# Confused about dressed states of atom

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Hello! Assume we have a 2 level system, with the ground state defined as the zero energy level and the excited state having an energy of ##\omega_0##. If we apply an oscillating electric field (assume dipole approximation and rotating wave approximation) of frequency ##\omega##, we have a time dependent hamiltonian. However, if we go to a frame rotating at the frequency of the laser, the explicit time dependence of the hamiltonian vanishes, and we are left with a time independent perturbation. Assuming the laser is on resonance i.e. ##\omega = \omega_0##, it can be shown that in this rotating frame, the energies of the eigenstates (which in the presence of the electric field are not the 2 levels of the atom anymore, but linear combinations of them) are ##\pm \frac{\Omega}{2}##, where ##\Omega## is the Rabi frequency. I understand it so far. However I am not sure what happens in the lab frame (where we actually care for experimental purposes). What I think happens, is that, if we go from the rotating frame back to the lab (non-rotating) frame, still with the laser on resonance, the energy of the lower eigenstate would be ## - \frac{\Omega}{2}## while the energy of the higher state would become ##\omega_0 + \frac{\Omega}{2}##, as now we add the ##e^{-i\omega_0 t}## term back to the upper state. However, this implies that in the lab frame we have 2 states with fixed (i.e. time independent) energy, which I am not sure it's right, given that the hamiltonian in the lab frame is explicitly time dependent, so I would imagine that its eigenstates would also be time dependent. Can someone help me understand how do we go back from the rotating frame with fixed energy levels to the lab frame? Thank you!

Twigg
Gold Member
Recall that the eigenstates in the rotating frame are superpositions of the ground and excited state. When you transform those states back to the lab frame, they will also be time-dependent. Those complex exponential factors cancel nicely.

Recall that the eigenstates in the rotating frame are superpositions of the ground and excited state. When you transform those states back to the lab frame, they will also be time-dependent. Those complex exponential factors cancel nicely.
Thanks for this. But what are the eigenvalues in the lab frame? Won't they also be time dependent?

Twigg
Gold Member
I had a derivation in mind when I wrote my last reply, but on reviewing it smells a bit fishy. It gives the right results, but only on resonance. If you'd still like to see it even with that caveat, let me know in a reply and I'll post it at my next chance.

Taking the eigenvalues in the lab frame is a dangerous game. How you would write the Hamiltonian in the lab frame, taking the RWA into account? There are a lot of pitfalls. That's why the AMO books (Foot, Metcalf) are so squirrely applying the RWA to the equations of motion rather than working with a Hamiltonian directly. If I get a chance this week, I'll take a look in Cohen-Tannoudji for a good solution.

Here's what I can say: when you do spectroscopy on a two-level system in the lab, you see a peak at the unperturbed resonance frequency ##\omega_0## plus the light shift that you calculate in the rotating frame. In almost every case, the behavior of the dressed states is what matters in experiment.

I had a derivation in mind when I wrote my last reply, but on reviewing it smells a bit fishy. It gives the right results, but only on resonance. If you'd still like to see it even with that caveat, let me know in a reply and I'll post it at my next chance.

Taking the eigenvalues in the lab frame is a dangerous game. How you would write the Hamiltonian in the lab frame, taking the RWA into account? There are a lot of pitfalls. That's why the AMO books (Foot, Metcalf) are so squirrely applying the RWA to the equations of motion rather than working with a Hamiltonian directly. If I get a chance this week, I'll take a look in Cohen-Tannoudji for a good solution.

Here's what I can say: when you do spectroscopy on a two-level system in the lab, you see a peak at the unperturbed resonance frequency ##\omega_0## plus the light shift that you calculate in the rotating frame. In almost every case, the behavior of the dressed states is what matters in experiment.
Thanks a lot for this! If you have time, sure, I would really appreciate your derivation.

Actually my question here came from thinking about light/AC Stark shift. In the derivations I found, the light shift i.e. the difference in energy for dressed states compared to the atom levels are calculated in the rotating frame (used for RWA). However, if we are to observe it, we observe it in the lab frame. For example, if I want to measure the resonant frequency of a transition for an atom trapped using lasers (for example in a MOT trap), the trapping laser will shift the resonant frequency (unless I am at a magic wavelength) and I need to correct for that in my experiment. So what confuses me is how do I go from the light shift calculated in the rotating frame to the shift I see in the lab. Based on what I read, I got the impression that they have the same values. However I am not totally sure how can that shift be a constant in the lab frame (as it is in the rotating frame), where the Hamiltonian is explicitly time dependent.

