\mid \tilde{E}_n \rangle^T = \left( \hat{A}^T_{(\vec{\mu},R,\theta,\beta_3)} \right)^n    \mid \tilde{0} \rangle_{T} =  \begin{cases} \varphi_<  (0, \hat{a}_2^\dagger) \mid n, 0 \rangle, \quad \text{if} \quad \beta_0 <2, \\
\varphi_\geq  (0, \hat{a}_1^\dagger) \mid 0, n \rangle, \quad \text{if}  \quad \beta_0 \geq 2 \end{cases}, \label{En-T}
\end{equation}

where some parameters have been absorbed by the  $\varphi$'s functions. Thus, by acting with the Hamiltonian on these states we get

\begin{equation}
\hat{H}^T_{(\beta_0,1)} \mid \tilde{E}_n \rangle^T =   \begin{cases}   n  \;   \varphi_<  (0, \hat{a}_2^\dagger) \;  \mid n, 0 \rangle + w_2 \;  \hat{a}_2^\dagger \;  \frac{\partial \varphi_< }{\partial \hat{a}_2^\dagger}  (0, \hat{a}_2^\dagger) \; \mid n,0\rangle , \quad \text{if} \quad \beta_0 <2, \\
w_1 \;  \hat{a}_1^\dagger \;  \frac{\partial \varphi_\geq }{\partial \hat{a}_1^\dagger}  (0, \hat{a}_1^\dagger) \;     \mid  0 , n \rangle +  n  \; \varphi_\geq  (0, \hat{a}_1^\dagger) \mid 0,n \rangle, \quad \text{if}  \quad \beta_0 \geq 2 \end{cases},  \label{H-T-En-E-En}
\end{equation}

which are consistent with the requirement of being eigenstates  of  $\hat{H}^T_{(\beta_0,1)}$  if and only if

\begin{equation}
\frac{\partial \varphi_< }{\partial \hat{a}_2^\dagger}  (0, \hat{a}_2^\dagger) =  c_{<} \; \;  \varphi_<  (0, \hat{a}_2^\dagger) \quad \text{ and} \quad   \frac{\partial \varphi_\geq }{\partial \hat{a}_1^\dagger}  (0, \hat{a}_1^\dagger) = c_{\geq} \; \; \varphi_\geq  (0, \hat{a}_1^\dagger) ,
\end{equation}

where $ c_{<}$ and $c_{\geq}$ are real constants, i.e,

\begin{equation}
\varphi_<  (0, \hat{a}_2^\dagger) = (\hat{a}_2^\dagger)^{c_{<}} \quad \text{and }
\quad  \varphi_\geq  (0, \hat{a}_1^\dagger) = (\hat{a}_1^\dagger)^{c_{\geq}} .
\end{equation}

Finally, inserting this last results into (<ref>) and  (<ref>), we get

\begin{equation}
\mid \tilde{E}_n \rangle^T =    \begin{cases} (\hat{a}_2^\dagger)^{c_{<}} \;  \mid n, 0 \rangle, \quad \text{if} \quad \beta_0 <2, \\
( \hat{a}_1^\dagger)^{c_{\geq}} \; \mid 0, n \rangle, \quad \text{if}  \quad \beta_0 \geq 2 \end{cases} \label{E-n-T-final}
\end{equation}

\begin{equation}
\hat{H}^T_{(\beta_0,1)} \mid \tilde{E}_n \rangle^T =   \begin{cases}   ( n   + w_2 \;  c_{<}  )  \mid \tilde{E}_n \rangle^T, \quad \text{if} \quad \beta_0 <2, \\
( w_1 \;  c_{\geq} +  n  )  \mid \tilde{E}_n \rangle^T, \quad \text{if}  \quad \beta_0 \geq 2 \end{cases}.
\end{equation}

We therefore observe that the energy spectrum in both situations, i.e., when $\beta_0 <2$ and when $\beta_0 \geq 2$ is isospectral to the one dimensional quantum oscillator up to a constant. In case that  $w_1$ and $w_2$ be integer or fractional numbers, the energy spectrum represents those of a commensurate 2-D quantum harmonic oscillator. As for the $c_{<}$ and $c_{\geq}$  constants, they should preferably be chosen as non-negative integers.  Finally, the energy eigenstates of the original Hamiltonian (<ref>) can be computed by acting with the unitary operator $\hat{T}_{(1,b,\beta_3,\theta)} $ on the states (<ref>), i.e.,
$\mid E_n \rangle = \hat{T}_{(1,b,\beta_3,\theta)}  \mid \tilde{E}_n \rangle^T.$

§.§ Generalized 2:1 commensurate quantum oscillator: case $ b^2=1$

When $ b^2= 4 \beta_+ \beta_- + \beta_3^2 =1,$  the $\alpha$ parameters could be taken different from $0$,  then various   situations can occur. Indeed, from the condition just  mentioned and from  equation (<ref>), we deduce that  when $\beta_3 =0,$

\begin{equation}
\beta_{\pm} = \frac{ e^{\pm i \theta} }{2} \quad \text{  and } \quad \alpha_{\pm} = \mp \; \frac{ e^{\pm  i \theta}}{2} \alpha_{3},
\end{equation}

when $\beta_{\pm} =0,$

\begin{equation}
\begin{cases} \alpha_- = \alpha_3 = 0, \quad \alpha_{+} \neq 0, \quad\text{ if} \quad  \beta_3 =1 \\  \alpha_+ = \alpha_3 = 0, \quad \alpha_{-} \neq 0, \quad\text{ if}  \quad \beta_3 =-1\end{cases} \label{beta-pm-conditions}
\end{equation}

and  when
$\beta_\pm  \neq 0$ and $\beta_3 \neq 0,$

\begin{equation}
\alpha_{+} = - \frac{\beta_+}{1-\beta_3} \alpha_3 \quad\text{ and}  \quad \alpha_{-} =  \frac{\beta_-}{1+\beta_3}  \alpha_3 .  \label{alpha-minus-plus}
\end{equation}

Let's start with the analysis of the second of these situations and then move on to the last one, because the first one can be obtained from the last one by making $\beta_3=0.$

§.§.§ Generalized two-dimensional  2:1  commensurate anisotropic quantum oscillator with linear coupling

When $\beta_{\pm}=0,$ then from the above condition for $b^2$ we have  $\beta_3= \pm 1.$  Then by inserting  the corresponding values of $\alpha$ and $\beta_3,$   which are summarized  in equation
(<ref>),  into  the algebraic equation system described in  (<ref>–<ref>), we get  30  different forms of expressing the set of $\mu$ and  $\nu$ type parameters in terms of the $\alpha$ and $\gamma$ parameters, for given values of $\beta_0$ and $\beta_3,$ 15 for each allowed value of $\beta_3.$ These relations are shown in appendix <ref>. In six of these relations,  three  for each allowed value of $\beta_3,$ the allowed value of $\beta_0$ is arbitrary but different from $\{\pm1,\pm3\}.$ This fact could give raise to different families of generalized  two-dimensional anisotropic  quantum oscillator systems, and  in particular,  of generalized two-dimensional fractional commensurate  anisotropic quantum oscillator systems. Unfortunately, the eigenstates of the corresponding generalized   lowering operator seem to be non-normalizable in the usual way.  We will therefore  focus on the particular case  described in equation (<ref>), which represents a 2:1 2-D commensurate anisotropic quantum oscillator with linear coupling, that is the case when $\beta_0=3$ and $\beta_3=1,$ with arbitrary nonzero $\gamma$ parameters:

\begin{eqnarray}
\hat{H}_{3, 1, \gamma_1,  \gamma_2} &=& 3 \hat{N} + \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1 +  \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2,  \label{H-A-17-2:1} \\
\hat{A}_{3,1,\gamma_1,\gamma_2, \alpha_+} & =&  \gamma_2^\ast \alpha_+  \hat{a}_1 + \mu_2 \hat{a}_2  + \frac{\gamma_1 \alpha_+}{2} \hat{a}_2^\dagger + \alpha_+ \hat{J}_-  +
\left( \gamma_2 \mu_2 +  \frac{\gamma_1 \gamma_2^\ast \alpha_+}{2}  \right)  \hat{I}. \label{A-A-17-2:1}
\end{eqnarray}

Let us now to compute the eigenstates of ladder operator ( <ref>), that is,

\begin{equation}
\hat{A}_{3,1,\gamma_1,\gamma_2, \alpha_+} \mid \lambda \rangle=  \lambda \mid \lambda \rangle, \quad
\lambda \in \mathbb{C}.  \label{A-3-1-gamma-1-gamma-2-alpha-plus}
\end{equation}

To solve this equation, let us first write it in the two-dimensional  Fock-Bargmann representation space of complex analytic functions, that is,

\begin{equation}
\alpha_+ \left( \gamma_2^\ast  +  \xi_2  \right)   \frac{\partial \psi}{\partial \xi_1} (\xi_1 , \xi_2) +  \mu_2  \frac{\partial \psi}{\partial \xi_2} (\xi_1 , \xi_2)   + \left(  \frac{\gamma_1 \alpha_+}{2} \xi_2  +  \gamma_2 \mu_2 +  \frac{\gamma_1 \gamma_2^\ast \alpha_+}{2}  -\lambda \right)  \psi(\xi_1,\xi_2)   =0, \quad \psi (0, \xi_2) = \phi_{2} (\xi_2), \label{A-lambda-psi=lambda-psi}
\end{equation}

where by convenience  we have added the initial condition $ \psi(0, \xi_2) = \phi_{2} (\xi_2),$ where $ \phi_{2} (\xi_2)$ is an arbitrary function of $\xi_2,$ which will influence directly  the form that the resulting fundamental state of the system will take.

The choice $\psi(0, 0) = 1,$   allows us express the analytic  function eigenstate   in the separable form $\psi (\xi_1, \xi_2) = \varphi_1 (\xi_1) \varphi_2 (\xi_2) \varphi_{2}.$ Then, by substituting this last expression into  (<ref>),  using the standard  techniques of resolution of linear partial differential equations, making  the following substitution during the development $\frac{\frac{d \varphi_1}{d \xi_1} (\xi_1)}{\varphi_{1} (\xi_1)}= c_1,$ where $c_1$ is an arbitrary complex  constant,  and then integrating the resulting first order linear differential equations in the variables $\xi_1$ and  $\xi_2 $, we get the function

\begin{equation}
\psi_{1} (\xi_1, \xi_2) = \exp\left[ c_1 \xi_1 \right] \exp\left[   \frac{1}{\mu_2}  \left( (\lambda-   \alpha_+ \gamma_2^\ast  c_1 )  -  (\gamma_2 \mu_2 + \frac{\gamma_1 \gamma_2^\ast \alpha_+} {2} ) \right) \xi_2  - \frac{\alpha_+}{2 \mu_2}  \left( c_1 \lambda + \frac{\gamma_1}{2}   \right) \xi_2^2   \right],
\end{equation}

whose form in the two-mode number representation is given by

\begin{equation}
\mid \lambda \rangle_1 = \mathcal{N}_1^{- \frac{1}{2}} \;  \exp\left[c_1 \hat{a}_1^\dagger \right] \exp\left[   \frac{1}{\mu_2}  \left( (\lambda -   \alpha_+ \gamma_2^\ast  c_1 )  -  (\gamma_2 \mu_2 + \frac{\gamma_1 \gamma_2^\ast \alpha_+} {2} ) \right)
\hat{a}_2^\dagger  - \frac{\alpha_+}{2 \mu_2}  \left( c_1 + \frac{\gamma_1}{2}   \right) {\hat{a}_2^\dagger}{}^2   \right] \mid 0,0\rangle,
\end{equation}

which represent a set of two-mode separable eigenstates of the ladder operator $ \hat{A}_{3,1,\gamma_1,\gamma_2, \alpha_+},$ having the  form of a canonical coherent state in the mode one and of a squeezed coherent state in the mode two. We observe that these states reduce to (<ref>) when all $\gamma$ parameters are equal $0.$

The fundamental state candidate, which we define as the eigenstate of  (<ref>) associated to the eigenvalue
$\lambda =0,$ is given by

\begin{equation}
\mid \tilde{0}  \rangle_1 = \mathcal{N}_1^{- \frac{1}{2}} \;  \exp\left[c_1 \hat{a}_1^\dagger \right] \exp\left[  -  \frac{1}{\mu_2} \left( \gamma_2^\ast  \alpha_+  c_1       +  (\gamma_2 \mu_2 + \frac{\gamma_1 \gamma_2^\ast \alpha_+} {2} ) \right)
\hat{a}_2^\dagger  - \frac{\alpha_+}{2 \mu_2}  \left( c_1 + \frac{\gamma_1}{2}   \right) {\hat{a}_2^\dagger}{}^2   \right] \mid 0,0\rangle,
\end{equation}

is also an eigenstate of  the Hamiltonian (<ref>), if and only if,  $c_1 = - \frac{\gamma_1}{2}. $
Thus, the true fundamental state of the system is

\begin{equation}
\mid \tilde{0}  \rangle_1 = \mathcal{N}_1^{- \frac{1}{2}} \;  \exp\left[- \frac{\gamma_1}{2} \hat{a}_1^\dagger \right] \exp\left[  - \gamma_2
\hat{a}_2^\dagger   \right] \mid 0,0\rangle,
\end{equation}

which assumes the normalized form

\begin{equation}
\mid \tilde{0}  \rangle_1 = \hat{D}_1 \left( - \frac{\gamma_1}{2} \right) \hat{D}_2 \left( - \gamma_2 \right) \mid 0,0\rangle, \label{2:1-linear-commensurate-separable-coherent-states}
\end{equation}

i.e., a set of  separable two-mode coherent states.

