If you expose an atomic 2-level system to a time dependent periodic electromagnetic wave, according to the AC-Stark effect (or Autler-Townes Effect) we experience an energy shift of the quantum system.

I always understood an energy shift to be a shift of the energies of the eigenstates of the system.

At the same time, we can calculate the time-evolution of the system. We then end up with Rabi oscillations, that is, the lower energy state $|1\rangle$ slowly becomes the higher energy state $ |2 \rangle$, and vice versa.

Isn't that a contradiction? If the time evolution isn't anymore given by $e^{-i\omega t}$, doesn't that mean that there aren't energy eigenstates anymore? What states do the energyshifts of the AC-Stark-effect then belong to? Or is there a special superposition of the states $|1\rangle$ and $|2\rangle$, which really is an eigenstate of the Hamiltonian, and thus has a time evolution with $e^{-i \omega t}$?


To be clear about that: I'm fully aware that you can also diagonalize a time dependent Hamiltonian, which will result in time dependent eivenvalues and eigenstates. However, the energyshifts from the AC-Stark effect are not time dependent, which leaves me still confused about where they come from.


2 Answers 2


At their heart, Rabi oscillations and Autler-Townes splittings describe exactly the same situation, but they do this from different perspectives, and they track different solutions of the same Schrödinger equation. When you're describing Rabi oscillations, you're looking at the time-dependent behaviour of an atomic eigenstate when you add an interaction hamiltonian into the mix; on the other hand, Autler-Townes looks at the same problem and decides that it only cares about the eigenstates of the total hamiltonian.

(They also describe fundamentally different experiments in terms of the timescales involved for your measurement device, but I'll get to those later.)

That said, the mention of instantaneous eigenstates has muddied the waters a bit, so let me explore in more depth exactly what we mean when we talk about eigenstates in the Autler-Townes perspective.

As a starting point, let's take the relevant section in Wikipedia, to try to unpack it:

In an appropriate rotating frame, and making the rotating wave approximation, $\hat {H}$ reduces to $$ \hat {H}=-\hbar \Delta |e\rangle \langle e|+{\frac {\hbar \Omega}{2}}\bigg(|e\rangle \langle g|+|g\rangle \langle e|\bigg). \tag{$*$}$$ Where $\Omega $ is the Rabi frequency, and $|g\rangle ,|e\rangle$ are the strongly coupled bare atom states.The energy eigenvalues are $$ E_{{\pm }}={\frac {-\hbar \Delta }{2}}\pm {\frac {\hbar {\sqrt {\Omega ^{2}+\Delta ^{2}}}}{2}}$$ and for small detuning, $$ E_{{\pm }}\approx \pm {\frac {\hbar \Omega }{2}}.$$

Everything here looks legit: the rotating-wave approximation (RWA) is generally plenty justified, and those are indeed the relevant eigenvectors and eigenvalues.

The catch, however, comes much earlier:

In an appropriate rotating frame,

which is where much of the confusion can stem from - in essence, from a shift to a suitable interaction picture. This setup means that when you're diagonalizing the Autler-Townes hamiltonian $(*)$, your basis states already encode a lot of time-evolution information, because relative to the static eigenstates $|0\rangle$ and $|1\rangle$, the rotating-frame basis states are given by $$ |g\rangle =|0\rangle \quad\text{and}\quad |e\rangle =e^{-i\omega t}|1\rangle $$ (where and this in turn means that the eigenstates $|\pm\rangle$ of the rotating-frame hamiltonian can be seen a satisfying the neat property that $\hat H|\pm\rangle = E_\pm |\pm\rangle = \hbar \omega_\pm|\pm\rangle$, but on a much more fundamental level they really encode two linearly independent solutions of the time-dependent Schrödinger equation: \begin{align} e^{-i\omega_+ t}|+\rangle &= e^{-i\omega_+ t}\left(A_+|g\rangle + B_+|e\rangle\right) = A_+e^{-i\omega_+ t}|0\rangle + B_+e^{-i(\omega_+ +\omega) t}|1\rangle ,\quad\text{and}\\ e^{-i\omega_- t}|-\rangle &= e^{-i\omega_- t}\left(A_-|g\rangle + B_-|e\rangle\right) = A_+e^{-i\omega_- t}|0\rangle + B_-e^{-i(\omega_- +\omega) t}|1\rangle . \end{align} In the rotating frame, these simplify into evolving eigenstates, but it's important to keep in mind that when seen from the original basis, they are actually rather complicated solutions.

Having said all of that, we can now kind of forget about the original basis, because the Rabi-oscillation formalism also tends to be worked from exactly the same rotating frame I developed above, so let's just accept the fact that $|e\rangle$ contains some hidden time dependence as a caveat, and just roll with it.