Twigg
Gold Member
I thought of a non-sketchy derivation. I'm going to be using conventions from Metcalf's "Laser Cooling and Trapping". Let me know if it's not clear how to translate the conventions. The big thing is the choice of the diagonal elements of the Hamiltonian, and that directly affects the form of the reference frame transformation. If you're getting into a project with MOTs, I recommend Metcalf's book anyways. If you're in a group, someone you know surely has it.

When you into the rotating frame, you apply a unitary transformation
$$U = \left[ \begin{matrix} e^{-i \delta t} & 0 \\ 0 & 1\end{matrix} \right]$$
To get the Hamiltonian from the rotating frame into the lab frame, do this:
$$H_{LF} = U^{\dagger} H_{RF} U$$
I may have gotten the order of ##U## and ##U^{\dagger}## mixed up. To double check, verify that ##\langle H_{LF} \rangle = \langle H_{RF} \rangle## given that ##|\psi_{RF} \rangle = U |\psi_{LF} \rangle##.

Starting with the rotating frame Hamiltonian,
$$H_{RF} = \frac{\hbar}{2} \left[ \begin{matrix} -2\delta & \Omega \\ \Omega & 0 \end{matrix} \right]$$

Note the eigenvalues are given by the characteristic equation: ##E*(E+\hbar\delta) - \left( \frac{\hbar\Omega}{2} \right)^2 = 0##

Now apply the rotating frame transformation:
$$H_{LF} = \frac{\hbar}{2} \left[ \begin{matrix} -2\delta & \Omega e^{+i\delta t} \\ \Omega e^{-i\delta t} & 0 \end{matrix} \right]$$

with characteristic equation ##E*(E+\hbar\delta) - \left( \frac{\hbar\Omega}{2} \right)^2 e^{+i\delta t} \times e^{-i\delta t} = E*(E+\hbar\delta) - \left( \frac{\hbar\Omega}{2} \right)^2 = 0##

The reason why ##\hbar \omega_0## doesn't appear in the eigen-energies is because this is the Hamiltonian for the dressed states: ##|1\rangle \otimes |1_S \rangle##, ##|2\rangle \otimes |0_S \rangle## where ##|1\rangle##, ##|2\rangle## are the atomic states and ##|N_S \rangle## is the state of the driving optical field with N photons in the optical mode S (contains polarization and wavevector information). Metcalf has a good discussion on this point in his second chapter on spontaneous emission. When you take the partial trace of this compound system, the ##\hbar \omega## will add back in and cancel the ##-\hbar\omega## from the factor of ##-\hbar \delta## leaving ##\hbar\omega_0##. So, just take the shift you get in the dressed picture and add it to ##\hbar \omega_0##.

Back to the real world, I know it was just an example you gave, but if you're thinking about light shifts in a run-of-the-mill MOT, they will (typically) be too small to measure underneath the Doppler broadened line widths.

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I thought of a non-sketchy derivation. I'm going to be using conventions from Metcalf's "Laser Cooling and Trapping". Let me know if it's not clear how to translate the conventions. The big thing is the choice of the diagonal elements of the Hamiltonian, and that directly affects the form of the reference frame transformation. If you're getting into a project with MOTs, I recommend Metcalf's book anyways. If you're in a group, someone you know surely has it.

When you into the rotating frame, you apply a unitary transformation
$$U = \left[ \begin{matrix} e^{-i \delta t} & 0 \\ 0 & 1\end{matrix} \right]$$
To get the Hamiltonian from the rotating frame into the lab frame, do this:
$$H_{LF} = U^{\dagger} H_{RF} U$$
I may have gotten the order of ##U## and ##U^{\dagger}## mixed up. To double check, verify that ##\langle H_{LF} \rangle = \langle H_{RF} \rangle## given that ##|\psi_{RF} \rangle = U |\psi_{LF} \rangle##.

