epsilon \rightarrow 0$ we will find that

$\Delta \mathcal{L}^{\infty}_{\phi^2} = \frac{m^2 \phi_a^2}{2}$

$\Delta \mathcal{L}^{\infty}_{\phi^4} = \frac{(\phi_a \phi_a)^2}{8N}$

and hence the total lagrangian is simply (in the leading order of $1/N$).

$\mathcal{L} = \frac{1}{2}(\partial_{\mu} \phi_a)^2$

This result seems to me highly suspicious... Did anybody do similar calculation? I looked all over the Internet and didn't find anything :( I will be very grateful for remarks and links to books or may be articles where similar problem is considered.

• I have never done explicit renormalization in a large N expansion, so I can't speak with authority, but the leading-order term in the $\frac{1}{N}$-expansion alone is frequently a bad approximation to the full theory, so "suspicious" results for it are nothing to worry too much about. – ACuriousMind Jul 6 '14 at 20:18
• @ACuriousMind: But if $N$ is large then it should still be a good approximation? – JeffDror Jul 7 '14 at 7:12
• Not sure if this is relevant, but if you take your original theory and send $N\rightarrow \infty$, you get a free theory, so then the loop corrections in the original theory in the $N\rightarrow \infty$ limit should reproduce the loop corrections in the free theory, ie, they should all be zero. Presumably to get a nontrivial answer you also need to send the coupling $\lambda\rightarrow \infty$ in such a way that the ratio $\lambda/N$ is fixed (you need to do a similar thing in large $N$ of yang mills, there the analogue of $\lambda/N$ is called the "'t Hooft" coupling") . – Andrew Jul 7 '14 at 14:46
• As explained here, the RG flow of the O(N) model is characterized by different fixed point, depending on the number of space-time dimensions considered. – Dilaton Jul 8 '14 at 11:03
• To me that step is not obvious either at present, but I am confident that he will explain it as soon as he sees your comment. – Dilaton Jul 8 '14 at 23:38