M3L12b.txt

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I want to give you a very brief derivation
of the canonical coupling between electromagnetic fields
and atoms.
I know this is covered in many textbooks,
and we have a very deep discussion
about it using the full QED formalism in 8.422.
But I feel course, 8.421 would not
be complete without a discussion like that.
And secondly, this is so important.
If you hear it twice from two different angles,
it's probably useful.
So I just want to remind you that if you
use classical electromagnetism and the Lagrangian formalism,
you find the very elegant result that the coupling
to the electromagnetic field can be introduced by modifying
the momentum that the canonical momentum, which appears
in the Lagrangian equation.
It's no longer the mechanical momentum which gives rise
to kinetic energy, but it is modified
by the vector potential.
So therefore the Hamiltonian can be-- at this point,
it's a classical Hamiltonian, but then we
use the same expression in quantum mechanics.
The Hamiltonian has kinetic energy and potential energy.
But the kinetic energy is no longer
p squared, because by p I mean now the canonical momentum.
We have to correct for the vector potential
to go from the canonical momentum
to the mechanical momentum, and if you squared it,
this gives the kinetic energy.
If you worry about the sum here and assume now
that the electron has a charge, the charge q of the electron
is minus e.
So that's why there's a plus sign here,
and not a minus sign.
We will use this fact that we have to substitute the momentum
operator the canonical momentum, which is no
longer the mechanical momentum.
We use it also in our quantum mechanical Hamiltonian
and the Schrodinger equation.
If you ask me, can you prove it?
No.
Nobody can prove that.
We can never prove what a quantum mechanical equation is.
We can just use physical understanding,
physical analogies-- and that's how quantum mechanics was
developed-- and say, based on the classical quantum
mechanical correspondence, this seems
to be a very reasonable assumption.
And ultimately, this very reasonable assumption
has now withstood the test of time for many, many decades,
or for almost 100 years.
So that's why we assume that doing the same substitution
in the Schrodinger equation for the momentum operator
gives the right result. But there is no way
how you can rigorously derive fundamental equations
in physics.
You can observe nature, postulate them,
and then define them.
So therefore, what is this line, which
was the classical Hamiltonian, by interpreting p
as an operator is now our quantum mechanical Hamiltonian.
And the one thing we have to consider now
is that once we have a quantum mechanical operator,
we have to be careful about commutator,
that certain quantities have to be ordered.
It matters which one you put first.
What is convenient, now, is to use the Coulomb gauge.
In the Coulomb gauge, divergence of A is 0.
But the del operator, the derivative operator,
is the operator of the canonical momentum.
And therefore, if you use the Coulomb gauge,
then we have commutativity between the canonical momentum,
p, and the vector potential.
And that's sort of nice.
Because if you take the square and square it out,
you get p dot A plus e dot P. But they are
the same in the Coulomb gauge.
So therefore, our Hamiltonian has now the following terms.

This is-- we call it H0 and we assume
that the Coulomb potential for the nucleus,
which eventually gives the electronic structure
of the atom, is included here.
So this is the Hamiltonian part, which describes
the structure of atoms.
And then we have two terms which couple
to the electromagnetic field.
So this is p dot A, or A dot p.
It doesn't matter in the Coulomb gauge.
We call this the interaction Hamiltonian.
And now we get a second-order term,
which is the A squared term.

Since it's second-order, we designate it
where we use the symbol H2.
In the following, we neglect the second-order term.
The argument is for weak fields, it doesn't matter.
It's not so important because it's higher-order.
However, in 8.422, we look at it more carefully.
And we can actually do a canonical transformation which
eliminates the A squared term.
So to drop it, it's not necessary,
you can actually show with a canonical transformation
that what I'm doing is actually more exact than just saying,
it's small and I eliminate it.
But anyway, I don't want to spend too much time on it.
All I wanted to remind you, what are the steps to obtain-- just
give me one second-- to obtain the interaction Hamiltonian?
And that's what I want to use for the remainder
of this class.
That we have the coupling of momentum
to the vector potential.
We assume that the vector potential--
we are not quantizing the electromagnetic field;
it's a semi-classical field-- is therefore a classical vector.
And we will investigate, what is now the coupling
between an atom through this Hamiltonian
to a plane wave of the electromagnetic field?
So therefore, we are introducing for the vector potential
the expression that the vector potential has an amplitude A0,
the polarization e hat, and then the plane wave vector e
to the iKr minus i omega t.
So this can be derived in a much more rigorous way,
but this is now the starting point of our discussion.
[INAUDIBLE]
OK.
OK.
So we want to talk about matrix element.
So the matrix element between two states
due to the interaction Hamiltonian for one plane wave
has the e to the minus i omega t dependence, which
is trivial in terms of we've assumed that there's
a monochromatic wave.
And what you want to discuss now is
this time-independent vector, Hba.

Let me get the vectors.
Let me get the polarization.
So the relevant operator, which connects
the two states a and b, is the momentum operator.