We observe that

\begin{equation}
\hat{H}_{3, 1, \gamma_1,  \gamma_2}  \mid \tilde{0}  \rangle_1 =  - \left(  \frac{\|\gamma_1\|^2}{2}  + \|\gamma_2\|^2 \right)   \mid \tilde{0}  \rangle_1,
\end{equation}

i.e, according to the freedom we have to choose the parameter $h_0$ in equation  (<ref>), which does not modify the general structure of the system in any way,  if we add   $  \left(\frac{\|\gamma_1\|^2}{2}  + \|\gamma_2\|^2  \right) \hat{I}$  to the Hamiltonian (<ref>), we realize that the quantum system governed by the Hamiltonian  $\hat{H}_{3, 1, \gamma_1,  \gamma_2} +  \left(\frac{\|\gamma_1\|^2}{2}  + \|\gamma_2\|^2 \right) \hat{I}, $ together with the set of ladder operators $\hat{A}_{3,1,\gamma_1,\gamma_2, \alpha_+} $ and $\hat{A}^\dagger_{3,1,\gamma_1,\gamma_2, \alpha_+} ,$  has a discrete energy spectrum which is identical to that of the canonical  quantum harmonic oscillator with associated energy eigenstates given by

\begin{equation}
\mid E_n \rangle_1 = \mathcal{N}_{n,1}^{-\frac{1}{2}} \left( \hat{A}^\dagger_{3,1,\gamma_1,\gamma_2, \alpha_+}\right)^n   \mid \tilde{0}  \rangle_1. \label{En-2:1-commensurate-linear-1}
\end{equation}

By writing the raisin operator $ \hat{A}^\dagger_{3,1,\gamma_1,\gamma_2, \alpha_+}$ in the form

\begin{equation}
\hat{A}^\dagger_{3,1,\gamma_1,\gamma_2, \alpha_+} = \hat{\Delta}_1 (\hat{a}_1^\dagger, \hat{a}_2^\dagger) +  \hat{\Delta}_2 (\hat{a}_1^\dagger, \hat{a}_2),
\end{equation}

\begin{equation}
\hat{\Delta}_1 (\hat{a}_1^\dagger, \hat{a}_2^\dagger) = \left( \gamma_2^\ast \mu_2^\ast + \frac{\gamma_1^\ast \gamma_2 \alpha_+^\ast}{2}  \right) \hat{I} +  \gamma_2 \alpha_{+}^\ast \hat{a}_1^\dagger + \mu_2^\ast \hat{a}_2^\ast, \quad \text{and} \quad
\hat{\Delta}_2 (\hat{a}_1^\dagger, \hat{a}_2) = \alpha_+^\ast \left(   \hat{a}_1^\dagger + \frac{ \gamma_1^\ast}{2}   \right) \hat{a}_2,
\end{equation}

satisfy the commutation relation

\begin{equation}
\left[  \hat{\Delta}_2 (\hat{a}_1^\dagger, \hat{a}_2),\hat{\Delta}_1 (\hat{a}_1^\dagger, \hat{a}_2^\dagger)   \right] = \alpha_+^\ast \mu_2^\ast \left(   \hat{a}_1^\dagger + \frac{ \gamma_1^\ast}{2}   \right) \hat{I},
\end{equation}

and adapting suitably the results of section <ref>, in particular equation (<ref>),  we get the normal ordering operator

\begin{equation}
\left( \hat{A}^\dagger_{3,1,\gamma_1,\gamma_2, \alpha_+}\right)^n = \sum_{k=0}^{[\frac{n}{2}]}    \frac{n!}{(n-2k)!} \frac{\left[
\alpha_+^\ast \mu_2^\ast \left(   \hat{a}_1^\dagger + \frac{ \gamma_1^\ast}{2}   \right)
\right]^k}{2^k \, k!}   \left( \hat{A}^\dagger_{3,1,\gamma_1,\gamma_2, \alpha_+}\right)^{n-2k}_{\lozenge}. \label{power-n-A-2:1-commensurate-linear-dagger-normal-ordering}
\end{equation}

Thus, by inserting this last result into equation  (<ref>), taking into account the structure of the
two-mode fundamental state  (<ref>) and using the standard properties of the harmonic oscillator unitary displacement  operators,  we obtain the following expression for the energy eigenstates of this system:

\begin{equation}
\mid E_n \rangle_1 = \mathcal{N}_{n,1}^{-\frac{1}{2}} =  \mathcal{N}_{n,1}^{-\frac{1}{2}} \hat{D}_1 \left( - \frac{\gamma_1}{2} \right) \hat{D}_2 \left( - \gamma_2 \right)  \sum_{k=0}^{[\frac{n}{2}]}   \frac{n!}{(n-2k)!} \frac{\left(
\alpha_+^\ast \mu_2^\ast   \hat{a}_1^\dagger\right)^k}{2^k \, k!}   \left(\mu_2^\ast \hat{a}_2^\dagger \right)^{n-2k} \mid 0,0 \rangle.
\end{equation}

Finally, using the results  (<ref> – <ref>), we get the normalized energy eigenstates

\begin{equation}
\mid E_n \rangle_1 =  \hat{D}_1 \left( - \frac{\gamma_1}{2} \right) \hat{D}_2 \left( - \gamma_2 \right) \frac{ \sum_{k=0}^{[\frac{n}{2}]}  \frac{1}{\sqrt{(n-2k)! k!} }       \left(\frac{\alpha^\ast_+  }{2 \mu^\ast_2 }\right)^k   \mid k, n-2 k\rangle}{\sqrt{  \sum_{k=0}^{[\frac{n}{2}]}   \frac{1}{(n-2k)! k!}  \left(\frac{ \|\alpha_+ \|  }{2 \|\mu_2\| }\right)^{2k} }}.
\end{equation}

The above procedure suggest us that there exist a unitary similarity transformation that can bring our Hamiltonian and ladder operators to a simpler structure. Indeed, we can show directly that

\begin{equation}
\hat{H}_{3,1,0,0} =    \hat{D}^\dagger_1 \left( - \frac{\gamma_1}{2} \right) \hat{D}^\dagger_2 \left( -\gamma_2 \right)  \left[\hat{H}_{3,1,\gamma_1,\gamma_2}  +  \left(\frac{\|\gamma_1\|^2}{2}  + \|\gamma_2\|^2 \right) \hat{I}   \right]  \hat{D}_1 \left( - \frac{\gamma_1}{2} \right) \hat{D}_2 \left( -\gamma_2 \right)  =  2 \hat{a}_1^\dagger \hat{a}_1 + \hat{a}_2^\dagger \hat{a}_2
\end{equation}

\begin{equation}
\hat{A}_{3,1,0,0,\alpha_+} =  \hat{D}^\dagger_1 \left( - \frac{\gamma_1}{2} \right) \hat{D}^\dagger_2 \left( -\gamma_2 \right)   \hat{A}_{3,1,\gamma_1,\gamma_2, \alpha_+}   \hat{D}_1 \left( - \frac{\gamma_1}{2} \right) \hat{D}_2\left( -\gamma_2 \right)  =  \mu_2 \hat{a}_2 + \alpha_+ \hat{J}_-  \label{A-two-mode-3-1-0-0-alpha-plus}
\end{equation}

i.e, the system reduces to  (<ref>).

There are many other solutions of the partial differential equation (<ref>) depending on the choice of the initial conditions. We will not try to find all them here, instead,  we will rather use the above fact to simplify the problem by taking advantage of the results of the previous sections. Doing that, we will obtain a particular set of two-mode   non-separable algebra eigenstates of the lowering operator $\hat{A}_{3,1,\gamma_1,\gamma_2, \alpha_+} $.    Thus, using the fact that the states  (<ref>) are eigenstates of the lowering operator  (<ref>), then the normalized eigenstates of  $\hat{A}_{3,1,0,0,\alpha_+}$ are given by

\begin{equation}
\mid  \lambda \rangle_{2} =   \frac{\hat{D}_1 \left( - \frac{\gamma_1}{2} \right) \hat{D}_2\left( -\gamma_2 \right)  \hat{D}_2 \left( \frac{\lambda}{\mu_2} \right)}{\sqrt{ \frac{1}{2}  + \frac{\| \mu_2 \|^2}{ \| \alpha_+ \|^2}   +  \frac{\| \lambda \|^2}{ \| \mu_2 \|^2}  +  \frac{\| \lambda \|^4 }{4  \| \mu_2\|^4}    }} \left[     \frac{ {\left( \hat{a}^{\dagger}_{2}  + \frac{\lambda^\ast}{ \mu_2^\ast}  \right)}^2 }{2} | 0, 0 \rangle  - \frac{\mu_2}{\alpha_+}
| 1 , 0  \rangle  \right].
\end{equation}

From this last equation we get the fundamental state of the system

\begin{equation}
\mid \tilde{0} \rangle_2 =   \frac{\hat{D}_1 \left( - \frac{\gamma_1}{2} \right) \hat{D}_2\left( -\gamma_2 \right)}{\sqrt{ \frac{1}{2}  + \frac{\| \mu_2 \|^2}{ \| \alpha_+ \|^2} }    } \left[     \frac{1 }{\sqrt{2}} | 0,  2 \rangle  - \frac{\mu_2}{\alpha_+}
| 1 , 0  \rangle  \right],
\end{equation}

from where we deduce that the action of the shifted  Hamiltonian on it is given by

\begin{equation}\left[\hat{H}_{3,1,\gamma_1,\gamma_2}  +  \left(\frac{\|\gamma_1\|^2}{2}  + \|\gamma_2\|^2 \right) \hat{I}   \right] \mid \tilde{0} \rangle_2 = 2 \mid \tilde{0} \rangle_2,
\end{equation}

i.e., the  energy  spectrum of the system is $E_n= n+2, \; n=0, 1, \ldots.$ Furthermore, using (<ref>), we get the energy eigenstates

\begin{eqnarray}
\mid E_n \rangle_2 &=& \mathcal{N}_{2,n}^{-\frac{1}{2}}         \;   \hat{D}_1 \left( - \frac{\gamma_1}{2} \right) \hat{D}_2\left( -\gamma_2 \right)   \;                \sum_{k=0}^{[\frac{n}{2}]}  \frac{1}{\sqrt{(n-2k)! k!} }       \left(\frac{\alpha^\ast_+  }{2 \mu^\ast_2 }\right)^k  \left[ \frac{\sqrt{(n-2k+1)!(n-2k+2)!}}{2} \mid k, n-2k+2 \rangle   \right. \nonumber \\ &+&
\left(\frac{\alpha^\ast_+}{\mu_2^\ast}\right)^2  \sqrt{\frac{(k+1)(k+2)}{2}}  \sqrt{(n-2k-1)!(n-2k)!} \mid k+2, n-2k-2 \rangle \nonumber \\ &+&  \left. \sqrt{2 (k+1)} \left(
\frac{\alpha^\ast_+}{\mu_2^\ast} (n-2k) - \frac{\mu_2}{\alpha_+}  \right) \mid k+1, n-2k\rangle
\right], \nonumber \\ \label{lambda-2-base-gen}
\end{eqnarray}

where, as before, $\mathcal{N}_{2,n}$ is a normalization constant given by $\mathcal{N}_{2,n} = {}_2\langle E_n \mid E_n \rangle_2.$

§.§.§ Interacting two-mode commensurate anisotropic quantum oscillators

When $b^2= 4 \beta_+ \beta_-  + \beta_3^2 =1 $ and $\beta_+ =\beta_{-}^\ast \neq 0$ and $\beta_3 \neq 0,$ the relations (<ref>) are satisfied. Then,  inserting them into  the algebraic equation system (<ref>–<ref>), we obtain

\begin{equation}
\begin{pmatrix}
\left( 1 - \frac{\beta_0 +\beta_{3}}{2} \right)& - \beta_+ \\
- \beta_-  &    \left( 1 - \frac{\beta_0 - \beta_{3}}{2} \right) \end{pmatrix} \begin{pmatrix}
\mu_1 \\ \mu_2
\end{pmatrix}
= \alpha_3 \; \begin{pmatrix}
- \left( \frac{\gamma_1^\ast}{2}  - \frac{\gamma_2^\ast \beta_+}{1-\beta_3} \right) \\
\left( \frac{\gamma_2^\ast}{2}  - \frac{\gamma_1^\ast \beta_-}{1+\beta_3}  \right)
\end{pmatrix}     \label{eq-mu-1-2}
\end{equation}

\begin{equation}
\begin{pmatrix}
\left( 1 + \frac{\beta_0 +\beta_{3}}{2} \right) &   \beta_- \\
\beta_+  &   \left( 1 + \frac{\beta_0 -\beta_{3}}{2} \right) \end{pmatrix} \begin{pmatrix}
\nu_1 \\ \nu_2
\end{pmatrix}
= \alpha_3 \; \begin{pmatrix}
\left( \frac{\gamma_1}{2}  + \frac{\gamma_2  \beta_-}{1 +\beta_3} \right) \\
- \left( \frac{\gamma_2}{2}  + \frac{\gamma_1  \beta_+}{1-\beta_3}  \right)
\end{pmatrix},     \label{eq-nu-1-2}
\end{equation}

where we have taken $\alpha_3$ as an non-zero independent parameter.

