M5L23n.txt

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I want to discuss it in this simplest possible scheme.
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 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 [INAUDIBLE] 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 the strong laser as a virtual level,
and you get the level scheme you want.
And in a dress state description,
you have sort of exactly the idealized level
scheme I'm presenting to you.
But therefore, with that justification,
let me just try to give you the simple scheme where
you can understand this absorption
cancellation by interference.
Later, in the end, we want to use lambda systems
with two stable ground states, 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
for 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
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 set up something like this.
So it must be fundamentally impossible to distinguish
photons emitted from 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 require a certain width.
And this is the 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 one,
through excited state out.
Then a photon comes out.
And eventually, we have now 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 we give you immediately an expression
where, in perturbation theory, we're not adding intensities.
We're adding the two amplitudes.
It's like a double-split 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 still in between.
And now we can scatter light.
We've talked about Rayleigh scattering and such.
And you know that, for an infinitely heavy atom,
the emitted radiation is the delta function at the incident
light.
And you do not know-- you can never
know-- which excited state was involved in the scattering.
So we assume that we have excited state 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 now
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 a product of two matrix elements,
or two photon matrix elements, which takes us from the ground
state through the light atom coupling to the excited state,
and then from the excited state via 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 it here for completeness,
we want to add the imaginary part.

That's more placeholder for a comment 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.
OK 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 two 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 cancellation if I'm detuning half in between.
But even if the matrix elements are different,
I will always find a detuning where this 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 [INAUDIBLE].
And therefore, I can neglect.
Any questions about that?

So OK, very trivially-- two excited states
which are indistinguishable-- it's
a 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 ingenious 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