so the Hilbert space can be decomposed into different $n$ manifolds spanned by the spin and cavity basis of $|-\frac{3}{2}\rangle |n\rangle $, $|-\frac{1}{2}\rangle
|n\rangle $, $|\frac{1}{2}\rangle |n\rangle $ and $|\frac{3}{2}\rangle
|n\rangle $. In the subspace containing only the $n$-th manifold, the Hamiltonian takes the form

$$\label{h0}
H^{^{0th}}=\left(
\begin{array}{cccc}
\omega n-\frac{9g^{2}}{4\omega } & \frac{\sqrt{3}}{2}\Delta G_{0}(n) & 0 & 0
\\
\frac{\sqrt{3}}{2}\Delta G_{0}(n) & \omega n-\frac{g^{2}}{4\omega } & \Delta
G_{0}(n) & 0 \\
0 & \Delta G_{0}(n) & \omega n-\frac{g^{2}}{4\omega } & \frac{\sqrt{3}}{2}%
\Delta G_{0}(n) \\
0 & 0 & \frac{\sqrt{3}}{2}\Delta G_{0}(n) & \omega n-\frac{9g^{2}}{4\omega }%
\end{array}%
\right).$$

The corresponding eigenvalues and eigenvectors are straightforwardly given by respectively $$\begin{aligned}
  \label{zeroenergy}
\varepsilon _{1,n} &=&\omega n-\frac{5g^{2}}{4\omega }-\frac{1}{2}%
B_{n}-2\chi _{1,n},  \notag  \label{zeroenergy} \\
\varepsilon _{2,n} &=&\omega n-\frac{5g^{2}}{4\omega }+\frac{1}{2}%
B_{n}-2\chi _{2,n},  \notag \\
\varepsilon _{3,n} &=&\omega n-\frac{5g^{2}}{4\omega }-\frac{1}{2}%
B_{n}+2\chi _{1,n},  \notag \\
\varepsilon _{4,n} &=&\omega n-\frac{5g^{2}}{4\omega }+\frac{1}{2}%
B_{n}+2\chi _{2,n},\end{aligned}$$and $$\begin{aligned}
|\varphi _{1,n}\rangle &\varpropto &\left(
\begin{array}{c}
-1 \\
K_{1,n} \\
-K_{1,n} \\
1%
\end{array}%
\right) ,|\varphi _{2,n}\rangle \varpropto \left(
\begin{array}{c}
1 \\
-K_{2,n} \\
-K_{2,n} \\
1%
\end{array}%
\right) ,  \notag \\
|\varphi _{3,n}\rangle &\varpropto &\left(
\begin{array}{c}
-1 \\
K_{3,n} \\
-K_{3,n} \\
1%
\end{array}%
\right) ,|\varphi _{4,n}\rangle \varpropto \left(
\begin{array}{c}
1 \\
-K_{4,n} \\
-K_{4,n} \\
