# Hamiltonian in CNOT gate implementation

I am studying the physical implementation of the CNOT gate from "Introduction to Quantum Computing" (Ray LaPierre) and I am confused about the order of operators in the tensor product of the calculation below - why is it swapped for the two different qubits?

The total Hamiltonian $$\hat{H}$$ is written as the sum of the following three terms: $$\hat{H}_1 = \frac{\hbar \omega_1}{2} \hat{\sigma}_{z1} \otimes \hat{I}$$ $$\hat{H}_2 = \hat{I} \otimes \frac{\hbar \omega}{2} \hat{\sigma}_{z2}$$ $$\hat{H}_{\mathrm{int}} = J \hat{\sigma}_{z1} \otimes \hat{\sigma}_{z2}$$
Each of these is a $$4 \times 4$$ operator since it acts on a 2-spins system and, as such, it is written as a tensor product of two $$2 \times 2$$ operators ($$\hat{\sigma}_{z1}$$, $$\hat{\sigma}_{z2}$$, and the identity $$\hat{I}$$). In each tensor product left $$\otimes$$ right, the left operator is acting on the first spin while the right operator is acting on the second spin. This is why for $$\hat{H}_1$$ and $$\hat{H}_2$$, the left and right parts are swapped with $$\hat{I}$$ whereas for $$\hat{H}_{\mathrm{int}}$$ both of them are different from $$\hat{I}$$ (representing the interaction Hamiltonian).