$\left\vert n\right\rangle $ to $\left\vert
n+1\right\rangle $. It is reasonable to set $\left\langle n+1\right\vert
R_{n+1,n}a^{\dagger }\left\vert n\right\rangle =\left\langle n+1\right\vert
F_{1}\left( a^{\dagger }a\right) a^{\dagger }\left\vert n\right\rangle $ by $$\begin{aligned}
R_{n+1,n} &=&\left\langle n+1\right\vert \sinh \left[ \frac{g}{\omega }%
\left( a^{\dagger }-a\right) \right] \left\vert n\right\rangle /\sqrt{n+1}
\notag \\
&=&\frac{1}{n+1}\frac{g}{\omega }e^{-\frac{g^{2}}{2\omega ^{2}}}L_{n}^{1}(%
\frac{g^{2}}{\omega ^{2}}).  \label{Rn}\end{aligned}$$Similarly, the term $aF_{1}\left( a^{\dagger }a\right) $ only has non-vanishing element $\langle n\left\vert aF_{1}\left( a^{\dagger }a\right) |n+1\right\rangle $. It follows that the term $F_{1}\left( a^{\dagger
}a\right) a^{\dagger }$ creates and $aF_{1}\left( a^{\dagger }a\right) $ eliminates a single photon of the cavity. The physics process is similar to that described in the RWA model, which facilitates the further analytic treatment.

The Hamiltonian now is $H^{1st}=H_{0}^{^{\prime }}+H_{1}^{^{\prime }}$: $$\begin{aligned}
H_{0}^{^{\prime }} &=&\omega a^{\dagger }a-\frac{g^{2}}{\omega }%
J_{z}^{2}+\Delta \beta J_{x}, \\
H_{1}^{^{\prime }} &=&\Delta J_{x}[G_{0}\left( a^{\dagger }a\right) -\beta
]+iJ_{y}\Delta \lbrack F_{1}\left( a^{\dagger }a\right) a^{\dagger
}-aF_{1}\left( a^{\dagger }a\right) ],  \notag\end{aligned}$$where $\beta =G_{0}\left( 0\right) =e^{-\frac{g^{2}}{2\omega ^{2}}}$.

Since the qubit and cavity in the noninteracting part $H_{0}^{^{\prime }}$ are decoupled, we apply a unitary transformation $S$ to diagonalize the qubit part in $H_{0}^{^{\prime }}$ $$S=\left(
\begin{array}{cccc}
-\frac{1}{C_{1}} & \frac{1}{C_{2}} & -\frac{1}{C_{3}} & \frac{1}{C_{4}} \\
\frac{K_{1}}{C_{1}} & -\frac{K_{2}}{C_{2}} & \frac{K_{3}}{C_{3}} & -\frac{%
K_{4}}{C_{4}} \\
-\frac{K_{1}}{C_{1}} & -\frac{K_{2}}{C_{2}} & -\frac{K_{3}}{C_{3}} 