Also in this case several situations can occur depending on the allowed values of the parameters. All  these situations are described in Appendices  <ref> and  <ref>,  where the generalized Hamiltonian with its associated lowering operator is detailed. According to what we have already seen in the previous sections, in the process of searching for normalizable eigenstates for the lowering operator, the use of unitary similarity transformations was very useful. Here we will use this tool  to bring  the generalized Hamiltonian (<ref>) and its corresponding associated lowering operator (<ref>) to assume a simpler form. In fact, by first performing a convenient unitary similarity transformation of type $su(2)$ and then a second transformation of the same characteristics but of type $h(2)$, that is, using the two-mode separable displacement operator associated with the 2-D quantum harmonic oscillator, it is possible to transform the two interacting quantum harmonic oscillator Hamiltonian with linear external coupling into a two non-interacting quantum oscillator Hamiltonain without external interaction.Moreover, the final structure of the Hamiltonian thus transformed will be identical to one of the basic forms already discussed in section <ref> with the only difference lying in the particular value of the new parameters.  In addition to all this, we have that the same successive unitary similarity transformations will lead the associated generalized lowering  operator to one of the  corresponding basic forms  of section   <ref>.

In general, a first $su(2)$ unitary similarity transformation of the generalized Hamiltonian (<ref>) with the passing operator

\begin{equation} \hat{T}_{(\epsilon,b, \beta_3, \theta)}= \exp\left[ - \arctan \left( \epsilon \sqrt{\frac{b -\epsilon \beta_3}{b+ \epsilon \beta_3}}  \right)  \left( e^{-i \theta} \hat{J}_+ - e^{i \theta} \hat{J}_-      \right)  \right],
\end{equation}

where $b^2= 4  \beta_+ \beta_- + \beta_3^2,$ and $\beta_\pm = R e^{\pm i \theta},$ will bring it to

\begin{equation}
\hat{H}^{T}_{(\beta_0, \epsilon,\beta_3, \theta, \vec{\tilde{\gamma}}_1, \vec{\tilde{\gamma}}_2)} =
\beta_0 \hat{N} + \epsilon b  \hat{J}_3 +  \tilde{\gamma}_{(1; \epsilon, b) }  \hat{a}_1^\dagger +  \tilde{\gamma}^\ast_{(1; \epsilon, b) } \hat{a}_1 +    \tilde{\gamma}_{(2; \epsilon, b, \beta_3) }   \hat{a}_2^\dagger  +   \tilde{\gamma}^\ast_{(2; \epsilon, b, \beta_3)}   \hat{a}_2  +  h_0  \hat{I}, \label{H-the-most-general-T}
\end{equation}

\begin{eqnarray}
\tilde{\gamma}_{(1; \epsilon, b, \beta_3) }  &=& \sqrt{\frac{b + \epsilon \beta_3}{2b}} \gamma_1 + \epsilon  e^{-i \theta}  \sqrt{\frac{b - \epsilon \beta_3}{2b}} \gamma_2, \\
\tilde{\gamma}_{(2; \epsilon, b, \beta_3) }  &=& \sqrt{\frac{b + \epsilon \beta_3}{2b}} \gamma_2  -  \epsilon  e^{i \theta}  \sqrt{\frac{b - \epsilon \beta_3}{2b}} \gamma_1.
\end{eqnarray}

A second unitary similarity  transformation using the passing operator

\begin{equation}
\hat{D}_{(1,2; \beta_0, \epsilon,b,\tilde{\gamma_1},\tilde{\gamma_2})} = \hat{D}_1 \left(-   \frac{ 2 \tilde{\gamma}_{(1; \epsilon, b,\beta_3) }  }{\beta_0 + \epsilon b} \right)  \hat{D}_2 \left(- \frac{2 \tilde{\gamma}_{(2; \epsilon, b, \beta_3) } }{\beta_0 - \epsilon b} \right),
\end{equation}

where $\hat{D}_1$ and $\hat{D}_2$ are the standard one-dimensional  displacement operators in the mode one and two, respectively,
brings the transformed Hamiltonian  (<ref>) to

\begin{equation}
\hat{H}^{T,D_{(1,2)}}_{(\beta_0, \epsilon,\beta_3, \theta, \vec{\tilde{\gamma}}_1, \vec{\tilde{\gamma}}_2)} =
\beta_0 \hat{N} + \epsilon b  \hat{J}_3  - 2 \frac{\|\tilde{\gamma}_{(1; \epsilon, b) }  \|^2}{\beta_0 + \epsilon b}   \hat{I}  -  2 \frac{\|\tilde{\gamma}_{(2; \epsilon, b) }  \|^2}{\beta_0 -  \epsilon b}     \hat{I} + h_0   \hat{I}, \label{H-the-most-general-T-D-1-2}
\end{equation}

that is, a two mode $(\beta_0 + \epsilon b) : (\beta_0 - \epsilon b)$  anisotropic quantum oscillator. The choice of $h_0 = 2 \frac{\|\tilde{\gamma}_{(1; \epsilon, b) }  \|^2}{\beta_0 + \epsilon b}  +   2 \frac{\|\tilde{\gamma}_{(2; \epsilon, b) }  \|^2}{\beta_0 -  \epsilon b}   $ would lead to a simple energy spectrum.

For example,  when  $b^2=1,$    $\frac{\gamma_1}{2} \neq  \frac{\gamma_2 \beta_-}{1-\beta_3},$  $\frac{\gamma_1}{2} \neq - \frac{\gamma_2 \beta_-}{1+\beta_3}$ and   $ \beta_0 \notin \{\pm1,\pm3 \},$ this composed unitary similarity transformation for the lowering (<ref>) operator leads    to

\begin{equation}
\hat{A}^{T, D_{(1,2)}}_{(\vec{0},\vec{0},\vec{\alpha},0)}  = - \epsilon \frac{\alpha_3 e^{i \epsilon \theta}}{2 \sqrt{1 - \beta_3^2} }  \hat{J}_{(-  \epsilon)},
\end{equation}

where $\hat{J}_{(-\epsilon)} = \hat{J}_{\mp}, $ when $\epsilon = \pm 1.$ Thus, when $\epsilon=1,$ by
using the results of section <ref>,  we get the energy eigenstates

\begin{equation}
\mid E_n \rangle =   \hat{D}_1 \left(- 2  \frac{ \tilde{\gamma}_{(1; 1, 1, \beta_3) }   }{\beta_0 +  1} \right)  \hat{D}_2 \left(- \frac{2 \tilde{\gamma}_{(2; 1, 1, \beta_3) } }{\beta_0 - 1} \right)  \hat{T}_{(1,1, \beta_3, \theta)} \mid 0, \kappa \rangle, \quad \kappa=1,2, \ldots,
\end{equation}

and the energy spectrum

\begin{equation}
E_n = (\beta_0 -1) \kappa +n. \quad n=0,1,\ldots, \quad \kappa= 1,2, \ldots.
\end{equation}

On the other hand,   under the same circumstances, when $\beta_0=3,$ under the  similarity transformation the  lowering operator
(<ref>)  becomes

\begin{equation}
\hat{A}^{T, D_{(1,2)}}_{(\vec{\mu},\vec{0},\vec{\alpha},0)}  = - \left[\sqrt{\frac{2}{1-\beta_3}}   \mu_1  e^{- i \theta}  + \frac{\alpha_3\tilde{\gamma}_{(2;1,1)}^\ast}{1 - \beta_3}   \right] \hat{a}_2  - \frac{\alpha_3 e^{i \theta}}{2 \sqrt{1 - \beta_3^2} }  \hat{J}_{-}, \label{annihilator-gen-1-2}
\end{equation}

when $\epsilon =1$ and

\begin{equation}
\hat{A}^{T, D_{(1,2)}}_{(\vec{\mu},\vec{0},\vec{\alpha},0)}  =  \left[\sqrt{\frac{2}{1-\beta_3}}   \mu_1   -  \frac{\alpha_3\tilde{\gamma}_{(1;-1,1)}^\ast}{1 - \beta_3}   \right] \hat{a}_1 + \frac{\alpha_3 e^{-i \theta} }{2 \sqrt{1 - \beta_3^2} }  \hat{J}_{+},
\end{equation}

when $\epsilon =-1.$

Thus, when $\epsilon=1,$ to find the eigenstates  of the transformed annihilator (<ref>), we can use the results of section  <ref> with the parameter  $\mu_2$ replaced by ${\tilde \mu_2} =  - \left[\sqrt{\frac{2}{1-\beta_3}}   \mu_1  e^{- i \theta}  + \frac{\alpha_3\tilde{\gamma}_{(2;1,1)}^\ast}{1 - \beta_3}   \right] $ and a the parameter  $\alpha_+$ replaced by
$\tilde{\alpha}_+ =   - \frac{\alpha_3 e^{i \theta}}{2 \sqrt{1 - \beta_3^2} }  .$ Hence, using  (<ref>) and (<ref>) we get

\begin{equation}
\widetilde{\mid  \lambda \rangle}_1 = \hat{D}_1 \left( c_1 \lambda \right) \hat{S}_2 \left( \tilde{\chi} \right)  \hat{D}_2 \left( \frac{\lambda}{\tilde{\mu}_2}  \cosh (\| \tilde{\chi} \|) \right) \mid  0 , 0 \rangle,
\end{equation}

$$\tilde{\chi}= \frac{\frac{ \lambda c_1 \tilde{\alpha}_+ }{\tilde{\mu}_2} } { \| \frac{ \lambda c_1 \tilde{\alpha}_+ }{\tilde{\mu}_2} \| }  \tanh^{-1} \left(   \| \frac{ \lambda c_1 \tilde{\alpha}_+ }{\tilde{\mu}_2} \|  \right), $$

which must verify   $\| \frac{ \lambda c_1 \tilde{\alpha}_+ }{\tilde{\mu}_2} \| < 1,$

\begin{equation}
\widetilde{\mid  \lambda \rangle}_2 = \frac{ \hat{D}_2 \left( \frac{\lambda}{\tilde{\mu}_2} \right)}{\sqrt{ \frac{1}{2}  + \frac{\| \tilde{\mu}_2 \|^2}{ \| \tilde{\alpha}_+ \|^2}   +  \frac{\| \lambda \|^2}{ \| \tilde{\mu}_2 \|^2}  +  \frac{\| \lambda \|^4 }{4  \| \tilde{ \mu}_2\|^4}    }} \left[     \frac{ {\left( \hat{a}^{\dagger}_{2}  + \frac{ \lambda^\ast}{ \tilde{\mu}_2^\ast}  \right)}^2 }{2} | 0, 0 \rangle  - \frac{\tilde{\mu}_2}{\tilde{\alpha}_+}
| 1 , 0  \rangle  \right].   \end{equation}

From these last equations, by making $\lambda=0,$   we  get the fundamental states

\begin{equation} \widetilde{\mid 0  \rangle}_1 =\mid 0 ,0 \rangle \quad \text{ and}  \quad \widetilde{\mid 0  \rangle}_2 =
\frac{  \left[     \frac{1}{\sqrt{2}} | 0, 2 \rangle  - \frac{\tilde{\mu}_2}{\tilde{\alpha}_+}
| 1 , 0  \rangle  \right]}{\sqrt{ \frac{1}{2}  + \frac{\| \tilde{\mu}_2 \|^2}{ \| \tilde{\alpha}_+ \|^2} }} \label{ground-states-commensurate-1-2-T-D}
\end{equation}

Hence, the eigenstates of the lowering operator operator   (<ref>)  are given by

\begin{equation}
\mid \lambda \rangle_1 =   \hat{D}_1 \left(-  \frac{ \tilde{\gamma}_{(1; 1, 1, \beta_3) }}{2} \right)  \hat{D}_2 \left(-  \tilde{\gamma}_{(2; 1, 1, \beta_3)}   \right)  \hat{T}_{(1,1, \beta_3, \theta)}  \widetilde{ \mid 0 \rangle}_1,
\end{equation}

\begin{equation}
\mid \lambda \rangle_2 =   \hat{D}_1 \left(-  \frac{ \tilde{\gamma}_{(1; 1, 1, \beta_3) }   }{2} \right)  \hat{D}_2 \left(-  \tilde{\gamma}_{(2; 1, 1, \beta_3) }  \right)  \hat{T}_{(1,1, \beta_3, \theta)}  \widetilde{ \mid 0 \rangle}_2.
\end{equation}

§ $SU(2)$ CHEN TYPE  $P:Q$ COMMENSURATE COHERENT STATES

In this section we will compute the energy eigenstates of the commensurate $p:q$ quantum oscillator,

\begin{equation}
\hat{\mathcal{H}}_{(p,q)} = \frac{\left[ p \; \hat{a}_1^\dagger \hat{a}_1 +  q  \;  \hat{a}_2^\dagger \hat{a}_2 \right]}{p \; q}, \label{H-p-q-commensurate}
\end{equation}

where  $p$ and $q$ are non-negative integer numbers.   To do this, we will use some results already obtained in previous section, in particular, those that relate to the  isotropic and to the  2:1 commensurate anisotropic 2D quantum oscillators. Then, we will generalize the structure of the lowering operators of these systems to include the case of the $p:q$ commensurate anisotropic quantum oscillator.

