# Cavity optomechanics Hamiltonian

In cavity optomechanics the radiation pressure exerted by light moves a mirror in a cavity. Because of that the resonance frequency of the cavity changes due to change in length of the cavity (cavity frequency, $$\omega_{cav} = n\pi c/L$$, $$L$$ is the length of the cavity). The Hamiltonian of the system is given by two harmonic oscillators i.e. the cavity mode and the mechanical mode coupled by the optomechanical Hamiltonian [as discussed in this review article, https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.86.1391, given by Eqs. (18)-(20)]:

$$H = \hbar \omega_{cav}a^\dagger a + \hbar \Omega_{m}b^\dagger b - \hbar g_0 a^\dagger a (b + b^\dagger)$$.

What I don't understand is that, since the cavity length $$L$$ is changed due to the radiation pressure, the cavity modes are now changed. So, should the modes not be represented by different creation and annihilation operators because the cavity modes are changing dynamically? How can we use the same annihilation (creation) operator '$$a$$' ('$$a^\dagger$$') for the optical mode in the Hamiltonian?

• isn't that the point? Once the optomechanical coupling $g_0$ becomes non-zero, the coherent state is no longer the eigenstate of the Hamiltonian. The eigenstate now becomes an admixture of optical and mechanical mode. – AmIAStudent Mar 6 at 3:27
• This paper might be of help. Therein a general formulation is developed which upon linearization (of cavity modes which depend on mirror position) gives the above hamiltonian. – Sunyam Mar 6 at 5:57