M3L14g.txt

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What I'm doing now is I really directly calculate for you
the rate of spontaneous emission.
And I'm not getting it through the back door
by treating absorption and saying, well, there's
n and n plus 1, or borrowing some argument from Einstein.
It's such an important quantity, we
should just hit the system with an Hamiltonian,
and out comes the spontaneous emission rate.
And this is what we are doing.

So the starting point is what we have discussed
at the beginning of the class.
We want to discuss Fermi's Golden Rule.
We want to use the rate, and to remind you
that the rate for process is the matrix element squared
by h plus square.
And then we have to multiply with the density of states.

So this is the density of states per polarization.
Actually, I made a few corrections to my notes,
because I realized I have to be very, very careful in telling
you what the states are, because this
is what this exercise is about.
And we have to be writing it down for one, mode by mode.
So the density of state is now per polarization--
we take care of polarizations later-- per unit frequency
interval.

This rate, if it's all of spontaneous emission,
is the Einstein A coefficient.
But there is one caveat, and this
is the emission of an atom, which
has a dipole moment, is not isotropic.
So I have to be a little bit more
careful with a solid angle.
I cannot just calculate a rate and assume everything is
isotopic.
If I would do that, I would save a few minutes,
but I would really expect something under the rug.
So what I'm calculating, first, is
the rate in a given unit angle.
And then, I do an integration over the unit angle.
And eventually, I will integrate over the dipolar pattern.
So therefore, the density of states, of the photon states,
is sort of photons with their K vectors going to all space.
But I wanted to have the density of states per unit angle.
And this quantity is, omega square 8 pi cube c cube times
V.
And of course, if you multiply this by 4 pi,
you get your normal density of states,
because the density of states isotropic.
But the rate which we calculate will not
be isotropic, because of the dipole matrix
element and the dipole pattern.
So therefore, we start with a differential formulation,
spontaneous emission per solid angle.
And then, when we integrate over the solid angle,
we have to take care of the sine square factor,
because of the dipole pattern.
Now, Fermi's Golden Rule takes us
from an excited state to ground state.
And since we use the fully quantized treatment,
our states are products states of atomic states and photon
states.
And so, we assume we start with an atom in the excited state.
And all modes, mode 1, mode 2, mode 3, are empty.
And the final state-- well, one photon is emitted,
and it can appear in any of the modes.
And we have to do an integral of all possibilities.

Good.
So we did all work with quantizing
the electromagnetic field, because we
want to calculate those matrix elements.
Let me just carry over the prefectors, the electric field
of a single photon.
Here, we have the dipole moment between the polarization
and the atomic dipole matrix element.
And now, since we are talking about an emission problem,
we have, from the matrix element squared,
as we just discussed, an n plus 1 factor.
But the population-- we start with zero
photons, so therefore, it's just 1.

So all the work we did on quantization
of the electromagnetic field is that, even without any photon
present, we have a coupling.
You can say it's a coupling caused by the vacuum, which
is like the coupling we would have gotten if we have
exactly one photon [INAUDIBLE].

So let me just write that out.

OK.
We have now taken care of the matrix element.
So we insert the matrix element now in our Fermi's Golden Rule
expression for dA, d omega.
Let me just keep track of all the factors.
We have an omega in the matrix element.
This comes from the electric field of a single photon.
We have a matrix element, dipole matrix element,
times polarization.
And the density of states gives us omega square.
So you see, already, we'll get all four spontaneous emission
omega cube expression.
One omega comes because the electric field
of a single a photon-- the electric field squared
of a single photon is proportional to omega.
And an omega square factor comes from the density of states.
It's really important to keep that apart.
The omega 3 dependence has two different sources.

It's always nice to see that we assumed an [INAUDIBLE] volume,
and it cancels out.
OK.
Let me just write it down.
And then we do the final step.
So if everything were isotropic, I could just
multiply with 4 pi, and the last factor would be dropped.
But if you want to go from the spontaneous emission
per solid angle to the total spontaneous emission,
we have to average it.
And what has angular factors is actually
the projection between the atomic dipole
moment and the polarization.

So this is the relevant term.
And now, we have to distinguish, there are two polarizations.
One polarization-- the polarization,
you have a dipole moment, and you
have light which is polarized in such a way the light goes here.
This is a dipole moment.
The light goes here.
And now, the light which propagates here
can have a polarization like this,
which has an projection of sine theta with a dipole moment.
And if the light goes there and the polarization is like this,
it's orthogonal to the dipole moment,
so the scalar product is zero.
So for one polarization, we have a sine theta factor.
For the second polarization, the scalar product
of the dipole moment is zero.
So therefore, this integration over the solid angle
boils down to that we can pull the matrix element out
of the integral.
And what is left is the projection factor,
sine square theta.
We have to integrate over the whole solid angle,
and this gives us 2/3.
So therefore, our final result is
that the microscopic expression for the A coefficient
has this factor of 3 in the denominator.
And this factor of 3 only comes because I correct the average
over the dipole pattern.
Well, then we have 4 pi epsilon0.
I mentioned the important depends on the frequency cube.

OK, so this is our final result for today.
And I will discuss next week what
are its units, how big is it, what
is the quality factor of the atomic oscillator.