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Application to the Ising Curie-Weiss model
The Ising Curie-Weiss model is the mean-field version of the famous Ising model (1920). Let $$n\geq 1 $$. Consider the complete graph on $$n$$ vertices, denoted by $$\mathcal{K}_n=\{1,\dots,n\}$$. For each site $$k\in\mathcal{K}_n$$, there is a discrete variable $$\sigma_k\in\{-1,1\}^n$$ representing the spin of site $$k$$. A spin configuration, $$\sigma=(\sigma_k)_{k\in \mathcal{K}_n}$$ is an assignment of spin value to each site. The energy of a configuration $$\sigma$$ is given by the Hamiltonian function


 * $$H_n(\sigma) = - \frac{1}{2n}\sum_{1\leq i,j\leq n} \sigma_i \sigma_j - \sum_{j} h_j\sigma_j$$,

where $$h_1,\dots,h_n$$ are external magnetic fields. The configuration probability is given by the Boltzmann distribution with temperature $$T>0$$:


 * $$\mathbb{P}_{n,T}(\sigma) ={e^{-H_n(\sigma)/T} \over Z_n(T)},$$

where the normalization constant


 * $$ Z_n(T) = \sum_{x\in \mathcal{K}_n} e^{-H_n(x)/T}$$

is the partition function. Let $$ S_n=\sigma_1+\dots+\sigma_n$$ for any $$n\geq 1$$.

Let us investigate on the asymptotic behaviour of $$S_n$$ when $$T\geq 1$$ and $$h_1=\cdots=h_n=0$$. Let $$\alpha \in [0,1[$$ and $$u\in \R$$.


 * $$ \begin{align}

E_{n,T}(u,\alpha)&:=\mathbb{E}_{n,T}\left[\exp\left(u\frac{S_n}{n^{1-\alpha}}\right)\right]=\frac{1}{Z_n(T)}\sum_{x \in \mathcal{K}_n} \exp\left(u\,\frac{x_1+\dots+x_n}{n^{1-\alpha}}+\frac{(x_1+\dots+x_n)^2}{2nT}\right)\\ &=\frac{1}{Z_n(T)}\sum_{x \in \mathcal{K}_n} \exp\left(u\frac{x_1+\dots+x_n}{n^{1-\alpha}}\right)\,\times\, \int_{\mathbb{R}}\exp \left( y\,(x_1+\dots+x_n)- \frac{nTy^2}{2}\right) \,\sqrt{\frac{nT}{2 \pi}}\, dy,\end{align}$$

where we used the Hubbard-Stratanovich transformation with $$ a=(nT)^{-1}$$. By interverting sum and integral, we get


 * $$ E_{n,T}(u,\alpha)=\frac{\sqrt{nT}}{Z_n(T)\sqrt{2\pi}}\int_{\mathbb{R}} \exp\left(- \frac{nTy^2}{2}\right)\sum_{x \in \mathcal{K}_n} \exp\left(\left(\frac{u}{n^{1-\alpha}}+y\right)(x_1+\dots+x_n)\right)\, dy.$$

Let $$ \varepsilon_1,\dots,\varepsilon_n$$ be $$n$$ i.i.d. random variables with distribution $$(\delta_{-1}+\delta_1)/2$$. We have


 * $$ \forall s\in \R\qquad \sum_{x \in \mathcal{K}_n} \exp\left(s\,(x_1+\dots+x_n)\right)=2^n\, \mathbb{E}\left[\exp\left(s\,(\varepsilon_1+\dots+\varepsilon_n)\right)\right]=2^n\, \mathbb{E}\left[\exp\left(s\,\varepsilon_1\right)\right]^n=2^n\,\mathrm{cosh}(s)^n.$$

As a consequence


 * $$ E_{n,T}(u,\alpha)=\frac{\sqrt{nT}\,2^n}{Z_n(T)\sqrt{2\pi}}\int_{\mathbb{R}} \exp\left(- \frac{nTy^2}{2}+n\mathrm{ln}\,\mathrm{cosh}\left(\frac{u}{n^{1-\alpha}}+y\right)\right)\, dy.$$

