We are interested in simulating the 1d Ising model Hamiltonian using a Quantum Circuit (QC). A similar question was posted before with no answers. Here we will assume, for simplicity, 3 lattice sites and $J=-1$.
Generically, the Hamiltonian is given as $$ H = -J \sum_{ij} \sigma_i^z \otimes \sigma_j^z.\tag{1} $$ For our case of interest this Hamiltonian becomes: $$ H = \sigma^z \otimes \sigma^z \otimes \mathbb{1} + \mathbb{1} \otimes \sigma^z \otimes \sigma^z.\tag{2} $$ Obviously I have not included any periodic boundary conditions. There are only three lattice sites so there are only two interaction terms. In what follows I will replace $\sigma^z$ with $Z$ implying the corresponding quantum gate.
The evolution operator corresponding to this Hamiltonian is given as $$\tag{3} U(t) = e^{-i (Z \otimes Z\otimes \mathbb{1} + \mathbb{1}\otimes Z \otimes Z)t}. $$ Should these operators not commute we would have to use the Trotter-Suzuki formula. However, they do commute and as a result there is no need to use it.
In each of the two summands there exists a unit operator which can be completely ignored from the circuit. Now, for the operator $Z \otimes Z$ the curcuit would read $$ \mathrm{CNOT} R_z(2t) \mathrm{CNOT}: $$
My question is whether as the generalization to the 3 lattice sites Hamiltonian is as simple forward as this:
Of course the $R_Z$ gate runs for $2t$ according the unitary $U(t)$. Finally, does this generalize as simply to the $n$ lattice site Ising model?