Can we diagonalize optomechanical hamiltonian? The optomechanical hamiltonian is given as 
$$\hat{H}=\hbar\Delta_{a}a^{\dagger}a+\hbar\omega_{m}b^{\dagger}b+\hbar g_{a}a^{\dagger}a(b^{\dagger}+b)$$
$a$ and $b$ are photonic and phononic operators and others are numbers. 
Can a Bogoliubov transformation be made to such a hamiltonian to find the normal frequencies?
Do we always have to linearize it? 
 A: Since the Hamiltonian in the question is  not atmost a quadratic in bosonic creation and annihilation operators, it is not possible to reduce it to uncoupled form by just a combination of Bogoliubov and displacement transformations. 
However it can be simplified by a generalized displacement transformation aka the Polaron transformation:
$$\hat{H} \rightarrow \hat{\tilde{H}}=\hat{U}_{}^{\dagger}\hat{H} \hat{U}_{}^{}$$
where 
$$\hat{U}_{}^{}=e_{}^{-\frac{g_{a}^{}}{\omega_{m}^{}}a_{}^{\dagger}a_{}^{}\left[b_{}^{\dagger}-b_{}^{}\right]}.$$
Using this generalized Polaron transformation, the transformed Hamiltonian acquires a diagonal form as given below
$$e_{}^{\frac{g_{a}^{}}{\omega_{m}^{}}a_{}^{\dagger}a_{}^{}\left[b_{}^{\dagger}-b_{}^{}\right]}\hat{H}e_{}^{-\frac{g_{a}^{}}{\omega_{m}^{}}a_{}^{\dagger}a_{}^{}\left[b_{}^{\dagger}-b_{}^{}\right]}=\hbar\Delta_{a}^{}a_{}^{\dagger}a_{}^{}-\hbar\frac{{g_{a}^{}}_{}^{2}}{\omega_{m}^{}}\left[a_{}^{\dagger}a_{}^{}\right]_{}^{2}+\hbar \omega_{m}^{}b_{}^{\dagger}b_{}^{}.$$
