The following content is provided
under a Creative Commons license.
Your support will help MIT OpenCourseWare continue
to offer high quality educational resources for free.
To make a donation or view additional materials
from hundreds of MIT courses, visit MIT OpenCourseWare
at ocw.mit.edu.
PROFESSOR: So, good afternoon.
Our subject for all of today and some part of Monday
is coherence in free-level systems.
And what I want to show you are there
are certain fundamentally new and, to some extent,
surprising processes which are possible when
you have three levels and not only two.
Because when you have three levels,
you have one excited state which is
coupled to two ground states.
And that gives us the possibility of interference.
And actually, all the special things
I will tell you about three-level system
are related to interference.
Certain phenomena will constructively interfere.
Other will destructively interfere.
And that leads to a number of important effects.
The first effect you want to talk about
is the possibility of dark states,
that there are superposition states which
are bombarded with two laser beams, and they're not excited.
Because the amplitudes to go to the excited state destructively
interfere.
When you have a state which is dark,
which cannot absorb the light, then you already understand
the possibility of doing lasing without inversion.
Because if a lot of the population is in a dark state,
you don't have to invert excited and ground state.
Because the dark state doesn't count.
Lasing without inversion is a little bit more intricate
than this, but just kind of the idea,
the notion of a dark state helps you already
to anticipate why something like lasing with inversion
may be possible.
The dark state is only dark in a very narrow frequency
interval of the two lasers.
So if you're a very narrow resonance,
then you have a very narrow resonance
where something goes dark.
But if something goes dark, it's not absorbing.
And therefore you have electromagnetically
induced transparency, which is a very, very sharp feature
in frequency space.
And when you have very, very sharp features in frequency,
you can change the group velocity of light,
because if transmission and absorption index of refraction
changes a functional frequency very,
very sharply, you have huge effects on propagation lights.
So kind of all of these different subjects which
I want to go through are different perspectives,
but you will find similar physics.
It's just we look at interference.
We look at dark states in various ways.
And each of them has created a cottage industry
and a subfield of itself.
So I started to remind you that you're
very familiar with the dark state in the situation where
we have optical pumping.
If you have two ground states in an excited state,
and you switch only one laser, sooner or later you
will just pump all the atoms into the state f
and the state f is dark, and vice versa.
So what I want to show you now is that even if both lasers are
on, then no matter in what kind of state or superposition state
you have the atoms, the atoms see
light, which is ready to excite them.
But there is-- and this is what I want to show you now--
a coherent superposition state, which is still the dark state.
So the last thing I did on Monday, I wrote down for you
our Hamiltonian, which is nothing
else in the well-known dipole Hamiltonian in the rotating
wave approximation.
And what is important now is that we have pathways
with Rabi frequency omega 1 and omega 2.
And we have one laser fields.
The photons are created, annihilated
by the a operator, and the second laser field,
where the operator is C and C dega.
So there are two ways how we go to the excited state.
We can couple from the ground state f to the excited state
and from the ground state to g to the excited state.
So when you shine on the light, atoms
may be in some initial state.
They will sort of spontaneously emit but scatter light.
But then after some transient, the state
which is stable against further illumination, is a dark state.
That's like optical pumping.
So you will optically pump into the following state,
which I write down for you.
And then we can inspect that indeed it is a dark state.
It is a coherent superposition of the two ground states.
And the coefficient involves the Rabi frequency,
omega 2 of laser 2, omega 1 of laser 1.
And we just normalize it by taking the quadrature some
of the two Rabi frequencies.
So this state is indeed a dark state.
So let me show that to you by looking
what happens when we expose it to this operator.
This operator v in the Hamiltonian
is the operator which couples the atoms to light, which
couples the atom to the excited state
and this causes spontaneous emission.
So what I want to show you that is a dark state,
since if you take this operator, apply it to the dark state
and ask, is there any possibility
how you can excite, how you can excite to the excited state?
And the answer is, it will be zero.
The dark state has a component on-- the dark state
has one component in f, and it has one component in g.
How did I want to do that?
Sometimes it's easier to say it in words
than to write down a long equation.
The light atom interaction couples the component
of the ground state to the excited state
with Rabi frequency omega 1 and the state f with omega 2.
However, in the construction of the dark state,
I've made sure that the amplitude of the component g
is omega 2.
So therefore when we apply this operator,
we have omega 1 from the operator and omega 2 here.
And here we also get omega 1 and omega 2.
And this minus sign means the two effects cancel.
So therefore, I've shown you that even
though in its full frame with second quantization, photon
operators and all this, this coupling is identically
0 and therefore we have a dark state.
So this phenomenon of having a dark state
and populating the dark state by light scattering,
you have spontaneous emission into the dark state
until the scattering light staggering stops and no matter
what the initial state of the atom
bar where the dark state is now 100% populated.