In previous sections we have seen that when $ p=q=1,$ i.e., when $\hat{H}_{(1,1)}$ represent the 2:D isotropic quantum oscillator, the associated lowering operator was given by $\hat{A}_{(1,1)}= \alpha_+ \hat{a}_1 + \alpha_-  \hat{a}_2, $ where we have renamed  the constants.  The normalized fundamental state in this case, for a given value of $\kappa= 0, 1, \ldots,$ in the number representation, was given by

\begin{equation}
\mid 0 \rangle^{(\kappa)}_{(1,1)} = \mathcal{N}^{- \frac{1}{2}}_{(\kappa)} \;  \left( \alpha_+  \hat{a}_2^\dagger -  \alpha_-  \hat{a}_1^\dagger  \right)^\kappa \mid 0,0 \rangle = \frac{\sum_{k=0}^{\kappa}  \sqrt{\binom{\kappa}{k}}  (-1)^\kappa  \alpha_-^k \alpha_+^{\kappa-k} \mid k, \kappa - k  \rangle }{\left(\| \alpha_+ \|^2 + \| \alpha_- \|^2\right)^{\frac{\kappa}{2}}}, \quad \kappa=0,1, \ldots, \label{kappa-1-1}
\end{equation}

and the associated chains of energy were computed using the corresponding raisin operator in the following way

\begin{equation}
\mid E_{n} \rangle^{(\kappa)}_{(1,1)} = \mathcal{N}^{- \frac{1}{2}}_{(n, \kappa)}  \; (\hat{A}^\dagger_{(1,1)})^n  \mid 0 \rangle^{(\kappa)}
\end{equation}

where $\mathcal{N}^{- \frac{1}{2}}_{(n, \kappa)}, \; n, \kappa=0,1,2, \ldots,$ are normalization constants.

We notice that for a fixed value of $\kappa,$ the energy spectrum was given by $E^{(\kappa)}_{n} = \kappa +n, \; n=0,1,2, \ldots.$ Then when $k=0,$ the  isotropic 2D quantum oscillator was isospectral to the 1-D canonical quantum oscillator. Indeed, in this case we had

\begin{equation}
[ \hat{\mathcal{H}}_{(1,1)} ,\hat{A}_{(1,1)}] = - \hat{A}_{(1,1)}, \quad  [\hat{\mathcal{H}}_{(1,1)} ,\hat{A}^\dagger_{(1,1)}] =  \hat{A}^\dagger_{(1,1)} \quad \text{and} \quad  [\hat{A}_{(1,1)} , \hat{A}^\dagger_{(1,1)}] = \left(\| \alpha_+ \|^2 + \| \alpha_- \|^2\right) \hat{I}. \label{algebra-standard-oscillator}
\end{equation}

Let us now introduce a new element to the discussion, that is,  the fact that the   fundamental states  (<ref>) form  a complete set of orthonormal states  which are also eigenstates of $\hat{\mathcal{H}}_{(1,1)}.$ Indeed,

\begin{equation}
\hat{\mathcal{H}}_{(1,1)} \mid 0 \rangle^{(\kappa)} = \kappa  \mid 0 \rangle^{(\kappa)}, \quad \kappa= 0,1, \ldots.
\end{equation}

Furthermore, we observe that the state $ \mid 0 \rangle^{(\kappa)}$ can be considered as the result of the action of the  raising operator

\begin{equation}
\hat{\mathcal{A}}^\dagger_{(1,1)} =   \alpha_+  \hat{a}_2^\dagger -  \alpha_-  \hat{a}_1^\dagger
\end{equation}

to the power $\kappa$ on the two-mode fundamental  state $\mid 0 , 0 \rangle,$ i.e.,

\begin{equation}
\mid 0 \rangle^{(\kappa)}_{(1,1)} = \left( \hat{\mathcal{A}}^\dagger_{(1,1)} \right)^\kappa \mid 0,0 \rangle.
\end{equation}

Moreover, we observe that the set of operators $\hat{\mathcal{H}}_{(1,1)},$  $\hat{\mathcal{A}}_{(1,1)}$  and   $\hat{\mathcal{A}}^\dagger_{(1,1)}$
form the algebra

\begin{equation}
[\hat{\mathcal{H}}_{(1,1)} ,\hat{\mathcal{A}}_{(1,1)}] = - \hat{\mathcal{A}}_{(1,1)}, \quad  [\hat{\mathcal{H}}_{(1,1)} ,\hat{\mathcal{A}}^\dagger_{(1,1)}] =  \hat{\mathcal{A}}^\dagger_{(1,1)} \quad \text{and} \quad  [\hat{\mathcal{A}}_{(1,1)} , \hat{\mathcal{A}}^\dagger_{(1,1)}] = \left(\| \alpha_+ \|^2 + \| \alpha_- \|^2\right) \hat{I},
\end{equation}

which is identical to (<ref>).

When $p=2$ and $q=1,$ we had essentially

\begin{equation}
\hat{\mathcal{H}}_{(2,1)} =  \frac{\left[ 2 \; \hat{a}_1^\dagger \hat{a}_1 +    \;  \hat{a}_2^\dagger \hat{a}_2 \right]}{2},
\quad  \text{and} \quad \hat{A}_{(2,1)} =  \alpha_-   \; \hat{a}_2  +   \alpha_+  \; \hat{a}_2^\dagger  \hat{a}_1 = \alpha_- \;  \hat{a}_2 + \alpha_+ \; \hat{J}_-  ,
\end{equation}

and from (<ref>) we had the ground states, that in the number representation look like

\begin{equation}
\mid 0 \rangle^{(k)}_{(2,1)} =  \mathcal{N}^{- \frac{1}{2}}_{(\kappa)}   \left( \frac{ \alpha_+  \;  (\hat{a}^\dagger_2)^2 - 2 \alpha_-  \; \hat{a}^\dagger_1}{2} \right)^{\kappa} \mid 0,0 \rangle = \frac{\sum_{k=0}^{\kappa}  \sqrt{\binom{\kappa}{k}} \sqrt{\frac{(2 (\kappa-k))! }{(\kappa-k)!}  }   (-1)^k  \alpha_-^k \left( \frac{\alpha_+}{2}\right)^{\kappa-k} \mid k, 2(\kappa - k)  \rangle }{\left(
\sum_{k=0}^{\kappa} \binom{\kappa}{k} \frac{(2(\kappa-k))!}{(\kappa-k)!}  \left(\frac{\|\alpha\|}{2} \right)^{2 (\kappa-k)} \|\alpha_-\|^{2k}\right)^{\frac{1}{2}}},
\end{equation}

where $ \kappa=0,1, \ldots.$
Again, we observe that the state $ \mid 0 \rangle^{(\kappa)}$ can be considered as the result of the action of the  raising operator

\begin{equation}
\hat{\mathcal{A}}^\dagger_{(2,1)} =    \frac{ \alpha_+  \;  (\hat{a}^\dagger_2)^2 - 2 \alpha_-  \; \hat{a}^\dagger_1}{2}
\end{equation}

to the power $\kappa$ on the two-mode fundamental  state $\mid 0 , 0 \rangle,$ i.e.,

\begin{equation}
\mid 0 \rangle^{(\kappa)}_{(2,1)} = \left( \hat{\mathcal{A}}^\dagger_{(2,1)} \right)^\kappa \mid 0,0 \rangle.
\end{equation}

Also, in this case, the state $ \mid 0 \rangle^{(\kappa)}_{(2,1)}, \; \kappa=1,2, \ldots,$ is an eigenstate of $ \hat{\mathcal{H}}_{(2,1)}$ with associated eigenvalue equal to $\kappa, \; \kappa=1,2,\ldots,$ then the energy  spectrum is given by $(E_{n}^{(\kappa)})_{(2,1)} = \kappa, \; \kappa=1,2, \ldots.$ Indeed, you can prove it directly or deduce that from  the commutation relations characterizing  the system, that is,

\begin{equation}
[\hat{\mathcal{H}}_{(2,1)} ,\hat{\mathcal{A}}_{(2,1)}] = - \hat{\mathcal{A}}_{(2,1)}, \quad  [\hat{\mathcal{H}}_{(2,1)} ,\hat{\mathcal{A}}^\dagger_{(2,1)}] =  \hat{\mathcal{A}}^\dagger_{(2,1)}
\end{equation}

\begin{equation}
[\hat{\mathcal{A}}_{(2,1)} , \hat{\mathcal{A}}^\dagger_{(2,1)}] = \| \alpha_+ \|^2   \hat{a}_2^\dagger \hat{a}_2 +  \frac{\| \alpha_+ \|^2  + 2   \| \alpha_- \|^2}{2 } \hat{I}.
\end{equation}

The previous results can be extended as follows:  if $p$ and $q$ are non-negative coprime integers, the special choice of the lowering operator

\begin{equation}
\mathcal{\hat{A}}_{(p,q)} = \frac{\alpha_{+}^\ast\; q \;  \hat{a}_2^p -  \alpha_{-}^\ast \; p \; \hat{a}_1^q}{p \; q}, \label{cal-A-p-q}
\end{equation}

assures us that the following commutation relation holds

\begin{equation}
[\mathcal{\hat{H}}_{(p,q)},\mathcal{\hat{A}}_{(p,q)}] = -\mathcal{\hat{A}}_{(p,q)} \label{H-cal-lowering-p-q}
\end{equation}

and consequently

\begin{equation}
[\mathcal{\hat{H}}_{(p,q)},\mathcal{\hat{A}}^\dagger_{(p,q)}] = \mathcal{\hat{A}}^\dagger_{(p,q)}. \label{H-cal-raising-p-q}
\end{equation}

This choice is truly inspired by the natural generalization of the corresponding form of the lowering operator when $p=q=1$ and when $p=2$ and $q=1,$ which for a more general case we estimate it could be

\begin{equation}
\hat{A}_{(p,q)} = \alpha_- \; ({\hat{a}^\dagger_1})^{(q-1)} \; \hat{a}_2 + \alpha_+ \; \hat{a}_1 \; ({\hat{a}^\dagger_2})^{(p-1)} , \label{lowering-A-generalized}
\end{equation}

and from the fact that its corresponding eigenstates associated  to the eigenvalue $\lambda=0$ are given by

\begin{equation}
\mid 0 \rangle^{(\kappa)}_{(p,q)} =  (\mathcal{\hat{A}}^\dagger_{(p,q)})^\kappa \mid 0,0 \rangle, \quad \kappa=0,1,2,\ldots. \label{kappa-0-p-q}
\end{equation}

Indeed, this last result follows directly from the fact that $[\hat{A}_{(p,q)},\mathcal{\hat{A}}^\dagger_{(p,q)}]=0, \; \kappa=0, 1,2, \ldots,$ and that $\hat{A}_{(p,q)} \mid 0,0 \rangle=0.$

Then, from all the above, we can say that a chain of orthonormalized eigenstates of the  $p:q$ commensurate anisotropic 2D quantum oscillator Hamiltonian  (<ref>) is given by the set of states (<ref>), which have the explicit normalized form

\begin{equation}
\mid 0 \rangle^{(\kappa)}_{(p,q)} =
\frac{\sum_{k=0}^{\kappa}  \sqrt{\binom{\kappa}{k}} \sqrt{\frac{(p (\kappa-k))! }{(\kappa-k)!}  }  \sqrt{\frac{(q k)!}{k!}} (-1)^k  \left(\frac{\alpha_-}{q}\right)^k \left( \frac{\alpha_+}{p}\right)^{\kappa-k} \mid q k, p(\kappa - k)  \rangle }{\left(
\sum_{k=0}^{\kappa} \binom{\kappa}{k} \frac{(p(\kappa-k))!}{(\kappa-k)!} \frac{(q k)!}{k!}
\left(\frac{\|\alpha_+\|}{p} \right)^{2(\kappa-k)} \left(\frac{\|\alpha_-\|}{q} \right)^{2k}
\right)^{\frac{1}{2}}}, \quad \kappa=0,1,2,\ldots. \label{p-q-energy-anisotropic-both}
\end{equation}

These states are certainty eigenstates of (<ref>) with  associated eigenvalues $E^{(\kappa)}_{(p,q)}= \kappa, \; \kappa=1,2, \ldots. $  Thus, we get suitable Chen $su(2)$ coherent states type[17]–[18], which have been obtained here on a simple way by using classical arguments of the theory of ladder operators.

There is something more, according to equation (<ref>),  the state $\mid 0,0 \rangle$ is not the only  eigenstate of $\hat{\mathcal{A}}$ with associated eigenvalue equal to $0.$ There are others,  depending on the values of $p$ and $q.$ Indeed, for fixed values of $p$ and $q,$ by using equation (<ref>), it is direct to check that

\begin{equation}
\mid 0  \rangle^{(k_1,k_2)}_{(p,q)} = \mid k_1, k_2 \rangle, \quad k_1 =0,\ldots, (q-1), \quad k_2 =0,\ldots, (p-1),
\end{equation}

are also eigenstates of $\hat{\mathcal{A}}$ with associated eigenvalue equal to $0.$ Then according to (<ref>) these states are also eigenstates of $\hat{\mathcal{H}}_{(p,q)}$ with associated eigenvalues equal to $E^{(k_1,k_2)}_{(0,p,q)}= \frac{k_1}{q}+\frac{k_2}{p}.$ Also, according to the commutation relations (<ref>) and (<ref>) the states defined as

\begin{equation}
\mid E \rangle^{(k_1,k_2)}_{(n, p,q)} = (\mathcal{N}^{(k_1,k_2)}_{(n, p,q)})^{- \frac{1}{2}}
(\hat{\mathcal{A}}^\dagger_{p,q})^n \mid k_1, k_2 \rangle, \quad k_1 =0,\ldots, (q-1), \quad k_2 =0,\ldots, (p-1),
\end{equation}

are eigenstates of $\hat{\mathcal{H}}_{(p,q)}$ with associated eigenvalues $ E^{(k_1,k_2)}_{(n, p,q)} = n + \frac{k_1}{q}+\frac{k_2}{p}, \; n=0,1,2,\ldots.$ Thus, again, by using classical arguments of the ladder operators theory, we obtain the right energy spectrum of the $p:q$ commensurate anisotropic quantum oscillator computed by Louck et Al. [14].

