Trace-class bounded-observable response

For a rank-one operator $R_{u,v}x=⟨x,v⟩u$, one has $\mathrm{\parallel }{R}_{u,v}{\mathrm{\parallel }}_1=\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$, and $\mathrm{\parallel }{R}_{u,v}-R_{u^{\mathrm{\prime }},v^{\mathrm{\prime }}}{\mathrm{\parallel }}_1\le \mathrm{\parallel }u-u^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }+\mathrm{\parallel }{u}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }v-v^{\mathrm{\prime }}\mathrm{\parallel }$. Strong continuity in [F2] therefore makes $t\mapsto U_0(t)R_{u,v}{U}_0(t{)}^{\ast }$ trace-norm continuous. Finite sums have the same property. By [F7], finite-rank SVD truncations $X_N$ of any trace-class $X$ satisfy $\mathrm{\parallel }X-X_N{\mathrm{\parallel }}_1\to 0$. The ideal estimate in [F3] and $\mathrm{\parallel }{U}_0(t)\mathrm{\parallel }=\mathrm{\parallel }{U}_0(t{)}^{\ast }\mathrm{\parallel }=1$ bound the conjugated tails uniformly by $\mathrm{\parallel }X-X_N{\mathrm{\parallel }}_1$, so $t\mapsto U_0(t)X{U}_0(t{)}^{\ast }$ is trace-norm continuous.

For finite-rank $X_N={\sum }_{j=1}^m{R}_{u_j,v_j}$, $B_t{X}_N={\sum }_j{R}_{B_t{u}_j,v_j}$ and $X_N{B}_t={\sum }_j{R}_{u_j,B_t{v}_j}$ because $B_t=B_t^{\ast }$. Strong continuity of $B_t$ makes both curves trace-norm continuous. For general trace-class $X$, take the finite-rank approximants from [F7] and use $\mathrm{\parallel }{B}_t(X-X_N){\mathrm{\parallel }}_1,\mathrm{\parallel }(X-X_N)B_t{\mathrm{\parallel }}_1\le \mathrm{\parallel }B\mathrm{\parallel }\mathrm{\parallel }X-X_N{\mathrm{\parallel }}_1$ uniformly in $t$. Thus $X\mapsto \frac{ih(t)}{\mathrm{\hslash }}[B_t,X]$ is strongly continuous on the trace-class Banach space and has norm bound $2\mathrm{\mid }h(t)\mathrm{\mid }\mathrm{\parallel }B\mathrm{\parallel }\mathrm{/}\mathrm{\hslash }$.

Apply [F6] on the trace-class Banach space with $K_t(X)=\frac{ih(t)}{\mathrm{\hslash }}[B_t,X]$ and $x_0=\rho$. By [F1] and [F8], $\rho$ is positive, has trace one, and satisfies $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1=1$; its Volterra solution has the claimed trace-norm linear approximation and remainder bound with $C_T=\frac{2}{\mathrm{\hslash }}{\int }_0^T\mathrm{\mid }h(s)\mathrm{\mid }\mathrm{\parallel }B\mathrm{\parallel }ds$. The Hilbert clause of [F6], applied to $K_t=ih(t)B_t\mathrm{/}\mathrm{\hslash }$ on $\mathcal{H}$, gives a unitary solution operator $W(t)=W_{ϵ}(t)$ satisfying $W^{\mathrm{\prime }}(t)u=ϵih(t)B_{t}W(t)u\mathrm{/}\mathrm{\hslash }$ for every $u\in \mathcal{H}$. For a rank-one $R_{u,v}$, $W(t)R_{u,v}W(t{)}^{\ast }=R_{W(t)u,W(t)v}$ is trace-norm differentiable and its derivative is $ϵK_t(W(t)R_{u,v}W(t{)}^{\ast })$, by the vector equation and skew-adjointness. Finite sums give the same integral equation. Approximate $\rho$ by finite-rank ${\rho }_N$ in trace norm using [F7]. Unitary invariance of trace norm follows from the ideal estimate [F3] applied to $W(t)$ and $W(t{)}^{\ast }$, both of norm one; hence $W(t){\rho }_{N}W(t{)}^{\ast }\to W(t)\rho W(t{)}^{\ast }$ uniformly, while the right-hand integral errors are at most $\mathrm{\mid }ϵ\mathrm{\mid }{C}_T\mathrm{\parallel }\rho -{\rho }_N{\mathrm{\parallel }}_1$. Thus $W(t)\rho W(t{)}^{\ast }$ solves the trace-class Volterra equation, and uniqueness in [F6] identifies it with ${\rho }_I(t)$.

The linear term in the expectation is $\frac{iϵ}{\mathrm{\hslash }}{\int }_0^{t}h(s)\mathrm{Tr}([B_s,\rho ]{\alpha }_t(A))ds$. By [F4], the trace integrand equals $\mathrm{Tr}(\rho [{\alpha }_t(A),B_s])$. The group law and stationarity give $[{\alpha }_t(A),B_s]={\alpha }_s([{\alpha }_{t-s}(A),B])$ and $\mathrm{Tr}(\rho {\alpha }_s(C))=\mathrm{Tr}(\rho C)$ for bounded $C$. Hence the integrand is $\mathrm{Tr}(\rho [{\alpha }_{t-s}(A),B])$, producing the stated convolution and response kernel.

The trace-norm remainder in step 1.3 changes the expectation by at most $\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }{\rho }_I(t)-\rho -ϵ{\int }_0^t{K}_s(\rho )ds{\mathrm{\parallel }}_1$ by [F3, F5]. This gives the stated bound. Since $A,B$ are self-adjoint and $\rho$ is positive, the commutator trace is purely imaginary by cyclicity; hence $R_{AB}(u)$ is real. No moment of $H$ occurs in any trace.