Metastability of Blume–Capel Model with Zero Chemical Potential and Zero External Field

Abstract

In this study, we investigate the metastable behavior of Metropolis-type Glauber dynamics associated with the Blume–Capel model with zero chemical potential and zero external field at very low temperatures. The corresponding analyses for the same model with zero chemical potential and positive small external field were performed in Cirillo and Nardi (J Stat Phys 150:1080–1114, 2013) and Landim and Lemire (J Stat Phys 164:346–376, 2016). We obtain both large deviation-type and potential-theoretic results on the metastable behavior in our setting. To this end, we perform highly thorough investigation on the energy landscape, where it is revealed that no critical configurations exist and alternatively a massive flat plateau of saddle configurations resides therein.

Appendix: Auxiliary Process

1.1 Original Auxiliary Process

Fig. 7

(Left) graph structure $({\widehat{V}}_{K},\,E({\widehat{V}}_{K}))$. (Right) graph structure $(V_{K},\,E(V_{K}))$

In this subsection, we define an auxiliary process which successfully represents the Metropolis dynamics on the edge typical configurations. For $K\ge 5$, we define a graph structure $({\widehat{V}}_{K},\,E({\widehat{V}}_{K}))$ (see Fig. 7 (left) for an illustration for the case of $K=5$). First, the vertex set ${\widehat{V}}_{K}\subseteq {\mathbb {R}}^{2}$ is defined by

$$\begin{aligned} {\widehat{V}}_{K}=\{(a,\,b)\in {\mathbb {R}}^{2}:0\le b\le a\le K\text { and }b\le 2\}\setminus \{(K,\,2)\}. \end{aligned}$$

(A.1)

Then, the edge structure $E({\widehat{V}}_{K})$ is inherited by the Euclidean lattice. We abbreviate by $0=(0,\,0)\in {\widehat{V}}_{K}$ and define

$$\begin{aligned} {\widehat{A}}_{K}=\{(a,\,b)\in {\widehat{V}}_{K}:a=K\text { or }b=2\}. \end{aligned}$$

Then, we define $\{{\widehat{Z}}_{K}(t)\}_{t\ge 0}$ as the continuous-time random walk on the aforementioned graph whose transition rate is uniformly 1. In other words, the transition rate ${\widehat{r}}_{K}:{\widehat{V}}_{K}\times {\widehat{V}}_{K}\rightarrow [0,\,\infty )$ is given by

$$\begin{aligned} {\widehat{r}}_{K}(x,\,y)={\left\{ \begin{array}{ll} 1 &{} \text {if }\{x,\,y\}\in E({\widehat{V}}_{K}),\\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Obviously, the process is reversible with respect to the uniform distribution on ${\widehat{V}}_{K}$.

We denote by ${\widehat{h}}_{\cdot ,\cdot }^{K}(\cdot )$ and $\widehat{\text {cap}}_{K}(\cdot ,\,\cdot )$ the equilibrium potential and capacity with respect to ${\widehat{Z}}_{K}(\cdot )$, respectively, in the sense of Definition 3.1. We define a constant ${{\mathfrak {c}}}_{K}>0$ by

$$\begin{aligned} {{\mathfrak {c}}}_{K}=|{\widehat{V}}_{K}|\widehat{\text {cap}}_{K}(0,\,{\widehat{A}}_{K}). \end{aligned}$$

(A.2)

Then, we have the following asymptotic lemma.

Lemma A.1

There exists a positive constant $\delta$ with $|\delta -0.435|<0.0001$ such that

$$\begin{aligned} \lim _{K\rightarrow \infty }{{\mathfrak {c}}}_{K}=\delta . \end{aligned}$$

Proof

We explicitly compute the equilibrium potential ${\widehat{h}}_{0,{\widehat{A}}_{K}}^{K}(\cdot )$. For simplicity, we write $h={\widehat{h}}_{0,{\widehat{A}}_{K}}^{K}$ and abbreviate by $h(a,\,b)=h((a,\,b))$ for $(a,\,b)\in {\widehat{V}}_{K}$. We define

$$\begin{aligned} a_{i}=h(K-i,\,0)\;\;\;\;;\;i\in \llbracket 0,\,K\rrbracket \;\;\;\;\text {and}\;\;\;\;b_{i}=h(K-i,\,1)\;\;\;\;;\;i\in \llbracket 0,\,K-1\rrbracket . \end{aligned}$$

