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  O(N) sigma model at large N

+ 6 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-07-07 10:43 (UCT), posted by SE-user user43283
asked Jul 6, 2014 in Theoretical Physics by user43283 (45 points) [ no revision ]
retagged Jul 7, 2014
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.

This post imported from StackExchange Physics at 2014-07-07 10:43 (UCT), posted by SE-user ACuriousMind
@ACuriousMind: But if $N$ is large then it should still be a good approximation?

This post imported from StackExchange Physics at 2014-07-07 10:43 (UCT), posted by SE-user JeffDror

2 Answers

+ 3 like - 0 dislike

This model was studied using RG and the epsilon-expansion (not at large N) by Wilson and Fisher. It shows interesting behaviour as a function of $d$. For d>4, the RG fixed point is the Gaussian one while for d<4, there is a new fixed point that is called the Wilson-Fisher fixed point. For N=1, it flows to the CFT associated with the Ising model at criticality. See here.

There is a brief and incomplete discussion on the large-N limit of the $O(N)$ model in the Erice lectures on large $N$ by Sidney Coleman that is reproduced in his book "Aspects of Symmetry".

answered Jul 8, 2014 by suresh (1,545 points) [ revision history ]
edited Jul 8, 2014 by suresh
+ 3 like - 0 dislike

I think that the derivation of the counterterms at n loops in the leading order in $1/N$ is essentially correct but I don't understand the result mentioned in the question for the resumation and the limit $\epsilon \rightarrow 0$.


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

then the sumation over n is indeed quite trivial (geometric serie) and one finds

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

which is non-trivial. In particular, the $\epsilon$ independant part is 

$- \frac{\lambda}{8N} (\phi^2)^2 \frac{1}{1- \frac{\lambda}{16 \pi^2}log(\mu/m)}$

which is a rather expected form (it is what we find when we resolve the renormalization group equation with a beta function computed at one loop, which as the effect of resum the logarithm coming from the one loop computation, and indeed at the leading order in $1/N$, the Feynman diagrams that remain are the chains of one-loops diagrams).

answered Jul 8, 2014 by 40227 (5,140 points) [ revision history ]
edited Jul 8, 2014 by 40227

@40227:  Thank you very much for your answer. The only thing I didn't get is how did you get  \(\epsilon\)-independent part of this expression:\(\Delta \mathcal{L}^{\infty}_{\phi^4} = -\frac{\lambda}{8N}(\phi^2)^2 \frac{1}{1-\frac{\lambda}{32 \pi^2 \epsilon}} \mu^{2\epsilon} \)

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