Ising Model with Curie–Weiss Perturbation

Abstract

Consider the nearest-neighbor Ising model on $\Lambda _n:=[-n,n]^d\cap {\mathbb {Z}}^d$ at inverse temperature $\beta \ge 0$ with free boundary conditions, and let $Y_n(\sigma ):=\sum _{u\in \Lambda _n}\sigma _u$ be its total magnetization. Let $X_n$ be the total magnetization perturbed by a critical Curie–Weiss interaction, i.e.,

$$\begin{aligned} \frac{d F_{X_n}}{d F_{Y_n}}(x):=\frac{\exp [x^2/\left( 2\langle Y_n^2 \rangle _{\Lambda _n,\beta }\right) ]}{\left\langle \exp [Y_n^2/\left( 2\langle Y_n^2\rangle _{\Lambda _n,\beta }\right) ]\right\rangle _{\Lambda _n,\beta }}, \end{aligned}$$

where $F_{X_n}$ and $F_{Y_n}$ are the distribution functions for $X_n$ and $Y_n$ respectively. We prove that for any $d\ge 4$ and $\beta \in [0,\beta _c(d)]$ where $\beta _c(d)$ is the critical inverse temperature, any subsequential limit (in distribution) of $\{X_n/\sqrt{{\mathbb {E}}\left( X_n^2\right) }:n\in {\mathbb {N}}\}$ has an analytic density (say, $f_X$) all of whose zeros are pure imaginary, and $f_X$ has an explicit expression in terms of the asymptotic behavior of zeros for the moment generating function of $Y_n$. We also prove that for any $d\ge 1$ and then for $\beta$ small,

$$\begin{aligned} f_X(x)=K\exp (-C^4x^4), \end{aligned}$$

where $C=\sqrt{\Gamma (3/4)/\Gamma (1/4)}$ and $K=\sqrt{\Gamma (3/4)}/(4\Gamma (5/4)^{3/2})$. Possible connections between $f_X$ and the high-dimensional critical Ising model with periodic boundary conditions are discussed.

Appendix: Limit Distribution of Lee–Yang Zeros When \(\beta <\beta _c(d)\)

In this appendix, we prove that when $\beta <\beta _c(d)$, the limiting distribution of Lee–Yang zeros has no mass in an arc containing $\exp (i0)$ of the unit circle. This is stated as a conjecture in Sect. 1.3 of [6]. As can be seen from below, the proof follows essentially from Theorem 1.2 of [12] and Theorem 1.5 of [23]. We present a slightly different and relatively self-contained proof here which might be better suited to the context. We came up with this proof before we knew of the existence of [12].

Let $\Lambda \subseteq {\mathbb {Z}}^d$ be finite. The Ising model on $\Lambda$ at inverse temperature $\beta \ge 0$ with free boundary conditions and external field $h\in {{\mathbb {R}}}$ is defined by the probability measure ${\mathbb {P}}_{\Lambda ,\beta ,h}$ on $\{-1,+1\}^{\Lambda }$ such that for each $\sigma \in \{-1,+1\}^{\Lambda }$

$$\begin{aligned} {\mathbb {P}}_{\Lambda ,\beta ,h}(\sigma ):=\frac{\exp \left[ \beta \sum _{\{u,v\}}\sigma _u\sigma _v+h\sum _{u\in \Lambda }\sigma _u\right] }{Z_{\Lambda ,\beta ,h}}, \end{aligned}$$

(127)

where the first sum is over all nearest-neighbor edges in $\Lambda$, and $Z_{\Lambda ,\beta ,h}$ is the partition function that makes (127) a probability measure. In this appendix, we will consider complex $h\in {\mathbb {C}}$. A famous result due to Lee and Yang [17] is that $Z_{\Lambda ,\beta ,h}\ne 0$ if $h\notin i{\mathbb {R}}$ where $i{\mathbb {R}}$ denotes the pure imaginary axis. So the fraction in (127) is well-defined if $h\notin i{\mathbb {R}}$ but it could take a complex value, and thus ${\mathbb {P}}_{\Lambda ,\beta ,h}$ is a complex measure. Let $\langle \cdot \rangle _{\Lambda ,\beta ,h}$ denote the expectation with respect to ${\mathbb {P}}_{\Lambda ,\beta ,h}$. For example, the magnetization at $u\in \Lambda$ is defined by

$$\begin{aligned} \langle \sigma _u\rangle _{\Lambda ,\beta ,h}:=\frac{\sum _{\sigma \in \{-1,+1\}^{\Lambda }}\sigma _u\exp \left[ \beta \sum _{\{u,v\}}\sigma _u\sigma _v+h\sum _{u\in \Lambda }\sigma _u\right] }{Z_{\Lambda ,\beta ,h}}. \end{aligned}$$

(128)

