M2L8d.txt

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We are now discussing the most important effect
due to the hyperfine structure.
And this is the fact that the nucleus has a magnetic moment,
and this magnetic moment couples to the magnetic field.
Even if you don't apply an external magnetic field--
we talked about that on Wednesday-- there
is an internal magnetic field created
by the electron with total angular momentum J.
So this is so to speak, the Zeeman Hamiltonian
of the nucleus in the magnetic field created by the electron.
And I will show you quickly in a simple derivation,
what the result of that is, but before I
do that, I also want to point out
that there's an alternative.
Right now, we said the nucleus experiences the magnetic field
created by the electron, but we can also
take the other approach.
The nucleus creates a vector potential
because of its magnetic moment, and the electron
which goes around the nucleus is not only
feeling the Coulomb potential, but also feeling a vector
potential.
And, of course, both different perspectives,
whether the electron moves in the magnetic field
of the nucleus or the nucleus experiences
the magnetic field of the electron,
both treatments have to agree.
I follow the more standard treatment
but the alternative treatment where the electron moves
in this electric and magnetic potential of the nucleus
is fully elaborated on the atomic physics Wiki.
So alternatively, electron moves in the potential
of the nucleus, which is the Coulomb potential.
We've discussed that, but then there is also
vector potential, created by the magnetic moment of the nucleus.
So you simply assume this is a potentially created
with a nucleus, and then you just
solve Schrodinger's equation.
And this approach is carried out on the Wiki.
However, since it's a little bit more standard
and there's an easy semi-classical derivation,
let me now discuss this one, because what
I like about it is it addresses one intuitive quantity,
namely the fact that there is an internal magnetic field.
We're not just solving the Schrodinger equation
for the whole system.
We're really estimating what is the magnetic field which
the electron creates at the position of the nucleus.
So let's now do a semi-classical derivation
of this internal magnetic field.
And I would immediately tell you this derivation
agrees quantitatively with a fully
quantum mechanical treatment.
So as often in these semi-classical derivations,
we have to separate two parts.
There are two ways how the electron
creates a magnetic field at the nucleus.
One is due to its orbital motion.
The electron is a ring current and creates a magnetic field.
But then the electron has magnetic moment
due to its spin.
This spin part is simply the potential of a magnetic dipole.
A unit vector where the magnetic dipole moment of the electron
is proportional to its spin, with a g-factor.
However-- and you can find that in all textbooks
on classical electrodynamics-- but there is one important term
which we have to add here which is also part of classical E&M,
and this is the delta function contribution.
You've probably seen it.
You'll find it in Jackson.
If you haven't, the model is that you
can assume that a magnetic moment is created by a ring
current and the ring current creates a magnetic moment
and you have the dipole potential
due to the magnetic moment.
However, if you have a ring current,
you can also ask what is the magnetic field
inside the current loop, and then
eventually do the transition where you allow
the current loop to go to zero.
That's how you make a point model of a magnetic dipole,
but what remains is sort of, as a delta function,
the location inside the loop.
I'm emphasizing it because it will be eventually
the delta function contribution, which
is important for S electrons.
And therefore, it is this contribution
which is the dominant effect in many situations.

OK.
So this is the magnetic field created.
It's a classical expression, but it's
valid in quantum mechanics, the expression
for the magnetic field created by the spin.
The second contribution is the orbital contribution.
And for that semi-classical, we just use Biot-Savart.

So Biot-Savart is usually the three dimensional integer
over the current density.
The volume integral or you can rewrite it as the current I d r
cos r over r cubed.
And eventually, if you now put in the electron charge
distribution, velocity cos r.
Well, velocity cos r means we get-- and that's what we want,
the orbital angular momentum.
The 1 over r cubed term means we have
to calculate an expectation value of the wave function
which is 1 over r cubed.
And the prefactor leads us-- it's
nothing else-- than 2 times the Bohr magneton.

