KMS state

Remarks

Every KMS state is invariant: $\omega \circ \tau_t = \omega$ for all $t$. Example: let $A = M_n(\C)$, let $H$ be a Hermitian matrix, and let $\tau_t(b) = e^{itH} b e^{-itH}$. The Gibbs state $\omega(a) = \tr(e^{-\beta H} a) / \tr(e^{-\beta H})$ satisfies the condition with $F_{a, b}(z) = \tr(e^{-\beta H} a e^{izH} b e^{-izH}) / \tr(e^{-\beta H})$. At $z = t + i \beta$, cyclicity of the trace turns this into $\omega(\tau_t(b) \, a)$; for this $A$ and $\tau$ the Gibbs state is the only KMS state. Non-example: if $H$ is not a multiple of the identity, the normalized trace $\frac{1}{n} \tr$ is not a $(\tau, \beta)$-KMS state for $\beta > 0$. For $a$ and $b$ entire analytic for $\tau$, the condition reads $\omega(a \, \tau_{i \beta}(b)) = \omega(b a)$.