This is called coherent population trapping.
Let me mention that this coherent population
trapping can be some good thing if you want to pump atoms
into a certain state and then they stop scattering light.
They're not heated up anymore.
But sometimes you want the atoms to scatter light.
For instance, in current laser cooling experiments
with molecules, there is a problem
that the multiplicity of the angular momentum states, which
are suitable for laser cooling in molecules
is such that you always have a dark state.
So if you irradiate the atoms with laser light,
they will not be on a cycling transition.
The cycling transition will stop.
So you have to play additional tricks.
One is to apply a transverse magnetic field
that the axis of the laser and the quantization axis given
by the magnetic field are in two different directions.
And then once the atom is pumped into a dark state,
it will actually precess, Zeeman precess out of the dark state.
Or magneto-optical traps for molecules
use a magnetic field gradient, which is rapidly switched,
because when you are in a dark state for one
value of the magnetic field, you just change the sign quickly.
And then the atoms are no longer in the dark state
for the new situation you've created.
So in other words, to have these wonderful dark state,
which is a major intellectual accomplishment
and eventually was exploited in many experiments,
for a number of experiments it's a real nuisance.
And you have to find ways around it
to make sure that the atoms are not stopping scattering light
and don't stop scattering light.
I gave you the simple example that the dark state
exists for resonant excitation.
But as you can immediately show, the dark state
also exists for detuning of resonance.
The only thing which is important as long as the two
lasers fulfill the two photon resonance condition.
So you want the difference between the two lasers
is the energy difference between the two states.
Also remember if you have two lasers which
are very different in power-- so let's say one Rabi frequency is
much smaller than the other one.
So then the dark state, the superposition
which is the dark state, the dark state
is now predominantly which one?
The state with the stronger coupling
or with the weaker coupling?
Well, it's the state with the weaker coupling.
The dark state has always a larger amplitude
in the state for which the coupling is weaker.
And that of course immediately makes a connection
to optical pumping.
For instance, if omega 1 the Rabi frequency is 0,
then the dark state is what we started out
with a trivial example, is a state g.
So you see that actually what I taught you
about coherent population trapping
is nothing else than extension of the concept
of optical pumping.
OK, so we've talked about the dark state
as a superposition of state g and f.
But if you have a two-dimensional Hilbert
space given by two ground states g and f,
we can now find a new basis.
We have defined the dark state.
But now we can get the orthogonal state,
which is the bright state.
It has a plus sign here and it has the omega 1 and omega 2,
the Rabi frequencies reversed.
So this state is now orthogonal.
And you can now visualize the preparation of the dark state
as follows.
We have the dark state.
We have the bright state.
The bright state, as you can immediately verify,
is strongly coupled by the laser field to the excited state.
So we have an excited state.
The laser fields-- well, its two laser fields strongly
couple the bright state to the excited state.
And then spontaneous emission-- two photon lights scattering--
can take you back to either of the two states.
But now you see that the concept of optical pumping,
which I started out with a trivial example,
is now applying in the new basis to the bright
and to the dark state.
And what happens is you would, by optical pumping,
populate the dark state and completely pump out
the bright state.
Just as a side remark, there are lots of subtleties.
Most of them you will probably see by yourself
by just inspecting the results.
But if you may ask yourself, what
happens when the Raman resonance is not
made, when the difference between the two lasers
is not exactly the difference between the two?
Well, then what you have is you have a superposition
of ground and excited state.
So we have the two ground states g and f.
But the two laser fields have a phase
which evolves different from the phase between g and f.
And suddenly, what you have is relative-- I mean,
everything is relative to the laser field, so to speak.
This plus sign turns into a minus sign.
So therefore, if you have pumped into the dark state,
but now the frequency difference of the two lasers
is slightly different from the frequency difference between g
and f, this mean in essence that as time
goes by, with the frequency, with the detuning away
from the two photon resonance, the dark state will now
precess into the bright state.
And it is only for the Raman resonance,
when it is exactly met, that you have a long-lived dark state,
a dark state which is a 2 eigenstate of this Hamiltonian.
Yes?
AUDIENCE: Would the precession also
change omega 1 into omega 2?
PROFESSOR: No, we have defined omega 1 and omega 2
as the Rabi frequencies of the two lasers.
Omega 1 is a Rabi frequency of the laser which
talks to, I think, the state g.
And omega two is the Rabi frequency
of the laser which talks to the state f.
So this doesn't change.
I'm just sort of telling you how you may want to think about it,
that if you put in all of the temporal phase factors,
on resonance, at least in some dressed atom picture,
all the phase factors are 0.
Because the e to the i omega t of the atomic wave function
is compensated by the light atom coupling.
And therefore phase factors just disappear,
and the state is stationary.
