Consider a quantum system described by an algebra $\mathcal{A}$ of local observables, which are supported on subsets of the lattice $\Lambda = \mathbb{Z}^{d}$. Given an observable $A$, let $A(t) = e^{i H t} A e^{-i H t}$ denote its time-evolution by the Hamiltonian $H$. Let us restrict in this question to short-range Hamiltonians, in the sense that $H$ can be written as $H = \sum_{X} h_{X}$, where the local term $h_{X}$ is supported on $X \subset \Lambda$, and $\parallel h_{X} \parallel$ is bounded and decays sufficiently rapidly with $\mathrm{diam}(X)$.

A state $\rho$ of this system is a linear functional on $\mathcal{A}$, whose action on an observable $A$ we will write as $\langle A \rangle_{\rho}$. We say that $\rho$ is a $\beta$-KMS state if for any local observables $A,B\in \mathcal{A}$ we have

$$ \langle A(t) B \rangle_{\rho} = \langle B A(t + i \beta) \rangle_{\rho}. $$

We further say that $\rho$ is an extremal $\beta$-KMS state if it cannot be written as a convex combination

$$\rho = \lambda \rho_{1} + (1-\lambda) \rho_{2},$$

with $0 < \lambda < 1$, where $\rho_{1}$ and $\rho_{2}$ are both $\beta$-KMS states.

Finally, we say that a state $\rho$ satisfies cluster decomposition if, for local observables $A_{X}$ and $B_{Y}$ respectively supported on regions $X,Y \subset \Lambda$, we have the decay to zero of the connected correlation function,

$$ \left| \langle A_{X} B_{Y} \rangle_{\rho} - \langle A_{X} \rangle_{\rho} \langle B_{Y} \rangle_{\rho} \right| \to 0, $$

as the distance between their supports $d(X,Y) \to \infty$.


Is it the case that all extremal $\beta$-KMS states of short-range Hamiltonians satisfy cluster decomposition? Does the answer depend on $d$ or $\beta$?

If this has not been proved, I would still be interested in knowing whether it is expected to be true, or if there are known counterexamples.


I am particularly interested in the low-temperature limit $\beta \to \infty$, where e.g. symmetry-breaking might be relevant. I also do not mind how rapidly the correlations go to zero.

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    $\begingroup$ You should look at Simon's The Statistical Mechanics of Lattice Gases (see, for instance, Corollary IV.1.8). $\endgroup$ Jun 4, 2021 at 5:56

1 Answer 1


Following Yvan Velenik's comment, the answer is affirmative for all finite temperatures, and can be found in Corollary IV.4.17 of Simon's The Statistical Mechanics of Lattice Gases. I will give the text of the corollary below, along with any relevant definitions.

Definition: Let $\mathcal{P}_{f}(\mathbb{Z}^{d})$ denote the set of all finite subsets of $\mathbb{Z}^{d}$.

Definition: Given the algebra $\mathcal{A}$ of quasilocal observables and a subset $X \subset \mathbb{Z}^{d}$, let $\mathcal{A}_{X}$ denote the subset of $\mathcal{A}$ consisting of observables supported on $X$.

Below will be made reference to a "strongly continuous one-parameter automorphism group" $\alpha$. One can have in mind time-evolution by a Hamiltonian, i.e. $\alpha = \{\alpha_{t}\}$, where $\alpha_{t}$ acts on an observable $A$ as $\alpha_{t}(A) = e^{i H t} A e^{-i H t}$.

Definition: Given a strongly continuous one-parameter automorphism group $\alpha$, let $K_{\beta}(\alpha)$ denote the set of $\beta$-KMS states with respect to $\alpha$.

Corollary IV.4.17: Let $\alpha$ be a strongly continuous one-parameter automorphism group, and let $\beta \neq 0$. Then $\rho \in K_{\beta}(\alpha)$ is an extreme point of that set if and only if, for all [quasilocal observables] $A \in \mathcal{A}$ and $\epsilon > 0$, there is a $\Lambda \in \mathcal{P}_{f}(\mathbb{Z}^{d})$ with $$| \langle A B \rangle_{\rho} - \langle A \rangle_{\rho} \langle B \rangle_{\rho} | \leq \epsilon \parallel \!\! B \parallel$$ for all $B \in \mathcal{A}_{\mathbb{Z}^{d} \setminus \Lambda}$.


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