PROFESSOR: OK.
Let's turn then to
the Dirac equation
and motivated, basically.
Through the Dirac equation
is the simplest way
to do perturbation theory
for the hydrogen atom.
It helps us derive what we
should think of perturbation.
So I'll discuss that quickly.
And there may be some of
that done in recitation.
So we're going to discuss
now the Dirac equation.
So the Dirac equation
begins with the observation
that we have E squared
minus p squared
c squared for a free particle
is m squared c to the fourth.
This dispersion relation that
rates E and p for any particle.
So if you wanted to
describe the dynamics
of a relativistic particle,
you could say, look,
the energy is the square
root of p squared c squared
plus m squared c to the fourth.
Therefore, I should take the
Hamiltonian to be that thing.
I should take H' to be that.
And work with a
Schrodinger equation
that has this H with
the square root.
Nobody does that, of course.
But you can do a
little bit with it.
If you notice that this is equal
to mc squared square root of 1
plus p squared over
m squared c squared.
And then expand.
In small velocities where
this is less than 1,
you get 1 plus 1/2
p squared over m
squared c squared minus
1/8 of this term squared.
So p squared p squared over m
to the fourth c to the fourth.
And then what do you get?
mc squared plus
1/2 2m p squared.
That's very nice.
Apart from the rest
energy, you get
the p squared over 2m
that you would expect.
And the next correction
is minus 18 p
squared p squared over m cube
c squared plus dot dot dot.
You know, that's one way you
could treat the Hamiltonian.
But it's very complicated.
Don't treat the
Hamiltonian too seriously
because the Schrodinger
equation would
have infinite number
of spatial derivatives.
p to the fourth, p to the
sixth, p to the eighth.
Never going to get
that understood.
But in terms of
perturbation theory,
we can think of p squared over
2m, your original Hamiltonian.
And here is a relativistic
correction that is small.
But Dirac was puzzled
by that square root.
He basically looked
at it and I think
he had sort of mathematical
inspiration in here.
He looked at this equation
and that square root
and he said, how
much would I wish
I could take that square root.
What can I do to take
that square root?
So he said, well, I could
take that square root.
This has p squared so c squared
plus m squared c to the fourth.
I could take that square
root if that would
be equal to something squared.
Which it's not because it's
missing the cross product
of the right amount.
And it's not there.
But if it would be
there, the cross product,
it would be a linear
function of p and mc squared.
So he says, OK, let me put
a linear function of p.
And the most general linear
function of p is obtained,
I want the c squared,
so c, by doing
the dot product of a vector
with p, a constant vector.
A constant vector times p is
the most general linear function
of p.
And here I'll output another
constant, mc squared.
And I hope I can
take the square root
because there will
exist some constants
and it somehow will work.
As of now, it cannot work
because it's always across
product.
But just follow these dots.
And then alpha dot p.
This is equal to c alpha
1 p1 plus c alpha 2 p2
plus c alpha 3 p3
plus beta mc squared.
So let's list all the
things that should
happen for this to work out.
Well, this square includes
the square of this,
the square of this,
the square of this,
and the square of this.
And those are what we want.
So we should have
that alpha 1 squared
is equal to alpha 2
squared is equal to alpha
3 squared is equal to beta
squared is equal to 1.
If that happens, well,
this term will give you
c squared p1 squared,
c squared p2 squared,
c squared p3 squared, and
m squared c to the fourth,
and I get everything to work.
But the cross products
must also work.
So for example, the cross
product of p1 with p2
should vanish.
And that, well, if you
ride these two factors like
that here you would say,
oh, what they need now
is that alpha i times alpha
j plus alpha j times alpha i
should be 0 for all
i different from j.
So for example, alpha 1,
alpha 2 plus alpha 2 alpha 1,
when you do the
product, should vanish.
And I kept the order
of these things there.
When I'm starting
to think that maybe
numbers is not going to work.
So i difference from j.