Both problems, then, are concerned with the hamiltonian $$ \hat {H}=-\hbar \Delta |e\rangle \langle e|+{\frac {\hbar \Omega}{2}}\bigg(|e\rangle \langle g|+|g\rangle \langle e|\bigg), \tag{$*$}$$ and now we can lay out the differences between them quite simply: the Rabi-oscillations formalism tends to look for solutions of the form $$ i\partial_t |\psi(t)\rangle = \hat H |\psi(t)\rangle, \quad \text{under }|\psi(0)\rangle = |g\rangle, $$ whereas Autler-Townes just solves for $$\hat H|\pm\rangle = \hbar \omega_\pm|\pm\rangle.$$ And, because they essentially just differ by a change of basis regarding which pair of solutions they track, you can obviously express the solutions of the Rabi problem as a superposition of Autler-Townes eigenstates, $$ |\psi(t)\rangle = \frac{1}{A_+B_--A_-B_+}\bigg( e^{-i\omega_+t}B_-|+\rangle - e^{-i\omega_-t}B_+|-\rangle \bigg), $$ and as the relative phase $e^{-i(\omega_+-\omega_-)t}$ between those two eigenstates evolves, $|\psi(t)\rangle$ goes from being along $|g\rangle$ to having as large a component along $|e\rangle$ as $|+\rangle$ does.

In practice, if you're looking for Rabi oscillations, you normally have some way of measuring population on the atomic basis, and you normally need this to be faster than the Rabi oscillation timescale $\sqrt{\Delta^2+\Omega^2} \geq \Omega$.

By contrast, Autler-Townes shifts normally show up if you're measuring the absorption spectrum (or similar) as you probe transitions from some additional state $|a\rangle$ in some other spectral region, and this normally requires the length of the probe to be longer than the Rabi period: resolving the Autler-Townes splitting requires a frequency resolution smaller than $$ \omega_+-\omega_- = \sqrt{\Delta^2+\Omega^2} $$ and this can only be done if your probe lasts longer than the Rabi period. As you can see, actual observations of Autler-Townes splittings and of Rabi oscillations describe fundamentally incompatible ways to measure the wavefunction.

OK, so finally, and just because I can, I'll include here a shameless plug to the fact that Autler-Townes splittings are now essentially measurable on real time, with the splittings appearing and disappearing as the light pulse strengthens and then decays, as compared with the pulse you use to observe the absorption, so you can now think about measuring things that look like this,

where you're seeing (a numerical simulation of) the splitting induced by a $\lesssim 10$-cycle-long pulse of near-IR, probed by a broadband extreme-UV pulse that's a good deal shorter than the period of the IR (yes, that's a thing that you can make and measure), scanning over a delay between the two pulses.

This is a growing field, known as attosecond transient absorption spectroscopy, and there's a lot we can say at the moment and even more that we'll be able to say in the coming years in terms of how things like Autler-Townes splittings grow and develop in the natural timescales of atoms.

The picture is of course more complicated (and for example it seems, in the picture above, that the RWA might not cut it at all even in regimes where the splitting is in the optical regimes (!)), so I'll just leave for further reading the paper where that figure came from,

and a second paper along the same lines,

  • Time-domain perspective on Autler-Townes splitting in attosecond transient absorption of laser-dressed helium atoms. M Wu et al. Phys. Rev. A 88, 043416 (2013).
  • $\begingroup$ This is a really nice exposition. I guess one pitfall here (and possibly part of the OP's motivation for this question) is that it is not always obvious whether you are measuring $|g/e\rangle$ or $|+/-\rangle$. In particular, AC stark shifts are described as level shifts of the $|+/-\rangle$ states, but when you actually see a little ball of optically trapped atoms it sure seems like you are probing the bare atoms themselves. $\endgroup$
    – Rococo
    Commented Apr 11, 2017 at 23:27
  • 1
    $\begingroup$ @Rococo Yeah, it's some pretty subtle ground, and it's easy to mis-step there. I suspect that the entire difference might just boil down to the length and bandwidth of the probe pulse you use to measure the dynamics, but I would need to think about it some more to give a categorical statement. In any case, it's not an obvious place where the bandwidth theorem is going to play a strong role, but it does lie close to the heart of the difference, so maybe that's where a lot of the unintuitiveness comes from. $\endgroup$ Commented Apr 12, 2017 at 9:39
  • $\begingroup$ @EmilioPisanty Pisanty Thank you for you elaborate Answer. I guess I understand the difference between the two pairs of states now. However, I'm still not sure about wether $|+\langle$ and $| - \langle$ are time dependent. Do you know a book that explains how the hamiltonian that you cited is derived? I fail doing it. $\endgroup$ Commented Apr 14, 2017 at 22:22
  • $\begingroup$ @Quantumwhisp In essence it is the Jaynes-Cummings hamiltonian, which is justified in any quantum optics text, $$H = \hbar\omega_0 |1⟩⟨1|+\frac12 \hbar\Omega \left(e^{-i\omega t}|1⟩⟨0|+e^{+i\omega t}|0⟩⟨1|\right).$$ To derive the rotating-frame hamiltonian, work out the TDSE for $\dot a$ and $\dot b$ in $|\psi(t)⟩ = a(t)|0⟩+b(t)e^{-i\omega t}|1⟩$. $\endgroup$ Commented Apr 14, 2017 at 22:28
  • $\begingroup$ As to whether $|\pm⟩$ are time-dependent or not, I guess it really depends on whether you think of the rotating-frame basis $|e⟩,|g⟩$ as "real" or not, and to what extent you buy that can depend on the specific system you're instantiating the Jaynes-Cummings hamiltonian in the first place. (cont.) $\endgroup$ Commented Apr 14, 2017 at 22:34

