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Mean field theory = large-N approximation?

+ 3 like - 0 dislike
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Wikipedia entry of 1/N expansion (or 't Hooft large-N expansion) mentions that

It (large-N) is also extensively used in condensed matter physics where it can be used to provide a rigorous basis for mean field theory.

I would like reference(s) that reviews this connection between MFT and large-N.


This post imported from StackExchange Physics at 2014-03-05 14:32 (UCT), posted by SE-user crackjack

asked Nov 5, 2013 in Resources and References by crackjack (110 points) [ revision history ]
recategorized Apr 24, 2014 by dimension10

1 Answer

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Any good book on statistical field theory and critical phenomena will have a chapter on large N. See for instance Zinn-Justin's book.

Keep in mind that the large N approximation is a special kind of mean-field theory (since it is both self-consistent and exact in large $N$). It is thus quite different from the usual mean-field theory use for example in the case of the Ising model.

It is called mean-field because one only have to minimize the action without computing any corrections (which are of order $1/N$). But the equation is self-consistent, contrary to the usual mean-field theory.

To summarize : for the $O(N)$ model, with $N\to\infty$, one could use the usual mean-field, or the "large $N$ mean-field". The results would not be the same, as the former is approximate but the latter is exact.

In principle, you can also use the large $N$ results to finite $N$, even though it is not exact in that case (but sometimes $3\gg1$). But it definitely does not work for the Ising model ($N=1$). Indeed, the large $N$ approach use the fact that the Goldstone modes (the $\pi_i$) dominate the physics compare to the fluctuation of $\sigma$ (because there is an infinite number of Goldstone modes compare to one $\sigma$). But in the Ising case, the only mode is always gapped (away from criticality), so the results of the large $N$ is non-sense.

This post imported from StackExchange Physics at 2014-03-05 14:32 (UCT), posted by SE-user Adam
answered Nov 5, 2013 by Adam (75 points) [ no revision ]
In Zinn-Justin's book, for example, I see a chapter on 1/N expansion of O(N) model and another chapter on MFT (steepest descent + corrections) of ferromagnetic Ising model, but not their equivalence. I want to see, for example, what 1/N expansion means for Ising model and its formal equivalence to MFT.

This post imported from StackExchange Physics at 2014-03-05 14:32 (UCT), posted by SE-user crackjack
@crackjack: I have edited my answer. Tell me if that answer your question.

This post imported from StackExchange Physics at 2014-03-05 14:32 (UCT), posted by SE-user Adam
Also, I dont understand the fundamental role of Goldstone modes in large-N. I thought, large-N expansion of 2D O(N) models is a prototype example of how the fields can be massive (through the non-perturbative large-N resummation effect) while their action naively appears to be massless.

This post imported from StackExchange Physics at 2014-03-05 14:32 (UCT), posted by SE-user crackjack
Thanks! It is much clearer now, except for last 4 lines: I was starting to warm up to the idea of an equivalence between large-N and MFT for Ising models. Zinn-Justin's book mentions (section 24.5 and A24.2) an expansion around MFT for Ising model, using $1/l$ as a parameter, with $ \beta \rightarrow l \beta $. Considering $N \sim l$ and $l \beta \sim N g^2$, I was reminded of an equivalence with 't Hooft large-N expansion. Also, with such a mean field expansion, the $\mathbb{Z}_2$ symmetry of Ising model becomes a spin variable with a gaussian distribution of spin values.

This post imported from StackExchange Physics at 2014-03-05 14:32 (UCT), posted by SE-user crackjack
The difference is that $l$ is an arbitrary parameter, useful to organize the calculation, but in fact equal to $1$. $N$ is a real parameter (though usually equal to $1,2$ or $3$), which represents the internal degrees of freedom of the spin field. For Ising, $N=1$, and the field has one internal degree of freedom, corresponding to the field $\sigma$ and $N-1=0$ $\pi$ field. The large $N$ expansion is based on the dominance of the Goldstone fields $\pi$, so it will only give non sense for Ising.

This post imported from StackExchange Physics at 2014-03-05 14:32 (UCT), posted by SE-user Adam
For instance, in $d=2$, the large $N$ predict no ordered phase at low temperature, which is perfectly fine for $N\geq 3$, but misses the BKT phase for $N=2$ and the ferromagnetic phase for $N=1$ (cf Onsager solution of the Ising model).

This post imported from StackExchange Physics at 2014-03-05 14:32 (UCT), posted by SE-user Adam

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