AC Stark Shift clarification

[kelly0303]

Hello! I think I am a bit confused about the AC Stark shift effect. Assuming we have a 2 level system with energy difference $\omega_0$ and we apply a laser of frequency $\omega$ and Rabi frequency $\Omega$. If we look at the problem in a frame rotating at the laser frequency, before we couple the laser to the atom, the 2 energy eigenvalues of the (atomic) system are 0 and $\Delta = \omega - \omega_0$. If we now couple the electric field to the atom, it can be shown that the eigenstates become $\frac{\Delta}{2} \pm \frac{\sqrt{\Omega^2+\Delta^2}}{2}$. Let's say that in an experiment we use a pulsed laser of fixed time length and scan the frequency of the laser, in order to find the frequency of the transition. If we ignore this AC Stark shift, we would get a normal Rabi spectrum of excitation probability vs detuning, from which we can extract $\omega_0$ as the center of that spectrum with a linewidth inversely proportional to the pulse length. I am not sure I understand what happens in practice, when we actually have to account for the AC Stark shift. If our laser has $\omega = \omega_0$, we would not be on resonance anymore, as the energy difference between the levels now is effectively $\omega_0 + \Omega$, which is different than the laser frequency. So what would be the actual resonant frequency that we would see in practice? Thank you!

 

[Twigg]

In practice, $\Omega << \omega_0$, so in most cases the AC Stark shift (sometimes called "light shift" in the context of optical atomic clocks) isn't a factor. Usually, as a rule of thumb, if you want a precise frequency measurement, you don't use a powerful field to excite the transition. It's a "less is more" type deal.

If you do see a circumstance where the AC Stark shift is significant, you will see a shifted resonance, where that shift varies roughly linearly with the applied field amplitude (square root of the applied power). Often times, the AC Stark shift is a source of noise for precision measurements because of amplitude noise on the exciting field. This is why "magic wavelength" transitions are useful for clocks and such. In short, magic wavelengths are sweet spots in the spectrum where light shifts from different excited states levels cancel out.

 

[kelly0303]

In practice, $\Omega << \omega_0$, so in most cases the AC Stark shift (sometimes called "light shift" in the context of optical atomic clocks) isn't a factor. Usually, as a rule of thumb, if you want a precise frequency measurement, you don't use a powerful field to excite the transition. It's a "less is more" type deal.

If you do see a circumstance where the AC Stark shift is significant, you will see a shifted resonance, where that shift varies roughly linearly with the applied field amplitude (square root of the applied power). Often times, the AC Stark shift is a source of noise for precision measurements because of amplitude noise on the exciting field. This is why "magic wavelength" transitions are useful for clocks and such. In short, magic wavelengths are sweet spots in the spectrum where light shifts from different excited states levels cancel out.

Thanks a lot for the reply! I guess that what confuses me is the equivalence between Rabi oscillations and dressed states. From what I read, they are the describing the same thing, in 2 different ways. Rabi oscillations use the eigenstates of the unperturbed atom while dressed states use the eigenstates of the atom interacting with the electromagnetic field. Basically it is just a change of basis for convenience and this leads to different energy levels for eigenstates used in the 2 basis. This energy difference is what is called the AC Stark Shift (right?). However, this seems to be just a mathematical change in energy not a physical one. More specifically, the formula for Rabi oscillations as a function of detuning, is symmetric with respect to a zero detuning i.e. based on that formula, the transition frequency that one would measure would be exactly $\omega_0$. However according to the AC stark shift, the measured transition frequency would be shifted with respect to $\omega_0$. From your explanation, I see that this shift is real, in the lab (although probably small for small electric fields) not just a mathematical shift due to basis change. So I am not sure why don't we see this in the formula for Rabi oscillations explicitly. Thank you!

 

[Twigg]

Been a while since I thought about these things. Had to shake a 1/4" of dust off the old book.

More specifically, the formula for Rabi oscillations as a function of detuning, is symmetric with respect to a zero detuning i.e. based on that formula, the transition frequency that one would measure would be exactly .

I'm a little confused by which formula you mean, because rabi oscillations occur at a frequency $\Omega$ not $\omega_0$. Can you give us the formula you're referring to?

As a rule of thumb, always think of the dressed states as the "real-er" states (I know there's no such thing, it's just a helpful gut instinct to have), and it'll make more sense. For Rabi oscillations, you are projecting a stationary dressed state onto a non-stationary unperturbed state, and that's why the population oscillates. Hope that helps!

 

[kelly0303]

Been a while since I thought about these things. Had to shake a 1/4" of dust off the old book.



I'm a little confused by which formula you mean, because rabi oscillations occur at a frequency $\Omega$ not $\omega_0$. Can you give us the formula you're referring to?