Then, we trivially have $a_{K}=h(0,\,0)=1$,

$$\begin{aligned} a_{0}=h(K,\,0)=0\;\;\;\;\text {and}\;\;\;\;b_{0}=h(K,\,1)=0. \end{aligned}$$

(A.3)

Moreover, by the Markov property, the following recurrence relations hold:

$$\begin{aligned}&3a_{i} =a_{i+1}+a_{i-1}+b_{i}\;\;\;\;;\;i\in \llbracket 1,\,K-1\rrbracket , \end{aligned}$$

(A.4)

$$\begin{aligned}&4b_{i} =b_{i+1}+b_{i-1}+a_{i}\;\;\;\;;\;i\in \llbracket 1,\,K-2\rrbracket , \end{aligned}$$

(A.5)

$$\begin{aligned}&2b_{K-1}=b_{K-2}+a_{K-1}\;\;\;\;\text {and}\;\;\;\;3a_{K-1}=1+a_{K-2}+b_{K-1}. \end{aligned}$$

(A.6)

Then, (A.4) and (A.5) induce the following relations:

$$\begin{aligned} a_{i+2}-7a_{i+1}+13a_{i}-7a_{i-1}+a_{i-2}&=0\;\;\;;\;i\in \llbracket 2,\,K-3\rrbracket , \end{aligned}$$

(A.7)

$$\begin{aligned} b_{i+2}-7b_{i+1}+13b_{i}-7b_{i-1}+b_{i-2}&=0\;\;\;;\;i\in \llbracket 2,\,K-3\rrbracket . \end{aligned}$$

(A.8)

Hence, we solve $t^{4}-7t^{3}+13t^{2}-7t+1=0$, which is equivalent to $(t+t^{-1})^{2}-7(t+t^{-1})+11=0$. This gives $t+t^{-1}=(7\pm \sqrt{5})/2$. Thus, we define the positive constants $\alpha _{1},\,\alpha _{2},\,\alpha _{3},\,\alpha _{4}>0$ such that $\alpha _{1}>\alpha _{2}$ are the solutions of $t+t^{-1}=(7+\sqrt{5})/2$ and $\alpha _{3}>\alpha _{4}$ are the solutions of $t+t^{-1}=(7-\sqrt{5})/2$. Then, there exist constants $p_{n}$ and $q_{n}$, $n\in \llbracket 1,\,4\rrbracket$ such that we have, for $i\in \llbracket 0,\,K-1\rrbracket$,

$$\begin{aligned} a_{i}=\sum _{n=1}^{4}p_{n}\alpha _{n}^{i}\;\;\;\;\text {and}\;\;\;\;b_{i}=\sum _{n=1}^{4}q_{n}\alpha _{n}^{i}. \end{aligned}$$

(A.9)

Based on the last formula, (A.3) implies

$$\begin{aligned} p_{1}+p_{2}+p_{3}+p_{4}=q_{1}+q_{2}+q_{3}+q_{4}=0, \end{aligned}$$

(A.10)

and (A.4) implies, for $i\in \llbracket 1,\,K-1\rrbracket$,

$$\begin{aligned} \sum _{n=1}^{4}\alpha _{n}^{i-1}\{p_{n}\alpha _{n}^{2}-(3p_{n}-q_{n})\alpha _{n}+p_{n}\}=0. \end{aligned}$$

