An adequate way to rewrite the following unitary superoperator Let us consider a set of superoperators: $X_1, \dots, X_8$ which acts on the density matrix $\rho$ as follows
\begin{equation} \label{eq:algebra} \tag{1}
\begin{array}{ll}
X_{1} \rho = a  \rho a^\dagger, & X_{2} \rho = a^\dagger \rho a,  \\ 
X_{3} \rho = b \rho b^\dagger,  & X_{4} \rho = b^\dagger  \rho b, \\ 
X_{5} \rho = a \rho  b^\dagger, & X_{6} \rho = b \rho a^\dagger, \\
X_{7} \rho = a^\dagger \rho b,  & X_{8} \rho = b ^\dagger \rho a,
\end{array}
\end{equation}
where $a$, $a^\dagger$, and $b$, $b^\dagger$ - bosonic operators.
I have to deal with the following-like expression
\begin{equation}  \label{prod}
\mathcal{F} = \prod\limits_{k=1}^9 \exp({\lambda_k}X_k) \rho, \tag{2}
\end{equation}
where the order of the exponents may be chosen arbitrary (coefficients $\lambda_i$ can be chosen arbitrary to simplify the formula, but not equal to zero).
Let us say that $\rho = |\alpha, \beta\rangle  \langle \alpha, \beta|$, where $|\alpha, \beta\rangle = \sum_{n,k} \alpha^n \beta^k/\sqrt{n!k!} |n,k\rangle$ - coherent state.
One can formally expand all of the exponents in \eqref{prod} and obtain
\begin{multline}  \label{bulky}
\mathcal{F} = \sum\limits_{\ell_1\dots,\ell_8}\frac{\lambda_{1}^{\ell_1}\dots \lambda_{8}^{\ell_8}}{\ell_1! \dots \ell_8!} 
(b^{\dagger})^{\ell_8}(a^{\dagger})^{\ell_7} b^{\ell_6} a^{\ell_5}(b^{\dagger})^{\ell_4} b^{\ell_3}(a^{\dagger})^{\ell_2} a^{\ell_1} |\alpha, \beta\rangle \times \\  \langle \alpha, \beta| 
(a^{\dagger})^{\ell_1} a^{\ell_2} (b^{\dagger})^{\ell_3} b^{\ell_4} (b^{\dagger})^{\ell_5} (a^{\dagger})^{\ell_6} b^{\ell_7} a^{\ell_8}
. \tag{3}
\end{multline}
The expression \eqref{bulky} is a quite bulky. I wonder if it is possible to rewrite \eqref{bulky} in form of action of some unitary operators from the left and ride sides: $U \rho U^\dagger$. Where $U$ - something like a chain of displacement or/and squeeze operators.
P.S. Once one had a deal with one-mode analog of the problem, where one had superoperators act as
\begin{equation}  \label{old}
Y_1\rho = a\rho, \quad Y_2\rho = a^\dagger\rho, \quad Y_3\rho = \rho a, \quad Y_4\rho = \rho a^\dagger,
\tag{4}
\end{equation}
In this case, it was possible to write action for $\rho = |\alpha\rangle  \langle \alpha|$ as follows
\begin{equation}  \label{good}
e^{\lambda Y_2}e^{-\lambda^* Y_1}e^{\lambda^* Y_3}e^{-\lambda Y_4} |\alpha\rangle  \langle \alpha| = e^{|\lambda|^2} D(\lambda)|\alpha\rangle  \langle \alpha| D^{\dagger}(\lambda),
\tag{5}
\end{equation}
where $D(\lambda)$ - displacement operator.
I would really like to obtain something like \eqref{good} (of course, more complicated and, probably, including squeeze operators), but for the \eqref{bulky} if it is even possible.
 A: First of all the product of operators you wrote is not unitary unless there are appropriate constraints on the $\lambda_i$'s, but let's not worry about this for the moment.
Since $\vert \alpha,\beta\rangle$ is a 2-mode coherent state, and since your sequence ${\cal F}$ can be done in any order, one way to proceed is to start with
$$
\exp(\lambda_1 X_1)\exp(\lambda_2 X_2)\exp(\lambda_5 X_5)\exp(\lambda_6 X_6)
$$
and use the property that coherent states are eigenstates of the annihilation operator $\hat a$ or $\hat b$ so that
$$
\exp(\lambda_1 X_1)\exp(\lambda_2 X_2)\exp(\lambda_5 X_5)\exp(\lambda_6 X_6)\rho
$$
will be proportional to itself.  You can then use the product of the remaining exponentials to act on each Fock state in
$$
\vert\alpha,\beta\rangle =\sum_{n_1n_2}\vert n_1 n_2\rangle\langle n_1n_2\vert \alpha,\beta\rangle
$$
or $\langle \alpha,\beta\vert$ and get a sum of Fock states.

Edit: Here are some details on how this would work.
\begin{align}
&\exp(\lambda_5 X_5)\exp(\lambda_3 X_3)\exp(\lambda_1 X_1)\vert\alpha\beta\rangle\langle \alpha\beta\vert \\ 
&\quad =\exp(\lambda_5 X_5)\exp(\lambda_3 X_3)\left[\sum_{k} \frac{ \lambda_1^k a^k}{k!} \vert \alpha\beta\rangle\langle \alpha\beta\vert (a^\dagger)^k\right]
\end{align}
but $a\vert\alpha\rangle=\alpha\vert \alpha\rangle$ and $\langle \alpha\vert a^\dagger = \alpha^* \langle\alpha\vert$ so we now get
\begin{align}
\exp(\lambda_5 X_5)&\exp(\lambda_3 X_3)\exp(\lambda_1 X_1)\vert\alpha\beta\rangle\langle \alpha\beta\vert \\
&= \exp(\lambda_5 X_5)\exp(\lambda_3 X_3)
\left[\sum_{k} \frac{ \lambda_1^k \alpha^k}{k!} \vert \alpha\beta\rangle\langle \alpha\beta\vert (\alpha^*)^k\right] \, ,\\
&= \exp(\lambda_5 X_5)\exp(\lambda_3 X_3)
\left[\sum_{k} \frac{ \lambda_1^k \vert \alpha\vert^k}{k!} \vert \alpha\beta\rangle\langle \alpha\beta\vert\right]\, ,\\
&= \exp(\lambda_5 X_5)\exp(\lambda_3 X_3) e^{\lambda_1 \vert \alpha\vert^2}
\vert \alpha\beta\rangle\langle \alpha\beta\vert \, ,\\
&=  e^{\lambda_1 \vert \alpha\vert^2}
\exp(\lambda_5 X_5)\exp(\lambda_3 X_3)
\vert \alpha\beta\rangle\langle \alpha\beta\vert \, .
\end{align}
You can redo the trick with the $\exp(\lambda_5 X_5)$ and $\exp(\lambda_3 X_3)$.
I'm not sure there's an easy way to handle the $X_2, X_4, X_7$ and $X_8$ terms except by brute force.  Maybe someone has a clever idea.
Contrary to what I initially suspected, the action of $\exp(\lambda_2X_2)\rho$ is not
$$
\exp(\lambda_2 a^\dagger)\rho \exp(\lambda_2 a)
$$
so
$$
\exp(\lambda_2 X_2)\exp(\lambda_1 X_1)\rho \ne D_1(\eta)\rho D_1^\dagger(\eta)
$$
Note that this is a borderline assignment question so I don't want to give additional details.