But if you write down what happens as a function of time
to the state, and you couple it as a function of time
to the laser field, you will find out
that there is an evolving relative phase
between the light field and the atomic state.
And the essence of that is that with these, there
will be a beat node at which the dark state becomes bright
and the bright state becomes dark.
OK.
One important application of the dark state
is the STIRAP technique.
STIRAP is adiabetic population transfer.
So STIRAP stands for stimulated-- what is the r?
Stimulated Rapid Adiabetic Passage?
AUDIENCE: I've seen both rapid and Raman.
PROFESSOR: So we have two choices.
It's both Raman and rapid.
A is adiabetic, p is passage.
And STI is stimulated, because it's
a coherent stimulated process.
The idea is the following.
With this concept of a dark state,
I can immediately explain to you how,
by changing the intensities of the laser beams,
you can make an adiabetic transfer from one state
g to the other state f.
And this is important because in many atomic physics
experiments, you start with one state
and then you want to prepare another state.
You always have the option of having some suitable pi pulse
and going from one state to the next.
But the pi pulse has to be exactly pi.
You have to be careful what the intensity
and duration of your pulse is.
But if you can adiabetically go from one stage
to the other one, this is terribly
robust against pretty much everything.
And I want to now use the concept of the dark state
to tell you how this population transfer works.
But before I even get into any explanation,
the picture is the following.
The dark state-- if you change laser parameters,
you can change which state is the dark state.
I gave you the simple example if one laser beam is on,
the state g is the dark state.
If the other laser beam is on, the state f is the dark state.
So if you switch one laser beam off and the other one on,
you've changed the definition of the dark state.
And as I want to show you is the dark state, if we can set it up
in the situation that the dark state is the absolute ground
state of the system, and you know when you change parameters
of your Hamiltonian, adiabeticity
tells you that you always stay in the ground state.
See now, even without any equations
you understand what adiabetic population transfer is.
But let's work it out.
So we want to understand-- and this is our adiabetic process--
what happens when the laser parameters, omega 1 and omega
2, change slowly.
So for pedagogical reasons, I will
drop all the assumptions I'm making now,
but let me make them.
So let's discuss degenerate states first.
So g and f are degenerate.
And let me also discuss blue-detuned light.
And you will see in a moment why.
So that would mean that-- and let's set up the situation.
We have the state g, f, and e.
We have a green laser beam and an orange laser beam.
And we know already, from what we have just learned,
that this should be transformed into--
and let me just be more specific here.
I said I want to use blue-detuned light.
So now we do what we just learned.
We express things as the dark state and the bright state.
And here we have the excited state.
And the beauty of degeneracy is that dark state
and bright state are superposition states,
but since they're superposition states
of two states with the same energy,
I can draw them without twisting my hand
or doing sort of a double line.
Anyway, but now what happens is the following.
The bright state is coupled to the excited
state by the two laser fields.
And the question is, what is the energy
of the bright state relative to the dark state?
Well, if I have a blue-detuned laser beam,
I get an ac-Stark shift, which increases the energy.
So therefore, the state which sees the light, which
scatters photon-- scattering photon
means there is an excited state admixture.
An excited state admixture means there's an ac-Stark effect.
And the ac-Stark effect is positive.
So therefore, the dark state is-- the uncoupled state
is the lowest state in the ground state manifold.
And even if you would not generalize it
to multiple levels and all that, if there is one dark state,
it's not upshifted by the ac-Stark effect.
Whereas every state which sees the light
is upshifted by the ac-Stark effect
which provides a blue shift.
So the dark state is the lowest quantum state.
So therefore, we can now use the general concept
in quantum mechanics of adiabetic state transfer
to achieve a perfect 100% transfer from one ground
state to the other one by tailoring as a function of time
our laser fields as follows.
We start out in a situation where the only beam which is on
is-- I want to change that to orange-- is the orange laser.
And when the orange laser is on, then the dark state is g.
Then we ramp down the orange laser.
But we ramp up the green laser.
And we have the situation that initially g
and eventually f is the dark state.
So this is a picture behind the rapid adiabetic passage
from state g to f.
I want to point out that the application of the two laser
pulse, first orange and then green
is called the counter-intuitive sequence.
Because if you ask yourself the problem,
you're starting out in state g and you
want to go over to state f, which laser
would you first switch on if you're in state g?
Well, if you want to get out of state g,
you would say first the green laser
to go up and then the orange laser to go down.
So what I just said, first the green and then the orange
is called the intuitive sequence.
But STIRAP works with the counter-intuitive sequence.
OK.
Any questions?
Yes?
AUDIENCE: So what's the intuition we have with this?
PROFESSOR: 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 adiabetic 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 adiabetic 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 discuss it 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 a way where you can go directly
without going through the excited state.
And this is done by the counter-intuitive 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 the
stimulate it down.
And this is not what STIRAP is about.
