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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.

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  • $\begingroup$ Does $\mathcal{L}'\rho = -i [\hat{H}',\rho]$ answer your question? If not, it's a bit unclear what you're looking for. $\endgroup$ Aug 15, 2018 at 14:34
  • $\begingroup$ I guess so. I am unsure if I could just do that or if I have to be cautious if superoperators are involved. I guess if dissipation of the form $\left(2 a \rho a^{\dagger} - a^{\dagger} a \rho - \rho ^{\dagger} a \right)$ is present I also just transform $a$ and $a^{\dagger}$? $\endgroup$
    – Timo
    Aug 15, 2018 at 14:43
  • $\begingroup$ To answer that you really need to add more context to your question. Is the displacement supposed to represent some physical change of the Hamiltonian due to e.g. external driving, or is it just a change of basis? $\endgroup$ Aug 15, 2018 at 15:02
  • $\begingroup$ I have added some context. The Hamiltonian is time-independent. $\endgroup$
    – Timo
    Aug 15, 2018 at 15:09

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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.

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  • $\begingroup$ Note the latest edit corrects a rather essential error in the final sentence of this answer. $\endgroup$ Aug 15, 2018 at 23:47

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