M5L23g.txt

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Let's not drop all the assumptions we've made.
So what happens when we change from
blue-detuned into red-detuned?
Well, for blue-detuning, we use the argument that adiabatically
we stay in the ground state, but for red-detuning it's
a higher state, and we adiabatically
stay in the higher state.
Same thing.
What happens if f and g are not degenerate?

Well, the argument is the following.
It's the same result. And I can quickly
show that to you by saying use the dressed state basis.
You have the state g with Na and Nb photons in laser
beams-- they're going to be consistent.
I called the operators Na and Nc.
And if you are now look at the state
f, which has one photon less here and one photon more here,
those two states are degenerate.
So I was just showing the situation
if g and f are not degenerate, here we
have a laser beam with Na photons,
here we have a laser beam with Nc photons.
If we include the photons in our description,
we have a degeneracy.
And that's what matters.
But what is relevant, of course, is
you have this degeneracy only when the energy
difference between g and f is the energy frequency difference
between the two laser beams.
So it is essential-- and you see that
again here-- that the two-photon Raman resonance
condition has to be met.
OK.
So we've dropped all the assumptions,
so we know you can-- in a general system,
you can transfer between two hyperfine states.
You can use red or blue-detuned laser.
Everything is robust.
OK.
So STIRAP goes by the name dark state transfer.
And what I want to address now is what
is the magic which is going on?
How can we transfer from state g to state f
without going to the excited state?
And the answer is-- I gave it already
to you-- it's not possible, because the only couplings we
have are the two couplings which I have shown.

But-- and this is sort of the gist you should carry away--
since the whole transfer is a coherent process--
and I will be more exact in a moment,
we're not building up a population
in the excited state, we're just building up a small amplitude.
And the population is a small amplitude squared.
It's, so to speak, epsilon squared.
And so you win one power of epsilon,
one power of the small amplitude,
if you do a coherent transfer as compared
to incoherent transfer, which relies
on shuffling over population.
And I want to sort of show you now how this pans out.
Jenny.
But in a limit of, you know, infinitely
adiabatic, infinitely slow, you still
would have the transfer happen without ever going
to the excited state, right?
Well I will show you that the integrated population
in the excited state goes as 1 over the transfer time.
So if you allow an infinite amount for the transfer,
you can make the population [INAUDIBLE] slow.
But ultimately, what I also want to show
you is that how much you put into the excited state
depends-- and I think this is a very satisfying result for me--
depends on your resources.
It depends on your laser power.
If you have infinite laser power,
you can keep the population infinitesimally small,
and if you have infinite time.
But if you go to the lab and do an experiment and say,
I want to be done in 24 hours, well,
this defines your time budget.
And the strongest laser you have in the laboratory
defines your laser power budget.
And given those two resources, there is a small amount,
and I want to give you an estimate for that.
And then we have to discuss, is that small amount
which we get in STIRAP, does that really
stand out over other methods to transfer population.
And the answer will be interesting.
OK.
So let me just say, no excited state
is needed as a stepping stone.