# Mean field theory = large-$N$ approximation?

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-$N$.

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I think Adam's answer is excellent, and I'd rather make this a comment but can't as I just signed up in order to answer.

While I agree with most of what Adam said, there are cases where large-N works well for $N=1$. The case I'm familiar with is the large-N expansion for the "Spin Ices" Dy$_2$Ti$_2$O$_7$ and Ho$_2$Ti$_2$O$_7$, which have Ising spins on a pyrochlore lattice. The best reference for this is Sergei Isakov's PhD thesis (University of Stockholm 2004), specifically Section 4.3 and Subsection 4.8.2.

It's noted in that thesis that we shouldn't expect large-N to work for pyrochlores below $N=3$; the expansion is singular at $N=2$ owing to order-by-disorder. Extensive checking against Monte Carlo simulations and experiment verify that the expansion is good for $N=1$ (references given in the thesis), but we don't know why this is.

I found myself needing to prove the equivalence of large-N and MFT at high temperatures. I apologise if it's bad etiquette to reference one's own work, but I provide a mathematical proof in Section 2.4 of my Master's thesis which can be found here (Perimeter Institute 2011).

I believe the main working difference between large-N and MFT is that MFT always contains a non-zero critical temperature $T_c$, but that $T_c\sim 1/N$ with $N$ the number of degrees of freedom at each lattice site. This means that large-N, with $N\rightarrow\infty$, has $T_c=0$ to zeroth order in the $1/N$ expansion (at which one generally works).

• Thanks for the comment. One question: in your case, are you talking about quantum Ising spins or classical spins ? – Adam Jul 14 '14 at 15:55

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.

• 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. – GuSuku Nov 5 '13 at 21:55
• 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. – Adam Nov 6 '13 at 14:34
• 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). – Adam Nov 6 '13 at 14:37