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For Hermitian matrices $H_1,\dots ,H_m$ and $t\ge 0$, let $S(t)={\prod }_{j=1}^m{e}^{-it{H}_j}$. ThenHere every norm is the spectral norm. A short proof starts with two summands. Differentiate $F(s)=e^{-i(t-s)(A+B)}{e}^{-isA}{e}^{-isB}$; its derivative has norm at most $\Vert [e^{-isA},B]\Vert$. Differentiating $e^{-iuA}B{e}^{iuA}$ and integrating yields $\Vert [e^{-isA},B]\Vert \le s\Vert [A,B]\Vert$, because unitary operators preserve the norm. Integrating $s$ from zero to $t$ gives $t^2\Vert [A,B]\Vert /2$. Inductively separate $H_1$ from the remaining sum, use the triangle inequality on their commutator, and apply the telescoping bound for products of operators to obtain the displayed many-term bound.

Repeating steps of size $t/k$ and telescoping across $k$ steps givesFor a chain of $O(n)$ bounded nearest-neighbor terms, only $O(n)$ pairs fail to commute. Consequently first-order product-formula Hamiltonian simulation has error $O(n{t}^2/k)$ and uses $O(nk)$ constant-size gates. The general bound is also given in Proposition 9 of Childs and collaborators' analysis of Trotter error.

The printed upper summation limit introduces $Z_n$ although only qubits $0,\dots ,n-1$ were defined. Literally the final term is undefined. Use the natural open-chain repairThis preserves the stated $n$-qubit system and agrees with the supplied sum-of-squares hint. If a cyclic convention $Z_n=Z_0$ was intended instead, it must be stated; the same argument works for its $n$ terms when $n\ge 2$. For $n=1$, the repaired open chain is empty and the target is the identity.

Here is a product-formula Hamiltonian simulation using exactly the two supplied lemmas. Let $L=n-1$ and choose an integer $r\ge L$. Every $h_i$ is a norm-one Hermitian matrix. One time slice is the product of two-qubit gatesTo compare a partial product with $e^{i(h_1+\cdots +h_j)/r}$, first propagate the previous error through the next unitary gate, which preserves the spectral norm, and then use Lemma A with $A=-(h_1+\cdots +h_{j-1})/r$, $B=-h_j/r$. Both norms are at most $j/r\le 1$. Its new error is at most $c(j/r{)}^2$ for a universal constant $c$. For an explicit choice, $c=4$ follows from the unitary Taylor bounds $\Vert e^{iA}-I-iA\Vert \le \Vert A{\Vert }^2/2$ and $\Vert e^{iA}-I\Vert \le \Vert A\Vert$. Induction and the triangle inequality therefore giveLemma B, the unitary product telescoping bound, now compares the $r$ repeated slices with $(e^{i{H}_{\mathrm{open}}/r}{)}^r=e^{i{H}_{\mathrm{open}}}$:Take, for example, $r=\max (1,L,\lceil 2c{\sum }_{j=2}^L{j}^2/ϵ\rceil )$. If the sum is zero the product is already exact; otherwise its error is at most $ϵ/2<ϵ$. There are $Lr$ two-qubit gates. This explicit lemma-based construction has fourth-degree dependence on $n$ for fixed precision:This is a sufficient polynomial, not an optimality claim. The polynomial-degree statement treats $ϵ$ as fixed; the inverse-precision dependence is displayed separately. No first-order term error is accumulated without the required repeated-slice factor.

The printed sum contains $Z_n$ although the stated labels end at $n-1$. We use the natural periodic convention $Z_n=Z_0$, with $n\ge 3$. If an open chain was intended, omitting the final term gives the same asymptotic bound.

Set $H_j=X_{j-1}{Z}_j$, with indices modulo $n$, and implement the product-formula Hamiltonian simulationEach factor acts on two qubits, so it is a two-qubit unitary operator. More explicitly, if $a=j-1$, $b=j$, and $C_{a\to b}$ is a controlled-NOT gate,where ${\mathsf{H}}_a$ is a Hadamard gate. This gives a constant number of one-qubit and two-qubit gates per factor; one-qubit gates can also be viewed as two-qubit gates tensored with the identity.

The spectral norm obeys the triangle inequality, submultiplicativity of the operator norm, and invariance under multiplication by unitary operators. In particular, the telescoping bound for products of operators gives $\Vert V^k-W^k\Vert \le k\Vert V-W\Vert$ for unitary $V,W$. The first-order unitary product-formula error bound consequently yieldsOnly neighboring terms can have a nonzero commutator, because all other supports are disjoint. There are $n$ neighboring unordered pairs, and $\Vert [H_j,H_l]\Vert \le 2\Vert H_j\Vert \Vert H_l\Vert =2$. Thus $\Vert U-\tilde{U}\Vert \le n/k$. Choosing $k>n/\epsilon$ givesThe coarser bound that counts all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairs also proves the often-used $O(n^3/\epsilon )$ construction. Locality improves that cubic estimate to the quadratic bound above; neither assertion is a lower bound on the best possible circuit.