2
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – SuperCiocia
    Commented Jul 7, 2022 at 23:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.