Why does Jaynes Cummings not describe an atom in free space? I'm a bit confused about the application of the Jaynes Cummings model and what exactly is meant by "one single mode": 
Usually, it is said that the Jaynes Cummings model describes a single atom in a high Q cavity. The atom then only interacts with a single mode of the light field and the Hamiltonian is written as: 
$$
H=H_0+H_I=\hbar \omega\sigma_+\sigma_-+\hbar\omega_La^{\dagger}a+\hbar g(\sigma_+a+\sigma_- a^{\dagger})
$$
Question 1: Why does this Hamiltonian only describe cavity QED and not e.g. an atom in free space? 
Question 2: Is it correct that this description is not valid for interaction with a laser field, since the coherent state is not an eigenstate of $a^{\dagger}a$, i.e. the above Hamiltonian describes interaction with a photon Fock state? 
I would appreciate some help, I'm a bit confused about these different models...
 A: I just recently learned about the Jaynes-Cummings model and will attempt to answer this question, despite that it was asked three years ago. I imagine that the questioner probably already have an answer, but I'll still provide some of my own thoughts.
To the first question, the most general interpretation of the Hamiltonian above is that it describes the interaction of an atom with a single mode field, not necessarily including a cavity. This means that the coupling term only has one mode of the photon creation and annihilation operators.
In the most general case, we would have some arbitrary $E$ field acting on an atom, and this $E$ field can be expanded as
$$E = \sum_{\lambda} \sqrt{\frac{\hbar \omega_k}{2\epsilon_0 V}} \left\{f(r)a_{\lambda} e^{i(k\cdot r - \omega_k t)} + H.C.\right\} \hat{\epsilon_\lambda}$$
Here, the summation is over $\lambda$, which is a tuple of indices $(\mu, k)$, where $\mu$ denotes the two polarizations of the light, and $k$ is the wave vector. $\hat{\epsilon_\lambda}$ is the polarization vector. The function $f(r)$ is the mode density, and it can be solved for different boundary conditions. Notice here that there is a pair of ladder operators $a$ and $a^\dagger$ for every mode.
With this general field, the JC Hamiltonian would have to be amended to
$$\hbar \omega \sigma_+ \sigma_- + \sum_{\lambda}\hbar \omega_k a^\dagger_\lambda a\lambda + \hbar\sum_{\lambda} g_\lambda (\sigma_+ a_\lambda + \sigma_- a^\dagger_\lambda)$$
In the case that the field is in free space, the wave vector can take any directions and any magnitude, and the summation over $k$ becomes an integral over the $k$ space. In the case where the field is inside a one dimensional cavity, the $k$ vector takes discrete values along the cavity axis, and continuous values along the two perpendicular axis.
With that, the short answer is, the JC Hamiltonian can describe an atom in free space, if you take the summation/integration over $\lambda$. But with the form you wrote down, where there is only a single mode interacting with the atom, we cannot create this type of coupling without a cavity, since the atom can essentially emit photon in all directions with any magnitude of the wave vector.
To the second question, the Hamiltonian is still valid for a laser interacting with atoms in a cavity, where the laser is represented by a coherent state. The JC Hamiltonian is essentially derived from a dipole approximation, in which the coupling term is $-d \cdot E$. Here the $E$ field is just an operator, it has nothing to do with which basis you choose to expand your state in. You can choose to solve the JC Hamiltonian in the Fock basis or in the coherent basis, since both are complete representations of the total Hilbert space (up to a factor of $\pi$ for the coherent basis). This is just like solving for the classic problem of a spin in a magnetic field, in which the Hamiltonian scales with $S_n$, for a magnetic field pointing along the vector $n$. You can solve it by quantizing in the $n$ direction, or you can choose a different quantization axis. In either cases, the Hamiltonian is always valid.
A: You need a cavity so that the modes are well separated in frequency.In  free space  there are arbitrarily close-in-frequency modes, and so it is impossible to let  the atom interact with only one of them.
