Finite unitary dynamics and Gibbs KMS identity

On a zero-dimensional Hilbert space the unique endomorphism is $I=0$, and its exponential is $I$, so the unitary and group identities hold directly. There is no trace-one density there; the remaining density and normalized-Gibbs clauses have no instance. On a nonzero space, diagonalize $H$ by [F1], with eigenvalues $E_j$. Then $U(t)$ has eigenvalues $e^{-it{E}_j\mathrm{/}\mathrm{\hslash }}$ of modulus one, so it is unitary. In the same eigenbasis, multiplying scalar exponentials gives $U(t+s)=U(t)U(s)$. The finite exponential series differentiates termwise and yields $i\mathrm{\hslash }{U}^{\mathrm{\prime }}(t)=HU(t)$.

In the eigenbasis of $H$, every matrix entry of $e^{izH\mathrm{/}\mathrm{\hslash }}B{e}^{-izH\mathrm{/}\mathrm{\hslash }}$ is a constant times $e^{iz(E_j-E_k)\mathrm{/}\mathrm{\hslash }}$, an entire scalar function. Thus $F_{A,B}$ is entire. For $z=t+iy$, $0\le y\le \beta \mathrm{\hslash }$, the real-time factors are unitary and the imaginary-time factors have norm at most $e^{\beta \mathrm{\parallel }H\mathrm{\parallel }}$ each. The trace against a density is bounded by operator norm, so $\mathrm{\mid }{F}_{A,B}(z)\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }{e}^{2\beta \mathrm{\parallel }H\mathrm{\parallel }}$ throughout the closed strip.

For finite matrices $C,D$, the coordinate trace gives $\mathrm{Tr}(CD)={\sum }_{i,j}{C}_{ij}{D}_{ji}=\mathrm{Tr}(DC)$ by interchanging the two finite indices. This cyclicity gives $\mathrm{Tr}(U\rho U^{\ast }A)=\mathrm{Tr}(\rho U^{\ast }AU)$. Unitary conjugation preserves positivity, trace, and eigenvalues; therefore it preserves the finite entropy $S(\rho )=-k_B{\sum }_{j}f(p_j)$, where $p_j$ are the density eigenvalues, $f(x)=x\log x$ for $x>0$, and $f(0)=0$. Thus $\mathrm{Tr}(\rho \log \rho )$ denotes $\mathrm{Tr}f(\rho )$, including singular densities; no value of $\log 0$ is used. Since ${\sigma }_{\beta }$ is a function of $H$, it commutes with $U(t)$, hence is stationary.

At $z=t+i\beta \mathrm{\hslash }$, ${\alpha }_z(B)=e^{-\beta H}{\alpha }_t(B)e^{\beta H}$. Hence $F_{A,B}(t+i\beta \mathrm{\hslash })=Z_{\beta }^{-1}\mathrm{Tr}(e^{-\beta H}A{e}^{-\beta H}{\alpha }_t(B)e^{\beta H})=Z_{\beta }^{-1}\mathrm{Tr}(A{e}^{-\beta H}{\alpha }_t(B))=\mathrm{Tr}({\sigma }_{\beta }{\alpha }_t(B)A),$ where the first equality uses the definition, the second uses finite trace cyclicity, and the third uses it again. The strip boundary identity follows.