I have a question regarding the operation of 2-qubit gates. Two transmon qubits are coupled via a capacitor, and a microwave is applied to one of the transmons. The Hamiltonian for this setup can be written as:
$$ \begin{align} H=\sum_{i=1}^2(\hbar(\omega_i-\omega_d)a_i^\dagger a_i-g(a_1^\dagger a_2+a_2^\dagger a_1)+A( a+ a^\dagger) \end{align} $$
I believe that by applying a SW transformation, i.e.,
\begin{align} e^{-iS}He^{iS}&\simeq H_0+g^2H_{diag}\\\\ S&=gS_1=gi\frac{a_1a_2^\dagger-a_2a_1^\dagger}{\hbar(\omega_1-\omega_2)}, \end{align} which is a unitary transformation, a 2-qubit gate can be generated. If $a_i$ were spin annihilation operators, interaction would arise from the microwave term because \begin{align} e^{-iS}a_1e^{iS}&\simeq a_1-\frac{g}{\hbar(\omega_1-\omega_2)}(2a_1^\dagger a_1-1)a_2. \end{align} However, in the case of boson annihilation operators, it seems that no interaction arises because \begin{align} e^{-iS}a_1e^{iS}&\simeq a_1+\frac{g}{\hbar(\omega_1-\omega_2)}a_2 \end{align} How should I interpret this difference? Does this mean that when we do not make the two-level system approximation, there is no interaction, but when we perform a low-level approximation of the bosonic Hamiltonian after the SW transformation, an interaction term emerges?
I have one more question. When I perform calculations considering the spin raising and lowering operators, after the SW transformation, I obtain four terms: ZI, IZ, XI, and ZX. \begin{align} e^{-iS}He^{iS}&\simeq \sum_{i=1}^2\frac{\hbar}{2}(\omega_i-\omega_d+g^2\delta_i)\sigma_i^z+A(\sigma_1^x-\frac{g}{\hbar(\omega_1-\omega_2)}\sigma_1^z\sigma_2^x ) \end{align} I believe that CR gates can be realized through the unitary evolution of ZX, but in this case, how do we eliminate the influence of the remaining three terms (ZI, IZ, XI)? (I speculate that IZ might be eliminated using microwave resonance.)