Starting with the rotating frame Hamiltonian,
$$H_{RF} = \frac{\hbar}{2} \left[ \begin{matrix} -2\delta & \Omega \\ \Omega & 0 \end{matrix} \right]$$

Note the eigenvalues are given by the characteristic equation: ##E*(E+\hbar\delta) - \left( \frac{\hbar\Omega}{2} \right)^2 = 0##

Now apply the rotating frame transformation:
$$H_{LF} = \frac{\hbar}{2} \left[ \begin{matrix} -2\delta & \Omega e^{+i\delta t} \\ \Omega e^{-i\delta t} & 0 \end{matrix} \right]$$

with characteristic equation ##E*(E+\hbar\delta) - \left( \frac{\hbar\Omega}{2} \right)^2 e^{+i\delta t} \times e^{-i\delta t} = E*(E+\hbar\delta) - \left( \frac{\hbar\Omega}{2} \right)^2 = 0##

The reason why ##\hbar \omega_0## doesn't appear in the eigen-energies is because this is the Hamiltonian for the dressed states: ##|1\rangle \otimes |1_S \rangle##, ##|2\rangle \otimes |0_S \rangle## where ##|1\rangle##, ##|2\rangle## are the atomic states and ##|N_S \rangle## is the state of the driving optical field with N photons in the optical mode S (contains polarization and wavevector information). Metcalf has a good discussion on this point in his second chapter on spontaneous emission. When you take the partial trace of this compound system, the ##\hbar \omega## will add back in and cancel the ##-\hbar\omega## from the factor of ##-\hbar \delta## leaving ##\hbar\omega_0##. So, just take the shift you get in the dressed picture and add it to ##\hbar \omega_0##.

Back to the real world, I know it was just an example you gave, but if you're thinking about light shifts in a run-of-the-mill MOT, they will (typically) be too small to measure underneath the Doppler broadened line widths.
Thank you so much for this! It was really helpful! One thing I am still confused about (and it might be more related to quantum mechanics in general, rather than the original question): based on your derivation, the eigenstates of the hamiltonian are the same in the rotating frame and lab frame (at least the shift due to the presence of the EM field is the same), which is basically (and I hope I got this right) why we talk about AC Stark shift in the lab frame, when all the derivations are made in the rotating frame. However, in the rotating frame, the hamiltonian is (by construction) time independent. In that case it makes sense to take the eigenstates of the hamiltonian as the energy of the two (dressed) levels and hence define the AC Stark shift. However, in the lab frame, the hamiltonian is explicitly time dependent, so doing the usual separation of variables in time and space doesn't work anymore and I am not sure if we can think of the eigenstates of the Hamiltonian still as energy levels. Can we still define eigenstates and eigenvalues in the case of a time dependent Hamiltonian, where in principle we would need to solve the full Schrodinger equation, not just an eigenvalue problem (which follows from the space-time separation of variables)?

Twigg
Gold Member
The rotating frame is a classic example of an "interaction picture," as opposed to a Schrödinger picture or Heisenberg picture. In any interaction picture, the observable energy levels will still be the eigenvalues of the Hamiltonian. This is a fundamental rule of quantum mechanics, and doesn't follow from how the Schrödinger equation is solved with separation of variables. But to the other side to your question, I believe you are correct that, in general, solving for the time evolution isn't as easy as solving eigenvalues and eigenstates. The Dyson series gives the general solution.

The rotating frame is a classic example of an "interaction picture," as opposed to a Schrödinger picture or Heisenberg picture. In any interaction picture, the observable energy levels will still be the eigenvalues of the Hamiltonian. This is a fundamental rule of quantum mechanics, and doesn't follow from how the Schrödinger equation is solved with separation of variables. But to the other side to your question, I believe you are correct that, in general, solving for the time evolution isn't as easy as solving eigenvalues and eigenstates. The Dyson series gives the general solution.
But then, how can we talk about energy levels (i.e. dressed states levels) in the lab frame? Of course we can diagonalize the hamiltonian containing the EM field, get the new eigenstates and eigenvalues, but are these physical? For example, assume we have a laser that we scan in order to find the right frequency, ##L_1## and a second laser ##L_2##, which is needed for the experimental procedure (can be for a MOT, or the laser used for ionization in a resonant ionization measurement, the actual implementation is not relevant for my question). If ##L_2## was not present, by scanning the frequency of ##L_1## (and ignoring other effects) we would find it centered at the actual atomic transition, ##\omega_0##. If we account for ##L_2##, we know that the resonant frequency will be shifted by a small amount ##\delta##, which is a function of the ##L_2## detuning ##\Delta## and the Rabi frequency of ##L_2##. So I guess my question is, if we scan ##L_1## now, (of course in the lab frame), will the resonant frequency be centered at ##\omega_0 + \delta##? I.e. will the eigenstates that ##L_1## sees be the eigenstates of the hamiltonian containing the atom-##L_2## interaction, even if that Hamiltonian is explicitly time dependent?

