- #1

BillKet

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- 29

Hello! I am reading about dressed states, and I am presented a situation in which we have a laser (the pump laser) on resonance with a 2 level (atomic) transition, and a second, weak laser (probe laser) that is scanned over a frequency range. The absorption spectrum of the probe laser, for different Rabi frequencies, ##\Omega## of the pump laser is shown in the attached figure. My question here is only for the case of ##\Omega = 0.3 \Gamma##. Given that we have atom laser interaction, the ground and excited state of the atom ##|g>## and ##|e>## are not eigenstates anymore. The eigenstates are, for zero detuning (##\Delta = 0##) ##|+> = \frac{|g>-|e>}{\sqrt{2}}## and ##|-> = \frac{|g>+|e>}{\sqrt{2}}##. So, the actual levels that the probe laser sees are ##|+>## and ##|->##. Given that the pump laser has a small power, the ground atomic state, ##|g>## will be more populated than the excited one ##|e>## (i.e. the population won't be split 50/50 as it would be for high laser power) in the steady state. However, the ##|+>## and ##|->## and equal linear combinations of ##|g>## and ##|e>##, so they should have equal populations. I also attached a little drawing with the levels and populations (populations are draw as circles, with bigger circles meaning bigger population). The 2 levels connected by the wiggly line represent 2 levels that can be connected by the probe laser, assuming the probe laser is on resonance. What confuses me is: why do we get an absorption at all (according to the first figure) when the probe laser is on resonance. According to the second figure, the 2 levels connected by the probe laser (wiggly line) have equal populations (even if ##|g>## and ##|e>## have different populations)? Shouldn't the absorption be zero, given that the amount of absorption and stimulated emission would cancel each other? Can someone help me understand what am I missing here? Thank you!