Let $\Lambda _n:=[-n,n]^d\cap {\mathbb {Z}}^d$ and $E_n$ be the set of all nearest-neighbor edges $\{u,v\}$ with $u,v\in \Lambda _n$. Note that

$$\begin{aligned} Z_{\Lambda _n,\beta ,h}&=\sum _{\sigma \in \{-1,+1\}^{\Lambda _n}}\exp \left[ \beta \sum _{\{u,v\}\in E_n}\sigma _u\sigma _v+h\sum _{u\in \Lambda _n}\sigma _u\right] \end{aligned}$$

(129)

$$\begin{aligned}&=\exp \left[ \beta |E_n|+h|\Lambda _n|\right] \sum _{\sigma \in \{-1,+1\}^{\Lambda _n}}\exp \left[ \beta \sum _{\{u,v\}\in E_n}(\sigma _u\sigma _v-1)+h\sum _{u\in \Lambda _n}(\sigma _u-1)\right] . \end{aligned}$$

(130)

Let $z=e^{-2h}$ throughout the appendix and write $Z_{\Lambda _n,\beta }(z):=Z_{\Lambda _n,\beta ,h}$. Then the outmost sum in (130) is a polynomial of z with degree $|\Lambda _n|$. So by the fundamental theorem of algebra, $Z_{\Lambda _n,\beta }(z)$ has exactly $|\Lambda _n|$ complex roots. The Lee–Yang circle theorem [17] says that these $|\Lambda _n|$ roots are on the unit circle $\partial {\mathbb {D}}$ where ${\mathbb {D}}:=\{z\in {\mathbb {C}}:|z|<1\}$ is the unit disk. So we may assume that these roots are

$$\begin{aligned} \exp (i\theta _{1,n}),\exp (i\theta _{2,n}),\dots ,\exp (i\theta _{|\Lambda _n|,n}) \text { where } 0<\theta _{1,n}\le \theta _{2,n}\le \dots \le \theta _{|\Lambda _n|,n}<2\pi .\nonumber \\ \end{aligned}$$

(131)

Note that we have used the fact that $Z_{\Lambda _n,\beta }(1)>0$. By the spin-flip symmetry, $Z_{\Lambda _n,\beta ,-h}=Z_{\Lambda _n,\beta ,h}$ for any $h\in {\mathbb {C}}$. As a result, those $|\Lambda _n|$ roots in (131) are symmetric with respect to the real-axis. Combining (130) and (131), we have

$$\begin{aligned} Z_{\Lambda _n,\beta }(z)=\exp [\beta |E_n|-(\ln z)|\Lambda _n|/2]\prod _{j=1}^{|\Lambda _n|}\left[ z-\exp (i\theta _{j,n})\right] ,~ z\in {\mathbb {C}}\setminus (-\infty ,0]. \end{aligned}$$

(132)

Here we take out $(-\infty ,0]$ from ${\mathbb {C}}$ so that $h=-(\ln z)/2$ is analytic in this slit domain. Therefore,

$$\begin{aligned} \frac{\partial \ln Z_{\Lambda _n,\beta }(z)}{\partial z}=-\frac{|\Lambda _n|}{2z}+\sum _{j=1}^{|\Lambda _n|}\frac{1}{z-\exp (i\theta _{j,n})},~\forall z\in {\mathbb {D}}\setminus (-1,0]. \end{aligned}$$

(133)

It is easy to see that

$$\begin{aligned} \left\langle \sum _{u\in \Lambda _n}\sigma _u\right\rangle _{\Lambda _n,\beta ,h}= & {} \frac{\partial \ln Z_{\Lambda _n,\beta ,h}}{\partial h}=\frac{\partial \ln Z_{\Lambda _n,\beta }(z)}{\partial z}\frac{\partial z}{\partial h}\nonumber \\&=-2z\frac{\partial \ln Z_{\Lambda _n,\beta }(z)}{\partial z},~\forall z\in {\mathbb {D}}\setminus (-1,0]. \end{aligned}$$

(134)

We define the average magnetization density in $\Lambda _n$ by

$$\begin{aligned} m_{\Lambda _n}(z):=\frac{\left\langle \sum _{u\in \Lambda _n}\sigma _u\right\rangle _{\Lambda _n,\beta ,h}}{|\Lambda _n|},~\forall z\in {\mathbb {C}}\setminus \partial {\mathbb {D}}, \end{aligned}$$

(135)

where we have dropped the $\beta$ dependence in $m_{\Lambda _n}(z)$. By (133) and (134), we have

$$\begin{aligned} m_{\Lambda _n}(z)= & {} 1-\frac{2z}{|\Lambda _n|}\sum _{j=1}^{|\Lambda _n|}\frac{1}{z-\exp (i\theta _{j,n})}\nonumber \\&=\frac{1}{|\Lambda _n|}\sum _{j=1}^{|\Lambda _n|}\frac{\exp (i\theta _{j,n})+z}{\exp (i\theta _{j,n})-z},~\forall z\in {\mathbb {D}}\setminus (-1,0]. \end{aligned}$$

