M1L3b.txt

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Let's start with a very, very short review
of this simple harmonic oscillator.
It just also connects me with the last lecture.
So how do we describe light in a single mode?
And single mode means it is perfectly monochromatic.

Well, we know that-- and that's what
we did last class-- that we have the Hamiltonian
for the electromagnetic field, which
is the quantum version of the electromagnetic energy,
e square, e square.
And from that, we derived the quantized Hamiltonian.
We introduced just as a reminder, vector potentials,
normal modes, and the normal mode operators
became the a degas.
And in second quantization, this is the Hamiltonian
for light, which is of course the Hamiltonian
for harmonic oscillator.
What I need today is, I have to rewrite the harmonic
oscillator, and we have to look at two variables, the momentum
and the position.
Actually, the momentum we later-- I
will show you that the momentum of this harmonic oscillator
is actually the operator of the electric field.
So we need that in a second.
And of course, I assumed from your knowledge
of the harmonic oscillator that position and momentum
are linear combinations, symmetric and antisymmetric
of a dagger a.
And the unit is the prefect or square
root of H [INAUDIBLE] omega over two,
or here it is h bar over 2 omega.
And of course in position and momentum operators,
the Hamiltonian is simply p square.
Should remind you of kinetic energy.
Omega square, q square should remind you of potential energy.

When we discuss light, we want to discuss the electric field.

I had shown you in one of those 100 equations, the operator
for the electric field.
The operator for the electric field
was of course related to the vector potential,
and the vector potential, we did a normal mode expansion
and then we put everything backward,
and we found that the electric field operator can actually
be written.
I have to fill in a few things, but essentially
is a minus a dagger.
And a minus a dagger is nothing else
than the momentum of the harmonic oscillator.
So here, the momentum has a special role
because it is the operator of the electric field of light
in a single mode.
It has a harmonic oscillator representation,
and the operator for the electric field
is-- I put in a few bells and whistles in a moment--
but is in essence, the momentum operator
of the mechanical analogous harmonic oscillator.

OK, but the electric field has a polarization.
We need that.

It has-- we have the photon energy and the quantization
volume.
So, what I'm writing down here is actually the electric field.
This square root is the electric field strength
of a single photon in the quantized volume of our cavity.
So v is the quantization volume.

Then, if you have a single mode of the electromagnetic field,
well, in contrast to harmonic oscillator,
we have a propagation kr plus omega t,
and here we have e to the minus i.

I hope I get the signs wrong here minus omega t.
So therefore-- but OK, we will mainly
be looking at an atom at the origin.
That's what we did in the dipole approximation.
So when I said that the electric field operator corresponds
to the momentum, I have to aid now
that it is at t equals zero, and at r equals zero.

We'll come back to that in a few moments.
OK, since I've given you the operator,
let me just put the operator symbol on it.
The operator of the electric field, I have to say now,
one word about Heisenberg operators
versus Schrodinger operators.
I assume you are all familiar with the two
different representations of quantum mechanics.
You often use the Schrodinger representation,
where the operators are time independent.
But in the Heisenberg representation,
the wave function, they state vector, is time independent,
and the time dependence of the evolution of the quantum system
leads to time dependence of the operator.
So what I've written down for you here
is actually the Heisenberg operator
because it has a time evolution.
But for a single mode, monochromatic electromagnetic
field, the time evolution is just e to the i omega t.

So if you want to obtain the Schrodinger operator
you have to eliminate the time dependence
by setting t equals zero.
At t equals zero, by definition, the Schrodinger operator
and the Heisenberg operator are the same.
But then in the Schrodinger picture,
the quantum state involves and the operator stays put.
In the Heisenberg picture, the quantum state stays constant
and you operator involves.
So, I am usually writing down for you
expressions for Heisenberg operators,
but if you simply set t equals zero in this expression,
you have the Schrodinger operator.

Questions?
Yes, Collin.
What happens if there's a factor of two that
shows up in the electric field of the single proton?
Like the [INAUDIBLE] I thought it looked like--
maybe it's just a factor of two--
Yes, you have to be careful what factors of 2.
I think I got it right here.
There is no extra factor of two.

But the question is-- OK, whenever you write down
equations for the electric field,
you want to use a and a degas photon number
or create photon numbers, so they are dimensionless.
So you always need a prefactor, which
has a dimension of the electric field.
And for the quantization of light, you have two choices.
One is I can take the electric field associated
with a single photon, that's one option.
The other option is I take, as a prefactor,
the electric field in the ground state, which you can say
is the zero point fluctuations of the electric field.
And the two differ by a factor of two
because the energy of the vacuum field
is one half h bar omega, whereas one quantum is h bar omega.
So that's one reason why you may sometimes find factors of 2
in some textbooks, or not in others,
but I think what-- I checked it.
I think as far as I know, it is consistent.
So [INAUDIBLE]

This is the Heisenberg operator for the electric field.
But it differs.
I mean at that point don't-- there is nothing to worry about
or get concerned about.
The question is for monochromatic light,
do we put the e to the i omega t into the state
or in the operator?
It's really just an e to the i omega
t factor has to be put either on the state, or on the operator.
These are the two choices.

Actually, my advice to you is don't even develop an intuitive
conceptual understanding, and whether you
are in the Schrodinger-- try to sort of see
the structure of the operator.
I sometimes like the omega t because it shows you
how the electric field evolves, and you see that explicitly.
But in the end, it's only when you do real calculation
that you have to think hard.
OK, which picture am I using now?
Which representation am I using now?
I'm probably not 100% rigorous in this course
when I write down quantum states and operators, whether I'm
always in the Schrodinger picture
or in the Heisenberg picture.
But by simply looking, does it have a time dependence?
Yes, it's a Heisenberg picture.
If I write down a quantum state which has a time dependence,
then I've chosen the Schrodinger representation.
It's as easy as that.
There was other?
There's a sign wrong up there because both a and a dagger
goes e to the i omega t, as you have written down.
Actually, I just wanted to say I was careful with my notes,
but I didn't carefully look at my notes.
This is what I have in my notes.
Does that satisfy you?
Yeah.
OK, good.
OK so, we want to talk about quantum states of light.
This simplest eigenstates are of course, the harmonic oscillator
eigenstates, but as I've already mentioned,
and we elaborate on it further, these are actually
non-classical states.
So quite often we don't want those states,
we want coherent states.
But we need those number of states as one representation.
So let me just introduce them.
So we label by n, the states with n photons,
and the energy of those states is n plus one half h bar omega.

The states are normalized.
Sometimes out of laziness, because it doesn't matter,
we can eliminate the one half.
It just make some equations more compact
and if I do that, I can write the Hamiltonian like this.
You are familiar, of course, with commutators
of a and a dagger and the only non-trivial matrix element
of a and a dagger is that it raises and lowers
the photon number by one.

OK, this was a sort review of the simple harmonic oscillator