KMS implies Gibbs in finite quantum systems

Statement

Let $\mathcal H$ be a finite-dimensional Hilbert space and let $H=H^\ast$ be a (bounded) Hamiltonian. Consider the Heisenberg time evolution

If a state $\omega$ on $\mathcal B(\mathcal H)$ satisfies the finite-volume KMS condition at inverse temperature $\beta>0$ for the dynamics $(\alpha_t)_{t\in\mathbb R}$, then $\omega$ is the quantum Gibbs state for $H$ at inverse temperature $\beta$, i.e.

where $\operatorname{Tr}(e^{-\beta H})$ is the quantum partition function .

Equivalently, if $\omega(A)=\operatorname{Tr}(\rho A)$ for a density operator $\rho$, then

This is the converse direction to Gibbs implies KMS .

Key hypotheses

• Finite-dimensional algebra $\mathcal B(\mathcal H)$ (so every state is normal and given by a density matrix).
• Dynamics is inner and generated by a self-adjoint $H$ via $\alpha_t(A)=e^{itH}Ae^{-itH}$.
• $\omega$ is a $\beta$-KMS state for some $\beta\in(0,\infty)$ in the sense of KMS .

Key conclusions

• $\omega$ must coincide with the Gibbs state for $H$ at inverse temperature $\beta$.
• In particular, for fixed $(H,\beta)$ the $\beta$-KMS state is unique on $\mathcal B(\mathcal H)$.
• Shifting $H\mapsto H+cI$ does not change $\omega$ (the constant cancels in normalization).

Proof idea / significance

Write $\omega(A)=\operatorname{Tr}(\rho A)$ for a density matrix $\rho$.

1. Apply the KMS identity with $A=I$ to get $\omega(\alpha_{i\beta}(B))=\omega(B)$ for all $B$ (analyticity is automatic in finite dimensions). In trace form this implies

   $$e^{\beta H}\rho e^{-\beta H}=\rho,$$

   hence $[\rho,H]=0$.

2. Use the full KMS relation

   $$\operatorname{Tr}(\rho\,A\,e^{-\beta H}Be^{\beta H})=\operatorname{Tr}(\rho\,BA)$$

   and cyclicity of the trace (together with $[\rho,H]=0$) to show that $\rho e^{\beta H}$ commutes with every matrix, hence $\rho e^{\beta H}=cI$ for some scalar $c$.

3. Therefore $\rho=c\,e^{-\beta H}$, and normalization $\operatorname{Tr}(\rho)=1$ fixes $c=\operatorname{Tr}(e^{-\beta H})^{-1}$.

Significance: in finite volume, “equilibrium = KMS” is equivalent to “equilibrium = Gibbs”. This equivalence motivates using the KMS condition as the definition of equilibrium in infinite-volume quantum statistical mechanics, where Gibbs density matrices may not exist.