frame, are:

$$\begin{aligned}
{\ddot {\hat x}}& = & - \omega^2_M {\hat x} - \frac{\hbar kA}{m}\sum_j
(\alpha_j^* {\hat a}_j+ \alpha_j {\hat a}_j^\dagger)\sin 2(kx_0-\phi_j)
-\Gamma_M {\dot {\hat x}} \nonumber\\
{\dot {\hat a}_j} & = & i \Delta_j^x {\hat a}_j - ikA \alpha_j {\hat x} \sin
2(kx_0-\phi_j) -\frac{\kappa}{2} {\hat a}_j.
\label{a2s}\end{aligned}$$

The resulting effective mechanical harmonic oscillator frequency is:

$$\begin{aligned}
\omega_M^2= \frac{2\hbar A k^2}{m} \sum_j |\alpha_j|^2 \cos{2(kx_0-\phi_j)}.
\label{freqs}\end{aligned}$$

We can restrict ourselves to real equilibrium fields. We take $\alpha_j ={\tilde
{\alpha}}_j e^{-i\theta_j}$ then transform ${\hat a}_j \to {\hat a}_j e^{i\theta_j}$. Thus $(\alpha_j^*
{\hat a}_j+ \alpha_j {\hat a}_j^*)\equiv {\tilde {\alpha}}_j( {\hat a}_j+ {\hat
a}_j^\dagger)$. We also rescale the mechanical oscillator coordinates ${\hat x} \to
\sqrt{2}X_{zpf}{\hat x}$ and ${\hat p} \to \sqrt{\hbar m\omega_M} {\hat p}$, where $X_{zpf}=\sqrt{\frac{\hbar}{2 m\omega_M}} $ is the zero-point fluctuation length scale. Hence, $\frac{{\hat P}^2}{2m}+\frac{1}{2} m \omega_M^2 {\hat x}^2 \to \frac{\hbar
\omega_M}{2}({\hat x}^2+{\hat p}^2)$.\
Below we drop the tilde so the equilibrium field values ${\tilde {\alpha}}_j 
\equiv \alpha_j$ are real. Using field operators ${\hat x}= ( {\hat b} +{\hat b}^\dagger)/\sqrt{2}$, the linearised dynamics for a two-optical mode system would correspond to an effective Hamiltonian:

$$\begin{aligned}
\frac{\hat H_{Lin}}{\hbar}&=&  -\Delta_1^x{\hat a}_1^\dagger {\hat a}_1 
-\Delta_2^x{\hat a}_2^\dagger {\hat a}_2
  + \omega_M(\Delta_1^x,\Delta_2^x){\hat b}^\dagger {\hat b}\nonumber\\
     & +&  