# Mean field theory = large-N approximation?

+ 3 like - 0 dislike
1022 views

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

recategorized Apr 24, 2014

+ 1 like - 0 dislike

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 (105 points)
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

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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.