(136)

By using the definition of $\langle \cdot \rangle _{\Lambda ,\beta ,h}$ (see (128)), we know that $m_{\Lambda _n}(z)$ is a rational function of z with poles on $\partial {\mathbb {D}}$, and thus (136) holds for each $z\in {\mathbb {D}}$. We define the empirical distribution

$$\begin{aligned} \mu _n:=\frac{1}{|\Lambda _n|}\sum _{j=1}^{|\Lambda _n|}\delta _{\exp (i\theta _{j,n})}, \end{aligned}$$

(137)

where $\delta _{\exp (i\theta _{j,n})}$ is the unit Dirac point measure at $\exp (i\theta _{j,n})$. Then

$$\begin{aligned} m_{\Lambda _{n}}(z)=\int _{\partial {\mathbb {D}}}\frac{e^{i\theta }+z}{e^{i\theta }-z}d\mu _{n}(e^{i\theta }),~\forall z\in {\mathbb {D}}. \end{aligned}$$

(138)

Since $\mu _n$’s live on the unit circle, $\{\mu _n:n\in {\mathbb {N}}\}$ is tight. So there is a subsequence of ${\mathbb {N}}$, $\{n_k:k\in {\mathbb {N}}\}$, such that $\mu _{n_k}\Longrightarrow \mu$ as $k\rightarrow \infty$ where $\mu$ is some probability measure on $\partial {\mathbb {D}}$. We will see later that $\mu$ is actually unique. Therefore,

$$\begin{aligned} m_{\Lambda _{n_k}}(z)=\int _{\partial {\mathbb {D}}}\frac{e^{i\theta }+z}{e^{i\theta }-z}d\mu _{n_k}(e^{i\theta })\rightarrow \int _{\partial {\mathbb {D}}}\frac{e^{i\theta }+z}{e^{i\theta }-z}d\mu (e^{i\theta }) \text { as }k\rightarrow \infty ,~\forall z\in {\mathbb {D}}. \end{aligned}$$

(139)

Note that

$$\begin{aligned} \Re m_{\Lambda _n}(z)=\int _{\partial {\mathbb {D}}}\Re \frac{e^{i\theta }+z}{e^{i\theta }-z}d\mu _{n}(e^{i\theta })=\int _{\partial {\mathbb {D}}}\frac{1-|z|^2}{|e^{i\theta }-z|^2}d\mu _n(e^{i\theta })>0,~\forall z\in {\mathbb {D}}, \end{aligned}$$

(140)

and $m_{\Lambda _n}(0)=1$. So $m_{\Lambda _n}$ is a Herglotz function. See Sect. 8.4 of [26] for more details. From (138), we know

$$\begin{aligned} |m_{\Lambda _n}(z)|\le \frac{1+r}{1-r}, \text { for any } z \text { satisfying } |z|\le r<1. \end{aligned}$$

(141)

So $\{m_{\Lambda _n}:n\in {\mathbb {N}}\}$ is locally uniformly bounded. It is well-known that

$$\begin{aligned} \lim _{n\rightarrow \infty }m_{\Lambda _n}(z)=\langle \sigma _0\rangle _{{\mathbb {Z}}^d,\beta ,h},~\forall z\in (0,1), \end{aligned}$$

(142)

where $\langle \cdot \rangle _{{\mathbb {Z}}^d,\beta ,h}$ is the expectation with respect to the unique infinite volume measure when $\beta \ge 0$ and $h>0$ (see, e.g., Proposition 3.29 and Theorem 3.46 of [10]). So by Vitali’s theorem (see, e.g., Theorem B.25 of [10]),

$$\begin{aligned} m(z):=\lim _{n\rightarrow \infty }m_{\Lambda _n}(z) \end{aligned}$$

(143)

exists locally uniformly on ${\mathbb {D}}$ and m is a Herglotz function. Comparing (139) and (143), we obtain

$$\begin{aligned} m(z)=\int _{\partial {\mathbb {D}}}\frac{e^{i\theta }+z}{e^{i\theta }-z}d\mu (e^{i\theta }),~\forall z\in {\mathbb {D}}. \end{aligned}$$

(144)

We define

$$\begin{aligned} m(z):=-m(1/z),~\forall z\in {\mathbb {C}}\setminus \bar{{\mathbb {D}}}. \end{aligned}$$

(145)

Since $\langle \sigma _x\rangle _{\Lambda _n,\beta ,h}=-\langle \sigma _x\rangle _{\Lambda _n,\beta ,-h}$ for any $x\in \Lambda _n$ and $h\in {\mathbb {C}}\setminus i{\mathbb {R}}$, we get

$$\begin{aligned} m(z)=\lim _{n\rightarrow \infty }m_{\Lambda _n}(z),~\forall z\in {\mathbb {C}}\setminus \bar{{\mathbb {D}}}. \end{aligned}$$

(146)

The following Stieltjes inversion formula on page 12 of [26] will be very import to our analysis of $\mu$.

