Volume-independent finite-range commutator path bound

Fix $B\in {\mathcal{A}}_Y$ and define $C_B^{\mathrm{\Lambda }}(S,s):={\sup }_{D\in {\mathcal{A}}_S,D\mathrm{\ne }0}\frac{\mathrm{\parallel }[{\alpha }_s^{\mathrm{\Lambda }}(D),B]\mathrm{\parallel }}{\mathrm{\parallel }D\mathrm{\parallel }}$ for every nonempty $S\subseteq \mathrm{\Lambda }$, with all operators viewed in ${\mathcal{A}}_{\mathrm{\Lambda }\cup Y}$. For $A\in {\mathcal{A}}_X$, put $H_X^{\mathrm{\Lambda }}={\sum }_{\mathrm{\varnothing }\mathrm{\ne }Z\subseteq X}\mathrm{\Phi }(Z)$ and $A_X(s)=e^{-is{H}_X^{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}A{e}^{is{H}_X^{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}$. The finite matrix exponential is unitary because its adjoint is $e^{is{H}_X^{\mathrm{\Lambda }}\mathrm{/}\mathrm{\hslash }}$ and the two commuting exponential series multiply to $I$; hence $A_X(s)$ ranges over all of ${\mathcal{A}}_X$ and preserves norm. Differentiating finite matrix exponentials and using the Jacobi identity, the terms contained in $X$ cancel and the terms disjoint from $X$ commute with $A_X(s)$. With $K_X(s)={\sum }_{Z\subseteq \mathrm{\Lambda }:Z\cap X\mathrm{\ne }\mathrm{\varnothing },Z\subseteq ̸X}{\alpha }_s^{\mathrm{\Lambda }}(\mathrm{\Phi }(Z))$, the remaining commutator $f(s)=[{\alpha }_s^{\mathrm{\Lambda }}(A_X(s)),B]$ satisfies $f^{\mathrm{\prime }}(s)=\frac{i}{\mathrm{\hslash }}[K_X(s),f(s)]-\frac{i}{\mathrm{\hslash }}{\sum }_{\begin{array}{c}Z\subseteq \mathrm{\Lambda }\\ Z\cap X\mathrm{\ne }\mathrm{\varnothing },Z\subseteq ̸X\end{array}}[{\alpha }_s^{\mathrm{\Lambda }}(A_X(s)),[{\alpha }_s^{\mathrm{\Lambda }}(\mathrm{\Phi }(Z)),B]]\mathrm{.}$ For the homogeneous equation, the finite matrix ODE $U^{\mathrm{\prime }}(s)=(i\mathrm{/}\mathrm{\hslash })K_X(s)U(s)$ has a solution on every compact interval: its Picard series converges uniformly because the $n$th iterated integral is bounded by $(T{\sup }_s\mathrm{\parallel }{K}_X(s)\mathrm{\parallel }\mathrm{/}\mathrm{\hslash }{)}^n\mathrm{/}n!$. Each $K_X(s)$ is self-adjoint, so differentiating $U(s{)}^{\ast }U(s)$ gives zero; since $U(0)=I$, $U(s{)}^{\ast }U(s)=I$. In finite dimension this makes $U(s)$ unitary, so conjugation by its propagator preserves the matrix norm. This argument also applies after reversing the time parameter, because the reversed homogeneous generator is still skew-adjoint. Set $\sigma =1$ if $t\ge 0$ and $\sigma =-1$ if $t<0$, and write $s=\sigma u$ for $0\le u\le \mathrm{\mid }t\mathrm{\mid }$. The equation for $f(\sigma u)$ has homogeneous commutator generated by $\sigma i{K}_X(\sigma u)\mathrm{/}\mathrm{\hslash }$ and a source multiplied by $\sigma$; the source sign does not change its norm. Variation of constants therefore gives the same norm estimate for either sign of $t$, with intermediate dynamics evaluated at signed times: $C_B^{\mathrm{\Lambda }}(X,t)\le C_B^{\mathrm{\Lambda }}(X,0)+\frac{2}{\mathrm{\hslash }}{\sum }_{\begin{array}{c}Z\subseteq \mathrm{\Lambda }\\ Z\cap X\mathrm{\ne }\mathrm{\varnothing }\end{array}}\mathrm{\parallel }\mathrm{\Phi }(Z)\mathrm{\parallel }{\int }_0^{\mathrm{\mid }t\mathrm{\mid }}{C}_B^{\mathrm{\Lambda }}(Z,\sigma u)du\mathrm{.}$ We enlarged the sum from crossing terms to all terms meeting $X$, which only increases its nonnegative right-hand side.

