What enduces transition in a two-state Quantum Mechanical system? Assume we have a two-state system, with an excited state, $ | e \rangle $, and a ground state, $|g\rangle$. The states have an energy difference of $E$. I want to talk about the transition frequency between the two, or the frequency we'd need to tune a laser to induce transition between the two.
In the Schrödinger picture, we can write $\langle e | H | g \rangle = \langle g | H | e \rangle = \langle g | H | g \rangle = 0$, and $\langle e | H | e \rangle = E$. However, this leaves no room to discuss a transition between the two states. Using the unitary time evolution operator, $U = \exp(-i\ t\ H\ /\ \hbar)$, does not change the ground state. In other words, $|\langle e|U|g\rangle|^2=0$ for all time. This is true no matter how we split of the energy via gauge transforms, the off diagonal terms will always be zero in this basis.
What is the minimum extra details we need to include in our model to talk about the transition frequency between the excited and ground state? I know we can heuristically say "use dipole operators" but I'm hoping for a more mathematically inspired answer than that.
Additionally, when we take a laser and tune it near $ \omega = E / \hbar $, where does this term actually enter in the Schrödinger picture - if at all?
 A: The crucial point is that it is not possible to neglect the presence of an electromagnetic field when studying transition processes in atoms. So even though the 2-level atomic system is in vacuum, you have to take into account the presence of a quantized electromagnetic radiation. Just to give you a more quantitative idea, the full Hamiltonian of the system is
$$ H = H_{atom} + H_{e.m.} + H_{int.}, $$
where $H_{atom}$ describes the atomic 2-levels system, and has $\left| g \right\rangle$ and $\left| e \right\rangle$ as eigenstates, $H_{e.m.}$ is the bare Hamiltonian of the electromagnetic radiation, and $H_{int.}$ is the interacting part that couples the atom to the radiation. The Hilbert space is now a tensor product of the atomic Hilbert space, and of the radiation Hilbert space, and a state can be written in the form $\left| a, \{n_{k,\lambda}\} \right\rangle $, where $a=g,e$ is the atomic part and $\{ n_{k,\lambda} \}$ is the number of photons in the state with momentum $k$ and polarization $\lambda$.
Now clearly because of $H_{int}$, the state where the atom is excited and the e.m. field is in the vacuum $\left| e, \{ 0 \} \right\rangle$ is no longer an eigenstate, and thus there is a probability of decaying in the $\left| g , 1_{k,\lambda} \right\rangle$ state (ground atomic state plus a photon emitted by the transition). This is a sketch of the spontaneous emission :)
A: You don't need to quantize the electromagnetic field to understand stimulated emission and absorption.  You just need to add time dependent off diagonal terms to your Hamiltonian that look like this: $V_0 sin(\omega t)$. If this is a laser, the frequency of the laser comes in here as $\omega$, and $V_0$ is proportional to the laser field strength.   (You do need quantum field theory to understand fully spontaneous emission though.)
For details on how to work this out and what the dynamics that result from this perturbation to the Hamiltonian look like, see Griffith's Introduction to Quantum Mechanics, Chapter 11 ("Quantum Dynamics"). This chapter addresses exactly the question you pose. Nearly every introductory quantum mechanics textbook covers this fundamental topic.