Moreover,  there are yet another set of eigenstates of (<ref>) with eigenvalue equal to $0,$ that is, the two-mode

\begin{equation}
\mid \tilde{0} \rangle_{(p,q)} = \frac{\frac{p}{\alpha^\ast_+ \; \sqrt{p!}} \mid 0 , p \rangle +   \frac{q}{\alpha^\ast_- \; \sqrt{q!}} \mid q , 0 \rangle}{ \left[  \frac{p^2}{|\alpha_+\|^2 \; p!    }        +     \frac{p^2}{|\alpha_-\|^2 \; q!    }         \right]^{\frac{1}{2}}  }          .
\end{equation}

It is direct to show that these states are also eigenstates of the $p:q$ commensurate anisotropic Hamiltonian (<ref>) with associated eigenvalue equal to $1$. Then, again, with the help of the raising operator $ \mathcal{\hat{A}}^\dagger_{(p,q)}$ we can generate a chain of eigenstates  of  (<ref>) in the usual way, i.e.,

\begin{equation}
\mid \tilde{E}_{n} \rangle_{(p,q)} = {\tilde{\mathcal{N}}}_{(p,q)}^{-\frac{1}{2}} \; (\mathcal{\hat{A}}^\dagger_{(p,q)})^{n} \; \mid \tilde{0} \rangle_{(p,q)}, \quad n=0,1, \ldots.
\end{equation}

By the way, the  energy spectrum of these states is given by $\tilde{E}_{n} = n+1, \; n=0,1, \ldots. $ The explicit  non-normalized form of these states is

\begin{eqnarray}
\mid \tilde{E}_{n} \rangle_{(p,q)} &=& {\tilde{\mathcal{N}}}_{(p,q)}^{-\frac{1}{2}} \;  \sum_{k=0}^{n} \binom{n}{k}  (-1)^k    \left(\frac{\alpha_-}{q}\right)^{k}   \left(\frac{\alpha_+}{p}\right)^{(n-k)} \nonumber \\ &\times& \left[  \frac{\sqrt{ (qk)! (p ((n+1)-k))!}}{\alpha^\ast_+ \; (p-1)!} \mid  qk, p((n+1) -k) \rangle +  \frac{ \sqrt{ (q(k+1))! (p (n-k)!}}{ \alpha^\ast_- \; (q-1)!  }   \mid  q(k+1), p(n -k) \rangle \right]. \nonumber\\
\end{eqnarray}

It might be worthwhile to find the generalized Hamiltonian $\hat{\mathbb{H}}_{(p,q)}$ that verifies $[\hat{\mathbb{H}}_{(p,q)}, \hat{A}_{(p,q)} ],$ where $\hat{A}_{(p,q)} $ is the lowering operator (<ref>). In general, this Hamiltonian should commute with the commutator $[\hat{A}_{(p,q)},\hat{A}_{(p,q)} ] $ and reduce to the isotropic or to the  $2:1$ commensurate anisotropic Hamiltonain, when $ p=1$ and $q=1$ and $p=2$ and $q=1,$ respectively.  One possibility is

\begin{equation}
\hat{\mathbb{H}}_{(p,q)} = \frac{p \;  \hat{a}_1^\dagger \hat{a}_1  + q \; \hat{a}_2^\dagger \hat{a}_2}{\left[ 1 - (p-1) (q-1)\right]}. \label{Hamiltonian-p-q-alternative}
\end{equation}

It is clear that the eigenstates of this Hamiltonian are given by the  states (<ref>). Moreover, its energy spectrum is $\mathbb{E}^{(\kappa)}_{(p,q)}= \frac{{p \; q  \; \kappa }}{1 -(p-1)(q-1)}, \; \kappa=0,1, \ldots. $ For example, when $p=q=1,$ it becomes the same as that of the canonical one-mode quantum oscillator, when $p > q $ and $q=1,$  $\mathbb{E}^{(\kappa)}_{(p,1)} = p \; \kappa, \;  \kappa= 1,2,\ldots,$ and when $p > q >1,$  $ \mathbb{E}^{(\kappa)}_{(p,q)} \leq 0, \forall  \kappa \in \mathbb{N}_{0}. $ In this last case, it could be interesting to define a new $p:q$ anisotropic 2D Hamiltonian as the negative of $\ref{Hamiltonian-p-q-alternative},$ which means that the old lowering operator become a the new raising operator and vice-versa.  Thus, the task   would be now to compute the eigenstates of

\begin{equation}
\tilde{\hat{A}}_{(p,q)} = \alpha^\ast_- \; ({\hat{a}_1})^{(q-1)} \hat{a}^\dagger_2 + \alpha^\ast_+ \;  \hat{a}^\dagger_1   ({\hat{a}_2})^{(p-1)},\label{lowering-tilde-A-generalized}
\end{equation}

which is harder to do than in the original case.  Indeed, the  corresponding  partial differential equation becomes more difficult to solve according to the increasing in the value of the parameters $p$ and $q.$

§ CONCLUSION

In this article we have  studied the conditions  under which  a quantum system, whose Hamiltonian is an element of the complex
$ \left\{ h (1) \oplus h(1) \right\} \roplus u(2)  $  Lie algebra, admits ladder operators which are also elements of this algebra.  As a realization of such a system we have chosen 2 interacting quantum oscillators with linear external coupling. We have proved that there are suitable unitary similarity transformations of the type $su(2)$ and $h(1) \oplus h(1)$ which  bring the Hamiltonian and its associated  ladder operators into a more fundamental  $u(2)$ form. The algebra eigenstates of the lowering operators   constructed  in this way have been computed, and we have found  known as well as new  structures of coherent and squeezed states for the $1:1,$ $2:1,$ and $1:2$ commensurate quantum oscillators. Furthermore, we have shown that the $1:2$ and $2:1$  commensurate anisotropic Hamiltonian are connected by a composed $su(2)$ similarity transformation. We have also shown that for a very special choice of the parameters the Hamiltonian and its associated ladder operators verify the  $su(2)$ Lie algebra. In this case, we have found an interesting truncated bosonic set of energy eigenstates and spectrum.  In the case that the characteristic  $su(2)$ parameter  $b \neq 1,$ we have found that not only the   isotropic  2D quantum oscillator is included in the set of $u(2)$ quantum systems  but also  the  fractional  commensurate and non-commensurate anisotropic  2-D oscillator systems. Finally, based on the experience gained in the treatment of the $1:1$ and $2:1$ systems, more precisely, on the particular form of the lowering operators and their respective eigenstates associated with the eigenvalue equal to $0,$ we have been able to  construct a valid extension of these operators for the   $p:q$ commensurate anisotropic case. We have found that the resulting energy eigenstates states in this case are comparable to the $SU(2)$ type coherent states proposed by Cheng. Something interesting we could do is to construct from these  latter the  corresponding commensurate anisotropic  2D  Schrödinger-type coherent states and then compute the  probability density of them  to finally see how the results differ from those obtained in  [16].

§ ACKNOWLEDGMENTS
The author would like  to thank Dr.  Artorix de la Cruz de Oña  of Dalhousie University for his valuable support and advice during the preparation of this article. He would also like to thank to his little purple doll Mary and to his beloved children by their encouragement  and great patience.

[1]  E. Schrödinger, Zum Heisenbergschen Unschärfeprinzip, Sitzungberichten der Preussischen
Akademie der Wissenschaften (Phys.-Math. Klasse), 19 (1932) ss. 296-323.

[2]  Heisenberg (1925). "Über quantentheoretische Umdeutung kinematischer und mechanischer Beziehungen". Zeitschrift für Physik (in German). 33 (1): 879–893. Bibcode:1925ZPhy...33..879H. doi:10.1007/BF01328377.

[3]  Dirac, Paul Adrien Maurice (1930). The Principles of Quantum Mechanics. Oxford: Clarendon Press.

[4]  Hilbert, David (2009). Sauer, Tilman; Majer, Ulrich (eds.). Lectures on the Foundations of Physics 1915–1927: Relativity, Quantum Theory and Epistemology. Springer. doi:10.1007/b12915. ISBN 978-3-540-20606-4. OCLC 463777694.

[5]  von Neumann, John (1932). Mathematische Grundlagen der Quantenmechanik. Berlin: Springer. English translation: Mathematical Foundations of Quantum Mechanics. Translated by Beyer, Robert T. Princeton University Press. 1955.

[6]  Weyl, Hermann (1950) [1931]. The Theory of Groups and Quantum Mechanics. Translated by Robertson, H. P. Dover. ISBN 978-0-486-60269-1.

[7] Klauder, John R (1960). "The action option and a Feynman quantization of spinor fields in terms of ordinary c-numbers". Annals of Physics.

[8] Barut, A. O.; Girardello, L. (1971). "New "Coherent" States associated with non-compact groups". Communications in Mathematical Physics. 21 (1)

[9] C. Brif Two-Photon Algebra Eigenstates: A Unified Approach to Squeezing  , Annals of Phys., 251:180-207,1996.

[10] C. Brif,   $SU(2)$ and $SU(1,1)$ Algebra Eigenstates : A Unified Analytic Approach to Coherent and Intelligent States, Int. J. Theor. Phys. 36:1651-1682,1997.

[11] N. Alvarez M.   and V. Hussin,  Generalized coherent and squeezed states based on the algebra $h(1) \oplus su(2)$,  J. Math. Phys. 43:2063, 2002.

[12] N. E. Alvarez M. $h(1) \oplus su(2)$ Vector algebra eigenstates with eigenvalues in the matrix domain,
arXiv:2301.10747 [quant-ph], 2023.

[13] Sunil Kumar;  C. L. Mehta, Time evolution operator for N interacting quantum harmonic oscillators,  J. Math. Phys.21, 2628–2631 , 1980.

[14] J. D. Louck, M. Moshinsky and  K. Wolf, Canonical transformations and accidental degeneracy . I. The anisotropic oscillator, J. Math. Phys. 14 692,  1973.

[15]  Sanjay Kumar, B. Dutta-Roy, Commensurate anisotropic oscillator, $SU(2)$ coherent states and the classical limit, J. Phys. A: Math. Theor. 41,  2008.

[16] J. Moran and V. Hussin, Coherent States for the Isotropic and Anisotropic 2D Harmonic Oscillators,Quantum Rep. 2019, 1(2), 260-270; arXiv:1911.06389v1.

[17] Chen Y F, Lan Y P, and Huang K F 2003 Phys. Rev. A 68 043803

[18] Chen Y F and Huang K F 2003 J. Phys. A: Math. Gen. 36 7751

[19] J. Moran, V. Hussin  and I. Marquette Non-linear ladder operators and coherent
states for the 2:1 oscillator, Journal of Physics A: Mathematical and Theoretical 54 (27), 275301, 2021.

[20] Schwinger J. Schwinger, Quantum Theory of Angular Mometum,  Biedenharn L.  C. and  Van Dam  (eds) , Academic Press, New York,1965, page 229.

[20] Bargmann, V. (1961), On a Hilbert space of analytic functions and an associated integral transform, Communications on Pure and Applied Mathematics, 14 (3): 187, doi:10.1002/cpa.3160140303, hdl:10338.dmlcz/143587.

[21] J. Moran and V. Hussin, A new method for constructing squeezed states
for the isotropic 2D harmonic oscillator , arXiv:2001.07652v1, 2020.

[22] Walter Wyss, Two Non-Commutative Binomial Theorems, arXiv:1707.03861v2, 2017.

