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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.

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Acknowledgements

The research of the second author was partially supported by NSFC grant 11901394 and that of the third author by US-NSF grant DMS-1507019. The authors thank two anonymous reviewers for useful comments and suggestions.

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

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. \(\square \)

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Camia, F., Jiang, J. & Newman, C.M. Ising Model with Curie–Weiss Perturbation. J Stat Phys 188, 5 (2022). https://doi.org/10.1007/s10955-022-02935-1

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Keywords

  • Ising model
  • Curie–Weiss interaction
  • Analytic density
  • Pure imaginary zeros
  • High dimensions
  • Periodic boundary conditions