# 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 $$$$\label{eq:algebra} \tag{1} \begin{array}{ll} X_{1} \rho = a \rho a^\dagger, & X_{2} \rho = a^\dagger \rho a, \\ X_{3} \rho = b \rho b^\dagger, & X_{4} \rho = b^\dagger \rho b, \\ X_{5} \rho = a \rho b^\dagger, & X_{6} \rho = b \rho a^\dagger, \\ X_{7} \rho = a^\dagger \rho b, & X_{8} \rho = b ^\dagger \rho a, \end{array}$$$$ where $$a$$, $$a^\dagger$$, and $$b$$, $$b^\dagger$$ - bosonic operators.

I have to deal with the following-like expression $$$$\label{prod} \mathcal{F} = \prod\limits_{k=1}^9 \exp({\lambda_k}X_k) \rho, \tag{2}$$$$ where the order of the exponents may be chosen arbitrary (coefficients $$\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 $$$$\label{old} Y_1\rho = a\rho, \quad Y_2\rho = a^\dagger\rho, \quad Y_3\rho = \rho a, \quad Y_4\rho = \rho a^\dagger, \tag{4}$$$$

In this case, it was possible to write action for $$\rho = |\alpha\rangle \langle \alpha|$$ as follows $$$$\label{good} e^{\lambda Y_2}e^{-\lambda^* Y_1}e^{\lambda^* Y_3}e^{-\lambda Y_4} |\alpha\rangle \langle \alpha| = e^{|\lambda|^2} D(\lambda)|\alpha\rangle \langle \alpha| D^{\dagger}(\lambda), \tag{5}$$$$ where $$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.

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.

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Jul 7, 2022 at 23:07