So with those two terms, we can now obtain our final expression
for the total magnetic field generated
by the electron at the origin.
And for that, I use the g-factor of 2
as it comes out of Dirac theory.
So now we have the total magnetic field.
We had this contribution L over r cubed.
If you inspect the dipole potential of this spin,
it has a contribution S over r cubed,
then it has this second contribution
to the dipole potential.
And finally, most importantly for the following discussion,
the delta function contribution which I discussed earlier.

If you have an S state, these first terms are 0, for L=0.
Because these are sort of terms which
have a dipole potential where positive and negative
contributions cancel out when you
do a spherical average and an S electron
performs a spherical average.
So it is 0 for L=0 due to the spherical average,
whereas the second part-- it would be 0 for L non-equals
to 0, because the probability for a non  Selectron to be
at the nucleus is 0.
So pretty much, this describes that.
So this describes the hyperfine structure.
Well, it describes the magnetic field created by the electron
and now we have to do the usual projection
in the following way: that the hyperfine structure is
the Zeeman Hamiltonian of the internal magnetic field
with a magnetic moment of the nucleus.
And the magnetic moment of the nucleus
is proportional to the angular momentum of the nucleus.
Sort of this argument even if it were not proportional,
it would rapidly precess and eventually be projected,
and the only direction which survives
is the direction of the angular momentum.
And similarly, the magnetic field--
you have a contribution of S and L, but S
and L rapidly precess around the resultant angular momentum J,
and therefore as a result, the internal magnetic field must
be, can only be, parallel to the angular momentum J.
If you do a fully quantum treatment,
it comes out immediately.
But if you do it semi-classically,
you calculate a magnetic field, you sort of
have to throw in this argument that you always
project on the axis of angular momentum.
And that means that the hyperfine interaction will
be the Hamiltonian [INAUDIBLE] or the operator will be the dot
product of our I dot J. For fine structure,
we had L dot S. For hyperfine structure, we have I dot J.
This is always how we couple angular momentum,
with a dot product.
The hyperfine constant goes by the letter, a.
And since, historically, a is measured
in frequency units in hertz, I have
to put in h Planck's quantum.
No, it's not h bar.
For historical reasons, it's h.
Yes?
Are I and J dimensionless, or will they
carry units in h bar?

Here, they are dimensionless.
Thanks for the question.
Because h is in frequency units.
It's in hertz and if you multiply with h,
we have an energy.
So therefore, I and J measure the angular momentum
in unit of h bar.
So it's not in that sense, it's a normalized angular momentum
operator.
The quantum numbers of I and J are not one half or one h bar.
It's just one half or one.
Other questions?

OK.
I can now take the expression with L and S and S dot r,
and evaluate it further, but I feel
I'm not providing any insight and you
can read about it on the Wiki.
So for non-S states, how to simplify this expression,
and get the final textbook result, I refer to the Wiki.
I want to discuss the most important part, namely
for S electrons, because hydrogen, all the alkalis,
have an S ground state.
So in that case, all we have to consider
is the delta function part, and if you
project the magnetic field onto the angular momentum axis,
we get the probability of the S electron to be at the origin.
And therefore, for S states, the hyperfine constant
is-- oh, I forgot one thing.
We have to parametrize the magnetic moment of the nucleus.
And that is done by using a nuclear magneton.
It's the same as the Bohr magnetron
where you have replaced the electron mass by the proton
mass.
And you have the nuclear g-factor.
Just as a reminder, the g-factor of the proton is 5.6.
The g-factor of the neutron is minus 3.8,
so the g-factor has nothing to do--
it's not even close to the factor 2,
which we obtained in the Dirac equation for the electron.
That just shows that the nucleons, protons and neutrons
are more complicated.
Well, they have quarks inside.
They have a complicated internal structure.
So therefore, the hyperfine constant involves now
the g-factor of the nucleus, the product of the nuclei magneton
with the Bohr magneton. And for hydrogen, this gives the famous result of 1420 MHz.