Conditional finite stationary response

Fix one cell $i$ of the preparation in Actually constructed Gibbs/KMS preparation at its assigned event $p_i\in O$, with $N_i=N_{\xi }(p_i)$, $T_i=T(p_i)$, and ${\mu }_i=\mu (p_i)$. Let $H_i=H_i^{\ast }$ be its proper Hamiltonian in joules, $Q_i$ its dimensionless number matrix, $G_i:=H_i-{\mu }_i{Q}_i$, and $\sigma :={\rho }_i$ its constructed density. Put ${\beta }_K:=(k_B{T}_K{)}^{-1}>0$ in joules${}^{-1}$ and ${\beta }_{\mathrm{l}\mathrm{o}\mathrm{c},i}:=N_i{\beta }_K=(k_B{T}_i{)}^{-1}$. Take Hermitian matrices $A,B\in \mathcal{B}({\mathcal{H}}_i)$ with $[A,Q_i]=[B,Q_i]=0$. Let $h:[0,S]\to \mathbb{R}$ be continuous for $S>0$ seconds, let $ϵ\in \mathbb{R}$ be dimensionless, and require $h(\tau )B$ to have units of joules. Define the local and Killing-time dynamics by

${\alpha }_{\tau }^{\mathrm{l}\mathrm{o}\mathrm{c}}(C)=e^{i{G}_i\tau \mathrm{/}\mathrm{\hslash }}C{e}^{-i{G}_i\tau \mathrm{/}\mathrm{\hslash }},{\alpha }_t^K(C)=e^{i{N}_i{G}_{i}t\mathrm{/}\mathrm{\hslash }}C{e}^{-i{N}_i{G}_{i}t\mathrm{/}\mathrm{\hslash }}\mathrm{.}$

Let $H_{K,i}:=N_i{H}_i+q{\mathrm{\Phi }}_i{Q}_i$ be the cell's physical Killing-energy operator, where $q$ is the common charge in coulombs and ${\mathrm{\Phi }}_i={\mathrm{\Phi }}_{\xi }(p_i)$ is in volts as in the preparation. Let $U_{\mathrm{l}\mathrm{o}\mathrm{c}}:[0,S]\to \mathcal{B}({\mathcal{H}}_i)$ and $U_K:[0,S\mathrm{/}{N}_i]\to \mathcal{B}({\mathcal{H}}_i)$ be the unique solutions of $i\mathrm{\hslash }\frac{d}{d\tau }{U}_{\mathrm{l}\mathrm{o}\mathrm{c}}(\tau )=(H_i-ϵh(\tau )B)U_{\mathrm{l}\mathrm{o}\mathrm{c}}(\tau ),U_{\mathrm{l}\mathrm{o}\mathrm{c}}(0)=I_{{\mathcal{H}}_i},$ $i\mathrm{\hslash }\frac{d}{dt}{U}_K(t)=(H_{K,i}-ϵN_{i}h(N_{i}t)B)U_K(t),U_K(0)=I_{{\mathcal{H}}_i}\mathrm{.}$ They are unitary propagators. Define the model expectations by $E_{\mathrm{l}\mathrm{o}\mathrm{c}}(\tau ):=\mathrm{Tr}\text{\hspace{0.17em}⁣}(\sigma U_{\mathrm{l}\mathrm{o}\mathrm{c}}(\tau {)}^{\ast }A{U}_{\mathrm{l}\mathrm{o}\mathrm{c}}(\tau )),E_K(t):=\mathrm{Tr}\text{\hspace{0.17em}⁣}(\sigma U_K(t{)}^{\ast }A{U}_K(t))\mathrm{.}$ Because $A$ and $B$ commute with $Q_i$, the second expectation is unchanged if $H_{K,i}$ is replaced by $N_i{H}_i$; that is the neutral-sector version of the Killing control. The exact finite-cell predictions are