§ ALLOWED VALUES OF THE PARAMETERS IN THE CASE $B=1$

When $b^2= 4 \beta_+ \beta_- + \beta_3^2 =1,$ and $\beta_{\pm}=0,$ the only values allowed for $\beta_3$ are $\pm 1 ,$
which in turn do not allow for non-zero $\alpha$ parameters other than $\alpha_{\pm}$, respectively. These values will split the entire parameter set giving rise to two different classes of generalized two-mode anisotropic Hamiltonian systems that can be obtained from each other by simply exchanging the index of the modes and $\alpha_+$ for $\alpha_-$ and $\gamma_1$ for $\gamma_2.$

§.§ The case $\beta_3=1$

When $\beta_{\pm}=0$ and $\beta_3=1,$ by  inserting the corresponding values of the $\alpha$ parameters  summarized  in equation
(<ref>)  into  the algebraic equation system (<ref>–<ref>), we obtain the following values for the vector $(\vec{\mu}, \vec{\nu}) \overset{\operatorname{def}}{=} (\mu_1,\mu_2,\nu_1,\nu_2) $  in terms of the remaining parameters:

* In case that $ \gamma_1=\gamma_2=0,$

\begin{equation}
(\vec{\mu}, \vec{\nu}) =\begin{cases}   (\mu_1,0,0,0), \quad \text{if} \quad \beta_0=1 \\ (0,\mu_2,0,0), \quad \text{if} \quad \beta_0=3   \\  (0,0,0, \nu_2), \quad \text{if} \quad \beta_0=-1   \\
(0,0,\nu_1,0), \quad \text{if} \quad \beta_0=-3   \\
(0,0,0,0),\quad \text{if} \quad \beta_0 \notin \{\pm1,\pm3\}
\end{cases}
\end{equation}

* In case that $ \gamma_1 \neq 0,\gamma_2=0  $

\begin{equation}
\\ (\vec{\mu}, \vec{\nu}) =\begin{cases}   (\mu_1,0,0,  \gamma_1 \alpha_+), \quad \text{if} \quad \beta_0 = 1  \\ (0,\mu_2,0,\frac{ \gamma_1 \alpha_+}{2}), \quad \text{if} \quad \beta_0=3 \\
(0,0,\nu_1,- \gamma_1 \alpha_+ ), \quad \text{if} \quad \beta_0=-3  \\
(0,0,0,\frac{2 \gamma_1 \alpha_+}{1 + \beta_0}),\quad \text{if} \quad \beta_0 \notin \{\pm1,\pm3\}
\end{cases}
\end{equation}

* In case that $  \gamma_1= 0, \gamma_2 \neq 0 $

\begin{equation}
(\vec{\mu}, \vec{\nu}) =\begin{cases}   (  \gamma_2^\ast  \alpha_+   ,\mu_2,0,0), \quad \text{if} \quad \beta_0 = 3  \\ ( -  \gamma_2^\ast  \alpha_+ ,0,0,\nu_2), \quad \text{if} \quad \beta_0=-1  \\
( - \frac{ \gamma_2^\ast  \alpha_+}{2},0,\nu_1,0 ), \quad \text{if} \quad \beta_0=-3  \\
( \frac{2 \gamma_2^\ast  \alpha_+}{\beta_0-1},0,0,0),\quad \text{if} \quad \beta_0 \notin \{\pm1,\pm3\}
\end{cases}
\end{equation}

* In case that $  \gamma_1  \neq 0 , \gamma_2 \neq 0  $

\begin{equation}
(\vec{\mu}, \vec{\nu}) =\begin{cases}   ( \gamma_2^\ast  \alpha_+  ,\mu_2,0,\frac{ \gamma_1 \alpha_+}{2}), \quad \text{if} \quad \beta_0 = 3  \\
( - \frac{ \gamma_2^\ast  \alpha_+}{2},0,\nu_1, -  \gamma_1 \alpha_+), \quad \text{if} \quad \beta_0=-3  \\
( \frac{2 \gamma_2^\ast  \alpha_+}{\beta_0-1},0,0,\frac{2 \gamma_1 \alpha_+}{1 + \beta_0}),\quad \text{if} \quad \beta_0 \notin \{\pm1,\pm3\}
\end{cases}
\end{equation}

§.§ The case $\beta_3=-1$

When $\beta_{\pm}=0$ and $\beta_3=-1,$ the same procedure leads to

* In case that $ \gamma_1=\gamma_2=0,$

\begin{equation}
(\vec{\mu}, \vec{\nu}) =\begin{cases}   (0,\mu_2, 0,0), \quad \text{if} \quad \beta_0=1 \\ (\mu_1,0,0,0), \quad \text{if} \quad \beta_0=3   \\  (0,0,\nu_1, 0), \quad \text{if} \quad \beta_0=-1   \\
(0,0,0,\nu_2), \quad \text{if} \quad \beta_0=-3   \\
(0,0,0,0),\quad \text{if} \quad \beta_0 \notin \{\pm1,\pm3\}
\end{cases}
\end{equation}

* In case that $ \gamma_1 \neq 0,\gamma_2=0  $

\begin{equation}
\\ (\vec{\mu}, \vec{\nu}) =\begin{cases}   (\mu_1, \gamma_1^\ast \alpha_-  ,0,  0), \quad \text{if} \quad \beta_0 = 3  \\
(0,- \gamma_1^\ast \alpha_-, \nu_1,0), \quad \text{if} \quad \beta_0=-1 \\
(0,  - \frac{\gamma_1^\ast \alpha_-}{2},0,\nu_2), \quad \text{if} \quad \beta_0=-3  \\
(0, \frac{2 \gamma_1^\ast  \alpha_-}{\beta_0 -1},0,0),\quad \text{if} \quad \beta_0 \notin \{\pm1,\pm3\}
\end{cases}
\end{equation}

* In case that $  \gamma_1= 0, \gamma_2 \neq 0 $

\begin{equation}
(\vec{\mu}, \vec{\nu}) =\begin{cases}   (0 ,\mu_2, \gamma_2 \alpha_-,0), \quad \text{if} \quad \beta_0 = 1
\\ (\mu_1,0, \frac{\gamma_2 \alpha_-}{2},0), \quad \text{if} \quad \beta_0=3  \\
(0,0, -\gamma_2 \alpha_-, \nu_2 ), \quad \text{if} \quad \beta_0=-3  \\
(0,0, \frac{2 \gamma_2  \alpha_-}{1+ \beta_0},0),\quad \text{if} \quad \beta_0 \notin \{\pm1,\pm3\}
\end{cases}
\end{equation}

* In case that $  \gamma_1  \neq 0 , \gamma_2 \neq 0  $

\begin{equation}
(\vec{\mu}, \vec{\nu}) =\begin{cases}   (\mu_1, \gamma_1^\ast \alpha_-, \frac{\gamma_2 \alpha_-}{2},0), \quad \text{if} \quad \beta_0 = 3  \\
(0, - \frac{\gamma_1^\ast \alpha_-}{2},- \gamma_2 \alpha_-, \nu_2 ), \quad \text{if} \quad \beta_0=-3  \\
( 0 , \frac{2 \gamma_1^\ast  \alpha_-}{\beta_0-1},\frac{2 \gamma_2 \alpha_-}{1 + \beta_0},0),\quad \text{if} \quad \beta_0 \notin \{\pm1,\pm3\}
\end{cases}
\end{equation}

§.§ The associated generalized Hamiltonian and lowering operator

The generalized Hamiltonian and lowering operator associated with the above parameter combination can be calculated directly using the equations (<ref>), (<ref>) and (<ref>). When the $\gamma$ parameters are nonzero, the generalized Hamiltonian (<ref>) can be viewed as two interacting quantum harmonic oscillators with additional linear coupling terms.  Labeling the generalized Hamiltonian and the generalized lowering operator with the minimal set of subscripts allowing a direct correspondence with the values of the parameters identifying each element of the above classes, that is, as $\hat{H}_{\beta_0, \beta_3, \gamma_1 , \gamma_2}$ and  $\hat{A}_{\beta_0,\beta_3,\gamma_1,\gamma_2,\alpha_{\mp}},$ we obtain the following lists:

\begin{eqnarray}
\hat{H}_{1, 1, 0, 0} &=& \hat{N} + \hat{J}_3, \nonumber \\
\hat{A}_{1,1,0,0,\alpha_+} & =&  \mu_1 \hat{a}_1 + \alpha_+ \hat{J}_-,  \label{HA-plus-basic-1}  \\
\hat{H}_{3, 1, 0 , 0} &=& 3 \hat{N} + \hat{J}_3,  \nonumber \\
\hat{A}_{3,1,0,0,\alpha_+}  &=& \mu_2 \hat{a}_2 +  \alpha_+ \hat{J}_- , \\
\hat{H}_{-1, 1, 0 , 0} &=&  -  \hat{N} + \hat{J}_3,  \nonumber \\
\hat{A}_{-1,1,0,0,\alpha_+}  &=& \nu_2 \hat{a}_2^\dagger +  \alpha_+ \hat{J}_- , \\
\hat{H}_{-3, 1, 0 , 0}  &=& -3 \hat{N} + \hat{J}_3, \nonumber \\
\hat{A}_{-3,1,0,0,\alpha_+} & =&  \nu_1 \hat{a}_1^\dagger + \alpha_+ \hat{J}_-, \\
\hat{H}_{\beta_0, 1, 0 , 0} &=& \beta_0 \hat{N} + \hat{J}_3, \nonumber \\
\hat{A}_{\beta_0,1,0,0,\alpha_+} & =&  \alpha_+ \hat{J}_- , \quad  \beta_0 \notin \{\pm 1, \pm 3\}. \label{HA-plus-basic-5}
\end{eqnarray}

\begin{eqnarray}
\hat{H}_{1, 1, \gamma_1  , 0} &=& \hat{N} + \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1, \nonumber \\
\hat{A}_{1,1,\gamma_1,0,\alpha_+} & =&  \mu_1 \hat{a}_1 + \gamma_1 \alpha_+ \hat{a}_2^\dagger + \alpha_+ \hat{J}_- +  \mu_{1} \gamma_1 \hat{I}, \\
\hat{H}_{3, 1, \gamma_1  , 0} &=& 3 \hat{N} + \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1,  \nonumber \\
\hat{A}_{3,1,\gamma_1,0,\alpha_+}  &=& \mu_2 \hat{a}_2 +  \frac{\gamma_1 \alpha_+}{2} \hat{a}_2^\dagger + \alpha_+ \hat{J}_- , \\
\hat{H}_{-3, 1, \gamma_1  , 0}  &=& -3 \hat{N} + \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1, \nonumber \\
\hat{A}_{-3,1,\gamma_1,0,\alpha_+} & =&  \nu_1 \hat{a}_1^\dagger -  \gamma_1 \alpha_+ \hat{a}_2^\dagger + \alpha_+ \hat{J}_-  -  \gamma_1^\ast \nu_1 \hat{I}, \\
\hat{H}_{\beta_0, 1, \gamma_1  , 0} &=& \beta_0 \hat{N} + \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1, \nonumber \\
\hat{A}_{\beta_0,1,\gamma_1,0,\alpha_+} & =&  \frac{2 \gamma_1 \alpha_+}{1+\beta_0} \hat{a}_2^\dagger +  \alpha_+ \hat{J}_- , \quad  \beta_0 \notin \{\pm 1, \pm 3\}.
\end{eqnarray}

\begin{eqnarray}
\hat{H}_{3, 1, 0,  \gamma_2} &=& 3 \hat{N} + \hat{J}_3 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2, \nonumber \\
\hat{A}_{3,1,0,\gamma_2, \alpha_+} & =&  \gamma_2^\ast \alpha_+  \hat{a}_1 + \mu_2 \hat{a}_2  + \alpha_+ \hat{J}_-  + \gamma_2 \mu_2 \hat{I}, \\
\hat{H}_{-1, 1, 0,\gamma_2} &=&  - \hat{N} + \hat{J}_3 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2,  \nonumber \\
\hat{A}_{-1,1,0,\gamma_2, \alpha_+}  &=& - \gamma_2^\ast  \alpha_+  \hat{a}_1 +  \nu_2 \hat{a}_2^\dagger + \alpha_+ \hat{J}_-   -     \gamma_2^\ast  \nu_2  \hat{I}, \\
\hat{H}_{-3, 1, 0,\gamma_2} &=& -3 \hat{N} + \hat{J}_3 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2, \nonumber \\
\hat{A}_{-3,1,0,\gamma_2, \alpha_+} & =&  - \frac{\gamma_2^\ast \alpha_+}{2} \hat{a}_1  +  \nu_1 \hat{a}_1^\dagger + \alpha_+ \hat{J}_- , \\
\hat{H}_{\beta_0, 1, 0,\gamma_2} &=& \beta_0 \hat{N} + \hat{J}_3 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2, \nonumber \\
\hat{A}_{\beta_0,1,0,\gamma_2, \alpha_+} & =&  \frac{2 \gamma_2^\ast }{\beta_0 -1} \hat{a}_1  +  \alpha_+ \hat{J}_-    \quad  \beta_0 \notin \{\pm 1, \pm 3\}.
\end{eqnarray}

\begin{eqnarray}
\hat{H}_{3, 1, \gamma_1,  \gamma_2} &=& 3 \hat{N} + \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1 +  \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2, \nonumber \\
\hat{A}_{3,1,\gamma_1,\gamma_2, \alpha_+} & =&  \gamma_2^\ast \alpha_+  \hat{a}_1 + \mu_2 \hat{a}_2  + \frac{\gamma_1 \alpha_+}{2} \hat{a}_2^\dagger + \alpha_+ \hat{J}_-  +
\left( \gamma_2 \mu_2 +  \frac{\gamma_1 \gamma_2^\ast \alpha_+}{2}  \right)  \hat{I}, \label{A-17} \\
\hat{H}_{-3, 1, \gamma_1,\gamma_2} &=& -3 \hat{N} + \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1 +  \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2,  \nonumber \\
\hat{A}_{-3,1,\gamma_1,\gamma_2, \alpha_+} & =&  - \frac{\gamma_2^\ast \alpha_+}{2} \hat{a}_1  +  \nu_1 \hat{a}_1^\dagger - \gamma_1 \alpha_+ \hat{a}_2^\dagger + \alpha_+ \hat{J}_-   -    \left( \gamma_1^\ast  \nu_1 - \frac{\gamma_1 \gamma_2^\ast \alpha_+ }{2} \right)  \hat{I}, \\
\hat{H}_{\beta_0, 1, \gamma_1,\gamma_2} &=& \beta_0 \hat{N} + \hat{J}_3 +  \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1 +  \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2, \nonumber \\
\hat{A}_{\beta_0,1,\gamma_1,\gamma_2, \alpha_+} & =&  \frac{2 \gamma_2^\ast \alpha_+}{\beta_0 -1} \hat{a}_1  +   \frac{2 \gamma_1 \alpha_+}{\beta_0 + 1} \hat{a}_2^\dagger
+  \alpha_+ \hat{J}_-  +  \frac{4 \gamma_1 \gamma_2^\ast \alpha_+}{\beta_0^2-1} \hat{I}, \quad  \beta_0 \notin \{\pm 1, \pm 3\},
\end{eqnarray}

