KMS Equilibrium States in Thermodynamic Limits

Finite-volume Gibbs states can converge to an infinite-volume KMS state even when the total partition function diverges. The essential input is convergence of both the states and their dynamics on a common observable algebra. This page proves that passage to the limit and explains the role of state extensions and boundary terms. The canonical KMS treatment explains the equilibrium interpretation and Gibbs derivation; the C*-dynamical KMS page develops analytic elements and closure for fixed dynamics.

Required background. Thermodynamic limits of local states supplies weak-star cluster points. The C*-dynamical KMS condition supplies the Gaussian analytic-element criterion used below.

Equilibrium limits involve changing dynamics

Let $\mathcal A$ be a unital C*-algebra. Suppose $\alpha^{(n)}$ and $\alpha$ are point-norm continuous automorphism groups such that, for every $B\in\mathcal A$ and $T<\infty$,

$$\sup_{|t|\leq T} \|\alpha_t^{(n)}(B)-\alpha_t(B)\| \longrightarrow0.$$

Fix $\beta>0$, use physical time, and put $\tau=\beta\hbar$. Let $\omega_n$ be $\beta$-KMS for $\alpha^{(n)}$, and suppose $\omega_n\to\omega$ weak-star. Then $\omega$ is $\beta$-KMS for $\alpha$.

This assertion differs from closure of the KMS set for one fixed group: the finite-volume automorphisms themselves depend on $n$. It does not assume norm convergence of the Hamiltonians generating them.

Gaussian smearing controls complex time uniformly

For $\epsilon>0$ define the entire analytic elements

$$B_\epsilon^{(n)} =\frac1{\sqrt{\pi\epsilon}} \int_{\mathbb R}e^{-t^2/\epsilon}\alpha_t^{(n)}(B)\,dt,$$

and similarly $B_\epsilon$ using $\alpha$. Real-time convergence and the isometric norm bound imply

$$B_\epsilon^{(n)}\longrightarrow B_\epsilon \quad\text{in norm}.$$

To justify this, first truncate the integral to $[-T,T]$, where convergence is uniform, and then bound both tails by $\|B\|$ times the Gaussian tail.

The analytic extension is

$$\alpha_z^{(n)}(B_\epsilon^{(n)}) =\frac1{\sqrt{\pi\epsilon}} \int_{\mathbb R} e^{-(t-z)^2/\epsilon}\alpha_t^{(n)}(B)\,dt.$$

For $z=i\tau$, the absolute value of its scalar kernel is $e^{\tau^2/\epsilon}e^{-t^2/\epsilon}$. The same argument therefore proves

$$\alpha_{i\tau}^{(n)}(B_\epsilon^{(n)}) \longrightarrow\alpha_{i\tau}(B_\epsilon) \quad\text{in norm}.$$

The bound is independent of $n$. No uniform estimate on the unsmeared operator $\alpha_{i\tau}^{(n)}(B)$ was assumed or derived.

Passing the analytic KMS identity to the limit

For every $A\in\mathcal A$, the finite-system KMS identity gives

$$\omega_n\!\left(A\alpha_{i\tau}^{(n)} (B_\epsilon^{(n)})\right) =\omega_n(B_\epsilon^{(n)}A).$$

Weak-star convergence permits fixed algebra elements as tests. The norm convergence just proved permits the elements in this equation to vary: if $C_n\to C$ in norm, then

$$|\omega_n(C_n)-\omega(C)| \leq\|C_n-C\|+|\omega_n(C)-\omega(C)| \longrightarrow0.$$

Consequently,

$$\omega(A\alpha_{i\tau}(B_\epsilon)) =\omega(B_\epsilon A).$$

Gaussian analytic elements are norm dense. The analytic-element criterion proved at the abstract KMS treatment now establishes the full strip condition for $\omega$. This completes the limit proof without repeating the fixed-dynamics closure argument.

The theorem also applies along a convergent subsequence or subnet. It establishes equilibrium of every cluster point, not uniqueness of the cluster point.

Finite-volume Gibbs states need a compatible extension

For finite-dimensional spin sites, let $H_{\Lambda_n}$ act inside an exhausting sequence of volumes and let

$$\rho_n= \frac{e^{-\beta H_{\Lambda_n}}} {\operatorname{Tr}e^{-\beta H_{\Lambda_n}}}.$$

On the whole quasi-local algebra define $\alpha_t^{(n)}=\operatorname{Ad}(e^{itH_{\Lambda_n}/\hbar})$. Its dynamics is trivial outside the volume. Extend $\rho_n$ by the product normalized trace on the exterior sites. The resulting state $\omega_n$ is KMS for this global group: finite-volume Gibbs cyclicity handles the interior, while the tracial state handles the trivial exterior dynamics.

An arbitrary exterior state would not suffice for this particular proof. For identity dynamics the KMS condition requires a trace, so a nontracial exterior state generally destroys the global KMS property. Nevertheless all exterior extensions have the same eventual values on each fixed local observable. Their local cluster points agree whenever the interior expectations converge.

Under the bounded finite-range lattice assumptions of Lieb–Robinson theory, finite-volume dynamics converges in the required compact-time norm sense on local observables and hence on the completion. More general interactions require that theorem’s summability hypotheses.

Boundary terms are admissible when the resulting finite-volume dynamics has the same limiting group. This condition must be verified, usually through distance-to-boundary estimates and uniform interaction bounds. Arbitrarily large or nonlocal boundary perturbations are not covered by calling them boundary conditions.

Independent spins give an explicit equilibrium limit

For $H_\Lambda=-h\sum_{x\in\Lambda}Z_x$,

$$Z_\Lambda=(2\cosh(\beta h))^{|\Lambda|}, \qquad \omega(Z_x)=\tanh(\beta h).$$

Every local Gibbs marginal stabilizes exactly. For a local observable the finite-volume dynamics also stabilizes once its support is inside the volume. The theorem gives a KMS state on the infinite quasi-local algebra, although $Z_\Lambda$ diverges as the volume grows.

This illustrates why a divergent global partition function does not exclude equilibrium. It also does not produce a density matrix in every preselected representation; macroscopic sector distinctions constrain normality.

Exercises

Why real-time convergence alone is insufficient without state convergence. Let a sequence have two distinct weak-star cluster points. What does the theorem conclude?

Solution

Apply the theorem separately along each convergent subsequence. Both limits are KMS for the same limiting dynamics and inverse temperature. It gives no equality between them. A uniqueness theorem is additional input, and multiple equilibrium limits can be physically meaningful.

An exterior-state counterexample. Set every finite-volume Hamiltonian equal to zero and choose a nontracial product state outside the volume. Is the extension KMS for the global identity dynamics?

Solution

Not at finite $n$: identity dynamics requires $\omega(AB)=\omega(BA)$. A nontracial exterior one-site state violates this for suitable matrix units. Using the normalized trace outside avoids the failure. The exterior choice still disappears from each eventual local marginal.

References

• O. Bratteli and D. W. Robinson, Operator Algebras and Quantum Statistical Mechanics 2: Equilibrium States, Models in Quantum Statistical Mechanics, 2nd ed., Springer, 1997, doi:10.1007/978-3-662-03444-6.
• B. Nachtergaele, R. Sims, and A. Young, “Quasi-locality bounds for quantum lattice systems. Part I. Lieb–Robinson bounds, quasi-local maps, and spectral flow automorphisms,” Journal of Mathematical Physics 60, 061101, 2019, doi:10.1063/1.5095769, author text.