M5L24m.txt

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Colin.
So the way you drew the level diagram,
you only have the f, I think, [INAUDIBLE].

Yes.
Actually I emphasized that we have an AC Stark shift,
and what I didn't say is-- when I discussed it here--
that the AC Stark shift pushes this level down
and pushes the other level up.
But since we are talking about a very broad resonance
in the excited state, for all practical matters,
the AC Stark shift doesn't matter.
Whereas for the narrow Raman resonance,
the AC Stark shift is important.
Also, you showed it on the plot that the shift of the excited
state [INAUDIBLE].
So maybe let me just-- sorry, let me just think.
So the AC Stark shift for that detuning
would shift this level up and would shift this level down.
So what's the second question?
You drew on that plot that the shift to the excitement state
was much lower-- or higher than the shift
to the ground [INAUDIBLE].

You mean those two shifts?
Yeah.
Oh, this shift is we have assumed-- we have
to now do the book-keeping.
We have assumed that the coupling
laser has a large detuning.
The coupling laser is very far away from resonance.
And if you want to hit the excited state,
we now need a Raman resonance, which is delta 2.
But the Raman resonance capital delta 2
means that we are smack on the single photon
resonance for the probe laser.
So this feature is pretty much we take the ground state
and go exactly to the excited state with a single photon.
Well, there is an AC Stark shift involved,
but it's not relevant here.
Whereas the other feature is the [INAUDIBLE] Raman feature.
And the one thing I wanted to point out in this context
is that there is actually a small energy
splitting between the two-photon Raman feature by the AC Stark
effect, whereas the EIT feature always
happens at Raman resonance delta equals 0.
Because-- just to emphasize that-- delta equals 0
is really you induce a coherence between g and f.
It's a dark state.
And when you have a dark state, in that situation,
you don't have an AC Stark effect.
So the EIT feature is at delta equals 0,
whereas the photon scattering features,
they suffer-- or they experience--
it may not be negative to suffer an AC Stark shift,
but they experience an AC Stark shift.

Further questions?
Yes.
[INAUDIBLE]?

Can you relate that to Doppler-free spectroscopy.
Actually, I don't think so, because I
would say for the whole discussion here,
let's assume we have an atom which has infinite mass which
is not moving at all.
We're really talking about internal coherences.
However-- and that's where it becomes related.
If you look at very narrow features
as a function of detuning, then of course Doppler shifts
play a role.
And if you have very, very, very narrow features,
you become sensitive to very, very, very small velocities.
And therefore, you have an opportunity to cool.
So if you can distinguish spectroscopically
an atom which moves a tiny little bit and an atom which
stands still, if you can by a narrow line
distinguish the two, then you can actually
laser cool this atom.
And the EIT feature can give you extremely high resolution.
I'm not discussing it here.
I've discussed the phenomenon of EIT, which is
coherent population trapping.
But there is an extension which is
velocity-selective coherent-- VSCPT,
velocity-selective coherent population trapping.
And VSCPT was a powerful method to cool atoms
below the recoil limit.
But I've not connected anything in-- of coherent population
trapping with a Doppler shift.
So for all this discussion, please
assume the atom is not moving.