Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-63/3/b/ii/solution

Write the final topological quantum order constants as and , to avoid confusing the allowed support diameter with the small time coefficient. For any initial operator of operator norm at most one,
and similarly for the partner state. The minus sign follows from and the Heisenberg picture convention . The Lieb-Robinson bound and its localization corollary apply to either time direction, using .
Choose , with , and localization buffer . The enlarged support obeys
The Lieb-Robinson localization by Haar twirling corollary supplies an operator approximating with
Do not assume that the approximate operator has norm at most one: it only has . Apply final-state local indistinguishability to . The two approximation errors then give
If the lattice has polynomially many sites in its diameter, , the error tends to zero uniformly in these supports. For sufficiently large , take . The initial pair is then indistinguishable within on supports of diameter at most . Together with the exact orthogonality from condition (i), this proves 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 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 prefactor merely because is large. This identifies exactly the assumption needed by the supplied localization proof.

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