$E_K(t)=E_{\mathrm{l}\mathrm{o}\mathrm{c}}(N_{i}t)(0\le t\le S\mathrm{/}{N}_i),$

and

$E_{\mathrm{l}\mathrm{o}\mathrm{c}}(\tau )=\mathrm{Tr}(\sigma {\alpha }_{\tau }^{\mathrm{l}\mathrm{o}\mathrm{c}}(A))+ϵ{\int }_0^{\tau }h(s)R_{\mathrm{l}\mathrm{o}\mathrm{c}}(\tau -s)ds+r(\tau ),$

where

$R_{\mathrm{l}\mathrm{o}\mathrm{c}}(u)=\frac{i}{\mathrm{\hslash }}\mathrm{Tr}\text{\hspace{0.17em}⁣}(\sigma [{\alpha }_u^{\mathrm{l}\mathrm{o}\mathrm{c}}(A),B]),C_S=\frac{2}{\mathrm{\hslash }}{\int }_0^S\mathrm{\mid }h(s)\mathrm{\mid }\mathrm{\parallel }B\mathrm{\parallel }ds,\mathrm{\mid }r(\tau )\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }\frac{(\mathrm{\mid }ϵ\mathrm{\mid }{C}_S{)}^2{e}^{\mathrm{\mid }ϵ\mathrm{\mid }{C}_S}}{2}\mathrm{.}$

Here $C_S$ is dimensionless and $R_{\mathrm{l}\mathrm{o}\mathrm{c}}$ has units $[A][B]\mathrm{/}(\mathrm{J}\mathrm{s})$. For the equilibrium correlations $C_{\mathrm{l}\mathrm{o}\mathrm{c}}(u)=\mathrm{Tr}(\sigma {\alpha }_u^{\mathrm{l}\mathrm{o}\mathrm{c}}(A)A)$ and $C_K(t)=\mathrm{Tr}(\sigma {\alpha }_t^K(A)A)$, let $S_{\mathrm{l}\mathrm{o}\mathrm{c}}$ and $S_K$ be their Fourier measures in the distributional convention $S(\omega )={\int }_{\mathbb{R}}{e}^{i\omega t}C(t)dt$, where $\omega$ is in seconds${}^{-1}$. Set $D_{N_i}(\omega )=N_i\omega$. Then

$S_K=(D_{N_i}{)}_{\ast }{S}_{\mathrm{l}\mathrm{o}\mathrm{c}},S_K(\omega )=N_i^{-1}{S}_{\mathrm{l}\mathrm{o}\mathrm{c}}(\omega \mathrm{/}{N}_i),$

the detailed-balance factor is unchanged because ${\beta }_K\mathrm{\hslash }(N_i\omega )={\beta }_{\mathrm{l}\mathrm{o}\mathrm{c},i}\mathrm{\hslash }\omega$, and any zero-frequency atom remains at zero and is kept separate from the $\coth$ formula. Finally, for the separately re-prepared finite Gibbs family ${\sigma }_{\lambda }\propto e^{-{\beta }_{\mathrm{l}\mathrm{o}\mathrm{c},i}(G_i-\lambda B)}$, where $\lambda B$ has units of joules,

${\frac{d}{d\lambda }\mathrm{Tr}({\sigma }_{\lambda }B)\mid }_{\lambda =0}={\beta }_{\mathrm{l}\mathrm{o}\mathrm{c},i}{\mathrm{Cov}}_{\mathrm{K}\mathrm{M},\sigma }(B,B)\ge 0,$

with ${\mathrm{Cov}}_{\mathrm{K}\mathrm{M},\sigma }(B,B)={\int }_0^1\mathrm{Tr}({\sigma }^{1-u}\delta B{\sigma }^u\delta B)du$ and $\delta B=B-\mathrm{Tr}(\sigma B)I$. It equals the ordinary variance when $[B,G_i]=0$; otherwise no such reduction is asserted. These are conditional model predictions, not reported observations or a theorem of relaxation.