Other questions?
Yes?
AUDIENCE: So if there is very strong coupling
to a fourth state from the excited state, because you need
to-- the population needs to travel through the excited
state in STIRAP, it's possible to spoil the whole process.
PROFESSOR: 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 you fourth state.
But by just calculating what is the time integrated population
in the excited state multiplied with gamma,
this is the amount of photons scattered.
And you can say if you try to do a STIRAP with the bosons
then condensate, every time you scatter, 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?
AUDIENCE: So are the two laser frequencies the same for that?
PROFESSOR: Yes.
I actually assumed that-- I come to that in a second--
that the two states are degenerate.
And for all this dark state busyness,
we have to fulfill the Raman resonance.
So if the two states are degenerate,
then the two laser beams.
AUDIENCE: So then what's the meaning
of increasing one of them?
We are satisfied--
PROFESSOR: 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, and 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 with the orange laser.
And we have two ways to accomplish it.
One is by frequency difference.
But if you don't have a frequency difference,
we need polarization selection rules.
Yes?
AUDIENCE: It's sort of a silly question.
In this particular case, if I have all my atoms in g,
can I turn on the green laser?
Can I optically pump everything to f?
Right?
And so in this particular case, is there
an advantage of doing STIRAP as opposed to optical pumping?
PROFESSOR: Yes there is, and we'll talk about that.
There is a difference between transferring population or only
transferring an amplitude.
And 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, 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 a PEC in one state after STIRAP transfer
is in another state, maybe if you
have the lasers in a counter-propagating way
back to zero 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 a 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, so lots of interesting questions.
First, let's now drop all the assumptions we've made.
So what happens when we change from blue detuning
to red detuning?
Well, for blue detuning, we use the argument
that adiabetically we stay in the ground state,
but for red detuning it's a higher state,
and we adiabetically 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.
Let me be consistent.
I call the operators c and Na and Nc.
And if you 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 made.
So we've talked 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've 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 are not building up population
in the excited state.
We're just building up a small amplitude.
And the population is the 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 show you now how this pans out.
Jenny?
AUDIENCE: But in the limit of infinitely adiabetic,
infinitely slow, you still would have the transfer happen
without ever going into the excited state, right?
PROFESSOR: Well, I will show you that the integrated population
in the excited state goes as one 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.
So let me just say, no.
Excited state is needed as a stepping stone.
I want to give you a quantitative touch
without doing too much math.
So let me just write down Schrodinger's equation.
We have the three amplitudes, the three states involved,
e, g, f.
Schrodinger's equation gives us the time derivative.
c dot f.
Since we assumed in the dressed state picture,
we can assume everything is degenerate.
But now we have couplings.
And the only couplings are-- and this
is what I was emphasizing-- between ground and excited
state with the Rabi frequency omega 1 and between the state f
and the excited state with Rabi frequency omega 2.
So the important part is-- I can write it down,
but the third equation is the one I want to focus,
that you can only build up population in the final state
by transferring with a Rabi frequency omega
two population from the excited state.
So let me just assume that Rabi frequency omega
1 equals omega 2 equals omega Rabi.
Let's have a symmetric situation so I can now completely focus
how can you get from the excited state to the final state,
but how you get from the ground state,
from the state g to the excited state is fairly symmetric here.
OK, so what we want is we want that at the end of the transfer
time, t transfer, the final state
amplitude is on the order of 1.
This would be 100% percent transfer.
So therefore, just looking at the differential equation
above, that means that this here has
to be 1 divided by the transfer time.
So therefore, we find that for this adiabetic transfer,
the excited state amplitude has to be--
and I neglect factors on the order of 2
now-- has to be 1 over the Rabi frequency times the transfer
time.
So the probability to be in the excited state
is the amplitude squared.
And if you are asking, which I think is a practical question
and I suggest to use this as a figure of merit,
when we discuss how well, how perfectly we have transferred
population, we can now ask, what is
the probability of spontaneous emission?
P spontaneous is the probability to be in the excited state
times gamma integrated over time.
So what we obtain for that is yes,
it's proportionate to gamma.
But we have the excited state amplitude squared.
We've multiplied with the transfer time.
So therefore, what we have in the denominator
is the laser power or the Rabi frequency
squared times the transfer time.
So therefore, this probability to go
to a spontaneous emission, which is also the integrated
probability of having populated the excited state,
it goes to 0 in the case that you have infinite laser power
or that you're infinitely patient.
So one question is, how long can the transfer time be?
And this is actually setting an limit to it.
We're talking about a coherent transfer.
So the transfer time has to be smaller
than the coherence time of your system.
For instance, we do that all the time in my lab.
If you go from on sublevel to another sublevel,
and we have magnetic field noise, if the magnetic field
noise is destroying the phase relationship between the two
hyperfine states, then the coherent transfer
gets interrupted, and then eventually we
are no longer talking about the amplitude.