As $K\ge 5$, this implies that

$$\begin{aligned} \begin{bmatrix}1 &{}\quad 1 &{}\quad 1 &{}\quad 1\\ \alpha _{1} &{}\quad \alpha _{2} &{}\quad \alpha _{3} &{}\quad \alpha _{4}\\ \alpha _{1}^{2} &{}\quad \alpha _{2}^{2} &{}\quad \alpha _{3}^{2} &{}\quad \alpha _{4}^{2}\\ \alpha _{1}^{3} &{}\quad \alpha _{2}^{3} &{}\quad \alpha _{3}^{3} &{}\quad \alpha _{4}^{3} \end{bmatrix}\begin{bmatrix}p_{1}\alpha _{1}^{2}-(3p_{1}-q_{1})\alpha _{1}+p_{1}\\ p_{2}\alpha _{2}^{2}-(3p_{2}-q_{2})\alpha _{2}+p_{2}\\ p_{3}\alpha _{3}^{2}-(3p_{3}-q_{3})\alpha _{3}+p_{3}\\ p_{4}\alpha _{4}^{2}-(3p_{4}-q_{4})\alpha _{4}+p_{4} \end{bmatrix}=0. \end{aligned}$$

As the square matrix is invertible (cf. Vandermonde matrix), we must have $p_{n}\alpha _{n}^{2}-(3p_{n}-q_{n})\alpha _{n}+p_{n}=0$ for all $n\in \llbracket 1,\,4\rrbracket$, which implies that

$$\begin{aligned} 3p_{n}-q_{n}=\frac{7+\sqrt{5}}{2}p_{n}\;\;\;\;;\;n=1,\,2,\;\;\;\;\text {and}\;\;\;\;3p_{n}-q_{n}=\frac{7-\sqrt{5}}{2}p_{n}\;\;\;\;;\;n=3,\,4. \end{aligned}$$

Hence, substituting (A.10) and the last display to (A.9) gives, for $i\in \llbracket 0,\,K-1\rrbracket$,

$$\begin{aligned} a_{i}= & {} p_{1}(\alpha _{1}^{i}-\alpha _{2}^{i})+p_{3}(\alpha _{3}^{i}-\alpha _{4}^{i}),\nonumber \\ b_{i}= & {} -\frac{1+\sqrt{5}}{2}p_{1}(\alpha _{1}^{i}-\alpha _{2}^{i})+\frac{-1+\sqrt{5}}{2}p_{3}(\alpha _{3}^{i}-\alpha _{4}^{i}). \end{aligned}$$

(A.11)

Substituting the last formula to (A.6) implies

$$\begin{aligned}&(2+\sqrt{5})p_{1}(\alpha _{1}^{K-1}-\alpha _{2}^{K-1})+(2-\sqrt{5})p_{3}(\alpha _{3}^{K-1}-\alpha _{4}^{K-1})\\&=\frac{1+\sqrt{5}}{2}p_{1}(\alpha _{1}^{K-2}-\alpha _{2}^{K-2})+\frac{1-\sqrt{5}}{2}p_{3}(\alpha _{3}^{K-2}-\alpha _{4}^{K-2}) \end{aligned}$$

and

$$\begin{aligned}&\frac{7+\sqrt{5}}{2}p_{1}(\alpha _{1}^{K-1}-\alpha _{2}^{K-1})+\frac{7-\sqrt{5}}{2}p_{3}(\alpha _{3}^{K-1}-\alpha _{4}^{K-1})\\&=1+p_{1}(\alpha _{1}^{K-2}-\alpha _{2}^{K-2})+p_{3}(\alpha _{3}^{K-2}-\alpha _{4}^{K-2}). \end{aligned}$$

Solving the last two equations, we can express $p_{1}$ and $p_{3}$ in terms of $\alpha _{1},\,\alpha _{2},\,\alpha _{3},\,\alpha _{4}$. Substituting these to the first equation of (A.11) for $i=K-1$, we deduce that $a_{K-1}$ equals

$$\begin{aligned} -\Big [(2-\sqrt{5})-\frac{1-\sqrt{5}}{2}\frac{\alpha _{3}^{K-2}-\alpha _{4}^{K-2}}{\alpha _{3}^{K-1}-\alpha _{4}^{K-1}}\Big ]+\Big [(2+\sqrt{5})-\frac{1+\sqrt{5}}{2}\frac{\alpha _{1}^{K-2}-\alpha _{2}^{K-2}}{\alpha _{1}^{K-1}-\alpha _{2}^{K-1}}\Big ] \end{aligned}$$

