Trace-class bounded-observable response — Examples

Finite-support sequences are dense in ${\mathrm{\ell }}^2$, so $D(H)$ is dense. The diagonal real operator is symmetric. If $y\in D(H^{\ast })$, testing the adjoint relation against each $e_n$ forces $(H^{\ast }y{)}_n=\mathrm{\hslash }\omega (n+\frac{1}{2})y_n$; this vector belongs to ${\mathrm{\ell }}^2$ exactly when ${\sum }_n(n+\frac{1}{2}{)}^2\mathrm{\mid }{y}_n{\mathrm{\mid }}^2<\mathrm{\infty }$. Thus $D(H^{\ast })=D(H)$ and $H^{\ast }=H$.

The Gibbs eigenvalues are $(1-q)q^n$, since ${\sum }_{n\ge 0}{e}^{-\beta \mathrm{\hslash }\omega (n+1\mathrm{/}2)}=e^{-\beta \mathrm{\hslash }\omega \mathrm{/}2}\mathrm{/}(1-q)$; hence $Z_{\beta }=e^{-\beta \mathrm{\hslash }\omega \mathrm{/}2}\mathrm{/}(1-q)$ is finite for every $\beta >0$ and [F1] applies. The Gibbs matrix is positive trace one and commutes with $U(t)$ because both are functions of $H$. The operator $B$ has norm $\mathrm{\mid }{b}_0\mathrm{\mid }$; its image lies in $\mathrm{span}\{e_0,e_1\}\subset D(H)$, so it maps $D(H)$ into itself. Its commutator $C=[H,B]$ is the finite-rank bounded operator $C{e}_0=\mathrm{\Delta }{b}_0{e}_1$, $C{e}_1=-\mathrm{\Delta }{b}_0{e}_0$, $C{e}_n=0$ for $n\ge 2$.

Since $B$ only connects $e_0$ and $e_1$, the unperturbed expectation $\mathrm{Tr}({\sigma }_{\beta }{\alpha }_t(B))$ is zero. Direct multiplication of the two nonzero matrix blocks gives $\mathrm{Tr}({\sigma }_{\beta }[{\alpha }_t(B),B])=-2i{b}_0^2(1-q{)}^2\sin (\omega t)$, so $R_{BB}(t)=2b_0^2(1-q{)}^2\sin (\omega t)\mathrm{/}\mathrm{\hslash }$. The constant-field convolution in [F2] is therefore $ϵh_0{\int }_0^t{R}_{BB}(u)du=\frac{2ϵh_0{b}_0^2(1-q{)}^2}{\mathrm{\Delta }}(1-\cos (\omega t))\mathrm{.}$ Here $C_T=2\mathrm{\mid }{h}_0\mathrm{\mid }\mathrm{\mid }{b}_0\mathrm{\mid }T\mathrm{/}\mathrm{\hslash }$; the response theorem also bounds the omitted term by $\mathrm{\mid }{b}_0\mathrm{\mid }(\mathrm{\mid }ϵ\mathrm{\mid }{C}_T{)}^2{e}^{\mathrm{\mid }ϵ\mathrm{\mid }{C}_T}\mathrm{/}2$.

The two-dimensional space $V=\mathrm{span}\{e_0,e_1\}$ and its orthogonal complement are invariant under the constant controlled Hamiltonian. Put $d=\mathrm{\Delta }\mathrm{/}2>0$, $g=ϵh_0{b}_0$, and $R=(d^2+g^2{)}^{1\mathrm{/}2}>0$. On $V$, let $X=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and $Z=\mathrm{diag}(1,-1)$. Then $H-ϵh_{0}B=\mathrm{\Delta }I-dZ-gX$ and $(dZ+gX{)}^2=R^{2}I$. Its exact propagator on this block is therefore $e^{-it\mathrm{\Delta }\mathrm{/}\mathrm{\hslash }}[\cos (Rt\mathrm{/}\mathrm{\hslash })I+i\sin (Rt\mathrm{/}\mathrm{\hslash })(dZ+gX)\mathrm{/}R]$. The initial diagonal block has entries $p_0=1-q$ and $p_1=(1-q)q$, with $p_0-p_1=(1-q{)}^2$; the complementary levels contribute zero to the expectation of $B$. Multiplying these two-by-two matrices gives the exact expectation $\mathrm{Tr}({\rho }_{ϵ}(t)B)=\frac{b_{0}gd(1-q{)}^2}{R^2}(1-\cos (2Rt\mathrm{/}\mathrm{\hslash }))\mathrm{.}$ For $ϵ=0$ this is zero, and its first derivative at $ϵ=0$ is the coefficient computed in step 3.1. The formula holds on every stated finite control interval. For the consistent extension using the same constant control for all $t\ge 0$, it is nonconstant and periodic whenever $ϵh_0{b}_0\mathrm{\ne }0$, so the exact expectation has no infinite-time limit; this conclusion follows from the exact block formula, not from extrapolating the finite-time remainder. This concerns the bounded projected transition observable, not oscillator position or an empirical measurement.