This must happen too
for this to be equal.
And moreover for
all i alpha i beta.
There shouldn't be cross
products between these ones
and this term either.
Plus beta alpha i is equal to 0.
And that's all
that should happen.
If that happens, we can
take the square root.
But if these things
are numbers, well, this
could be phases or 1 or minus 1.
And that's never going to work.
This is twice this product.
So [INAUDIBLE] OK,
they're matrices.
But that's not bad,
if they're matrices,
we'll have like spinors,
like Pauli was doing.
And we'll work on spinors.
But it turns out that this
one, the simplest solution
is something with
4 by 4 matrices.
So the solution is alpha
is 0 sigma sigma 0.
And beta is 1 minus 100.
This is the simplest
solution, 4 by 4 matrices.
So that Dirac Hamiltonian, H
Dirac, is equal to the energy.
And the energy was the
square root of this.
So it's just this thing.
So it's c alpha dot
p plus b mc squared.
That's our Dirac Hamiltonian.
But we need to deal with spinors
that now are four dimensional.
These matrices,
alpha and beta, are
four dimensional because
the segments that are inside
are two dimensional.
And these are 2 by 2 matrices.
So when you write the Dirac
equation as i H bar d psi dt
equal H Dirac psi, psi is a
four component thing, a chi
and a phi.
And what people
discover is that chi
behaves like the Pauli spinor.
And this one is a small
thing, a small correction.
So there is a whole analysis of
this system in which, in order
to put the magnetic
field, you can put in here
for the Dirac
Hamiltonian, H Dirac,
coupled to electromagnetism
has a c alpha p plus e over
ca plus beta mc squared.
And you still have to
add the potential, V
of r, that comes from
[INAUDIBLE] minus e squared
over r.
It's the value of
the electron charge
that is the potential of r.
The scalar potential phi of r.
So this is the
Dirac Hamiltonian.
And you have these things.
And this is something
you can read
if you're interested
in [? Shankar. ?] There
is a derivation of the Pauli
equation for the electron.
Which is the
spinors that you are
familiar with by eliminating
in a recursive expansion phi.
And it may be that
Professor Metlitski
is going to go through
that in recitation as well.
So what happens?
What you get is that you
get an H chi equal E chi.
And H is what?
Here is the grand prize.
From the Dirac equation, we
can rewrite the Hamiltonian
of the hydrogen atom in a more
accurate way, a more complete
Hamiltonian.
And it has p squared
over 2m plus V of r,
which is what we've called H0.
It's there.
It's the first term.
That's what you would expect.
Then there is that term
that we anticipated.
Some relativistic
correction here.
So it comes like minus p to the
fourth over 8m cubed c squared.
And we call this
delta H relativistic.
And then you get this
term that corresponds
to spin orbit coupling
that also shows up.
2m squared c squared 1 over r
dv dr. You've studied this thing
and you gave a heuristic
explanation for it.
That's another term.
It's kind of difficult
to get the right value
by a simple argument.
You always get it
off by a factor of 2
that is associated
to Thomas precession.
But here it comes out directly
with the right number.
This is called spin
orbit coupling.
Delta H of spin orbit.
And there's one more term.
Plus H squared to this
[INAUDIBLE] 8 m squared
c squared Laplacian of V.
Which is called of this V of r.
It's called the Darwin term.
Delta H Darwin.
Not the biologist.
Some Darwin.
So the job is set for us.
We have to do
perturbation theory.
Now you remember that all
the terms associated with H0
were of the form energies
with alpha squared mc squared.
This was like H0.
If you look at the units
of all of these terms,
we looked at them in the notes.
Delta H. They all go like
alpha to the fourth mc squared.
So there is a difference
of alpha squared there.
1/19000 smaller.
That is the fine structure
of the hydrogen atom.
And our task next time will be
to compute the effect of this
and see what happens with all
the levels of the hydrogen
atom.
This is our
[? final ?] structure.
So we'll do that next time.