Iterate this integral inequality. All intermediate times retain sign $\sigma$; in absolute-time parameters the iterated region is $0\le u_n\le \cdots \le u_1\le \mathrm{\mid }t\mathrm{\mid }$, whose volume is $\mathrm{\mid }t{\mathrm{\mid }}^n\mathrm{/}n!$. Thus the same factorial weights apply for either sign of time. The zero-length term is at most $2\mathrm{\parallel }B\mathrm{\parallel }{\mathbf{1}}_{X\cap Y\mathrm{\ne }\mathrm{\varnothing }}$. A length-$n$ contribution, $n\ge 1$, is a chain $Z_1,\dots ,Z_n\subseteq \mathrm{\Lambda }$ with $Z_1\cap X\mathrm{\ne }\mathrm{\varnothing }$, $Z_{j+1}\cap Z_j\mathrm{\ne }\mathrm{\varnothing }$, and $Z_n\cap Y\mathrm{\ne }\mathrm{\varnothing }$, weighted by ${\prod }_{j=1}^n\mathrm{\parallel }\mathrm{\Phi }(Z_j)\mathrm{\parallel }$ and $(2\mathrm{\mid }t\mathrm{\mid }\mathrm{/}\mathrm{\hslash }{)}^n\mathrm{/}n!$. The first interaction sum is at most $\kappa \mathrm{\mid }X\mathrm{\mid }$; each later sum is at most $\kappa \mathrm{\mid }{Z}_j\mathrm{\mid }\le \kappa M$, so the total length-$n$ weight is at most $\kappa \mathrm{\mid }X\mathrm{\mid }(\kappa M{)}^{n-1}$. After $N$ iterations, the unresolved remainder is bounded by $2\mathrm{\parallel }B\mathrm{\parallel }(2\mathrm{\mid }t\mathrm{\mid }\mathrm{/}\mathrm{\hslash }{)}^{N+1}\kappa \mathrm{\mid }X\mathrm{\mid }(\kappa M{)}^N\mathrm{/}(N+1)!$; it tends to zero. Hence no infinite path expansion is assumed without convergence.

If $X\cap Y=\mathrm{\varnothing }$, any contributing length-$n$ chain satisfies ${\mathrm{dist}}_{\mathrm{\infty }}(X,Y)\le nR$: choose a site in each successive nonempty intersection and apply the triangle inequality, using ${\mathrm{diam}}_{\mathrm{\infty }}{Z}_j\le R$ at each of the $n$ steps. Thus its indicator is at most $e^{\mu (nR-{\mathrm{dist}}_{\mathrm{\infty }}(X,Y))}$. Since $M\ge 1$, $\kappa \mathrm{\mid }X\mathrm{\mid }(\kappa M{)}^{n-1}\le \mathrm{\mid }X\mathrm{\mid }(\kappa M{)}^n$. Summing the resulting exponential series gives $C_B^{\mathrm{\Lambda }}(X,t)\le 2\mathrm{\parallel }B\mathrm{\parallel }\mathrm{\mid }X\mathrm{\mid }{e}^{-\mu {\mathrm{dist}}_{\mathrm{\infty }}(X,Y)}(e^{2\kappa M{e}^{\mu R}\mathrm{\mid }t\mathrm{\mid }\mathrm{/}\mathrm{\hslash }}-1)\mathrm{.}$ If $X\cap Y\mathrm{\ne }\mathrm{\varnothing }$, then the distance is zero and the zero-length term together with the same series is bounded by $2\mathrm{\parallel }B\mathrm{\parallel }\mathrm{\mid }X\mathrm{\mid }{e}^{2\kappa M{e}^{\mu R}\mathrm{\mid }t\mathrm{\mid }\mathrm{/}\mathrm{\hslash }}$. In either case, multiplying by $\mathrm{\parallel }A\mathrm{\parallel }$ proves the stated estimate.