Specified finite GNS detailed balance — Examples

Since $Z$ commutes with $\sigma$ and $Z^2=I$, the map $A\mapsto ZAZ$ is self-adjoint in the $\sigma$-weighted inner product and is an involution. Hence $D_{\varphi }$ is self-adjoint and nonpositive, annihilates diagonal matrices, and sends $X,Y$ to $-{\gamma }_{\varphi }X,-{\gamma }_{\varphi }Y$. Adding it to the Gibbs-jump dissipator preserves stationarity and the scalar kernel: the sum of its two nonpositive quadratic forms can vanish only in the scalar kernel of the Gibbs-jump part. Thus [F2] and [F3] apply to this full generator. The two-state population equation is $\dot{z}=-{\gamma }_1(z-z_{\mathrm{e}\mathrm{q}})$, while the off-diagonal equation is ${\dot{\rho }}_{01}=-({\gamma }_2+i\omega ){\rho }_{01}$, by direct multiplication of the two jump matrices and the dephaser. The dual observables satisfy $G(X)=-{\gamma }_{2}X-\omega Y$ and $G(Y)=\omega X-{\gamma }_{2}Y$, so $e^{tG}X=e^{-{\gamma }_{2}t}(X\cos \omega t-Y\sin \omega t)$; also $e^{tG}(Z-z_{\mathrm{e}\mathrm{q}}I)=e^{-{\gamma }_{1}t}(Z-z_{\mathrm{e}\mathrm{q}}I)$.

Applying [F3] with $A=B=X$, the commutator identity $[Y,X]=-2iZ$ and $z_{\mathrm{e}\mathrm{q}}=-\tanh (\beta \mathrm{\Delta }\mathrm{/}2)$ give $R_{XX}(t)=2\tanh (\beta \mathrm{\Delta }\mathrm{/}2)e^{-{\gamma }_{2}t}\sin (\omega t)\mathrm{/}\mathrm{\hslash }$. Since ${\int }_0^{\mathrm{\infty }}{e}^{-{\gamma }_{2}t}\sin (\omega t)dt=\omega \mathrm{/}({\gamma }_2^2+{\omega }^2)$, the displayed ${\chi }_{\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{n}}(0)$ follows.

For $q\in \mathbb{R}$, $H(q)=\mathrm{\Delta }Z\mathrm{/}2-qX$ satisfies $H(q{)}^2=((\mathrm{\Delta }\mathrm{/}2{)}^2+q^2)I$. Put $\lambda (q)=\sqrt{(\mathrm{\Delta }\mathrm{/}2{)}^2+q^2}$. The finite spectral calculus gives $e^{-\beta H(q)}=\cosh (\beta \lambda (q))I-\frac{\sinh (\beta \lambda (q))}{\lambda (q)}H(q)$, hence the partition function $\mathcal{Z}(q):=\mathrm{Tr}(e^{-\beta H(q)})=2\cosh (\beta \lambda (q))$ and $\mathrm{Tr}(e^{-\beta H(q)}X)\mathrm{/}\mathcal{Z}(q)=q\tanh (\beta \lambda (q))\mathrm{/}\lambda (q)$. Differentiating at $q=0$ gives $2\tanh (\beta \mathrm{\Delta }\mathrm{/}2)\mathrm{/}\mathrm{\Delta }=2\tanh (\beta \mathrm{\Delta }\mathrm{/}2)\mathrm{/}(\mathrm{\hslash }\omega )$; [F4] identifies this derivative as the static Gibbs susceptibility.

Using $X^2=Y^2=Z^2=I$, $YX=-iZ$, $\mathrm{Tr}(\sigma Z)=z_{\mathrm{e}\mathrm{q}}$, and $\mathrm{Tr}(\sigma X)=\mathrm{Tr}(\sigma Y)=0$ in step 1.1 gives the stated $C_{XX}$ and $C_{ZZ}$. Integrating their real parts yields ${\mathrm{\Gamma }}_X={\int }_0^{\mathrm{\infty }}{e}^{-{\gamma }_{2}t}\cos (\omega t)dt={\gamma }_2\mathrm{/}({\gamma }_2^2+{\omega }^2)$ and ${\mathrm{\Gamma }}_Z=(1-z_{\mathrm{e}\mathrm{q}}^2){\int }_0^{\mathrm{\infty }}{e}^{-{\gamma }_{1}t}dt=(1-z_{\mathrm{e}\mathrm{q}}^2)\mathrm{/}{\gamma }_1$, consistent with [F2].

The two susceptibilities differ because their denominators are respectively ${\gamma }_2^2+{\omega }^2$ and ${\omega }^2$, with ${\gamma }_2>0$. The dynamic formula used [F3]'s fixed-dissipator perturbation, whereas the static formula differentiates a Gibbs state whose Hamiltonian changes with $q$. Therefore no equilibrium identity equating them is asserted for this model.

The Gibbs weights are ${\pi }_0=b\mathrm{/}(a+b)$ and ${\pi }_1=a\mathrm{/}(a+b)$: their ratio is $b\mathrm{/}a=e^{-\beta \mathrm{\Delta }}=e^{-\beta (E_0-E_1)}$, and they sum to one. Thus ${\pi }_{0}a={\pi }_{1}b$ and $\pi (v)=0$. The two-state generator has centered eigenvector $Z-z_{\mathrm{e}\mathrm{q}}I$ with eigenvalue $-(a+b)$; therefore $e^{sW}v=e^{-(a+b)s}v$ and $⟨v,v{⟩}_{\pi }=u^2(1-z_{\mathrm{e}\mathrm{q}}^2)$.

Substituting this covariance into [F1] gives

Evaluating the integral gives the displayed finite-time formula. Dividing by $2T$ and letting $T\to \mathrm{\infty }$ gives $u^2(1-z_{\mathrm{e}\mathrm{q}}^2)\mathrm{/}(a+b)$.