[@rev]). Then its dynamical equations read $$\begin{aligned}
\dot{\hat{a}}&=&-\kappa\hat{a}-i\Delta \hat{a}+ig(\hat{b}+\hat{b}^\dagger)\hat{a}+E +\sqrt{2\kappa}\hat{\xi}_c(t),\nonumber\\
\dot{\hat{b}}&=&-\gamma_m\hat{b}-i\omega_m\hat{b}+ig\hat{a}^\dagger\hat{a}+\sqrt{2\gamma_m}\hat{\xi}_m(t),
\label{e2}\end{aligned}$$ which directly uses the system-reservoir coupling Hamiltonian in Eq. (\[couple1\]).

II. LINEARIZED DYNAMICS OF WEAKLY COUPLED QUANTUM OPTOMECHANICAL SYSTEMS
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For the total Hamiltonian $H(t)=H_e(t)+H_{om}(t)+H_{sr}(t)$ in our used rotating frame, we take another interaction picture with respect to the part of $H_e(t)$, to have the cavity mode $\hat{a}$ displaced by the unitary operation $U_0(t)={\cal T}\exp\{-i\int_0^t d\tau H_e(\tau)\}$ to $\hat{a}-i\frac{e^{i\Delta t}-1}{\Delta}E$. Then the remaining part of the Hamiltonian $H(t)$ will be transformed to $$\begin{aligned}
H^{in}(t)&=& U_0^\dagger(t)\big(H_{om}(t)+H_{sr}(t)\big)U_0(t)\nonumber\\
&=&\underbrace{ig E\frac{e^{i\Delta t}-1}{\Delta}e^{-i\omega_mt} \hat{a}^\dagger \hat{b}-ig  E \frac{e^{-i\Delta t}-1}{\Delta}e^{i\omega_mt} \hat{a} \hat{b}^\dagger
+ig E\frac{e^{i\Delta t}-1}{\Delta}e^{i\omega_mt} \hat{a}^\dagger \hat{b}^\dagger-ig E \frac{e^{-i\Delta t}-1}{\Delta}e^{-i\omega_mt} \hat{a} \hat{b}}_{H_{couple}(t)}\nonumber\\
&-&g\underbrace{\big|\frac{e^{i\Delta t}-1}{\Delta}E\big|^2(e^{-i\omega_mt}\hat{b}+e^{i\omega_m t}\hat{b}^\dagger)}_{H_{displace}(t)}
+i\sqrt{2\kappa}\{(\hat{a}^\dagger +i\frac{e^{-i\Delta t}-1}{\Delta}E)\hat{\xi}_p(t)
-(\hat{a}-i\frac{e^{i\Delta t}-1}{\Delta}E)\hat{\xi}_p^\dagger(t)\}\nonumber\\
&+&i\sqrt{2\gamma_m }\{\hat{b}^{\dagger}\hat{\xi}_{m}(t)-\hat{b}\hat{\xi}_{m}^\dagger (t)\}
-\underbrace{g\hat{a}^\dagger\hat{a}(e^{-i\omega_mt}\hat{b}+e^{i\omega_m t}\hat{b}^\dagger)}_{H_{nl}(t)}.
\label{Hi}\end{aligned}$$ Among this transformed Hamiltonian $H^{in}(t)$, the effect of the cubic term $H_{nl}(t)$ is insignificant compared with the others. Because it is non-commutative with the rest of the Hamiltonian, its action will modify the mode operators in the rest part of $H^{in}(t)$ by the unitary action [@s-laser] $$\begin{aligned}
U_{nl}(t,\tau)&=&{\cal T}e^{i\int_\tau^t dt' g\hat{a}^\dagger\hat{a}(e^{-i\omega_m t'}\hat{b}+e^{i\omega_m t'}
\hat{b}^\dagger)}\nonumber\\
&=& e^{i\int_\tau^t dt' g\hat{a}^\dagger\hat{a}\hat{b}e^{-i\omega_m t'}}e^{i\int_\tau^t dt' g\hat{a}^\dagger\hat{a}\hat{b}^\dagger e^{i\omega_m t'}}
e^{i\int_\tau^t dt'\frac{g^2}{\omega_m}(1-e^{i\omega_m(t'-\tau)})(\hat{a}^\dagger\hat{a})^2},
\label{NL}\end{aligned}$$ as it can be seen from the expectation value of any system operator 