Hamiltonian in CNOT gate implementation

Question

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?

Answer 1

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).