We are talking about population in the excited state.
OK, so we have this result, which at least demystifies what
many people say-- actually I've heard
truly experienced researchers in the field to say,
we can do the transfer without ever populating
the excited state.
So the truth is what I just derived for you.
But now we want to compare it to what
would happen if you use an incoherent picture.
If you use an incoherent picture,
let's assume we would use a pi pulse,
put all the population in the excited state,
and then we would stimulate it down.
So we have a pi pulse.
So we build our population.
And then we use a second pi pulse.
Well, we would then have about unity population in the excited
state.
It would emit light at a rate gamma.
And it would take the time of the inverse Rabi frequency,
the Rabi period, to stimulate it down.
Now you can say, what happens if-- maybe that's
dumb to have a pi pulse and put all the population in, take
a pi over 20 pulse, put a little bit population in, stimulate it
down.
The next batch, the next batch, the next batch, but you
get exactly the same result, because each population which
you've put into the excited state
had to wait for a Rabi period to be stimulated down.
So therefore in the incoherent picture,
if you sort of transfer batches of population,
you realize that the probability for spontaneous emission
or for populating a four state is independent of the transfer
time, because it cancels out.
If you do small pieces of population,
you have a smaller rate of emission,
but you integrate for longer and the two temporal factors
cancel out.
And now you realize that the coherent transfer
is more favorable by another power of omega Rabi times t
transfer in the denominator.
So that's why people prefer STIRAP.
Questions about that?
Can you suggest another method how
we could transfer population from one state to the next?
I gave you two examples.
One was this magic, or not so magic, dark state transfer.
The other one is a method where we
put population, stimulated it down, next batch of population,
stimulate it down.
What is another process which we can do to transfer population?
AUDIENCE: Landau-Zener?
PROFESSOR: Pardon?
AUDIENCE: Landau-Zener?
PROFESSOR: Landau-Zener?
Actually, the STIRAP is a Landau-Zener.
The dark state is always the lowest state.
There is a crossing sort of in between,
but it's an avoided crossing.
Actually, in a way, you may want to think about it.
The STIRAP is actually a Landau-Zener transfer
between where the lower state is a dark state,
the upper state is a bright state.
Will?
AUDIENCE: I still have a question.
So Landau-Zener [INAUDIBLE] But the STIRAP scheme
is sensitive to the run-on resonance.
PROFESSOR: Absolutely it is.
AUDIENCE: So how robust is it against small violations
[INAUDIBLE]?
PROFESSOR: Well, I gave you already the hint
when I discussed that if you are detuned from the Raman
resonance, this detuning, at the detuning frequency
would be between dark and bright state.
So this picture what is dark and what is bright
is scrambled up by that.
So in other words, as long as you're
inverse detuning is smaller than the transfer time, you're OK.
Because the dark state is the dark state.
But if during the transfer time you turn the dark state
into a bright state, you've completely
messed up the scheme.
So in other words, I know people in this hallway, in my group
want to STIRAP transfer between molecules.
And we have to stabilize our lasers.
So we have automatically little bit laser fluctuations.
But if you do the transfer in a millisecond,
it is sufficient to have the laser frequency under control
at the kilohertz level.
So that's a requirement.
But back to my question.
We have discussed sort of an incoherent transfer
and we've discussed STIRAP.
Can you suggest another method how
you can go from one count state to the next?
AUDIENCE: We talked about optical pumping.
PROFESSOR: Optical pumping?
Yes, but optical pumping means we
really go with spontaneous emission.
And then we have 100% spontaneous emission.
So we want a process where we go from one momentum state
to another momentum state.
Well--
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yes.
Optical pumping is a two-photon process.
Actually, everything's a two-photon process here.
But optical pumping is actually a process
where we have a Raman to photon transition.
The first leg is stimulated.
The second one is spontaneous.
But I would suggest now what [INAUDIBLE] said,
let's have a two photon Raman transition.
So we want to go from state g to state f with two photon Raman
transition.
Let's assume we have a detuning delta.
So now, do you have any expectations
whether these-- we can do a pi pulse.
Two photon Raman transition is a two photon Rabi frequency.
I'll remind you in a second of it.
And if you do a pi pulse in that,
we have half a Rabi oscillation, which
takes us from cg to f and we have 100% transfer.
So do you have an expectation if this Raman
process how will it turn out to be in our figure of merit,
how many photons are emitted from the excited
state during the transfer?
Will it be the same as STIRAP?
Will it be the same as in population transfer?
Will it be something else?
AUDIENCE: You have to scale.
PROFESSOR: It will scale with the detuning,
but-- OK, what I will do is actually-- I will in a moment
say, OK.
We have a certain laser.
What I'm going to say?
