Displacement transformation of Liouvillian superoperator

Question

The displacement operator $D(\alpha)$ has the property $D^{\dagger}(\alpha) \hat{a} D(\alpha) = \hat{a} + \alpha$. We obtain the Hamiltonian $\hat{H}'$ in the displaced frame from the transformation $\hat{H} \rightarrow D^{\dagger}(\alpha) \hat{H} D(\alpha) = \hat{H}'$. The evolution of the density operator is given by the von Neumann equation $$ \dot{\rho} = -i \left[ \hat{H}, \rho \right] + \kappa \left (2 \hat{a} \rho \hat{a}^{\dagger} - \hat{a}^{\dagger} \hat{a} \rho - \rho \hat{a}^{\dagger} \hat{a} \right) = \mathcal{L} \rho \tag{1} $$ with $\mathcal{L}$ the Liouvillian super operator. What is the transformation I have to apply to $\mathcal{L}$ to obtain it in the displaced frame? I would like to transform $(1)$ with a displacement transformation by $\alpha$ where $\alpha$ is obtained from mean-field equation therefore obtaining equations for the quantum fluctuations.

Answer 1

In general, consider a change of basis described by a time-independent unitary operator $U$, which transforms the density operator as $\rho \to \rho' = U\rho U^\dagger$. Let the equation of motion for the density operator in the old basis be $$ \dot{\rho} = -i[H,\rho] + \kappa\mathcal{D}[a]\rho,$$ where $\mathcal{D}[a]\rho = 2a\rho a^\dagger - a^\dagger a \rho - \rho a^\dagger a $ is a Lindblad dissipator. Then the new equation of motion is clearly $$ \dot{\rho}' = \frac{\partial}{\partial t}\left( U\rho U^\dagger\right) = U \dot{\rho} U^\dagger,$$ where we used the fact that $\dot{U} = 0$ by assumption. Plugging away and making liberal use of $U U^\dagger = 1$, one obtains $$ \dot{\rho}' = -i[H',\rho'] + \kappa \mathcal{D}[a']\rho',$$ where $H' = U H U^\dagger$ and $a' = U a U^\dagger$. Now follow the same steps to show that, if $U(t)$ is time-dependent, then the Hamiltonian is instead modified as $H' = U H U^\dagger + i \dot{U} U^\dagger$ (hint: you need the identity $\partial_t (U U^\dagger) = \dot{U}U^\dagger + U \dot{U}^\dagger = 0$).

In your example, $a' = a+\alpha$ and you can use the handy shift-invariance of the Lindblad equation, i.e. the combined transformation $$a\to a+\alpha \qquad H\to H +i \kappa( \alpha a^\dagger -\alpha^* a )$$ leaves the Lindblad equation invariant.