Both ways of description are equivalent and equally valid, but one may be more convenient in certain situations. One can use either eigenfunctions of time-independent $H_0$ or eigenfunctions of time-dependent $H(t)$, since both are self-adjoint and have complete set of eigenfunctions. The choice usually depends on the frequency of the external field.

Level shifts due to external field are considered when the external field changes slowly enough (e.g. microwave frequencies) that one can probe those level shifts with another radiation of higher frequencies (optical, UV).

Mathematically, this is supported by the fact that if there is no other external disturbance, the $\psi$ function can be with good accuracy expressed as sum

$$ \psi(t) = \sum_m a_m \Phi_m(t) $$ where coefficients $a_m$ are constant in time and $\Phi_m(t)$ are eigenfunctions of the total Hamiltonian: $$ H(t)\Phi_m(t) = E_m(t) \Phi_m(t). $$ The eigenvalues $E_m$ change in time which can be described as energy levels changing.

If the external field changes too quickly (frequency comparable to resonance frequency $(E_1-E_0)/\hbar$or higher), the approximation of time-independent $a_m$'s is no longer accurate and both eigenfunctions and coefficients change in time. In such a case it is more customary to express the $\psi$ function as

$$ \psi(t) = \sum_m c_m(t) \Phi^{0}_m $$

where $\Phi^0_m$ are eigenfunctions of the main time-independent Hamuiltonian: $$ H_0\Phi^0_m = E^0_m \Phi^0_m. $$ Then both eigenfunctions and eigenvalues are fixed in time and the time-independencecan be kept, without loss of generality, only in the coefficients $c_m$.

  • $\begingroup$ So you are saying that energy shifts (like the AC stark effect) do only occur with frequencies far lower than the resonance frequency? I doubt that to be true, because setups like a dipole trap make use of the AC stark effect, and use laser-light as electromagnetic wave, which has roundabout the same frequency as the transition in the atom. $\endgroup$ Commented Apr 8, 2017 at 12:46
  • $\begingroup$ No, energy shifts are purely artefact of the manner of expansion of $\psi$ into linear combination of eigenfunctions. If one uses eigenfunctions of $H_0$, there are no changes of levels. If one uses eigenfunctions of $H(t)$, there are. $\endgroup$ Commented Apr 8, 2017 at 19:22
  • $\begingroup$ So, you say, if I choose a proper linear combinatioin of the states $| 1 \rangle$ and $|2 \rangle$, then this state will be stationary (time evolution given by the energy eigenvalue), although states $|1 \rangle$ and $|2 \rangle$ perform rabi oscillations? $\endgroup$ Commented Apr 8, 2017 at 19:27
  • $\begingroup$ Rabi oscillations happen to $\psi$, not to eigenfunctions of Hamiltonian. What do you mean by $|1\rangle,|2\rangle$? $\endgroup$ Commented Apr 8, 2017 at 19:34
  • $\begingroup$ State $|1\rangle$ and 2 are supposed to be the the eigenstates of $H_0$. If $|\Psi \rangle$ initially is $|1\rangle$, then it will perform Rabi oscillations, the same for 2. Those are obviously not the eigenstates of $H(t)$, but a linear combination of those states COULD be the eigenstate of $H(t)$. I'm writing could, because I'm not sure wether $H(t)$ has Eigenstates at all. I'm doubting it, because I find it unintuitive that a time dependent operator has time independent eigenvalues. $\endgroup$ Commented Apr 8, 2017 at 19:41

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