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(I will not use the “hat” ($$\hat{}$$) for the operators unless it is strictly necessary)

The Rabi Hamiltonian for 2 qubit is $\label{eqn:H2R} H_{2q}=\hbar \omega_0{a^{\dagger}}a + \hbar \frac{\omega_1}{2}\sigma^{(1)}_z + \hbar \frac{\omega_2}{2}\sigma^{(2)}_z + \hbar(\Gamma_1\sigma^{(1)}_x + \Gamma_2\sigma^{(2)}_x)({a^{\dagger}}+ a)$.

Using the following displacement operator properties $\begin{split} &{D^{\dagger}(\alpha)}aD(\alpha)=a + \alpha \\ &{D^{\dagger}(\alpha)}{a^{\dagger}}D(\alpha)={a^{\dagger}}+ \alpha^*\\ &{D^{\dagger}(\alpha)}D(\alpha)=D(\alpha){D^{\dagger}(\alpha)}={\mathbb{1}}\end{split}$

the Hamiltonian is easily rewritten in term of the displacement operator $H_{2q}=\hbar\omega_0 D^{\dagger}(\omega^{-1}_0 \hat{\gamma}^{\dagger}){a^{\dagger}}a D(\omega^{-1}_0 \hat{\gamma}) + \hbar \frac{\omega_1}{2}\sigma^{(1)}_z + \hbar \frac{\omega_2}{2}\sigma^{(2)}_z - \omega^{-2}\hat{\gamma}^{\dagger}\hat{\gamma}$

where $\hat{\gamma}=\Gamma_1\sigma^{(1)}_x + \Gamma_2\sigma^{(2)}_x$.

Taking $$\Gamma_i\in\mathbb{R}$$ and remembering $$\sigma_x^{\dagger}=\sigma_x$$ one has $$\hat{\gamma}^{\dagger}=\hat{\gamma}$$.

The eigenstates of Hamiltonian (\ref{eqn:H2R}) are the displaced Fock states....: $|N_{nm}nm>={D^{\dagger}(\omega^{-1}\gamma_{nm})}|N>|n>|m>$ with $$N\in\mathbb{N}$$, $$n,m=\{+,-\}$$ and $$\gamma_{nm}=n\Gamma_1+m\Gamma_2$$ is the eigenvalue of the operator $$\hat{\gamma}$$ on the 2-qubits bases $$\{|nm>\}=\{|++>,|+->,|-+>,|-->\}$$.

Negletting the costant therm of the Hamiltonian it becomes

$H_{2q}=\hbar\omega_0 D^{\dagger}(\omega^{-1}_0 \hat{\gamma}^{\dagger}){a^{\dagger}}a D(\omega^{-1}_0 \hat{\gamma}) + \hbar \frac{\omega_1}{2}\sigma^{(1)}_z + \hbar \frac{\omega_2}{2}\sigma^{(2)}_z - 2\hbar\frac{\Gamma_1\Gamma_2}{\omega^2}\sigma^{(1)}_x\otimes\sigma^{(2)}_x$

Calculate now the matrix’s elements in the bases $$\{|N_{nm}nm\}$$ in order to find later the eigenenergies of the system $\begin{split} &<M_{st}st|H_1|N_{nm}nm>=\epsilon_N\delta_{sn}\delta_{tm}<M_{st}|N_{nm}>\\ &<M_{st}st|H_2|N_{nm}nm>=\epsilon_1(1-\delta_{sn})\delta_{tm}<M_{st}|N_{\bar{n}m}>+\epsilon_2\delta_{sn}(1-\delta_{tm})<M_{st}|N_{n\bar{m}}>\\ &<M_{st}st|H_3|N_{nm}nm>=-\epsilon_{12}(nm)\delta_{sn}\delta_{tm}<M_{st}|N_{nm}> \end{split}$ where $$\bar{n}$$ is the negation of $$n$$, $$H_1=\hbar\omega_0 D^{\dagger}(\omega^{-1}_0 \hat{\gamma}){a^{\dagger}}a D(\omega^{-1}_0 \hat{\gamma})$$, $$H_2=\hbar \frac{\omega_1}{2}\sigma^{(1)}_z + \hbar \frac{\omega_2}{2}\sigma^{(2)}_z$$, $$H_3=- 2\hbar\frac{\Gamma_1\Gamma_2}{\omega^2}\sigma^{(1)}_x\otimes\sigma^{(2)}_x$$, $$\epsilon_N=\hbar\omega_0 N$$, $$\epsilon_1=\hbar \frac{\omega_1}{2}$$, $$\epsilon_2= \hbar \frac{\omega_2}{2}$$ and $$\epsilon_{12}=2\hbar\frac{\Gamma_1\Gamma_2}{\omega^2}$$. One can easily identify two diagonal terms and an off-diagonal one.