I am struggling to interpret the Langevin equation for the outputs of an optical cavity. I am using Quantum Noise by Gardiner & Zoller, and Quantum Optics by Walls & Milburn as primary guides. I have seen Wolpertinger's solution to a similar question. (Link.) However it doesn't quite cover the problems I have.
The Langevin equation for a cavity of $n$ partially-transmitting mirrors is
$$\frac{d}{dt}\hat{a}_{c}(t)=\frac{i}{\hbar}[\hat{H}_{c},\hat{a}_{c}(t)]-\gamma\hat{a}_c(t) +\sum_n\sqrt{\gamma_n}\hat{a}_{in}(t).$$
Where $\hat{a}_{c}(t)$ is the (Heisenberg-picture) annihilation operator for the cavity mode, $\hat{H}_{c}$ is the Hamiltonian for the cavity mode which in general can be non-linear, and $\gamma = \frac{1}{2}\sum_n\gamma_n$ are the constants that couple the cavity mode to the driving fields $\hat{a}_{in,n}(t)$. The input operators are obtained by integrating over the continuous mode operators for the external fields - see the question linked above for a derivation.
A typical method of solving these equations after substitution of a suitable Hamiltonian is to take the Fourier transform and in doing so define a frequency decomposition of the annihilation operator as
$\tilde{a}_{c}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dte^{i\omega t}\hat{a}_{c}(t)$.
The input-output relations can then be found algebraically using the boundary condition
$\hat{a}_{in,n}(t)+\hat{a}_{out,n}(t) = \sqrt{\gamma_n}\hat{a}_{c}(t)$.
In the simplest case of an empty linear cavity the Hamiltonian used is $\hat{H}_{c} = \hbar\omega_{c}(\hat{N}+\frac{1}{2})$, which brings me to the problems I have: Why can we treat the cavity field as though it is a single mode? The cavity must have some non-zero line-width and be partially resonant at frequencies other than $\omega_{c}$ - shouldn't these terms also contribute to the Hamiltonian? Furthermore, I would expect the annihilation operator to have a decomposition like
$\hat{a}_{c}(t) = \int d\omega \zeta(\omega) \hat{a}(t,\omega) $,
$\zeta(\omega)$ being some suitable lineshape function, and $\hat{a}(t,\omega)$ being continuous-mode operators. This would give $\hat{a}_{c}(t)$ the same units of $\sqrt{\text{Hz}}$ as the driving fields. However it is clear from looking at the units in the boundary condition that $\hat{a}_{c}(t)$ must not have units, as $\gamma_n$ has units of $\text{Hz}$. So from both the units and the Hamiltonian it seems to me that the cavity is treated as having single frequency. But if this was true then $\tilde{a}_c(\omega)$ would not be interesting.
I assume that what's happening here is a narrow-band assumption: If the cavity line-width is small compared to the resonant frequency (MHz vs THz) then we can get away with treating the cavity as effectively a single mode system. Then $\tilde{a}_c(\omega)$ tells us the RF spectrum that we will see after detection centers the noise around DC. (In this sense we can have our cake and eat it too.) But I only ever see this assumption briefly discussed, if at all, so any further explanation or suggested reading on this topic would be appreciated.