We make the change of variables $$z=n^{\alpha}(y+u/n^{1-\alpha})$$ :


 * $$ \begin{align} E_{n,T}(u,\alpha)&=\underbrace{\frac{\sqrt{nT}\,2^n}{Z_n(T)\sqrt{2\pi}\,n^{\alpha}}}_{C_{n,T}(\alpha)}\,\exp\left(-\frac{Tu^2}{2n^{1-2\alpha}}\right)\,\int_{\mathbb{R}} \exp\left(Tuy-\frac{nT}{2}\left(\frac{z}{n^{\alpha}}\right)^2+n\mathrm{ln}\,\mathrm{cosh}\left(\frac{z}{n^{\alpha}}\right)\right)\, dz\\

&=C_{n,T}(\alpha)\,\exp\left(-\frac{Tu^2}{2n^{1-2\alpha}}\right)\,\int_{\mathbb{R}} \exp\left(Tuz-nG_T\left(\frac{z}{n^{\alpha}}\right)\right)\, dz,\end{align}$$ where $$ G_T(s)= Ts^2/2-\mathrm{ln}\,\mathrm{cosh}(s)$$ for any $$s\in \R$$. The function $$G_T$$ is continuous on $$\mathbb{R}$$ and satisfies
 * $$|G_T(s)|$$ goes to $$0$$ when $$n$$ goes to $$+\infty$$,
 * $$G_T$$ has a unique minimum at $$0$$,
 * in the neighbourhood of $$0$$, $$ G_T(s)=\left(T-1\right)s^2/2-s^4/12+o(s^4)$$.

As a consequence, Laplace's method gives us
 * $$\int_{\mathbb{R}} \exp\left(Tuz-nG_T\left(\frac{z}{n^{\alpha}}\right)\right)\, dz \quad\underset{n\to+\infty}{\longrightarrow}\quad

\left\{\begin{array}{cl} \displaystyle{\int_{\mathbb{R}} \exp\left(Tuz-(T-1)\frac{z^2}{2}\right)\, dz} & \mbox{if}\quad T>1,\quad \alpha=1/2,\\ \displaystyle{\int_{\mathbb{R}} \exp\left(uz-\frac{z^4}{12}\right)\, dz} & \mbox{if}\quad T=1,\quad \alpha=1/4.\\ \end{array}\right. $$ By taking $$u=0$$, we get
 * $$\frac{1}{C_{n,T}(\alpha)} \quad\underset{n\to+\infty}{\longrightarrow}

\quad\left\{\begin{array}{cl} \displaystyle{\int_{\mathbb{R}} \exp\left(-(T-1)\frac{z^2}{2}\right)\, dz=\sqrt{\frac{2\pi}{T-1}}} & \mbox{if}\quad T>1,\quad \alpha=1/2,\\ \displaystyle{\int_{\mathbb{R}} \exp\left(-\frac{z^4}{12}\right)\, dz} & \mbox{if}\quad T=1,\quad \alpha=1/4.\\ \end{array}\right. $$ Lévy's continuity theorem yields
 * $$\frac{S_n}{\sqrt{n}}\quad \overset{\mathcal{L}}{\underset{n\to+\infty}{\longrightarrow}} \quad\mathcal{N}\left(0,\frac{T}{T-1}\right)\quad\mbox{if}\quad T>1,$$

and
 * $$\frac{S_n}{n^{3/4}} \quad\overset{\mathcal{L}}{\underset{n\to+\infty}{\longrightarrow}} \quad\frac{\displaystyle{\exp\left(-\frac{x^4}{12}\right)\, dx}}{\displaystyle{\int_{\mathbb{R}} \exp\left(-\frac{z^4}{12}\right)\, dz}}\quad\mbox{if}\quad T=1.$$