divided by

$$\begin{aligned}&\Big [(2+\sqrt{5})-\frac{1+\sqrt{5}}{2}\frac{\alpha _{1}^{K-2}-\alpha _{2}^{K-2}}{\alpha _{1}^{K-1}-\alpha _{2}^{K-1}}\Big ]\Big [\frac{7-\sqrt{5}}{2}-\frac{\alpha _{3}^{K-2}-\alpha _{4}^{K-2}}{\alpha _{3}^{K-1}-\alpha _{4}^{K-1}}\Big ]\\&-\Big [(2-\sqrt{5})-\frac{1-\sqrt{5}}{2}\frac{\alpha _{3}^{K-2}-\alpha _{4}^{K-2}}{\alpha _{3}^{K-1}-\alpha _{4}^{K-1}}\Big ]\Big [\frac{7+\sqrt{5}}{2}-\frac{\alpha _{1}^{K-2}-\alpha _{2}^{K-2}}{\alpha _{1}^{K-1}-\alpha _{2}^{K-1}}\Big ]. \end{aligned}$$

As $\alpha _{1}>\alpha _{2}$ and $\alpha _{3}>\alpha _{4}$, we have $(\alpha _{2}/\alpha _{1})^{K-1}\rightarrow 0$ and $(\alpha _{4}/\alpha _{3})^{K-1}\rightarrow 0$ as $K\rightarrow \infty$. Thus, we may calculate

$$\begin{aligned} \lim _{K\rightarrow \infty }a_{K-1}&=\frac{-\Big [(2-\sqrt{5})-\frac{1-\sqrt{5}}{2}\frac{1}{\alpha _{3}}\Big ]+\Big [(2+\sqrt{5})-\frac{1+\sqrt{5}}{2}\frac{1}{\alpha _{1}}\Big ]}{\Big [(2+\sqrt{5})-\frac{1+\sqrt{5}}{2}\frac{1}{\alpha _{1}}\Big ]\Big [\frac{7-\sqrt{5}}{2}-\frac{1}{\alpha _{3}}\Big ]-\Big [(2-\sqrt{5})-\frac{1-\sqrt{5}}{2}\frac{1}{\alpha _{3}}\Big ]\Big [\frac{7+\sqrt{5}}{2}-\frac{1}{\alpha _{1}}\Big ]}\\&=\frac{2\sqrt{5}-\frac{1+\sqrt{5}}{2}\alpha _{2}+\frac{1-\sqrt{5}}{2}\alpha _{4}}{5\sqrt{5}+\frac{3-5\sqrt{5}}{2}\alpha _{2}+\frac{-3-5\sqrt{5}}{2}\alpha _{4}+\sqrt{5}\alpha _{2}\alpha _{4}}. \end{aligned}$$

In the second equality, we used that $\alpha _{1}\alpha _{2}=\alpha _{3}\alpha _{4}=1$. By substituting the exact values of $\alpha _{i}$, this is asymptotically 0.5649853624. Moreover, as $0=(0,\,0)\in {\widehat{V}}_{K}$ is connected only to $(1,\,0)\in {\widehat{V}}_{K}$, we have by [18, (7.1.39)] that

$$\begin{aligned} \widehat{\text {cap}}_{K}(0,\,{\widehat{A}}_{K})=\frac{1}{|{\widehat{V}}_{K}|}[h(0,\,0)-h(1,\,0)]=\frac{1-a_{K-1}}{|{\widehat{V}}_{K}|}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \lim _{K\rightarrow \infty }{{\mathfrak {c}}}_{K}=\lim _{K\rightarrow \infty }|{\widehat{V}}_{K}|\widehat{\text {cap}}_{K}(0,\,{\widehat{A}}_{K})=1-\lim _{K\rightarrow \infty }a_{K-1}\approx 0.4350146376, \end{aligned}$$

which concludes the proof. $\square$

Remark 1

In the periodic boundary case, we need a completely different auxiliary process to estimate the structure of edge typical configurations. Namely, the desired process is a Markov chain on the collection of subtrees of a $K\times 2$-shaped ladder graph with semi-periodic boundary conditions (i.e., open on the horizontal boundaries and periodic on the vertical ones). In this case, we deduce an upper bound for the corresponding capacity, which is sufficient to obtain the bound (9.3). We refer to [24, Proposition 7.9] for more information on this estimate.

