= Solution
Write the final <topological quantum order> constants as $a>0$ and $\varepsilon<1$, to avoid confusing the allowed support diameter with the small time coefficient. For any initial operator $A_X$ of <operator norm> at most one,
$$
\langle\psi_0|A_X|\psi_0\rangle=\langle\psi_1|A_X(-t)|\psi_1\rangle,
$$
and similarly for the partner state. The minus sign follows from $|\psi_0\rangle=e^{itH}|\psi_1\rangle$ and the <Heisenberg picture> convention $A_X(t)=e^{itH}A_Xe^{-itH}$. The <Lieb-Robinson bound> and its localization corollary apply to either time direction, using $|t|$.
Choose $\operatorname{diam}(X)\leq aL/4$, $t=\tau L$ with $v\tau\leq a/8$, and localization buffer $l=aL/8$. The enlarged support $X'$ obeys
$$
\operatorname{diam}(X')\leq\operatorname{diam}(X)+2(v|t|+l)\leq3aL/4<aL.
$$
The <Lieb-Robinson localization by Haar twirling> corollary supplies an operator $B_{X'}$ approximating $A_X(-t)$ with
$$
\delta_L:=\|A_X(-t)-B_{X'}\|\leq\mu v\tau L|X|e^{-\mu aL/16}.
$$
Do not assume that the approximate operator has norm at most one: it only has $\|B_{X'}\|\leq1+\delta_L$. Apply final-state <local indistinguishability> to $B_{X'}/(1+\delta_L)$. The two approximation errors then give
$$
\left|\langle\psi_0|A_X|\psi_0\rangle-\langle\widetilde\psi_0|A_X|\widetilde\psi_0\rangle\right|\leq\varepsilon(1+\delta_L)+2\delta_L.
$$
If the lattice has polynomially many sites in its diameter, $N(L)=O(L^p)$, the error tends to zero uniformly in these supports. For sufficiently large $L$, take $\delta_L\leq(1-\varepsilon)/[2(\varepsilon+2)]$. The initial pair is then indistinguishable within $\varepsilon'=(1+\varepsilon)/2<1$ on supports of diameter at most $aL/4$. Together with the exact orthogonality from condition (i), this proves \b[the initial state is topologically ordered for sufficiently small linear-time coefficient]. The constants for initial and final order need not be identical.
The <backward preservation of local indistinguishability> has an explicit finite-system qualification: it holds whenever the displayed $L|X|e^{-\mu aL/16}$ error is small enough. The conventional bounded-density, fixed-dimensional lattice interpretation supplies that condition. The question leaves growth control implicit; for an arbitrary collection of qudits one cannot discard the $|X|$ prefactor merely because $L$ is large. This identifies exactly the assumption needed by the supplied localization proof.
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