We have to compare apples with apples.
And here we have the detuning as a parameter.
And what I want to replace is I will--
what will replace the detuning by the time it
takes to do the pi pulse.
So the detuning is determined by the condition
that we want to do the transfer in the same time as the STIRAP.
So this will eliminate the detuning.
So we want to use the same resources.
Our resources are time and laser power.
And we want to use our resources of time and laser power
in the same way for the two photon transition.
So the question is what happens to-- what
is the answer to the question, how much excited state is
involved in the two photon Raman process as compared to STIRAP?
Do you have an expectation?
Who of you is doing STIRAP in the laboratory?
A few people, yeah.
Actually, I know that 99-- my observation
is that 99% of the people say STIRAP is special
because it's a dark state transfer.
And when I started to confront famous people in the field
and say the two photon Raman process does exactly the same,
they were completely surprised.
Anyway I want to show you in one equation
that the two photon Raman process has
exactly the same integrated population in the excited
state, and therefore it is as good-- the two photon Raman
process performs exactly as well as the so-called dark state
transfer, where you never go through the excited state.
And this for me completely demystifies it,
because the two photon Raman process
has in common with the STIRAP that everything is coherent.
And if everything is coherent, it
means we are not building a population.
We are only building up amplitude in the excited state.
So let me just show you in one equation
that indeed it works out in that way.
It's also a nice way to quickly recapitulate
what we learned in the previous chapter
about two photon transitions.
So the two photon Rabi frequency is nothing else
than the product of the two Rabi frequencies divided
by the detuning.
And of course as-- if you use studied larger and larger
detuning as you said, everything will slow down.
But we know that the transfer time
will be the inverse of the two photon Rabi frequency.
And this is delta over omega square.
So therefore, we will eliminate delta from our equations
by replacing it by the time it will
take to perform a pi pulse between the two ground states.
OK, so what is the probability to be in the excited state?
Well, we've done the theory of the ac-Stark effect.
We have beaten perturbation theory to death.
You know the first order admixture
is omega Rabi over delta and the probability
is omega Rabi squared over delta squared.
So therefore, the probability to spontaneously emit a photon
is omega Rabi squared over delta squared times gamma times t
transfer.
And this is nothing else than gamma divided by omega Rabi
squared times t transfer.
So if you use a Raman pulse to transfer the population
coherently, the integrated population
of the excited state, which is a measure for heating and measure
for winding up in the wrong state, emission
to a false state is inversely proportional to your laser
power and inversely proportional to the time
you can afford to do the transfer.
Exactly the same as for STIRAP.
And the limit to the transfer time in both cases
is set by the coherence time.
Try to tell people who do STIRAP that a two photon Raman
pulse would do exactly the same as STIRAP,
at least in populating the excited state.
They will be surprised.
Because here, you explicitly go through the excited state.
We don't have a counter-intuitive sequence.
We switch on both lasers simultaneously.
But I think this for me illustrates or completely
demystifies what people have said about the dark state
transfer and the STIRAP process.
Any question?
Cody?
AUDIENCE: Why do people use STIRAP?
What's the advantage?
PROFESSOR: Well, why are you using Landau-Zener sweeps when
you go from one hyperfine state to the next,
and why don't you use a pi pulse with your f generator,
because it's more robust.
AUDIENCE: [INAUDIBLE].
As you said [INAUDIBLE].
PROFESSOR: But similar, if you detune--
if you detune your-- the condition is the same.
If you have your two Raman laser and you detune them,
you no longer have resonant Rabi oscillations.
You have off-resonant Rabi oscillations.
And you will find if you want to do the transfer in half
a period, you have to make sure that you don't
get one cycle of the beat node or one
cycle of the detuning in your transfer time.
So the condition for the frequency stability for the two
photon Raman and the STIRAP I think
are pretty exactly identical.
So I think it's pretty much the robustness.
And you have to decide what you need and what you want.
For instance, if you have laser beam and the laser beam has
an inhomogeneous profile, you cannot meet the pi pulse
condition for the atoms in the middle and at the edge
of your laser beam.
But in STIRAP you just provide plenty of extra power
and then everything is robust against the laser beam profile.
So there are clearly advantages like that.
But all what I point out is the advantage
is not the population of the excited state.
It's the same.
Questions?
OK.
So this is dark state and STIRAP.
Our next topic is lasing or gain without inversion.
So in other words, I want to show you
what was actually rather recent accomplishment
that this common belief that you need population in the excited
state, which is larger in the ground state,
for optical gain and lasing is not true.
Well, it's true for two level system,
but it's not true when you have more than two levels.
And I discuss it here for the case of a three-level system.
And this is really insight which came as recent as in the '80s.
One reference is the person who pioneered a lot of work
on three-level systems and provided a lot of insight
was Steve Harris at Stanford.
