Is the total number of excitation conserved with Rabi Hamiltonian?

From quantum optics textbooks, I have learned about Jaynes-Cummings Hamiltonian for a description of atom and field interaction, as an approximated model under the so-called rotating wave approximation. Under the approximation, it drops out the counter-rotating terms that are non-energy-conserving terms, often said textbooks. This makes sense because the counter-rotating terms do not commute with the free-evolution Hamiltonian $\hat{H}_{0}$ of atom and field, i.e., they do not preserve the total number of excitation in the whole system. Thus, I tend to drop out the non-energy-conserving terms whatsoever. Otherwise, the solution of the Schrödinger equation might not be normalized or may be unphysical.

Nevertheless, when the rotating wave approximation is not valid, e.g., in the ultra strong coupling regime, one has to consider the Rabi model, the original model that contains the counter rotating terms, in order to correctly and quantitatively describe actual physical phenomena. Here my question arises.

The Rabi-model has the non-energy-conserving terms, the counter-rotating terms. How can it be physical? For example, if one starts with a certain initial state $\vert \psi(t=0)\rangle$ under the Rabi-model, the total energy of the state $\vert \psi(t)\rangle$ changes in time. Does this make sense?

• Can you write down what you refer to as the 'Rabi model'? – Harry Levine Jan 26 '18 at 22:43
• @HarryLevine The Rabi mode that I referred to is like $g(\hat{a}+\hat{a}^{\dagger})(\hat{\sigma}_{-}+\hat{\sigma}_{+})$, where $\hat{a}$ and $\hat{a}^{\dagger}$ are the annihilation and creation operator of a single-mode field, and $\hat{\sigma}_{-}=\vert g\rangle\langle e\vert$ and $\hat{\sigma}_{+}=\vert e\rangle\langle g\vert$ with $\vert g\rangle$ and $\vert g\rangle$ being the ground and excited state of two-level atom. – Veteran Jan 29 '18 at 5:17
• @HarryLevine I thought it's more or less obvious what I was referring to as the Rabi model from the context where the rotating wave approximation, Jaynes-Cummings Hamiltonian, etc are mentioned. – Veteran Jan 29 '18 at 5:19
• Thank you for posting the explicit form of the model. Can you additionally say what you mean by 'the total energy of state $|\psi(t)\rangle$ changes in time'? If $|\psi(t=0)\rangle$ is an eigenstate of $H$, then it's energy does not change in time. If it is not an eigenstate, then one can consider the expectation value of its energy $\langle\psi | H | \psi \rangle$, and this value also does not change in time. – Harry Levine Jan 29 '18 at 20:46
• @HarryLevine The expectation value of the energy $\langle \psi\vert\hat{H}\vert\rangle$ changes in time due to the counter rotating terms $\hat{a}\sigma_{-}$ and $\hat{a}^{\dagger}\sigma_{+}$ that don't commute with the free-evolution Hamiltonian $\hat{H}_{0}$. This is my question. – Veteran Jan 30 '18 at 12:43