Weighted interaction commutator bound

For a finite nonempty $S\subseteq \mathrm{\Lambda }$, define $C_B^{\mathrm{\Lambda }}(S,u)={\sup }_{D\in {\mathcal{A}}_S,D\mathrm{\ne }0}\mathrm{\parallel }[{\alpha }_u^{\mathrm{\Lambda }}(D),B]\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }D\mathrm{\parallel }$. Let $U_S(u)$ solve $U_S^{\mathrm{\prime }}(u)=-i{H}_S(u)U_S(u)\mathrm{/}\mathrm{\hslash }$, $U_S(0)=I$, where $H_S(u)={\sum }_{\mathrm{\varnothing }\mathrm{\ne }Z\subseteq S}{\mathrm{\Phi }}_u(Z)$. Put ${\tilde{\alpha }}_u^S(D)=U_S(u)D{U}_S(u{)}^{\ast }$ and $f(u)=[{\alpha }_u^{\mathrm{\Lambda }}({\tilde{\alpha }}_u^S(D)),B]$. Product differentiation and the Jacobi identity give $f^{\mathrm{\prime }}(u)=\frac{i}{\mathrm{\hslash }}[K_S(u),f(u)]+\frac{i}{\mathrm{\hslash }}{\sum }_{\begin{array}{c}Z\subseteq \mathrm{\Lambda }\\ Z\cap S\mathrm{\ne }\mathrm{\varnothing },Z\subseteq ̸S\end{array}}[{\alpha }_u^{\mathrm{\Lambda }}({\tilde{\alpha }}_u^S(D)),[B,{\alpha }_u^{\mathrm{\Lambda }}({\mathrm{\Phi }}_u(Z))]],$ where $K_S(u)={\sum }_{Z\subseteq \mathrm{\Lambda }:Z\cap S\mathrm{\ne }\mathrm{\varnothing },Z\subseteq ̸S}{\alpha }_u^{\mathrm{\Lambda }}({\mathrm{\Phi }}_u(Z))$ is self-adjoint. Indeed, the local propagator cancels the terms contained in $S$, and terms disjoint from $S$ commute with $D$ by [F3]. The homogeneous equation $g^{\mathrm{\prime }}=(i\mathrm{/}\mathrm{\hslash })[K_S(u),g]$ preserves norm because its propagator is unitary; existence follows from the uniformly convergent finite-dimensional Picard series. Variation of constants, $\mathrm{\parallel }[R,Q]\mathrm{\parallel }\le 2\mathrm{\parallel }R\mathrm{\parallel }\mathrm{\parallel }Q\mathrm{\parallel }$, and enlargement of the crossing-support sum to all terms meeting $S$ therefore give $C_B^{\mathrm{\Lambda }}(S,u)\le C_B^{\mathrm{\Lambda }}(S,0)+\frac{2}{\mathrm{\hslash }}{\sum }_{\begin{array}{c}Z\subseteq \mathrm{\Lambda }\\ Z\cap S\mathrm{\ne }\mathrm{\varnothing }\end{array}}{\int }_0^u\mathrm{\parallel }{\mathrm{\Phi }}_v(Z)\mathrm{\parallel }{C}_B^{\mathrm{\Lambda }}(Z,v)dv\mathrm{.}$ The same inequality applies to the inverse path, which is generated over a reversed parameter interval by the interaction $-{\mathrm{\Phi }}_{u-v}$ and has the same norm bound.

At $u=0$, disjointness gives $C_B^{\mathrm{\Lambda }}(S,0)=0$ when $S\cap Y=\mathrm{\varnothing }$, while $C_B^{\mathrm{\Lambda }}(S,u)\le 2\mathrm{\parallel }B\mathrm{\parallel }$ for all supports and times by the commutator inequality and isometry of the dynamics. Iteration of step 1.1 produces time-ordered integrals over $0\le v_n\le \cdots \le v_1\le u$ and chains $Z_1,\dots ,Z_n$ with $Z_1\cap S\mathrm{\ne }\mathrm{\varnothing }$, $Z_{j+1}\cap Z_j\mathrm{\ne }\mathrm{\varnothing }$ and $Z_n\cap Y\mathrm{\ne }\mathrm{\varnothing }$, with product ${\prod }_{j=1}^n\mathrm{\parallel }{\mathrm{\Phi }}_{v_j}(Z_j)\mathrm{\parallel }$. All support sums are finite. For the iteration remainder, put $Q_{\mathrm{\Lambda }}={\sum }_{\mathrm{\varnothing }\mathrm{\ne }Z\subseteq \mathrm{\Lambda }}{\sup }_{v\in [0,T]}\mathrm{\parallel }{\mathrm{\Phi }}_v(Z)\mathrm{\parallel }<\mathrm{\infty }$, which is finite by norm continuity and compactness. Its absolute size after $n$ iterations is at most $2\mathrm{\parallel }B\mathrm{\parallel }(2Q_{\mathrm{\Lambda }}u\mathrm{/}\mathrm{\hslash }{)}^n\mathrm{/}n!$, so it tends to zero. This volume-dependent number is used only to discard the remainder. For the chain terms, define the time-dependent kernel $K_v(x,y)={\sum }_{Z\subseteq \mathrm{\Lambda }:x,y\in Z}\mathrm{\parallel }{\mathrm{\Phi }}_v(Z)\mathrm{\parallel }$. At each fixed time tuple, summing over all endpoint and overlap sites overcounts each nonnegative interaction-chain term at least once. Thus the order-$n$ contribution for $S=X$ is at most $2\mathrm{\parallel }B\mathrm{\parallel }(2\mathrm{/}\mathrm{\hslash }{)}^n{\int }_{0\le v_n\le \cdots \le v_1\le u}{\sum }_{x\in X,y\in Y}(K_{v_1}\cdots K_{v_n})(x,y)d{v}_n\cdots d{v}_1\mathrm{.}$ Kernel products here sum only over sites of the finite volume. No supremum over time has been moved inside an interaction sum for this bound.

By [F1], for each $v\in [0,T]$ separately, $K_v(x,y)\le M_{G}G(d_{\mathrm{\infty }}(x,y))$. Induction using [F2] gives $(K_{v_1}\cdots K_{v_n})(x,y)\le M_G^n{C}_G^{n-1}G(d_{\mathrm{\infty }}(x,y))(n\ge 1):$ the intermediate-site convolution over the finite volume is bounded by the full nonnegative lattice convolution. This bound is uniform in every time tuple. Its integral over the time-ordered simplex is $u^n\mathrm{/}n!$ times that bound. Substituting into step 2.1 and summing the convergent exponential series gives $C_B^{\mathrm{\Lambda }}(X,u)\le \frac{2\mathrm{\parallel }B\mathrm{\parallel }}{C_G}(e^{2C_G{M}_{G}u\mathrm{/}\mathrm{\hslash }}-1){\sum }_{x\in X,y\in Y}G(d_{\mathrm{\infty }}(x,y))$. The finite-volume remainder justifies the iteration limit; alternatively the displayed factorial majorant dominates every partial chain sum. Multiplication by $\mathrm{\parallel }A\mathrm{\parallel }$ proves the path estimate, including $A=0$. Applying the same time-ordered argument to the reversed interaction path, whose norm at every time is bounded by $M_G$, proves the inverse estimate. For a constant interaction take $u=\mathrm{\mid }t\mathrm{\mid }$; reversal changes its sign but not its norm. All constants in the resulting estimates depend only on $C_G,M_G$, not on the volume-dependent remainder bound.