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Put $\sqrt{p}=\sin \theta$ and $\sqrt{1-p}=\cos \theta$, where $0<\theta <\pi /2$. In the ordered basis $(|b⟩,|g⟩)$, the product of the two reflections isThe leading minus sign is a physically irrelevant global phase. The remaining rotation matrix rotates the good-bad plane through $2\theta$, whereEquivalently, one iterate changes the initial angle $\theta$ to $3\theta$.

The operators $X_i$ act on different qubits and therefore commute, soSince $X=HZH$,The scalar $e^{-it}$ is a global phase. Thus each factor uses two Hadamard gates and one phase gate, and applying the factors in parallel or sequentially gives an exact quantum circuit of $\boxed{3n}$ elementary gates.

Each summand of $J_X$ acts on one qubit, while each summand $Z_i{Z}_j$ of $J_Z$ acts on two. Therefore $J=J_X+J_Z$ is a 2-local Hamiltonian. We haveSplit $t$ into $r$ steps of length $\delta =t/r$ and use the second-order product formulaFor one step, $\Lambda =O(\delta n^2)$ in the stated estimate, so the spectral-norm error is $O({\delta }^3{n}^6)$. The error bound for a product of unitary operators makes the total errorIt is therefore enough to choosewith $r$ also large enough that the small-step estimate applies.

All $Z_i{Z}_j$ terms commute. A factor $e^{-i\delta Z_i{Z}_j}$ uses a constant-size circuit of two controlled-NOT gates and one phase gate, up to a global phase, so one product-formula step costs $O(n^2)$ gates. The complete Hamiltonian simulation consequently has size