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

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 $O_Y$.

Let $U(t)=e^{-itH}$. Define the second initial state by $|{\tilde{\psi }}_0⟩=U(t{)}^{†}|{\psi }_2⟩$, while $|{\psi }_0⟩=U(t{)}^{†}|{\psi }_1⟩$. The unitary time evolution preserves their inner product:Both initial states are normalized and evolve to the chosen final pair. This supplies the orthogonal partner needed for topological quantum order. The remaining requirement is local indistinguishability, which is proved in the next condition's solution.

Write the final topological quantum order constants as $a>0$ and $\epsilon <1$, to avoid confusing the allowed support diameter with the small time coefficient. For any initial operator $A_X$ of operator norm at most one,and similarly for the partner state. The minus sign follows from $|{\psi }_0⟩=e^{itH}|{\psi }_1⟩$ and the Heisenberg picture convention $A_X(t)=e^{itH}{A}_X{e}^{-itH}$. The Lieb-Robinson bound and its localization corollary apply to either time direction, using $|t|$.

Choose $\mathrm{diam}(X)\le aL/4$, $t=\tau L$ with $v\tau \le a/8$, and localization buffer $l=aL/8$. The enlarged support $X^{\prime }$ obeysThe Lieb-Robinson localization by Haar twirling corollary supplies an operator $B_{X^{\prime }}$ approximating $A_X(-t)$ withDo not assume that the approximate operator has norm at most one: it only has $\Vert B_{X^{\prime }}\Vert \le 1+{\delta }_L$. Apply final-state local indistinguishability to $B_{X^{\prime }}/(1+{\delta }_L)$. The two approximation errors then giveIf 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 ${\delta }_L\le (1-\epsilon )/[2(\epsilon +2)]$. The initial pair is then indistinguishable within ${\epsilon }^{\prime }=(1+\epsilon )/2<1$ on supports of diameter at most $aL/4$. 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 $L|X|e^{-\mu aL/16}$ 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.