Atom-field interaction for two level system: decomposition of the dipole moment on $|0\rangle$ and $|1\rangle$

Question

On page 145 of Exploring the quantum, atom cavities and photons by Serge Haroche and Jean-Michel Raimond, we have a two level system (an atom) interacting with a classical field. From what I understood, we can see the problem as a nucleus with one electron having two orbitals of eigenstates $|0\rangle$ and $|1\rangle$. We thus write the Hamiltonian:

$$ H = \hbar \omega_0 |1\rangle \langle 1| - \vec{d}\cdot\vec{E} $$

Where $\vec{d}=e \vec{r}$ and $\vec{r}$ is the position of the electron relative to the atom. On this page, the author rewrites $\vec{d}=\langle 0 | \vec{r} | 1 \rangle \ |0\rangle \langle 1 | + \langle 1 | \vec{r} | 0 \rangle \ |1\rangle \langle 0 | $. The diagonals terms of this operator vanish for parity reasons (I'm ok with this as fundamental is even and excited state is odd).

I don't understand how we can assume that with the interaction, we still can assume that our atom still has two eigenstates $|0\rangle$ and $|1\rangle$. The interaction could add energy levels for example.

If it is an approximation, what are the requirements to ensure that it is true?

Answer 1

For me the interaction could add energy levels for example.

It can't. The dimension of the state space that describes the atom is fixed, and adding an interaction cannot change it.

On the other hand, your construction does rest on the assumption that it is only those two states that are relevant to the evolution. If that assumption fails (which can be because e.g. there's another state that's closer to resonance with the driver, or which has a much stronger coupling, among a myriad possible reasons) then the other states obviously also need to be included, in both the dipole coupling as well as the atomic hamiltonian.