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In a telescoping local-shell decomposition, split the spectral filter integral at a time proportional to the shell distance. A Lieb-Robinson bound controls short times, while the two-sided almost-exponential spectral filter tail controls long times. Polynomial shell-weight growth then gives the displayed decay. The estimate is asymptotic at large distance; small shells retain the elementary operator-norm bound. The evenization of a nonnegative bandlimited filter can supply the two-sided tail from a one-sided existence statement.

Undo a short evolution of two orthogonal locally indistinguishable states. Orthogonality is preserved exactly. A Lieb-Robinson bound localizes every backward-evolved observable within an expanded region and bounds its norm error. After normalizing the approximate observable, the initial expectation difference is at most $\epsilon (1+\delta )+2\delta$. Polynomial volume growth in the system diameter controls the localization prefactor and yields preservation for sufficiently small linear-time coefficient.

A Duhamel comparison can bound truncation error using commutators evolved under the full Hamiltonian. This avoids assuming that a black-box Lieb-Robinson bound for $H$ also holds for every truncation. Polynomial shell weights and exponential distance suppression make the sum over omitted shells convergent. Comparing two truncations through the full evolution bounds their difference.

Iterating the integral commutator inequality gives ordered interaction chains with factors $(2|t|{)}^n/n!$. A per-site interaction weight at most $s{e}^{-\mu }$ bounds each successive chain extension by $ks{e}^{-\mu }$. No chain shorter than the interaction distance connects separated supports. Reversing chains supplies the smaller support prefactor, producing the Lieb-Robinson bound. Overlapping supports require retaining the equal-time commutator term.

Apply Haar twirling conditional expectation to the complement of a neighbourhood around the initial operator support. A Lieb-Robinson bound on commutators with arbitrary complement-supported unitaries then bounds the approximation error without summing over sites. For a bound proportional to $e^{-\mu d}(e^{2kst}-1)$, choosing $v=4ks/\mu$ gives an error at most $\mu vt|X|\Vert A_X\Vert e^{-\mu l/2}$ outside radius $vt+l$.

The operator support is the set of subsystems on which an operator may act nontrivially; it acts as the identity outside that set. Unitary time evolution can enlarge the support, while a Lieb-Robinson bound controls the error of replacing the evolved operator by one with smaller support. The support diameter and the number of sites in the support are different quantities.

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 $O_Y$.

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.

Put $\lambda =2ks$. For $t\ge 0$, apply a valid Lieb-Robinson bound to each commutator in the supplied comparison of the two evolutions. For a disjoint shell at support distance $d$,In the usual uniform interaction-norm setting, truncation preserves the strength bound, so the same constants apply to $H_d$. The given polynomial shell-weight hypothesis and $e^{\lambda \tau }-1\le e^{\lambda t}$ now giveConsequentlyThe question explicitly supplies a bound for the full $H$; such a bound alone should not be assumed to transfer to $H_d$. A full-evolution comparison for truncated dynamics avoids that additional assumption. The Duhamel comparison formula givesNow use only the full-Hamiltonian bound, and sum the shell weights. Since ${\sum }_{q\ge d}{q}^{\alpha }{e}^{-\mu q}=O(d^{\alpha }{e}^{-\mu d})$, the errors for $j=d$ and $j=d-1$ are each bounded by $O(\Vert h_Z\Vert |t|d^{\alpha }{e}^{-\mu d+\lambda |t|})$. The triangle inequality comparing both evolutions to ${\tau }_t^H(h_Z)$ proves the same boxed estimate from the stated full-system assumption. Negative times follow by the reverse-time Duhamel formula; the commutator bounds use $|t|$.

For the interaction-term shells used to enforce $H_0=h_Z$, shell $d$ corresponds to support distance $d-1$ for distinct terms. Replace the displayed commutator estimate by the coarser bound with $e^{\lambda \tau }$, retaining the equal-time contribution for overlaps. Its distance factor is $e^{-\mu (d-1)}=e^{\mu }{e}^{-\mu d}$, and the fixed factor $e^{\mu }$ is absorbed by the big-O constant. This supplies the same claimed shell estimate without incorrectly using a zero equal-time bound for overlapping terms.

Let $R=X(l)$ and $C$ be its complement, with Hilbert-space dimension $d_C$. Use Haar twirling conditional expectation to defineThis is an operator supported on $R$. The normalization $1/d_C$ is essential: the partial trace alone would not fix an operator already supported on $R$.

Subtract the integrand from $A_X(t)$ and use the operator norm triangle inequality. Since $A-UA{U}^{†}=[A,U]U^{†}$ and every unitary operator has norm one, the assumed Lieb-Robinson bound, applied to a unitary supported on the entire complement, givesThe factor $\min (|X|,|C|)$ was bounded by $|X|$; there is no sum over sites and hence no volume-dependent prefactor.

ChooseFor $t,l\ge 0$, $e^{2kst}-1\le 2kst{e}^{2kst}$, and thereforeThus $A_{X(l)}(t)=A_R(t)$ is the desired Lieb-Robinson localization by Haar twirling. The displayed choice assumes the usual positive constants $\mu ,s$; if the interaction bound has $s=0$, the error is identically zero and any positive $v$ works. For negative times the same argument uses $|t|$.

Letwhere the two signs describe the Heisenberg antiferromagnet and Heisenberg ferromagnet. A standard Lieb-Robinson bound is obtained by iterating the Heisenberg picture equation and bounding nested commutators by operator norms. Its constants depend on the interaction only through quantities such asChanging $J$ to $-J$ changes neither the supports nor the norms $\Vert \Phi (X)\Vert$. Every term in the nested-commutator estimate acquires at most an irrelevant sign before its absolute value is taken. Therefore both chains obey exactly the same estimatewith the same $C,\mu$, and Lieb-Robinson velocity $v_{\mathrm{LR}}$. No unitary equivalence of the two Hamiltonians is required.

Consider a depth-$D$ circuit whose gates have range at most $r$. Under backward Heisenberg evolution, a one-site observable has a backward light cone of a local quantum circuit of radius at most $rD$. Choose sites $i,j$ separated by $L>2rD$. Their two backward light cones are disjoint. Since the input is a product state, expectations factorize, and hence every connected correlation function between the two output observables vanishes.

For the GHZ statehowever,at every separation. The light cones must therefore overlap, which forcesFor opposite ends of a one-dimensional chain, $L=\Theta (N)$, so $D=\Omega (N)$ and no constant-depth local circuit can prepare the GHZ state. The continuous-time version follows directly from the Lieb-Robinson bound, with preparation time at least $L/(2v_{\mathrm{LR}})$ up to exponentially small tails.

This is the GHZ-state circuit-depth lower bound. Finite-depth local circuits define equivalence within a gapped phase, so a state with this long-range order cannot lie in the same circuit phase as a product state. The GHZ state is the finite-size cat state associated with spontaneous symmetry breaking; it is not a unique short-range-entangled ground state. The persistent distant correlation is precisely the obstruction.