Direct multiplication gives $X^2=Y^2=Z^2=I$, $YX=-iZ$, and $[Z,X]=2iY$. The Gibbs matrix is $\sigma =(I-\tanh (x)Z)\mathrm{/}2$, so $\mathrm{Tr}(\sigma Z)=-\tanh (x)$ and $\mathrm{Tr}(\sigma X)=0$.

Conjugation by $e^{-itH\mathrm{/}\mathrm{\hslash }}$ yields ${\alpha }_t(X)=X\cos (\omega t)-Y\sin (\omega t)$. Hence $C_{XX}(t)=\mathrm{Tr}(\sigma {\alpha }_t(X)X)=\cos (\omega t)-i\tanh (x)\sin (\omega t)$ and $R_{XX}(t)=2\tanh (x)\sin (\omega t)\mathrm{/}\mathrm{\hslash }$. The spectral measure has its two transition atoms at $\pm \omega$ and satisfies $S_{XX}(-\omega )=e^{-\beta \mathrm{\Delta }}{S}_{XX}(\omega )$, in agreement with [F2].

For comparison, the static Gibbs family for $H(h)=H-hX$ has $Z(h)=2\cosh (\beta \sqrt{(\mathrm{\Delta }\mathrm{/}2{)}^2+h^2})$. By [F3], ${\mathrm{\partial }}_h\log Z(h)=\beta \mathrm{Tr}({\sigma }_{h}X)$, so $\mathrm{Tr}({\sigma }_{h}X)=\frac{h}{\sqrt{(\mathrm{\Delta }\mathrm{/}2{)}^2+h^2}}\tanh \text{\hspace{0.17em}⁣}\left(\beta \sqrt{(\mathrm{\Delta }\mathrm{/}2{)}^2+h^2}\right)$ and ${{\mathrm{\partial }}_h\mathrm{Tr}({\sigma }_{h}X)\mid }_{h=0}=\frac{2}{\mathrm{\Delta }}\tanh (x)$. Since ${\mathrm{Var}}_{\sigma }(X)=1$ and $\tanh (x)<x$ for $x>0$ (the derivative of $x-\tanh x$ is ${\tanh }^{2}x>0$), this static susceptibility is strictly below $\beta {\mathrm{Var}}_{\sigma }(X)=\beta$. Thus the noncommuting Gibbs derivative is not replaced by $\beta$ times ordinary variance.

Integrating the kernel for constant $h_0$ gives $\delta ⟨X(t)⟩=\frac{2ϵh_0}{\mathrm{\Delta }}\tanh (x)(1-\cos (\omega t))+r_{ϵ}(t)$, with $\mathrm{\mid }{r}_{ϵ}(t)\mathrm{\mid }\le (\mathrm{\mid }ϵ\mathrm{\mid }{C}_T{)}^2{e}^{\mathrm{\mid }ϵ\mathrm{\mid }{C}_T}\mathrm{/}2$ and $C_T=2\mathrm{\mid }{h}_0\mathrm{\mid }T\mathrm{/}\mathrm{\hslash }$, because $\mathrm{\parallel }X\mathrm{\parallel }=1$.

When $ϵh_0\mathrm{\ne }0$, the leading response oscillates without a long-time limit; the finite-time theorem does not imply relaxation to the static Gibbs derivative. This is an exact model calculation, not an observation.