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If $H={\sum }_j{h}_j$ and all local terms commute, thenexactly. Approximating each fixed-size factor to error at most $\epsilon /m$ and applying the telescoping bound for products of operators gives total operator norm error at most $\epsilon$.

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 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.

Write the local Hamiltonian as $H={\sum }_{j=1}^m{h}_j$. Since its terms commute, their matrix exponentials factor exactly:Each factor acts on at most two qubits and can be compiled over a fixed universal quantum gate set to operator norm error at most $\epsilon /m$. The telescoping bound for products of operators then bounds the total error by the sum of the factor errors, at most $\epsilon$. Because $m$ is polynomial in $n$ and the Solovay--Kitaev theorem gives gate count polynomial in $\log (m/\epsilon )$ for each fixed-dimensional factor, this is an efficient commuting local Hamiltonian simulation. Finally, the eigenvalue equation $H|\Psi ⟩=\lambda |\Psi ⟩$ impliesso $|\Psi ⟩$ remains an eigenstate and its eigenvalue is $e^{-i\lambda t}$.