Duhamel derivative and finite Kubo–Mori response

On a compact interval of $s$, the matrix norms of $M(s)$ and $M^{\mathrm{\prime }}(s)$ are bounded. The exponential power series and its differentiated series converge uniformly there, since the derivative of $M^n$ is ${\sum }_{j=0}^{n-1}{M}^j{M}^{\mathrm{\prime }}{M}^{n-1-j}$, whose norm is at most $n\mathrm{\parallel }M{\mathrm{\parallel }}^{n-1}\mathrm{\parallel }{M}^{\mathrm{\prime }}\mathrm{\parallel }$. Expanding both exponentials in the integral, integrating term by term, and using ${\int }_0^1(1-u{)}^j{u}^{k}du=j!k!\mathrm{/}(j+k+1)!$ (obtained by integration by parts) gives the same differentiated series. This proves the Duhamel formula.

For finite matrices $C,D$, $\mathrm{Tr}(CD)={\sum }_{i,j}{C}_{ij}{D}_{ji}=\mathrm{Tr}(DC)$ by interchanging finite indices. Take the trace in step 1.1. This cyclicity makes every integrand trace equal to $\mathrm{Tr}(e^M{M}^{\mathrm{\prime }})$. Since $Z=\mathrm{Tr}{e}^M>0$, the scalar chain rule gives $(\log Z{)}^{\mathrm{\prime }}=\mathrm{Tr}(e^M{M}^{\mathrm{\prime }})\mathrm{/}Z=\mathrm{Tr}(\rho M^{\mathrm{\prime }})$. Setting $M^{\mathrm{\prime }}=X_a$ proves the first control derivative.

Differentiate $\mathrm{Tr}(\rho X_b)$ in the $X_a$ direction using step 1.1 and the quotient rule. The result is ${\int }_0^1\mathrm{Tr}({\rho }^{1-u}{X}_a{\rho }^u{X}_b)du-\mathrm{Tr}(\rho X_a)\mathrm{Tr}(\rho X_b)\mathrm{.}$ Because ${\rho }^{1-u}{\rho }^u=\rho$, cyclicity shows that replacing each $X$ by its centered version $\delta X$ changes the integral by exactly the subtracted product, giving the stated Hessian.

Diagonalize the positive definite $\rho$ with eigenvalues $p_i>0$. For Hermitian $X$, the quadratic form equals ${\sum }_{i,j}\mathrm{\mid }\delta X_{ij}{\mathrm{\mid }}^{2}L(p_i,p_j)$, where $L(a,b)={\int }_0^1{a}^{1-u}{b}^{u}du>0$. Thus it is real and nonnegative. For $a\mathrm{\ne }b$, direct integration gives $L(a,b)=(a-b)\mathrm{/}(\log a-\log b)$, and $L(a,a)=a$. Concavity of $\log$ gives $(1-u)\log a+u\log b\le \log ((1-u)a+ub)$, strictly when $a\mathrm{\ne }b$ and $0<u<1$. Exponentiating and integrating yields $L(a,b)\le (a+b)\mathrm{/}2$, strictly when $a\mathrm{\ne }b$. Substituting the matrix entries yields the upper bound by $\mathrm{Tr}(\rho (\delta X{)}^2)={\mathrm{Var}}_{\rho }(X)$. If $X$ commutes with $\rho$, its entries between distinct eigenspaces vanish, so equality holds. Conversely, if $X$ does not commute with $\rho$, some $\delta X_{ij}\mathrm{\ne }0$ has $p_i\mathrm{\ne }{p}_j$; its term is strictly below the arithmetic-mean bound while every other term is at most that bound, so the total inequality is strict. Thus equality holds exactly in the commuting case.

For $H(h)=H_0-hB$, put $M=-\beta H(h)$. Then $M^{\mathrm{\prime }}=\beta B$, so step 3.1 gives ${\mathrm{\partial }}_h\mathrm{Tr}({\rho }_{h}B)=\beta {\mathrm{Cov}}_{KM,{\rho }_h}(B,B)\ge 0$. Since $E\mapsto e^{-\beta E}\mathrm{/}Z(h)$ is injective on the real eigenvalues of $H(h)$, $H(h)$ and ${\rho }_h$ have the same eigenspaces; hence $[H(h),B]=0$ if and only if $[{\rho }_h,B]=0$. Step 4.1 therefore gives equality with $\beta {\mathrm{Var}}_{{\rho }_h}(B)$ exactly in the commuting case, and strict inequality otherwise. No ordinary-variance identity is inferred without this condition.