M3L14f.txt

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Of course, all I need to come back
to spontaneous emission, stimulated emission,
are the matrix elements of these operators a and a dagger.
And this is where, of course, stimulated
and spontaneous emission and all that comes in.
The non-vanishing matrix elements
in this description of the electromagnetic field
are the ones where a annihilates a photon
and the matrix element is square root n.
Or where a creates a photon, adds a photon
to n photons already present.
And then the matrix element is n plus 1.

OK.
So we went from A to Q and p, and we went to a and a dagger
But A is also related to the electric field
by taking the time derivative of the vector potential.

So now, of course, we can go from our expressions of a
and a dagger all the way back.
Just substitute, substitute, substitute,
and find an expression for the electric field
in terms of a and a dagger.

The result is that we have a and here we have a dagger,
the [? repolarization ?] vector.
We have the plane wave vector and the complex conjugate.
And the complex conjugate of a is a dagger,
so the electric field is a superposition
of a and a dagger.
The electric field becomes an operator
which is the sum of creation and annihilation operator.

So with that, we can go back to our Hamiltonian.
Our Hamiltonian for the interaction
between light and atoms, in the simplest possible case,
was the dipole Hamiltonian, which involves a dipole matrix
element.
The charge of the electron is negative.
That's why the minus sign has disappeared.
And now, all we do is from our treatment
before in the Schrodinger equation, where
the electric field was an external field-- now
the electric field becomes the operator acting
on the quantum state of the electromagnetic field.

So, by the way, this prefactor here
is-- because the rest of it is just
dimensionless-- this prefactor has-- so we have the matrix
element here.
This prefactor is an electric field.
And something you should always know-- this electric field
is actually the electric field of a single photon.
This is the correct normalization.
If you want to factor out the volume, the frequency, and all
that, you combine these factors in such a way
that it's an electric field of a single photon.
Then we have the dipole moment.
And then we have an expression with creation and annihilation
operator over here.
I assume now that the atom sits at r equals 0,
so why should I carry forward an e to the iKr term?
We conveniently place the atoms at r equals 0.
But I have to say a word or two about the e to the i omega
t factor.
I have been deliberately cavalier
about my formulation in quantum mechanics
whether I use the Schrodinger or the Heisenberg picture.
And you know in the Schrodinger picture of a wave function
is time-dependent, not the operator.
In the Heisenberg picture, it's the other way around.
And I have to tell you, every time I
do a calculation and look at [? a group, ?]
I'm getting confused about the two pictures.
So anyway, trust me that in this case, when
I want to discuss the Schrodinger picture,
the time-dependent factor should not be present.
But it's a little bit-- you really
have to look at the derivation and carefully
realize the two are connected with a unitary transformation.
You really have to figure out in which representation you are.
But I want to not focus on the formality here.
But I'm not caring for what this factor,
because I want to discuss the Schrodinger picture.
OK.
Yes.
So now we can look at the matrix elements of our interaction
Hamiltonian.

And just to be clear, we have written down
this Hamiltonian for just a single mode of the radiation
field.

Depending on what we're interested in,
we may have to sum over many, many modes.
So we are looking at transitions from an initial state, which
may be an excited state, to a final state which
may be a ground state.
And since we have quantized the magnetic field,
we also have to specify the state of the quantum field.
And we assume that the uncoupled Hamiltonian, of course,
has simply number states as eigenstates,
n photons or n prime photons.

So the only non-vanishing matrix elements are the following.

e is the charge.
e hat is the polarization.
Epsilon 1 is the electric field of a single photon.
And, of course, we only have a coupling
by the fully quantized Hamiltonian
when we have a dipole matrix element connecting
state a and b.
And these are all things we have already
discussed in another context.
But now, the a's and a daggers, which
only act on the photon field, give rise to two couplings.
One is absorption, and one is emission.

Absorption takes place when we look at the matrix element
when the final state has one more photon.
And emission takes the other way around.
When the final state has one more-- which way do we go?

Let me just write it down and then read it off.

I think I've inverted it.
But anyway, initial and final state
can differ by plus 1 photon or minus 1 photon.
In one case, it's absorption; the other case, it's emission.
And the matrix element is n or n plus 1.

So one is absorption, and one is emission.

So finally, if you ask, what are the rates
of absorption and emission?

When we assume we have a situation
where-- and we have now discussed a matrix element,
and this matrix element could become the basis of Fermi's
Golden Rule.
We just have to specify time-dependent perturbation
theory.
But in any case, whatever we do when we talk about a rate,
it will involve the matrix element squared.
So now we can ask what happens when we couple
ground and excited states.

And let's assume we have an excited state
and we sum over all possible photon occupation
numbers of the ground state.
Well, when we go from the excited state to the ground
state, there will be only one term contributing to the sum
where we have one photon more, because it has been emitted.
So therefore, because of the square root n and n
plus 1 dependence of the matrix element,
we find that for the processes where a photon is emitted,
where the atomic system gives away a photon,
the sum of all the possible rates becomes simply n plus 1.
And in the case of absorption, it becomes n.

So in other words, we have now done with field quantization
what Einstein pulled out of a thermodynamic equilibrium
argument.
Namely, that if you have a system
that the rate of emission versus the rate of absorption
is n plus 1 over n.
But we did not assume any spectral distribution.
We know this n plus 1 over n applies to every single mode
of the electromagnetic field.

Questions about that?

I also want to tell you, just as a side remark, a lot of people
think that when emission is n plus 1,
the plus 1 is different from n.
That this plus 1 is sort of a spontaneously emitted photon
which has maybe some random phase.
And the n, which is stimulated photons,
they go in the same mode as-- they're
identical to the photons already present.
I don't see any of that in that treatment.

So a spontaneously emitted photon
is identical to the photon which would
be emitted in a stimulated way.

We just have n plus 1.
This is the matrix element for coupling to this mode.

At some point, spontaneous emission
can happen in many modes.
But if it goes to many modes, then there
is some integral or some summation involved,
and this can cause a certain randomness.
But at the level of a single mode,
I do not see any difference between the one photon
and the n photons at this level of discussion.
Just keep that in mind.
And actually, we'll discuss micromasers
You can have put an excited atom in the cavity,
and you have a fully reversible exchange.
So you spontaneously emit; you absorb.
You spontaneously emit; you absorb.
You have Rabi oscillations, which involve a single photon,
and they involve spontaneous emission.
Fully reversible, completely unitary time evolution.