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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?

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  • $\begingroup$ 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. $\endgroup$
    – wcc
    Mar 6, 2019 at 3:27
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    $\begingroup$ 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. $\endgroup$
    – Sunyam
    Mar 6, 2019 at 5:57

2 Answers 2

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This is beacause, it is based on the assumption that only one optical and mechanical mode interact. Each optical cavity supports in principle an infinite number of modes and mechanical oscillators have more than a single oscillation/vibration mode. The validity of this approach relies on the possibility to tune the laser in such a way, that it populates a single optical mode only Furthermore, scattering of photons to other modes is supposed to be negligible, which holds if the mechanical (motional) sidebands of the driven mode do not overlap with other cavity modes, i.e. if the mechanical mode frequency is smaller than the typical separation of the optical modes. I hope this will somehow clear your doubt.

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  • $\begingroup$ Welcome to Physics! We do have MathJax here to format equations. You can see the notation page in help center for details, if you aren't familiar with it (though it's similar to LaTeX). $\endgroup$
    – Kyle Kanos
    Sep 9, 2019 at 11:21
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  • As it was correctly pointed in the other answer, only one mode is considered coupled to the oscillator, which is why one need not use additional creation/annihilation operators.
  • The frequency of this mode does change as the mirror moves! The wavelength of the cavity mode is however huge compared to the displacement of the mirror, so this change can be neglected. Yet, it is sometimes included - you will probably meat such Hamiltonians, although this seriously complicates the math.
  • Finally, cavity-mirror system is only one realization of such a Hamiltonian. In some cases this problem does not arise at all.
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