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Malamala

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Malamala

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- #2

DrClaude

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Yes, a laser resonant with a transition will lead to Rabi cycling between the ground and the excited state, but you have to compare the Rabi frequency with the duration of the experiment. If the laser is not intense, you will probably never reach a the point where the population of the excited state is being lowered by the laser.

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f95toli

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However, if you saturate an

If you are working with a

(*) Assuming the lifetime of the excited state is still much longer than the pulse.

- #4

Malamala

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@f95toli @DrClaude thank you for your replies! I am actually new to this, so I am not totally sure I know what I mean either. Sorry for that! So one thing I think is getting me confused is that, from the derivations that I have seem for Rabi oscillations, it looks like the state being populated (assuming a 2 level system) is oscillating with time, so the final position depends on the time you stop the pulse. However from time dependent perturbation theory (from my QM class, Griffiths book) I remember that if you send a laser towards a 2 level system, you reach a steady state, where the rate of going to the upper state is equal to the rate of going to the lower state, so the number of electrons in the upper state will become (on average) constant. But as far as i understand, in the case of Rabi oscillations, you move all the electrons at once i.e. they are all in the upper state then lower state and so on, so you never reach a steady state. I am not totally sure what am I missing, or how to combine the 2 pictures (which seem to be contradicting). Thank you!

However, if you saturate anensembleof two-level systems (which is the simplest case) you will indeed end up in a situation where on average 50% of the population is in the ground state. However, this still leaves the other 50% so you will still see a signal.

If you are working with asingletwo-level system (say a qubit or a single ion) you will indeed have Rabi oscillations and "saturation" is not really a thing, the state of the system when you turn off the laser will depend on total areas of the pulse (amplitude*time) you've applied, regardless of how long the pulse is (*) . Hence, you could indeed be unlucky and not see anything.

(*) Assuming the lifetime of the excited state is still much longer than the pulse.

- #5

f95toli

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One thing to keep in mind is that QM is all about probabilities. If you measure a single system only once you will always find it in

The "basic" Rabi model is for a

A real TLS will spontaneously relax (transition back to the ground state) with some characteristic time T1 and de-phase (time T2). This also means that the amplitude of real Rabi oscillations will decay with a time constant T1 (which btw how T1 is measured).

Now, if you apply lots of power you can make sure that the system is immediately excited again once it has decayed. If you plot the power dependence you will find that you then reach a point where the dependence becomes nearly flat and this is what is known as saturation. If you work through the math you will find that the amount of power that is needed for saturation depends on T1*T2.

But again, this makes no sense in the basic Rabi model since there it is assumed that T1 and T2 are infinite.

This is probably the source of your confusion.

In lasers and similar systems you are typically dealing with ensembles of TLS but math is similar. However, instead of measuring the same system many times, you are now usually measuring lots of identical systems.

Now, you

- #6

hutchphd

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The answer is yes!

But this is not the population dictated by "thermal" interactions with the rest of the world. In equilibrium the population of the higher E state (in an ensemble of atoms) will be fewer by the Boltzmann factor e

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