Theorem A

(Stieltjes Inversion Formula) Let $\gamma :=\{e^{it}: a<t<b\}$ be an open arc on $\partial {\mathbb {D}}$ with endpoints $e^{ia}$ and $e^{ib}$, $0<b-a<2\pi$. Then

$$\begin{aligned} \mu (\gamma )+\frac{1}{2}\mu (\{e^{ia}\})+\frac{1}{2}\mu (\{e^{ib}\})=\lim _{r\uparrow 1}\frac{1}{2\pi }\int _a^b \Re m(re^{i\theta }) d\theta . \end{aligned}$$

(147)

In particular, Theorem A implies the $\mu$ that we obtained from the subsequential limit is unique, that is, we have $\mu _n\Longrightarrow \mu$ as $n\rightarrow \infty$. We call $\mu$ the limiting distribution of Lee–Yang zeros. Note that $\mu$ is actually a function of $\beta$. Let $\beta _c(d)$ be the critical inverse temperature of the Ising model on ${\mathbb {Z}}^d$. We are ready to prove the main result about $\mu$.

Theorem B

For any $d\ge 1$ and any $\beta \in [0,\beta _c(d))$, there is $\epsilon >0$ (which only depends on $\beta$ and d) such that

$$\begin{aligned} \mu \left( \{e^{it}:-\epsilon<t<\epsilon \}\right) =0. \end{aligned}$$

(148)

Proof

By Theorem A,

$$\begin{aligned}&\mu \left( \{e^{it}:-\epsilon<t<\epsilon \}\right) +\frac{1}{2}\mu \left( \{e^{i\epsilon }\}\right) +\frac{1}{2}\mu \left( \{e^{i\epsilon }\}\right) =\lim _{r\uparrow 1}\frac{1}{2\pi }\int _{-\epsilon }^{\epsilon } \Re m(re^{i\theta }) d\theta \end{aligned}$$

(149)

$$\begin{aligned}&\quad =\lim _{r\uparrow 1}\frac{1}{2\pi }\left[ \int _{-\epsilon }^{0} \Re m(re^{i\theta }) d\theta +\int _{0}^{\epsilon } \Re m(re^{i\theta }) d\theta \right] \end{aligned}$$

(150)

$$\begin{aligned}&\quad =\lim _{r\uparrow 1}\frac{1}{2\pi }\left[ \int _{0}^{\epsilon } \Re m(re^{-i\theta }) d\theta +\int _{0}^{\epsilon } \Re m(re^{i\theta }) d\theta \right] . \end{aligned}$$

(151)

By Theorem 1.5 of [23], we have that m(z) is complex analytic in a neighborhood of $z=1$. So we may pick $\epsilon$ small such that m is analytic in $D(1,2\epsilon ):=\{z\in {\mathbb {C}}:|z-1|<2\epsilon \}$. Then both $\Re m(re^{-i\theta })$ and $\Re m(re^{i\theta })$ are bounded if $re^{-i\theta }$ and $re^{i\theta }$ are in $D(1,\epsilon )$. The dominated converge theorem implies that

$$\begin{aligned} \mu \left( \{e^{it}:-\epsilon<t<\epsilon \}\right) +\frac{1}{2}\mu \left( \{e^{i\epsilon }\}\right) +\frac{1}{2}\mu \left( \{e^{i\epsilon }\}\right) \end{aligned}$$

(152)

$$\begin{aligned} =\frac{1}{2\pi }\int _{0}^{\epsilon }\left[ \lim _{r\uparrow 1}\Re m(re^{-i\theta }) +\lim _{r\uparrow 1} \Re m(re^{i\theta })\right] d\theta . \end{aligned}$$

(153)

By (145) and (146), and continuity of m in $D(1,2\epsilon )$, we have

$$\begin{aligned} \lim _{r\uparrow 1}\Re m(re^{-i\theta })=- \lim _{r\uparrow 1}\Re m(r^{-1}e^{i\theta })=-\lim _{r\uparrow 1}\Re m(re^{i\theta }). \end{aligned}$$

(154)

Plugging this into (153), we get

$$\begin{aligned} \mu \left( \{e^{it}:-\epsilon<t<\epsilon \}\right) +\frac{1}{2}\mu \left( \{e^{i\epsilon }\}\right) +\frac{1}{2}\mu \left( \{e^{i\epsilon }\}\right) =0, \end{aligned}$$

(155)

which concludes the proof of the theorem. $\square$