As a rule of thumb, always think of the dressed states as the "real-er" states (I know there's no such thing, it's just a helpful gut instinct to have), and it'll make more sense. For Rabi oscillations, you are projecting a stationary dressed state onto a non-stationary unperturbed state, and that's why the population oscillates. Hope that helps!

I am sorry for not being clear. The formula I was talking about is the probability of transition as a function of detuning, for a pulse of fixed length of time, $T$: $$\frac{1}{1+(\frac{\Delta}{\Omega})^2}sin^2(\frac{T}{2}\sqrt{\Omega^2+\Delta^2})$$ If you plot this as a function of $\Delta$, you get this. As you can see in this plot, the peak appears when the detuning is exactly zero. So as far as I understand, based on this formula, if you scan the frequency of your laser, with the interaction time and Rabi frequency fixed, you would get the peak of that plot (or the symmetry axis) to be exactly when $\Delta = 0$, which implies that the peak appears when the laser frequency is exactly equal to the transition frequency. However, in this formula, the magnitude of the Rabi frequency, doesn't affect the center of the plot i.e. even for huge electric fields, the center is at zero detuning. However, if there is an AC Stark shift, and I scan my laser, I should see in my data the transition frequency shifted (and in later doing the analysis I would have to correct that value to get back the real $\omega_0$). I am not sure what I am missing (or what I missunderstand here), but my question is, why don't we see this intensity dependent shift in the formula above i.e. why can't I see the AC Stark shift directly from that formula? Thank you!

 

[Twigg]

Ok, I'm not 100% confident here, but my understanding is that what you wrote actually does include the AC Stark shift. Here's my argument:
Take the rotating frame Hamiltonian ($\hbar = 1$ because I'm a lazy chump): $$H = \bigl( \begin{smallmatrix} -\delta & \Omega/2\\ \Omega/2 & 0 \end{smallmatrix} \bigr)$$
Diagonalize: $$E_\pm = \frac{1}{2}[-\delta \pm \sqrt{\delta^2 + \Omega^2}]$$ $$\Delta E = E_+ - E_- = \sqrt{\delta^2 + \Omega^2}$$
and that's the (full, non-approximate) AC stark shifted precession rate

To elucidate, looking at the Rabi rate is different from doing fluoerescence spectroscopy, because the mixing of excited and ground states is built into Rabi flopping whereas it is a complicating factor for fluoresence spectroscopy. When I said you'd see the AC stark shift move your resonance earlier, I was talking about a fluorescence-like measurement.

 

[kelly0303]

Ok, I'm not 100% confident here, but my understanding is that what you wrote actually does include the AC Stark shift. Here's my argument:
Take the rotating frame Hamiltonian ($\hbar = 1$ because I'm a lazy chump): $$H = \bigl( \begin{smallmatrix} -\delta & \Omega/2\\ \Omega/2 & 0 \end{smallmatrix} \bigr)$$
Diagonalize: $$E_\pm = \frac{1}{2}[-\delta \pm \sqrt{\delta^2 + \Omega^2}]$$ $$\Delta E = E_+ - E_- = \sqrt{\delta^2 + \Omega^2}$$
and that's the (full, non-approximate) AC stark shifted precession rate

To elucidate, looking at the Rabi rate is different from doing fluoerescence spectroscopy, because the mixing of excited and ground states is built into Rabi flopping whereas it is a complicating factor for fluoresence spectroscopy. When I said you'd see the AC stark shift move your resonance earlier, I was talking about a fluorescence-like measurement.

Ohhh! So what you mean is that if the lifetime is big compared to the interaction time (i.e. pulse length), we would see Rabi oscillations which already include the AC Stark shift (or at least what we would call the AC Stark shift in the case of fluorescence measurement). So our experimental data (ignoring other practical complications) would be a described by that $sin^2$ function, centered exactly at $\omega_0$ i.e. no correction needed for the transition frequency from the actual measured one. If on the other hand the interaction time is longer compared to the lifetime, we won't see Rabi oscillations, and we get in a regime where the data is described by a Lorentz profile (instead of the $sin^2$) and it is in this case where the effect of the AC Stark shift is explicit i.e. the peak of this Lorentz profile is not at $\omega_0$, but shifted depending on the laser intensity and we need to correct the measured transition frequency in order to get the actual one. Is my understanding correct?

 

[Twigg]

I guess that's what I'm saying, hadn't thought of it that way. Now I have a bunch of questions. Dangit!

Again, I'm not 100% confident. I feel like I need to read some clock papers now.

 

[kelly0303]

I guess that's what I'm saying, hadn't thought of it that way. Now I have a bunch of questions. Dangit!

Again, I'm not 100% confident. I feel like I need to read some clock papers now.

Hahaha! Well to be honest your comment really pointed me in a direction I haven't thought of before (i.e. that it makes sense to talk about AC Stark shift only in that specific regime), so thank a lot for that! I will read more into it. Please let me know if you find something else while reading.