AC Stark Shift clarification

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  • Thread starter kelly0303
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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!
 

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