Does the AC Stark Effect in Multilevel Systems depend on whether a state is populated or not?

Question

For the sake of simplicity, assume a 3 level system, subject to a Hamiltonian $H_0$. The Hilbert space is then completely spanned by 3 states with energies $E_1$, $E_2$ and $E_3$. I'd like to understand what happens in the following experiment:

• I prepare the system in state $|1\rangle$
• Now I illuminate the system with monochromatic light-field of frequency $(E_2 - E_3)/\hbar = \omega_{23}$. This light is kept shining during the next steps
• Because $\omega_{23}$ is far from the frequencies $\omega_{13}$, the system will still be in state $|1\rangle$
• Now I do spectroscopy: I illuminate the system with a 2nd light field, with frequencies around $\omega_{13}$, or $\omega_{12}$. In a detector, I track whether the light of the 2nd source is absorbed from the system

Now what will happen? Will I see a peak in the absorption or the 2nd light field for the unperturbed frequency $\omega_{13}$? That is, the 2nd light field excites the system from the state $|1\rangle$ to the unperturbed state $|3\rangle$?

Or will I see two peaks in the absorption for the frequencies $\omega_{13} \pm \tilde{\Omega}$? That is, the 2nd light field excites the system from the state $|1 \rangle$ to a state $|+\rangle$ or $|-\rangle$, which are eigenstates of the old hamiltonian $H_0$ + the interaction with light field 1. If light field 1 is treated as a quantum field, then $|+\rangle$ and $-\rangle$ would be what is usually called the "dressed states".

For this question, I proposed a simple experimental setup, because I hope that the question "what will happen" can be answered unambiguously, and in past question attempts I wasn't able to ask proper questions about the theoretical side of the model. Because of that I also want to, for the purpose of the question, disregard any effects of broadening, or as well as aspects of the actual implementations of the experiment.

The reason I'm asking the question is that I usually only see calculations of the AC-Stark-Shift in 2-level systems, where the system already is in one of the 2 states that are mixed by the lightfield (for example when it undergoes rabi oscillations).

The seperate case that light mixes and shifts eigenenergies of two states that the system is NOT in, I don't see this case in standard literature on the topic.

Answer 1

I believe that the effect does not depend on whether the level is populated or not.

The experiment you are proposing is basically an upside-down Autler-Townes effect. There, you strongly pump two levels and then use a weak probe to measure the absorption from one of the levels into an unpumped state. It is found that the absorption line is split into two lines separated by the Rabi frequency $\Omega$ (assuming zero detuning for convenience, as in your question).

To explain this effect, it is perhaps easier to consider the fully quantum description, in which we have to diagonalize the Hamiltonian of the coupled light and atom. In the absence of coupling, the eigenstates are found to be $\vert 3,n \rangle$ and $\lvert 2,n+1 \rangle$ (assuming that level $3$ lies above level $2$) where $n$ labels the number of pump photons, forming an infinite ladder of states spaced apart by $\omega_{23}$. The coupling then causes the two states in each rung of the ladder to be mixed into dressed states, separated by $\Omega$, which is the AC Stark shift.

There has been no mention of any population anywhere, and all we have done is find the eigenstates of some coupled Hamiltonian. In the Autler-Townes effect, the dressed states are initially populated, and the probe excites them into the unpumped state. It should be clear however that, in the case of your reversed experiment, the probe can just as well excite from the unpumped state (level $1$) into the dressed states, as they exist regardless of whether or not there is population in them. So to answer your question, you should see absorption lines at $\omega_{12}\pm\Omega/2$ as well as $\omega_{13}\pm\Omega/2$.

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You might ask, why do only two of the rungs give rise to absorption lines? The answer is that the probe beam can only couple states in which the number $n$ of pump photons is unchanged, so its only non-zero matrix elements are $\langle 1,n \rvert V \lvert 2,n \rangle$ and $\langle 1,n \rvert V \lvert 3,n \rangle$. Given that the dressed states in the $n$th rung are formed of equal superpositions of $\lvert 2,n+1\rangle $ and $\lvert 3,n\rangle$, we couple only to the $(n-1)$th and $n$th rung, the former being responsible for $\omega_{12}\pm\Omega/2$, and the latter for $\omega_{12}+\omega_{23}\pm\Omega/2=\omega_{13}\pm\Omega/2$.

Answer 2

If I understood you correctly, the experiment you proposed is an example of an electromagnetically induced transparency (EIT) setup with cold/ultracold matter. In this case, you would prepare your sample in a well-defined state and perform spectroscopy using a weak probe beam while the stronger coupling field is present. As a result, you would see a typical spectroscopy peak with a dip in the middle. You can find plenty of calculations of that effect in the literature for $\Lambda$, ladder, or $V$ configuration. I can recommend checking out Daniel Steck's lecture notes.

I would also like to point out that the resonance condition you mentioned is a crucial component of this setup. In this case, I prefer to use the name "Autler-Townes effect," which indicates that the outcome is more complicated than a simple level shift for two strongly coupled states. However, if you're talking about a far-detuned light (like in optical dipole traps), I don't have any issue with using AC Stark shift as the name. Again, please keep in mind that this is just my personal opinion.