And now, for the plane wave, we want
to do an expansion in orders of Kr.
The leading term will be the dipole approximation,
and the next-order term will give rise
to magnetic dipole and electric quadrupole transitions.
So first, the fact that I can do this extension
requires that K dot a is much smaller than 1.
So it's a long-wavelength approximation,
which you can see is valid as long
as the atom is smaller than the wavelengths
we are talking about.
And I will show you later when I discuss
in its higher-order term that for atoms in the ground state,
this is actually also an expansion of alpha.
So again we retrieve the fine structure constant.
So when we do this expand of this plane wave exponential,
every term is smaller than the previous term
by the fine structure constant, alpha, which is 1/137.
The leading term is the 1, and this gives rise
to the dipole approximation.
So what I want to show you now in one or two
minutes is that this is indeed the dipole approximation.
It leads to a matrix element, which
is electric field times r, the dipole moment.
What we have right now is A dot p, the vector potential
times the momentum.
So in other words, I want to show you
that the A dot p matrix element is equivalent to an e dot r
matrix element.
This can be done with a canonical transformation, which
we do in 8.422.
But here let me just show you the elementary discussion.
You want to replace the vector potential
by the electric field.
So therefore, we use the fact that the electric field
is the derivative of the vector potential.
And the derivative of the vector potential
yields, since we have an e to the minus i omega
t dependence, that we simply multiply
with a factor of omega, of the frequency.

And what we obtain here is, then, the amplitude
of the electric field E0.
So with that, we have a matrix element
which now involves the electric field, but still
the matrix element of the momentum operator.

But we can easily go from the momentum operator
to the precision operator by taking the fact
that the momentum operator is nothing
else than the commutator of r with the Hamiltonian H0.

The kinetic energy in H0 is p squared.
And the commutator of r with p squared is simply p.
So this is what I'm using here.
And these are the prefactors.

So therefore, if you take the momentum operator between two
states, we have this relation between the momentum operator
and the precision operator.
But we have the momentum operator with the commutator.
And r, H0 gives us r, but when H0 acts on b,
which gives the energy Ea.
And for the other part of the commutator, which is H0, r,
you can have H0 act on the left hand side of b.
And this gives us the energy Eb.

Dividing by H bar, what we get it
is the energy difference Eb minus Ea,
or in frequency units, omega ba.
So this is now how we have implemented the commutator.

So that by inserting this into this equation,
we now finally obtain our result for the matrix element
in the dipole approximation.
We have e, the electric field amplitude polarization,
the matrix element b r a, omega ba.
And here we have 1 over omega in the denominator,
which came from the derivative of the vector potential, dE dt.

So this here is the matrix element of the dipole operator.

So in other words, I have derived for you
that the interaction matrix element
in the dipole approximation is the dipole operator
times the electric field as long as I
make the approximation that this frequency factors
on the order of unity.
Now this is clearly the case near resonance.

So you should be safe.

However, if you do the more rigorous derivation
of the dipole Hamiltonian using canonical transformation,
this factor of omega ba over omega does not appear.
So again, the dipole approximation
is better than I presented today.
The true approximations, dropping the a squared term
and approximating this ratio of frequencies with 1,
is not necessary if you use the other methods using
canonical transformation.

Any questions about that?

Let me make one comment.
I've shown you in this derivation
that the A dot p term, the A dot p
interaction within the dipole approximation,
is identical to the e dot r interaction.
So these are the same operators and if you would exactly
calculate an atomic structure calculation,
you would get the same coupling and the matrix element.
However, in practice, there are important differences.
Because the e dot r operator, the r operator,
has a lot of weight where the electron is
far away from the nucleus, because it's multiplied with r.
Whereas the p operator emphasizes the derivative
of the wave function, and this is usually
strong at close distances.
So in the end, the results have to be the same.
But if you make an approximation to your wave function,
one formulation may be numerically much more accurate
than the other one.
But fundamentally, the two terms are the same,
and they are related the way how I derived it for you.
Yes?
When we say it's the dipole approximation,
does it mean that it's a long wavelength or the small field
or everything together?
So what's the absolute fundamental assumption?
The absolute fundamental assumption
of the dipole approximation is that we
assume-- let me just scroll up.

In the equation above, we had an expression
for the vector potential.
And the dipole approximation means
that we can neglect the position dependence of the vector
potential A of r.
So we approximate A of r as being A evaluated
at the origin of the atom.
And although they extend over the size of the atom,
we do not have to consider the spatial dependence
of the vector potential.
So this is sufficient.
The other thing's that A squared is small,
and that when near resonance, I have
to make those assumptions if I want to use
this elementary derivation.
But with the canonical transformation
as is discussed in atom-photon interactions,
you do not need those additional assumptions.
So the only assumption behind the dipole approximation
is that one.
And this means that the extent of the atom
is much smaller than the optical wavelengths.