\begin{eqnarray}
\hat{H}_{1, -1, 0, 0} &=& \hat{N} - \hat{J}_3, \nonumber \\
\hat{A}_{1,-1,0,0,\alpha_-} & =&  \mu_2 \hat{a}_2 + \alpha_- \hat{J}_+,  \label{HA-minus-basic-1}\\
\hat{H}_{3, -1, 0 , 0} &=& 3 \hat{N} - \hat{J}_3,  \nonumber \\
\hat{A}_{3,-1,0,0,\alpha_-}  &=& \mu_1 \hat{a}_1 +  \alpha_- \hat{J}_+, \\
\hat{H}_{-1, -1, 0 , 0} &=&  -  \hat{N} - \hat{J}_3,  \nonumber \\
\hat{A}_{-1,-1,0,0,\alpha_-}  &=& \nu_1 \hat{a}_1^\dagger +  \alpha_- \hat{J}_+ , \\
\hat{H}_{-3, -1, 0 , 0}  &=& -3 \hat{N} - \hat{J}_3, \nonumber \\
\hat{A}_{-3,-1,0,0,\alpha_-} & =&  \nu_2 \hat{a}_2^\dagger + \alpha_- \hat{J}_+, \\
\hat{H}_{\beta_0, -1, 0 , 0} &=& \beta_0 \hat{N} - \hat{J}_3, \nonumber \\
\hat{A}_{\beta_0,-1,0,0,\alpha_-} & =&  \alpha_- \hat{J}_+ , \quad  \beta_0 \notin \{\pm 1, \pm 3\}. \label{HA-minus-basic-5}
\end{eqnarray}

\begin{eqnarray}
\hat{H}_{3, -1, \gamma_1  , 0} &=& 3 \hat{N} - \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1, \nonumber \\
\hat{A}_{3,-1,\gamma_1,0,\alpha_-} & =&  \mu_1 \hat{a}_1 + \gamma_1^\ast \alpha_-\hat{a}_2  + \alpha_- \hat{J}_+  +  \gamma_1  \mu_{1} \hat{I}, \\
\hat{H}_{-1, -1, \gamma_1  , 0} &=&  - \hat{N} - \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1,  \nonumber \\
\hat{A}_{-1,-1,\gamma_1,0,\alpha_-}  &=&  \gamma_1^\ast \alpha_-  \hat{a}_2  +  \nu_1 \hat{a}_1^\dagger + \alpha_- \hat{J}_+  -   \gamma_1^\ast \nu_1 , \\
\hat{H}_{-3, -1, \gamma_1  , 0} &=& -3 \hat{N} - \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1, \nonumber \\
\hat{A}_{-3,-1,\gamma_1,0,\alpha_-} & =&  - \frac{\gamma_1^\ast \alpha_-}{2} \hat{a}_2 + \nu_2 \hat{a}_2^\dagger  + \alpha_- \hat{J}_+, \\
\hat{H}_{\beta_0, -1, \gamma_1  , 0}  &=& \beta_0 \hat{N} -  \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1, \nonumber \\
\hat{A}_{\beta_0,-1,\gamma_1,0,\alpha_-} & =&  \frac{2 \gamma_1^\ast  \alpha_-}{\beta_0 - 1} \hat{a}_2 +  \alpha_- \hat{J}_+   , \quad  \beta_0 \notin \{\pm 1, \pm 3\}.
\end{eqnarray}

\begin{eqnarray}
\hat{H}_{1, -1, 0,  \gamma_2} &=&  \hat{N} - \hat{J}_3 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2, \nonumber \\
\hat{A}_{1,-1,0,\gamma_2, \alpha_-} & =&  \mu_2  \hat{a}_2  +  \gamma_2 \alpha_- \hat{a}_1^\dagger + \alpha_- \hat{J}_+  + \gamma_2 \mu_2 \hat{I}, \\
\hat{H}_{3, -1, 0,\gamma_2} &=&  3 \hat{N} - \hat{J}_3 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2,  \nonumber \\
\hat{A}_{3,-1,0,\gamma_2, \alpha_-}  &=& \mu_1 \hat{a}_1 +  \frac{\gamma_2 \alpha_-}{2} \hat{a}_1 ^\dagger + \alpha_- \hat{J}_+, \\
\hat{H}_{-3, -1, 0,\gamma_2} &=& -3 \hat{N} - \hat{J}_3 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2, \nonumber \\
\hat{A}_{-3,-1,0,\gamma_2, \alpha_-} & =&  -  \gamma_2 \alpha_- \hat{a}_1^\dagger  +  \nu_2 \hat{a}_2^\dagger + \alpha_- \hat{J}_+ -\gamma_2^\ast \nu_2  \hat {I}, \\
\hat{H}_{\beta_0, -1, 0,\gamma_2} &=& \beta_0 \hat{N} - \hat{J}_3 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2, \nonumber \\
\hat{A}_{\beta_0,-1,0,\gamma_2, \alpha_-} & =&  \frac{2 \gamma_2 \alpha_- }{\beta_0 + 1} \hat{a}_1^\dagger  +  \alpha_- \hat{J}_+,    \quad  \beta_0 \notin \{\pm 1, \pm 3\}.
\end{eqnarray}

\begin{eqnarray}
\hat{H}_{3, -1, \gamma_1,  \gamma_2} &=& 3 \hat{N} - \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1 +  \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2, \nonumber \\
\hat{A}_{3,-1,\gamma_1,\gamma_2, \alpha_-} & =&  \mu_1 \hat{a}_1  +  \gamma_1^\ast \alpha_-  \hat{a}_2 + \frac{\gamma_2 \alpha_-}{2} \hat{a}_1^\dagger + \alpha_- \hat{J}_+  +
\left(\gamma_1 \mu_1 + \frac{\gamma_1^\ast \gamma_2  \alpha_-}{2}  \right)  \hat{I}, \\
\hat{H}_{-3, -1, \gamma_1,\gamma_2} &=& -3 \hat{N} - \hat{J}_3 + \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1 +  \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2,  \nonumber \\
\hat{A}_{-3,-1,\gamma_1,\gamma_2, \alpha_-} & =&  - \frac{\gamma_1^\ast \alpha_-}{2} \hat{a}_2  - \gamma_2 \alpha_-  \hat{a}_1^\dagger  +  \nu_2 \hat{a}_2^\dagger + \alpha_- \hat{J}_+   -    \left(\gamma_2^\ast  \nu_2 - \frac{\gamma_1^\ast  \gamma_2  \alpha_- }{2} \right)  \hat{I}, \\
\hat{H}_{\beta_0, -1, \gamma_1,\gamma_2} &=& \beta_0 \hat{N} -  \hat{J}_3 +  \gamma_1 \hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1 +  \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2, \nonumber \\
\hat{A}_{\beta_0,-1,\gamma_1,\gamma_2, \alpha_-} & =&  \frac{2 \gamma_1^\ast \alpha_-}{\beta_0 -1} \hat{a}_2  +   \frac{2 \gamma_2 \alpha_-}{\beta_0 + 1} \hat{a}_1^\dagger
+  \alpha_- \hat{J}_+  +  \frac{4 \gamma_1^\ast \gamma_2 \alpha_-}{\beta_0^2-1} \hat{I}, \quad  \beta_0 \notin \{\pm 1, \pm 3\},
\end{eqnarray}

§ $ \LEFT\{ H (1) \OPLUS H(1) \RIGHT\} \ROPLUS U(2) $   HAMILTONIAN AND ASSOCIATED LOWERING OPERATOR

When $b^2= 4\beta_+ \beta_- + \beta_3^2 =1$ and $\beta_\pm \neq 0,$ according to equations  (<ref>) and  (<ref>),  several situations may arise  depending on the values of the parameters involved.  In such a case, it is pertinent to start  the analysis by fixing  the values of the  parameters $\gamma$ and then to analyze the conditions that the remaining parameters must satisfy for the resulting algebraic equation to have nonzero solutions.

§.§ Case $b^2=1,$ $\beta_{0} \notin  \{1,3,-1,-3\}$ and $\vec{\gamma}=0$

In this case, the generalized Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(\beta_0,\vec{\beta},\vec{0}, \vec{0})} =   \beta_0 \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} + h_0 \hat{I}
\label{H-beta-0-beta}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{equation}
\hat{A}_{(\vec{0},\vec{0},\vec{\alpha},0)}  = \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right). \label{A-vec-alpha}
\end{equation}

We notice that by performing a  similarity transformation with the help of the unitary operator $\hat{T}_{(\epsilon,b,\beta_3,\theta)},$ described in equation ( <ref>), these operators  take the form

\begin{equation}
\hat{H}^{T_{\epsilon}}_{(\beta_0,\beta_3,\vec{0}, \vec{0})} = \beta_{0} \hat{N} + \epsilon \hat{J_3} +  h_0  \hat{I}= \frac{(\beta_0 + \epsilon)}{2}
\hat{a}_1^\dagger \hat{a}_1 + \frac{(\beta_0 - \epsilon)}{2}
\hat{a}_2^\dagger \hat{a}_2 +  h_0  \hat{I}  \label{H-beta-0-beta-3}
\end{equation}

\begin{equation}
\hat{A}^{T_{\epsilon}}_{(\vec{0},\vec{0},\alpha_3,0)}  =  \begin{cases}   - \frac{ \alpha_3 e^{i \theta}}{2R} \hat{J}_-, \quad \text{if} \quad \epsilon =1 \\    \frac{ \alpha_3 e^{- i \theta}}{2R} \hat{J}_+, \quad \text{if} \quad \epsilon = - 1,
\end{cases}    \label{A-vec-alpha-3},
\end{equation}

respectively, i.e., they show the structure of a $(\beta_0 + \epsilon): (\beta_0 -\epsilon)$ rate anysotropic Hamiltonian together with its associated $su(2)$ canonical lowering operator.

§.§ Case $b^2=1,$ $\beta_{0} \in  \{1,3\}$ and $\vec{\gamma}=0$

When all the parameters $\gamma $ are equal to $0,$ and $\beta_0 \neq -1$ and $\beta_0 \neq -3,$  we have  $\nu_1=\nu_2=0,$ and therefore we   fall in the case already discussed at the beginning of this article, i.e., we must analyze under what conditions the parameters  $\mu_1$ and $\mu_2$ are nonzero, and we find that it is possible if and only if  $\beta_0=1$ or $\beta_0 =3.$ The expression of the generalized Hamiltonians and its associated lowering operators in this case are shown in equations   ( <ref>– <ref>).

§.§ Case $b^2=1,$ $\beta_{0} \in  \{-1,-3\}$ and $\vec{\gamma}=0$

On the other hand, when all the parameters $\gamma $ are equal to $0$  and  $\beta_0 \neq  1$ and $\beta_0 \neq 3,$  we have  $\mu_1=\mu_2=0.$  Then from  (<ref>)  we deduce  $\nu_2 = - \frac{2 \beta_+ \nu_1}{2 + \beta_0 -\beta_3}.$ Hence, the  generalized Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(\beta_0,\vec{\beta},\vec{0}, \vec{0})} =   \beta_0 \hat{N} +  \vec{\beta} \cdot \vec{\hat{J}} + h_0 \hat{I}
\label{H-nu-beta-0-beta}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{equation}
\hat{A}_{(\vec{0},\vec{\nu},\vec{\alpha},0)}  =   \nu_1 \left(   \hat{a}_1^\dagger  -   \frac{2 \beta_+ }{2 + \beta_0 -\beta_3} \hat{a}_2^\dagger \right)  +  \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}  \hat{J}_+  - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  +  \hat{J}_3 \right). \label{A-nu-vec-alpha}
\end{equation}

§.§ Case $b^2=1,$  and  $\frac{\gamma_1}{2}= \frac{\gamma_2 \beta_-}{1-\beta_3}$

In this case, depending on the value of $\beta_0,$ we  distinguish the following subcases:

* When $\beta_0 =1,$ the  Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(1,\vec{\beta},\vec{\gamma_1}, \vec{\gamma_2})} =   \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} +  \frac{ 2 \gamma_2 \beta_-}{1-\beta_3}  \hat{a}_1^\dagger + \frac{ 2 \gamma_2^\ast \beta_+}{1-\beta_3}  \hat{a}_1  + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2 +   h_0 \hat{I}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{equation} \hat{A}_{(\vec{\mu},\vec{\nu},\vec{\alpha},a_0)}  =   \mu_1 \left(   \hat{a}_1  +  \frac{2 \beta_- }{1 + \beta_3} \hat{a}_2 \right) +  \alpha_3  \gamma_2  \left(  \frac{2 \beta_-}{1 -\beta_3^2}  \hat{a}_1^\dagger - \frac{ 1}{1-\beta_3}  \hat{a}_2^\dagger \right) + \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right) + \frac{\gamma_2 \mu_1}{\beta_+}.
\end{equation}

