Proper clock two-level response

The eigenvectors of $Z$ have energies $+\mathrm{\Delta }\mathrm{/}2$ and $-\mathrm{\Delta }\mathrm{/}2$. Therefore $\mathrm{Tr}(e^{-{\beta }_{\mathrm{l}\mathrm{o}\mathrm{c}}{H}_{\mathrm{l}\mathrm{o}\mathrm{c}}})=e^{-{\beta }_{\mathrm{l}\mathrm{o}\mathrm{c}}\mathrm{\Delta }\mathrm{/}2}+e^{{\beta }_{\mathrm{l}\mathrm{o}\mathrm{c}}\mathrm{\Delta }\mathrm{/}2}=2\cosh ({\beta }_{\mathrm{l}\mathrm{o}\mathrm{c}}\mathrm{\Delta }\mathrm{/}2)$, and $\mathrm{Tr}(\sigma Z)=p_{+}-p_{-}=-\tanh ({\beta }_{\mathrm{l}\mathrm{o}\mathrm{c}}\mathrm{\Delta }\mathrm{/}2)$.

With $\theta =\mathrm{\Delta }\tau \mathrm{/}\mathrm{\hslash }$, direct multiplication using [L1] gives $e^{i{H}_{\mathrm{l}\mathrm{o}\mathrm{c}}\tau \mathrm{/}\mathrm{\hslash }}X{e}^{-i{H}_{\mathrm{l}\mathrm{o}\mathrm{c}}\tau \mathrm{/}\mathrm{\hslash }}=X\cos \theta -Y\sin \theta$. Thus $[{\alpha }_{\tau }^{\mathrm{l}\mathrm{o}\mathrm{c}}(X),X]=-\sin \theta [Y,X]=2iZ\sin \theta$, so the Kubo definition gives $R_{\mathrm{l}\mathrm{o}\mathrm{c}}(\tau )=\frac{i}{\mathrm{\hslash }}\mathrm{Tr}(\sigma [{\alpha }_{\tau }^{\mathrm{l}\mathrm{o}\mathrm{c}}(X),X])=\frac{2}{\mathrm{\hslash }}\tanh ({\beta }_{\mathrm{l}\mathrm{o}\mathrm{c}}\mathrm{\Delta }\mathrm{/}2)\sin \theta$.

The only nonzero matrix elements of $X$ in the $Z$ eigenbasis exchange the two energy eigenvectors, so the two transition frequencies are $-\mathrm{\Delta }\mathrm{/}\mathrm{\hslash }$ from the $+$ state and $+\mathrm{\Delta }\mathrm{/}\mathrm{\hslash }$ from the $-$ state. Their Gibbs weights are $p_{+}$ and $p_{-}$ from step 1.1; hence the stated two-line spectral measure follows from the finite spectral expansion. The cell theorem, invoked for the assigned geometric cell under its inherited assumption [F2], sends each frequency $\omega$ to $N\omega$ without changing its weight, giving the Killing-time lines and periods in the Example.

Put $R(\lambda )=\sqrt{(\mathrm{\Delta }\mathrm{/}2{)}^2+{\lambda }^2}$. Since $H_{\mathrm{l}\mathrm{o}\mathrm{c}}-\lambda X=(\mathrm{\Delta }Z\mathrm{/}2-\lambda X)$ and $ZX+XZ=0$ by [L1], its square is $R(\lambda {)}^{2}I$, so $e^{-{\beta }_{\mathrm{l}\mathrm{o}\mathrm{c}}(H_{\mathrm{l}\mathrm{o}\mathrm{c}}-\lambda X)}=\cosh ({\beta }_{\mathrm{l}\mathrm{o}\mathrm{c}}R(\lambda ))I-\frac{\sinh ({\beta }_{\mathrm{l}\mathrm{o}\mathrm{c}}R(\lambda ))}{R(\lambda )}(H_{\mathrm{l}\mathrm{o}\mathrm{c}}-\lambda X)\mathrm{.}$ Taking traces and using $\mathrm{Tr}(H_{\mathrm{l}\mathrm{o}\mathrm{c}}X)=0$ and $\mathrm{Tr}(X^2)=2$ from [L1] gives $\mathrm{Tr}({\sigma }_{\lambda }X)=\lambda \frac{\tanh ({\beta }_{\mathrm{l}\mathrm{o}\mathrm{c}}R(\lambda ))}{R(\lambda )}\mathrm{.}$ Differentiating this explicit scalar expression at $\lambda =0$ yields the displayed static susceptibility. It is a derivative of a newly normalized Gibbs family, whereas the finite-time Kubo response in step 2.1 keeps $\sigma$ fixed.

Since $R_K(t)$ is a nonzero sinusoid when ${\beta }_{\mathrm{l}\mathrm{o}\mathrm{c}}>0$ and $\mathrm{\Delta }>0$, it is periodic and does not decay as $t\to \mathrm{\infty }$. The theorem supplies no absolute-integrability hypothesis for a Green–Kubo limit, so the oscillatory finite response cannot be relabelled as a dc transport coefficient.