Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 63 / 3 / a / Solution

First take disjoint supports $X,Y$, the setting in which the displayed Lieb-Robinson bound can hold with an $e^{2kst}-1$ factor. For overlapping supports a nonzero equal-time commutator is possible, whereas that factor vanishes at $t=0$.

Iterate the given integral inequality for $C_A(Y,t)$. At time zero, $C_A(Y,0)=0$ if $Y$ misses $X$, and in general $C_A(Y,0)\le 2\Vert A_X\Vert$. Define the positive interaction-chain weightsThe Lieb-Robinson interaction-chain expansion givesThe ordered time integrations produce $|t{|}^n/n!$. For a finite system the iterative remainder tends to zero, since the total interaction weights are finite and the factorial dominates repeated integrations.

For the interaction distance convention counting the fewest interacting hyperedges needed to connect disjoint supports, $a_n=0$ when $n<d(X,Y)$. The per-site interaction bound givesDropping the final endpoint restriction consequently bounds $a_n\le |Y|(ks{)}^n{e}^{-\mu n}$. Reversing the chains gives the same estimate with $|X|$. ThereforeMultiply by $\Vert B_Y\Vert$ and sum the exponential series to obtainFor overlapping supports, retain the initial term. A valid general version adds $2\Vert A_X\Vert \Vert B_Y\Vert 1_{X\cap Y\ne \varnothing }$ to the right side. A coarser bound with $e^{2ks|t|}$ in place of $e^{2ks|t|}-1$ is also sufficient for later shell estimates, with the conventional support distance zero on overlaps. Thus the printed bound needs disjoint supports or an equal-time term. The printed $C(Z,s)$ and $O_Z$ inside a supremum over $O_Y$ are also inconsistent labels; the iteration uses $C_A(Z,\tau )$ and $O_Y$.