And this was a paper in the late '80s which
enunciated the possibility of having
lasing without inversion.
And when you are a graduate student like me,
who was educated earlier, you say, gee whiz.
That's not what I learned in class.
But of course it's often the small sprint.
It's the assumption for two-level system.
So the secret behind lasing without inversion
is that you can have cancellation of absorption
by interference.
In other words, this belief that you need inversion
comes that you want that the stimulated emission is
stronger than the absorption.
But if you have a three-level system, you have to be careful.
We have interference effect.
And what I want to show you is that we can create a situation
where we have destructive interference for absorption,
but we don't have destructive interference
for stimulated emission.
So therefore, we can now afford to have more population
in the ground state, because the absorption out of the ground
state does not have to be suppressed by minimizing
the population in the ground state.
We have an interference effect, which suppresses absorption
from the ground state population.
And therefore, the ground state population
can be larger than in a two-level system.
I could simply tell you you can hide population
in the dark state, end of discussion,
and you get the basic idea.
But the concept of lasing without inversion
is richer than that.
It's connected to the dark state, but there is more to it.
So that's why I want to spend the rest of today's class
in discussing this concept.
So I want to discuss it in the simplest possible scheme.
So I want to discuss first the simple system.
In the end, this is not the system
in which it has been accomplished.
But I think what you realize is I draw you
up a very simple scheme.
I make all assumptions, and there is maybe
no atomic, no suitable atom where you can realize it.
But then I will do what I've done so often in this class.
I turn around and say, but if you go to a tri-state basis,
we can exactly engineer that.
So in other words, if you need a certain level scheme
and you don't get it, you can just take another level,
put it there with a strong laser as a virtual level,
and you get the level scheme you want.
And in a dress state description,
you have exactly the idealized level
scheme I'm presenting to you.
So therefore with that justification,
let me just try to give you the simple scheme where
you can understand this absorption
cancellation by interference.
I mean, later in the end, we want
to use lambda systems with two stable ground
state, two hyperfine states.
That's how all the research is done.
But now I assume we have a V system.
So we have two excited states and we have one ground state.
And I want to assume that by coupling through radiation,
spontaneous emission, those excited states
couple to the same continuum.
I want to say that this is an important assumption.
If you have two states, hyperfine states,
n equals plus 1, n equals minus 1,
which would emit photons of different circular
polarization, no.
They are not coupling to the same continuum.
They would be distinguishable, and then certain interference
effects will not happen.
So eventually, the magic of quantum interference
comes if it is absolutely indistinguishable
which excited state has emitted the photon.
So we have to really be-- set up something like this.
So it must be fundamentally impossible to distinguish
photons emitted through one state or the other.
So we know already if we treat the system,
we couple the atoms to the continuum.
That means in the end, that those states
acquire a certain width.
And yes, this is a situation we want to discuss.
So what we have set up now is a situation
where we have, with these two states, two
indistinguishable paths to scatter photons.
So let's say we take the ground state
and we shine laser light on it.
It's a two photon process.
A photon goes in, a photon comes out.
And we can now go through excited state 1,
through excited state out then a photon comes out
and eventually we have now the coupled through this two photon
process the ground state to continuum of modes.
But those two processes to go through e1 and e2
are indistinguishable.
So I will give you immediately an expression
where in perturbation theory, we are not adding intensities.
We are adding the two amplitudes.
It's like a double-slit experiment.
Or in other words, the situation I want to discuss now
is the following.
We have all of the population in the ground state.
We have the two excited states.
And our laser is tuned in between.
And now we can scatter light.
We've talked about really scattering and such.
And you know that for an infinitely heavy atom,
the emitted radiation is a delta function at the incident light.
And you will not know, you can never
know which excited state was involved in the scattering.
So we assume that we have excited
state-- which one is the lower one?
I think e1.
This is e2.
And our laser beam is detuned.
We have a detuning of delta 1 and delta 2 respectively.
So now we have to add amplitudes when we do perturbation theory.
So in second order perturbation theory,
for the light scattering, if you want
to derive Fermi's golden rule as we have done a number of times
here, you remember, the critical part--
I'm not writing down the whole expression-- the critical part
is sort of a product of two matrix element or two photon
matrix element which takes us from the ground state
through the light atom coupling to the excited state.
And then from the excited state why are the coupling eventually
back to the ground state, which is now a continuum of modes.
And we sum over all possible modes.
And what we have here is the detuning denominator.
Well, if you want to do a little bit better,
I'm just mentioning here for completeness,
we want to add the imaginary part.
That's more placeholder for a command I want to make later.
It's not really essential.
But the new thing I want to discuss now
is that since we have two excited states and not one,
we have to sum over the two excited states,
and it's a coherent sum.
So if I detune between-- if my laser detuning is tuned
in between, then delta 1 is positive
and delta 2 is negative.
