Commutation of field and atomic ladder operators in the Dicke model Consider the Dicke model, whose Hamiltonian is (in the rotating wave approximation)
\begin{equation}
\hat H=\omega_c \hat a^\dagger \hat a+\omega_0 \hat \sigma_z + g(\hat a^\dagger \hat \sigma + \hat a \hat \sigma^\dagger)
\end{equation}
Here $\hat a^\dagger$ and $\hat a$ are the creation and annihilation operators for a photon in the cavity mode and $\hat \sigma^\dagger$ and $\hat \sigma$ are the atomic raising and lowering operators. Take for simplicity the case of a single atom, so $\hat \sigma=|g \rangle \langle e |$ and $\hat \sigma^\dagger= |e \rangle \langle g |$.
Do the field and atom operators commute? That is, is it the case that $[\hat a^\dagger, \hat \sigma]=[\hat a, \hat \sigma^\dagger]=0$?
I believe that they do from considerations of dissipative evolution. Consider a free atom subject only to a dissipation due to spontaneous emission, neglecting the Hamiltonian terms. Then the master equation is 
\begin{equation}
\frac{\partial \hat \rho}{\partial t} = -\gamma(\hat \rho \hat \sigma^\dagger \hat \sigma + \hat \sigma^\dagger \hat \sigma \hat \rho - 2 \hat \sigma \hat \rho \hat \sigma^\dagger)
\end{equation}
The expectation value of the lowering operator changes according to 
\begin{equation}
\frac{\partial \langle \hat \sigma \rangle}{\partial t} = Tr\{\hat \sigma \frac{\partial \hat \rho}{\partial t} \} = -\gamma \big [ \langle \hat\sigma^\dagger \hat\sigma \hat\sigma \rangle+\langle \hat\sigma \hat\sigma^\dagger \hat\sigma \rangle -2\langle \hat\sigma^\dagger \hat\sigma \hat\sigma \rangle \big ] = -\gamma \langle \hat \sigma \rangle
\end{equation}
which has as its solution
\begin{equation}
\langle \hat \sigma \rangle(t) = \langle \hat \sigma \rangle_0 e^{-\gamma t}
\end{equation}
If the field operators and atomic operators commute, then with the previous process we find
\begin{equation}
\frac{\partial \langle \hat a \rangle}{\partial t} = Tr\{\hat a \frac{\partial \hat \rho}{\partial t} \} = -\gamma \big [ \langle \hat\sigma^\dagger \hat\sigma \hat a \rangle+\langle \hat a\hat\sigma^\dagger \hat\sigma \rangle -2\langle \hat\sigma^\dagger \hat a \hat\sigma \rangle \big ] = 0,
\end{equation}
which makes sense as there is no cavity decay in the master equation. On the other hand, if they do not commute, then
\begin{equation}
\frac{\partial \langle \hat a \rangle}{\partial t}  \neq 0
\end{equation}
despite the lack of dissipative processes for the cavity field.
This argument is handwavy at best as the master equation neglects Hamiltonian terms for the cavity and atomic energies, but those should not cause any non-imaginary time evolution like the decay in the non-commuting case.
To repeat the initial question, do the field and atom operators commute? That is, is it the case that $[\hat a^\dagger, \hat \sigma]=[\hat a, \hat \sigma^\dagger]=0$? 
 A: Good question. Yes, they do commute. In constructing the many-body Hilbert space for two different systems $A$ and $B$, the direct product of the individual spaces 
$$\mathcal H = \mathcal H_A \otimes \mathcal H_B$$
has to be taken to describe the combined system. This is the only way consistent with fulfilling the requirements of QM and all possible measurements on the subsystems if these are uncoupled and don't know of each other. 
Any operator $K_A$ defined on subsystem $A$ is promoted to $K \mapsto K_A \otimes \mbox {Id}_B$ which then acts on the combined space (and analogously $L_B \mapsto   \mbox{Id}_A  \otimes L_B$). 
With this construction, operators $K_A$, $L_B$  initially acting on different subspaces automatically commute
$$
[K_A \otimes \mbox{Id}_B,\; \mbox{Id}_A  \otimes L_B ]=0
$$
since 
$$
(K_A \otimes \mbox{Id}_B)(\mbox{Id}_A  \otimes L_B)= K_A \otimes L_B =(\mbox{Id}_A  \otimes L_B)(K_A \otimes \mbox{Id}_B).
$$
This commutation also holds for fermions operators of different fermions type, e.g. electrons and muons or different fermionic atom species.
