that $
X_{zpf}=\sqrt{\hbar/(2m\omega_M)}$ here too depends on the detunings via $\omega_M$.

For triple mode hybridization, $\omega_M \sim \Delta_{1} \sim \Delta_{2} $. For convenience, we also take $\phi_1=0, \phi_2=\pi/4$. Then, since $\tan 2kx_0 = {|\alpha_2|^2}/{|\alpha_1|^2}$, we can re-write Eq. \[freqs\]:

$$\begin{aligned}
\omega_M^2= \frac{2\hbar A k^2}{m \cos{2kx_0}}|\alpha_1|^2.
\label{freq2}\end{aligned}$$

We consider near symmetric driving of the two optical modes for which $R \sim 1$ and thus $kx_0 \approx \phi/2=\pi/8$ so $\omega_M \sim  \left({\frac{2\hbar A
k^2}{m}}\right)^{1/2} n_{1,2}^{1/2} $. Hence,

$$\begin{aligned}
{\tilde g}_{1,2} \sim \left( {\frac{\hbar k^2}{4}}\right)^{1/4} \left( \frac{A^3
n_{1,2}}{ m}\right)^{1/4}.\end{aligned}$$

Since the optomechanical coupling increases only very slowly with cavity photon number the most effective means to reach strong coupling regimes is to increase the nanosphere size to the maximum practical value ($r\sim 200-250$nm).

For the ideal case where the nanosphere radius $r$ is small [@Zoller] ie. $\lambda \gg r$, then $A(r\ll \lambda) \equiv A_0(r)$ where the small nanosphere coupling takes the form: $$\begin{aligned}
A_0(r) = \frac{3}{2} \frac{\epsilon_r-1}{\epsilon_r-2} \frac{V_s}{V_c} \omega_L
\label{ideal}\end{aligned}$$ where $V_s= 4/3\pi r^3$ is the sphere volume (and hence $m=V_s \rho$ where the density $\rho=2000$Kgm$^{-3}$ for silica). In turn, $V_c=\pi (w/2)^2 L$ is the cavity volume, where $w\approx 40\,\mu$m is the cavity waist and $L \simeq 0.5 - 1$cm is the cavity 