So therefore, we do a sum here with opposite signs.
And if I can neglect the gamma, assuming the gamma
is small or much smaller than the detuning,
then I'm just adding up two numbers with opposite sign.
And depending now on the matrix element, if they're identical
I get a cancellation if I'm detuning half between.
But even if the matrix element are different,
I will always find a detuning where the sum is 0.
So to the extent that I can neglect the imaginary part,
the gamma in the denominator, I have now this situation
that this is 0.
It vanishes for a certain laser frequency, omega 0,
tuned between the levels e1 and e2.
And what I've assumed here is that the detunings are larger
than the decay widths, and therefore I can neglect that.
Any questions about that?
So OK.
Very trivially, two excited states
which are indistinguishable.
It's the modern version of Feynman's
double-slit experiment.
We add up the amplitudes and by necessity, we get 0.
But that sounds so trivial, but it took a Steve Harris,
some genius person to invent it, to realize
that when we have light at omega 0, it would be absorbed by e1.
It would be absorbed by e2.
But it is not absorbed in the situation
we have with e1 and e2.
But now we have to make the connection that the statement
that we need inversion for lasing was related
to absorption, but now we have cancelled absorption.
So now what we want to do is we want
to create a situation where we have lasing.
We want to put population into the excited state.
But the population in this excited state
will be much, much smaller than the population in the ground
state.
So you're not even close to inversion.
So let me draw the diagram.
We have lots and lots and lots and lots
of population in the ground state.
We have 0 population in e1.
And in e2, we just have a little bit.
And now you would say, well, you don't have inversion.
How can you get lasing?
But if we have now a cavity at frequency omega 0,
then you can say the Lorentzian profile of the excited state e2
overlaps with omega 0.
So at least you have some possibility for the excited
state e2 to decay into the cavity mode.
And therefore, a very weak field in the cavity mode
will be amplified by stimulated emission
without suffering any absorption.
So therefore it will be amplified and we have gain.
So if you have a cavity at omega 0,
we have gain because we have some stimulated emission
at omega 0 but no absorption.
Of course, if you would tune your cavity
to be in a resonance with e2, you
would have more stimulated emission,
but you would have much more absorption
and you would not be able to get a net gain for a small probe
field.
So you would not have lasing.
So you can say that what we have achieved here
in a three-level system, we have accomplished
destructive interference for absorption.
But the way how we've put population just in e2,
we have not any destructive interference
for the stimulated gain.
So we have tweaked on the lasing equation,
on the laser equation the absorption part,
but not the stimulated emission part by interference
in a three-level system.
So how can this be realized?
Let me just give you the first possible realization
and then I think we should stop.
We could, in principle, realize it
with hydrogen and a DC electric field.
Remember, hydrogen has the 1s level, has the 2s level,
and the 2p state.
And you all had a nice, little nice homework assignment
that when you have a DC field, you actually couple 2s and 2p.
And if your DC field is stronger than the Lamb shift,
you actually create a superposition
of 2s and 2p states.
And now, if you tune your laser right in between,
you have a dark resonance.
So the situation is if I would now
plot the absorption versus frequency,
you would find that if you're in resonance with the upper state
you absorb, but then even in between,
there is this notch where you have zero absorption.
So for this simple case with hydrogen
in a DC electric field, the dark resonance
occurs when you have equal detuning.
You have an equal detuning for the atom with respect
to the two states.
And I like to sort of connect it to classical physics.
You know that absorption is related.
When you absorb, you have light scattering.
The photons are not disappearing.
They're just scattered into other modes.
That's what absorption is about.
And classically, light scattering
comes because you have an oscillating dipole moment.
But what happens is if you have this coherent superposition
and you're right in between, you have a positive detuning
with one harmonic oscillator and a negative detuning
with the other harmonic oscillator.
So if you drive an harmonic oscillator-- and this is one
and this is the other-- with positive detuning,
with red detuning, the dipole moment
is in phase with the electric field.
In the other case, it's out of phase.
So now you have the two excited states
and you're driving two equal but opposite dipole moments
in those two states.
And therefore, the two dipole moments add up to 0.
Therefore the laser, just using our understanding
of a simple harmonic oscillator, is not
creating any dipole moment.
Therefore, we're not scattering any light
and therefore we're not absorbing any light.
So this is how you can now realize in the hydrogen atom
using DC magnetic field and creating this level structure,
you can create lasing without inversion.
But I will tell you on Wednesday-- no,
today is Wednesday-- next Monday,
next week that we don't have to deal
with the difficulties of hydrogen atoms
in the laboratory.
We can just use our favorite alkalize
and dress the alkalize up with laser beams
and eventually create the same situation
with an atom which is experimentally
much simpler to address.
Any questions?
OK, one more week to go.
See you on Monday.