M5L23f.txt

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OK.
Any questions?
Yes.
[INAUDIBLE]

Well, I gave you the intuition.
I said you want to have a dark state.
The dark state is the ground state, the lowest state
of the manifold.
So you want to first make sure that g is the ground state.
And then you turn [? opposite ?] your Hamiltonian,
and eventually f is the ground state.
And this is exactly the situation.
f is the ground state with the green light on.
I've given you the intuitive picture.
Now, I will actually come back to that.
There is-- the kind of-- it's called the magic
of the adiabatic state transfer that, if you always
stay in the dark state-- if you always
stay in the dark state-- you never populate the excited
state.
So it looks like magic that you can
go from the ground state to the final state
without having ever any population in the excited
state.
And of course-- can that be true?
No.
Because there is no coupling [? direct. ?] So somehow,
in this picture of adiabatic transfer
always staying in the dark and never populating
the excited state, maybe we have over-emphasized something.
We need a little bit of excited state.
Otherwise, we cannot go there, because our Hamiltonian does
not allow the direct path from g to f.
So a lot of people think that STIRAP
is a method where you can go from g to f without going
through the excited state.
I have many, many people talking about it.
So I want to sort of discuss that in the next few minutes,
and to some extent demystify it.
But since you're asking me, now, in a black and white picture,
STIRAP is the way-- I'm correcting myself in a moment,
but STIRAP is the way where you can go directly without going
through the excited state.
And this is done by the counterintuitive sequence where
you just want to keep atoms in the dark state,
whereas the intuitive sequence is really based on the fact you
want to take population here, put population here, and then
stimulate it down.
And this is not what STIRAP is about.
Other questions?
Will.
So if there is very strong coupling
to [INAUDIBLE] from the excited state,
because you need to [INAUDIBLE] travel
through the excited state during STIRAP,
it's possible to spoil the whole process.
We will talk about it.
And actually, in the next minute,
I want to sort of simply calculate how many photons
are emitted during the passage.
And if you now say the photons have a branching ratio,
this could be a branching ratio to your [? fourth ?] state.
But by just calculating what is the time-integrated population
in the excited state, I multiply it with gamma,
this is the amount of photons scattered.
And you can see, if you try to do
a STIRAP with the Bose-Einstein condensate, every time you
scatter you reach-- you heat up, you reach momentum states
outside the condensate, and those momentum states
are your fourth state.
So actually, by discussing how much light scattering happens
in this transfer, I will pretty much
talk indirectly about how much population
is going into a fourth state.
So that's what you want to understand now,
how true is the magic of the dark state transfer where
we can go from one state to the next
without ever going through the excited state.
Yes?
So are the two laser frequencies the same, or [INAUDIBLE]?
Yes.
I actually assumed that-- I'll come to that in a second--
that the two states are degenerate.
And for all this dark state business,
we have to fulfill the [? Riemann ?] resonance,
so if the two states are degenerate,
then the two laser beams.
So then what's the meaning of increasing
the level [INAUDIBLE].
OK.
One possibility is that this is an M equals 0 state,
this is an M equals minus 1 state, this is M equals plus 1,
and this is sigma plus, and this is sigma minus [? radiation. ?]
So yes, this was actually very important
for the whole discussion of three-level coherence,
that this level can only talk to the green laser,
this level can only talk to the orange laser,
and we've two ways to accomplish it.
One is by frequency difference.
But if you don't have a frequency difference,
we need polarization [INAUDIBLE].

Yes.
[INAUDIBLE] question [INAUDIBLE] if I [? have all ?] [INAUDIBLE]
and I turn on the green laser, and I optically pump everything
[? to that. ?] And so in this particular case,
is there an advantage of doing [INAUDIBLE] optical pumping
[INAUDIBLE]?
Or is it--

Yes, there is.
And we'll talk about that.
There is a difference between transferring population or only
transferring an amplitude.

So it depends what you want.
If all you want is to get the atoms from g to f
and you don't care how they are heated up or so, that's OK.
But the STIRAP process is a stimulated process.
So you have momentum k, and here you have the momentum k2.
So you take the atoms from an initial state
and put them in a very well-defined momentum state.
So [? BEZ ?] in one state after STIRAP transfer
is in another state maybe if you have
the lasers in a counter-propagating way
back to 0 momentum.
Whereas if you do optical pumping, which
involves spontaneous emission, then you always
have dissipation.
You create entropy.
And because spontaneous emission goes into many modes,
you have many different recoils.
And eventually, your final state is no longer a single state,
it has many different momentum states involved.
So that's why we've been talking here
about the coherent transfer, sort
of assuming that optical pumping with spontaneous emission
and the many modes involved is not what
you want because of heating.

OK.