1.2 Projected Auxiliary Process

Based on the original auxiliary process defined in the preceding subsection, we define a projected auxiliary process which is obtained by simply projecting the elements in ${\widehat{A}}_{K}$ to a single element ${{\mathfrak {d}}}$. Rigorously, we define a graph structure $(V_{K},\,E(V_{K}))$ (see Fig. 7 (right) for an illustration for the case of $K=5$). The vertex set $V_{K}\subseteq {\mathbb {R}}^{2}$ is defined by

$$\begin{aligned} V_{K}=({\widehat{V}}_{K}\setminus {\widehat{A}}_{K})\cup \{{{\mathfrak {d}}}\}. \end{aligned}$$

(A.12)

Then, the edge structure $E(V_{K})$ is inherited by $E({\widehat{V}}_{K})$; we have $\{x,\,y\}\in E(V_{K})$ for $\{x,\,y\}\in E({\widehat{V}}_{K})$, $x,\,y\in {\widehat{V}}_{K}\setminus {\widehat{A}}_{K}$, and we have $\{x,\,{{\mathfrak {d}}}\}\in E(V_{K})$ for

$$\begin{aligned} x\in {\widehat{V}}_{K}\setminus {\widehat{A}}_{K}\;\;\;\;\text {satisfying}\;\;\;\;\exists y\in {\widehat{A}}_{K}\;\;\;\;\text {with}\;\;\;\;\{x,\,y\}\in E({\widehat{V}}_{K}). \end{aligned}$$

Then, we define $\{Z_{K}(t)\}_{t\ge 0}$ as the continuous-time Markov chain on $(V_{K},\,E(V_{K}))$ whose transition rate $r_{K}$ is defined by $r_{K}(x,\,y)={\widehat{r}}_{K}(x,\,y)$ if $x,\,y\ne {{\mathfrak {d}}}$ and

$$\begin{aligned} r_{K}(x,\,{{\mathfrak {d}}})=r_{K}({{\mathfrak {d}}},\,x)=\sum _{y\in {\widehat{A}}_{K}}{\widehat{r}}_{K}(x,\,y). \end{aligned}$$

This process is reversible with respect to the uniform distribution on $V_{K}$.

We denote by $h_{\cdot ,\cdot }^{K}(\cdot )$, $\text {cap}_{K}(\cdot ,\,\cdot )$, $D_{K}(\cdot )$ the equilibrium potential, capacity, and Dirichlet form with respect to $Z_{K}(\cdot )$, respectively, in the sense of Definition 3.1. Then, by the strong Markov property, it is immediate from the definition that

$$\begin{aligned} h_{0,{{\mathfrak {d}}}}^{K}(x)={\widehat{h}}_{0,{\widehat{A}}_{K}}^{K}(x)\;\;\;\;;\;x\in {\widehat{V}}_{K}\setminus {\widehat{A}}_{K}\;\;\;\;\text {and}\;\;\;\;h_{0,{{\mathfrak {d}}}}^{K}({{\mathfrak {d}}})={\widehat{h}}_{0,{\widehat{A}}_{K}}^{K}(y)=0\;\;\;\;;\;y\in {\widehat{A}}_{K}. \end{aligned}$$

Therefore, by (3.4), we have (cf. (A.2))

$$\begin{aligned} |V_{K}|\text {cap}_{K}(0,\,{{\mathfrak {d}}})=|{\widehat{V}}_{K}|\widehat{\text {cap}}_{K}(0,\,{\widehat{A}}_{K})={{\mathfrak {c}}}_{K}. \end{aligned}$$

(A.13)