Photon number for a multimode cavity Let's assume I have a cavity with one main frequency and two different sideband frequencies characterized by the annihilation operators $a$, $a_1$ and $a_2$. If I want to calculate the number of photons in the cavity, should it be ($a+ a_1+a_2$)$^\dagger$($a+a_1+a_2$)? If so, how do I understand this?
 A: For a single mode $\hat{a}$ the number of photons $N_\mathrm{ph}$ is given by the 
$$N_\mathrm{ph} = \langle \hat{a}^\dagger \hat{a}\rangle $$
This is to be understood as the photon population of the mode.
How to go to multiple modes? You are counting photons in different degrees of freedom, which is modes are. So you should add the photon numbers of each mode. In the OP's case
\begin{align}N_\mathrm{ph} &= \langle \hat{a}^\dagger \hat{a}\rangle + \langle \hat{a}_1^\dagger \hat{a}_1\rangle + \langle \hat{a}_2^\dagger \hat{a}_2\rangle \\ &= \langle \hat{a}^\dagger \hat{a} + \hat{a}_1^\dagger \hat{a}_1 + \hat{a}_2^\dagger \hat{a}_2\rangle \,,\end{align}
which is not equal to the expression suggested by the OP
$$N_\textrm{ph}\neq\langle (\hat{a}^\dagger  + \hat{a}_1^\dagger + \hat{a}_2^\dagger)(\hat{a} +  \hat{a}_1 + \hat{a}_2)\rangle.$$
The reason is very simple and can also be understood classically. When you are counting photons, you are counting intensities, not amplitudes.
An example for where this is important is the Hamiltonian of the electromagnetic field (see e.g. wiki). In its simplest form
$$H = \sum_\lambda E_\lambda \hat{a}_\lambda^\dagger\hat{a}_\lambda.$$
So $\langle H \rangle$ is simply the sum over the energies of each mode times the number of photons in each mode, as we would expect.
A: I assume the sideband frequencies you mention are must smaller than the free spectral range of the cavity.
In this limit it is best to think about this problem in the following terms. There is only ONE TEM00 optical mode supported by the cavity*.
The sidebands that you mention represent a temporal oscillation of the amplitude of that particular mode at the modulation frequency. Note that this is operating in a Heisenberg picture for the photon operator because we are imagining a time-dependent operator. However, you were already working with a time-dependent operator when you specified that the carrier tone has sidebands.
The instantaneous photon number would then be calculated as
\begin{align}
\hat{n}(t) =& \hat{a}^{\dagger}(t)\hat{a}(t)\\
\langle \hat{n}(t) \rangle =&\langle \hat{a}^{\dagger}(t)\hat{a}(t)\rangle = n_0 + \eta \cos(\Omega t + \phi)
\end{align}
Where $n_0$ is the average carrier photon number, $\eta$ is the sideband modulation depth, $\Omega$ is the modulation frequency and $\phi$ is the modulation phase.
Note that here $\eta$ is represented in units of photon number so if you wish you could interpret this somehow as the number of photons in the sidebands but I'll reiterate that that might be a little misleading as those aren't actually separate modes of the optical cavity and that I much prefer the single time-dependent mode perspective.
*Two modes if you are considering two different polarizations of light