Twigg
Gold Member
but are these physical
Absolutely. Note that, as I showed in my derivation, expectation values are the same in the lab frame and the rotating frame. What I'm trying to get at is that energy levels are energy levels, in any frame so long as the frame is only a unitary transformation away. Edit: even if that unitary transformation is time-dependent.

As for the example, we're neglecting the effect of L1 on resonance, right? Just checking.

If so, then yes, L1 will see a resonance at ##\omega_0 + \delta## where ##\delta## is the light shift due to L2.

Sorry, I feel like I keep missing your underlying confusion. One suggestion is that you might try to solve as an exercise the Rabi problem of a spin-1/2 particle in a magnetic field with a large, time-independent ##B_z## and a small oscillating ##B_{x} = B_0 \cos(\omega t)##. You'll find it very familiar, except without the obfuscation of the RWA.

Sorry for the scatter brained reply. Typing from my phone.

Absolutely. Note that, as I showed in my derivation, expectation values are the same in the lab frame and the rotating frame. What I'm trying to get at is that energy levels are energy levels, in any frame so long as the frame is only a unitary transformation away. Edit: even if that unitary transformation is time-dependent.

As for the example, we're neglecting the effect of L1 on resonance, right? Just checking.

If so, then yes, L1 will see a resonance at ##\omega_0 + \delta## where ##\delta## is the light shift due to L2.

Sorry, I feel like I keep missing your underlying confusion. One suggestion is that you might try to solve as an exercise the Rabi problem of a spin-1/2 particle in a magnetic field with a large, time-independent ##B_z## and a small oscillating ##B_{x} = B_0 \cos(\omega t)##. You'll find it very familiar, except without the obfuscation of the RWA.

Sorry for the scatter brained reply. Typing from my phone.
Thanks a lot! Actually the first paragraph clarified it for me now! However I am now a bit confused about this: "As for the example, we're neglecting the effect of L1 on resonance, right? Just checking.". Does the ##L_1## laser has any influence on itself? I.e. if we had just ##L_1##, without ##L_2##, wouldn't we measure the resonant frequency at ##\omega_0## without any shift?

Twigg
Gold Member
if we had just , without , wouldn't we measure the resonant frequency ##\omega_0## at without any shift?
Even the probe laser will light shift the resonance, it's just a tiny effect. In a MOT, it would surely be below the Doppler linewidth. Light shifts most commonly arise due to optical dipole traps, but in theory any laser will do it.

Does the laser has any influence on itself?
Remember, what we're discussing the system that contains both the laser field(s) and the atom. L1 doesn't "influence itself", it perturbs the atom. But yes, even doing spectroscopy with a single laser, you will (idealistically) see a light shift.

(Edit: added strikethrough to erroneous claims)

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Even the probe laser will light shift the resonance, it's just a tiny effect. In a MOT, it would surely be below the Doppler linewidth. Light shifts most commonly arise due to optical dipole traps, but in theory any laser will do it.

Remember, what we're discussing the system that contains both the laser field(s) and the atom. L1 doesn't "influence itself", it perturbs the atom. But yes, even doing spectroscopy with a single laser, you will (idealistically) see a light shift.
Wait, in an experiment with a single laser (assuming the transition is a delta function), won't we measure ##\omega_0## exactly (ignoring the effect coming from the counter-rotating wave which I think is called Bloch-Siegert effect and does shift the resonance even for a single laser)? In the rotating wave approximation the dressed state formalism (which leads to the light shift) and Rabi oscillations formalism (which uses the unperturbed atomic levels as the basis) are equivalent. And in the Rabi approach, the transition frequency appears exactly at ##\omega_0##. Shouldn't they both either predict a shift or not predict it?

• Twigg
Twigg
Well, I'm second guessing myself now. I believe you're right about not seeing a light shift on a single laser spectroscopy experiment. My bad. Hold on a moment, I need to extract my foot from my mouth. • 