* When $\beta_0 =3,$ the  Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(3,\vec{\beta},\vec{\gamma_1}, \vec{\gamma_2})} =  3  \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} +  \frac{ 2 \gamma_2 \beta_-}{1-\beta_3}  \hat{a}_1^\dagger + \frac{ 2 \gamma_2^\ast \beta_+}{1-\beta_3}  \hat{a}_1  + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2 +   h_0 \hat{I}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{equation} \hat{A}_{(\vec{\mu},\vec{\nu},\vec{\alpha},0)}  =   \mu_1 \left(   \hat{a}_1  +  \frac{2 \beta_- }{\beta_3 -1}  \hat{a}_2 \right) +  \alpha_3  \gamma_2 \left(  \frac{\beta_-}{1 -\beta_3^2}  \hat{a}_1^\dagger - \frac{1}{2(1-\beta_3)}  \hat{a}_2^\dagger \right) + \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right).
\end{equation}

* When $\beta_0 =-3,$ the  Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(-3,\vec{\beta},\vec{\gamma_1}, \vec{\gamma_2})} =  -3  \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} +  \frac{ 2 \gamma_2 \beta_-}{1-\beta_3}  \hat{a}_1^\dagger + \frac{ 2 \gamma_2^\ast \beta_+}{1-\beta_3}  \hat{a}_1  + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2 +   h_0 \hat{I}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{equation} \hat{A}_{(\vec{0},\vec{\nu},\vec{\alpha},a_0)}  =  \nu_1  \hat{a}_1^\dagger + 2  \left( \frac{ \beta_+ \nu_1}{1+\beta_3} + \frac{\gamma_2 \alpha_3}{1-\beta_3^2}  \right) \hat{a}_2^\dagger + \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right) - 2 \gamma_{2}^\ast \left(\frac{2 \beta_+ \beta_3 \nu_1 + \gamma_2 \alpha_3 }{1- \beta_3^2}\right).
\end{equation}

* When $\beta_0  \notin \{1,3,-1,-3 \},$ the  Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(\beta_0,\vec{\beta},\vec{\gamma_1}, \vec{\gamma_2})} =  \beta_0  \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} +  \frac{ 2 \gamma_2 \beta_-}{1-\beta_3}  \hat{a}_1^\dagger + \frac{ 2 \gamma_2^\ast \beta_+}{1-\beta_3}  \hat{a}_1  + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast \hat{a}_2 +   h_0 \hat{I}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{equation} \hat{A}_{(\vec{0},\vec{\nu},\vec{\alpha},0)}  =  \frac{ 2 \alpha_3 \gamma_2 (3 +\beta_0)}{(1-\beta_3^2)\left( (2 + \beta_0)^2 -1  \right)     }  \left( 2 \beta_-  \hat{a}_1^\dagger    -  (1 + \beta_3) \hat{a}_2^\dagger \right) + \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right).
\end{equation}

§.§ Case $b^2=1,$  and  $\frac{\gamma_1}{2}= - \frac{\gamma_2 \beta_-}{1+\beta_3}$

In this case, depending on the value of $\beta_0,$ we  distinguish the following sub-cases:

* When $\beta_0 =-1,$ the  Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(-1,\vec{\beta},\vec{\gamma_1}, \vec{\gamma_2})} =  - \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} -  \frac{  2 \gamma_2 \beta_-}{1+\beta_3}  \hat{a}_1^\dagger - \frac{  2 \gamma_2^\ast \beta_+}{1+\beta_3}  \hat{a}_1 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast  \hat{a}_2 +   h_0 \hat{I}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{equation} \hat{A}_{(\vec{\mu},\vec{\nu},\vec{\alpha},a_0)}  =  \alpha_3  \gamma_2^\ast  \left(  \frac{2 \beta_+}{1 -\beta_3^2}  \hat{a}_1 + \frac{ 1}{1+\beta_3}  \hat{a}_2 \right) +  \nu_1 \left(   \hat{a}_1^\dagger  -  \frac{2 \beta_+ }{1 - \beta_3} \hat{a}_2^\ast \right) +  \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right) + \frac{\gamma_2^\ast \nu_1}{\beta_-}.
\end{equation}

* When $\beta_0 =-3,$ the  Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(-3,\vec{\beta},\vec{\gamma_1}, \vec{\gamma_2})} =  - 3 \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} -  \frac{  2 \gamma_2 \beta_-}{1+\beta_3}  \hat{a}_1^\dagger - \frac{  2 \gamma_2^\ast \beta_+}{1+\beta_3}  \hat{a}_1 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast  \hat{a}_2 +   h_0 \hat{I}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{equation} \hat{A}_{(\vec{\mu},\vec{\nu},\vec{\alpha},0)}  =    \alpha_3  \gamma_2^\ast  \left(  \frac{\beta_+}{1 -\beta_3^2}  \hat{a}_1 + \frac{1}{2(1+\beta_3)}  \hat{a}_2 \right) +  \nu_1 \left(   \hat{a}_1^\dagger  +  \frac{2 \beta_+ }{1 + \beta_3}  \hat{a}_2^\dagger \right) + \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right).
\end{equation}

* When $\beta_0 =3,$ the  Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(3,\vec{\beta},\vec{\gamma_1}, \vec{\gamma_2})} =  3 \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} -  \frac{  2 \gamma_2 \beta_-}{1+\beta_3}  \hat{a}_1^\dagger - \frac{  2 \gamma_2^\ast \beta_+}{1+\beta_3}  \hat{a}_1 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast  \hat{a}_2 +   h_0 \hat{I}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{equation} \hat{A}_{(\vec{\mu},\vec{0},\vec{\alpha},a_0)}  =  \mu_1  \hat{a}_1 - 2  \left( \frac{ \beta_- \mu_1}{1-\beta_3} + \frac{\gamma_2^\ast \alpha_3}{1-\beta_3^2}  \right) \hat{a}_2 + \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right) + 2 \gamma_{2} \left(\frac{2 \beta_- \beta_3 \mu_1 - \gamma_2^\ast \alpha_3 }{1- \beta_3^2}\right).
\end{equation}

* When $\beta_0  \notin \{1,3,-1,-3 \},$ the  Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(\beta_0,\vec{\beta},\vec{\gamma_1}, \vec{\gamma_2})} =  \beta_0 \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} -  \frac{  2 \gamma_2 \beta_-}{1+\beta_3}  \hat{a}_1^\dagger - \frac{  2 \gamma_2^\ast \beta_+}{1+\beta_3}  \hat{a}_1 + \gamma_2 \hat{a}_2^\dagger + \gamma_2^\ast  \hat{a}_2 +   h_0 \hat{I}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{equation} \hat{A}_{(\vec{\mu},\vec{0},\vec{\alpha},0)}  =  \frac{ 2 \alpha_3 \gamma_2^\ast (3 -\beta_0)}{(1-\beta_3^2)\left( (2 - \beta_0)^2 -1  \right)     }  \left( 2 \beta_+  \hat{a}_1  +  (1 - \beta_3) \hat{a}_2 \right) + \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right).
\end{equation}

§.§ Case $b^2=1,$    $\frac{\gamma_1}{2} \neq  \frac{\gamma_2 \beta_-}{1-\beta_3}$ and  $\frac{\gamma_1}{2} \neq - \frac{\gamma_2 \beta_-}{1+\beta_3}$

* When $\beta_0 =3,$ the  Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(3,\vec{\beta},\vec{\gamma_1}, \vec{\gamma_2})} =  3  \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} +   \gamma_1
\hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1  + \gamma_2 \hat{a}_2^\dagger +  \gamma_2^\ast \hat{a}_2 +   h_0 \hat{I}
\label{H-2-D-gen-3-gamma-all-diff}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{eqnarray} \hat{A}_{(\vec{\mu},\vec{\nu},\vec{\alpha},a_0)} & =&   \left[ \mu_1  \hat{a}_1  + \left( \frac{\alpha_3}{\beta_+}  \left(
\frac{\gamma_1^\ast}{2} - \frac{ \gamma_2^\ast \beta_+}{1-\beta_3}
\right)    -   \frac{2 \mu_1 \beta_- }{1-\beta_3}  \right)\hat{a}_2 \right] +
\frac{\alpha_3}{2} \left(
\frac{\gamma_2}{2} + \frac{\gamma_1 \beta_+}{1-\beta_3} \right)  \left[ \frac{2 \beta_-}{1+\beta_3}  \hat{a}_1^\dagger - \hat{a}_2^\dagger   \right]
\nonumber \\ &+&  \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right) \nonumber \\ &+&
2 \mu_1   \left(
\frac{\gamma_1}{2} - \frac{ \gamma_2  \beta_-}{1-\beta_3}
\right) + \frac{\gamma_2 \alpha_3}{\beta_+}   \left(
\frac{\gamma_1^\ast}{2} - \frac{ \gamma_2^\ast \beta_+}{1-\beta_3}
\right) + \alpha_3   \left(
\frac{\gamma_2}{2} + \frac{ \gamma_1 \beta_+}{1-\beta_3}
\right)  \left(
\frac{\gamma_2^\ast}{2} - \frac{ \gamma_1^\ast \beta_-}{1+\beta_3}
\right). \label{A-2-D-gen-3-gamma-all-diff}
\end{eqnarray}

* When $\beta_0 =-3,$ the  Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(-3,\vec{\beta},\vec{\gamma_1}, \vec{\gamma_2})} = - 3  \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} +   \gamma_1
\hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1  + \gamma_2 \hat{a}_2^\dagger +  \gamma_2^\ast \hat{a}_2 +   h_0 \hat{I}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{eqnarray} \hat{A}_{(\vec{\mu},\vec{\nu},\vec{\alpha},a_0)} & =&
\frac{\alpha_3}{2} \left(
\frac{\gamma_2^\ast}{2} - \frac{\gamma_1^\ast \beta_-}{1+\beta_3} \right)  \left[ \frac{2 \beta_+}{1- \beta_3}  \hat{a}_1 + \hat{a}_2   \right]  +  \left[ \nu_1  \hat{a}_1^\dagger  +  \left( \frac{\alpha_3}{\beta_-}  \left(
\frac{\gamma_1}{2} +  \frac{ \gamma_2  \beta_-}{1+\beta_3}
\right)    +   \frac{2 \nu_1 \beta_+}{1+\beta_3}  \right)\hat{a}_2^\dagger \right]
\nonumber \\ &+&  \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right) \nonumber \\ &-&
2 \nu_1   \left(
\frac{\gamma_1^\ast}{2} + \frac{ \gamma_2^\ast  \beta_+}{1+\beta_3}
\right) - \frac{\gamma_2^\ast \alpha_3}{\beta_-}   \left(
\frac{\gamma_1}{2} + \frac{ \gamma_2  \beta_-}{1+\beta_3}
\right) + \alpha_3   \left(
\frac{\gamma_2^\ast}{2} - \frac{ \gamma_1^\ast \beta_-}{1+\beta_3}
\right)  \left(
\frac{\gamma_2}{2} + \frac{ \gamma_1  \beta_+}{1-\beta_3}
\right).
\end{eqnarray}

* When $\beta_0  \notin \{1,3,-1,-3 \},$ the  Hamiltonian  (<ref>) becomes

\begin{equation}
\hat{H}_{(\beta_0,\vec{\beta},\vec{\gamma_1}, \vec{\gamma_2})} =  \beta_0  \hat{N} + \vec{\beta} \cdot \vec{\hat{J}} +   \gamma_1
\hat{a}_1^\dagger + \gamma_1^\ast \hat{a}_1  + \gamma_2 \hat{a}_2^\dagger +  \gamma_2^\ast \hat{a}_2 +   h_0 \hat{I} \label{gen-gen-hamiltonian-beta-0-diff}
\end{equation}

and the lowering operator  (<ref> )  reduces to

\begin{eqnarray} \hat{A}_{(\vec{\mu},\vec{\nu},\vec{\alpha},a_0)}  &=&  \frac{2  \alpha_3}{ 1-\beta_0}
\left( \frac{\gamma_2^\ast}{2}  - \frac{\gamma_1^\ast \beta_-}{1+\beta_3} \right) \left( \frac{2 \beta_+}{1-\beta_3}  \hat{a}_1 + \hat{a}_2     \right)   \nonumber \\&+  &\frac{2  \alpha_3}{1+\beta_0}
\left( \frac{\gamma_2}{2}  +  \frac{\gamma_1 \beta_+}{1-\beta_3} \right) \left( \frac{2 \beta_-}{1+\beta_3}  \hat{a}_1^\dagger - \hat{a}_2^\dagger     \right) \nonumber \\&+&   \alpha_3 \left(  \frac{\beta_-}{1+\beta_3}   \hat{J}_+   - \frac{\beta_+}{1-\beta_3}   \hat{J}_-  + \hat{J}_3 \right) \nonumber \\ &+& \frac{8 \alpha_3}{1-\beta_0^2} \left( \frac{\gamma_2^\ast}{2}  - \frac{\gamma_1^\ast \beta_-}{1+\beta_3} \right)  \left( \frac{\gamma_2}{2}  +  \frac{\gamma_1 \beta_+}{1-\beta_3} \right). \label{gen-gen-lowering-beta-0-diff}
\end{eqnarray}