PROFESSOR: Today we have to
continue with our discussion
of the hydrogen atom.
We had derived or
explained how you
would derive the corrections
to the original Hamiltonian,
the Hamiltonian you've studied
already in a couple of courses,
this h0 Hamiltonian
for the hydrogen atom
that has a kinetic term
and a potential term.
This Hamilton we've
studied for a while.
And what we've said was that the
Dirac equation provided a way
to get systematically all the
corrections, the first order
of that Hamiltonian, and
that's what we have here.
That was the end product of that
discussion in which the better
Hamiltonian or the perturbed
Hamiltonian including more
effects was the original one
plus a relativistic correction
because this is not the
relativistic kinetic energy.
Then there was a
spin orbit coupling
in which you couple the
spin of the electron
to the orbital angular
momentum of the vector.
And finally, there
is a Darwin term.
It's a surprising term.
This is Charles Gaston
Darwin, a British physicist,
that discovered this term.
And all these terms were
suppressed with respect to h0,
in the sense that
h0 had energies.
The energies
associated to h0 went
like alpha squared mc squared.
The fine structure constant,
that's for h0 veera.
But for h1 or for
delta h, the energies
go like alpha to the
fourth mc squared,
where m is the mass
of the electron,
and that's about 20,000,
19,000 times smaller.
So this is our
perturbation, and that's
what you have to understand.
So we'll begin with
the Darwin term.
Then we'll turn to
the relativistic term,
then to the spin orbit
term, and, by the end
of today's lecture, we'll
put it all together.
So that's our plan.
So let's start with
the Darwin term.
What is this term?
So Darwin term, I will
try to evaluate it.
So it depends on the
potential energy.
It has the Laplacian of
the potential energy.
So the potential energy
is minus e squared over r.
So how much is the Laplacian
of the potential energy?
Well, the Laplacian of v
would be minus e squared
times the Leplacian of 1 over r.
And that's minus e squared times
minus 4 pi delta function of r.
That's something you study in
in E and M. The Laplacian of 1
over r is related
to a delta function.
It's the charge density
of a point particle
that produces that potential.
So our result here is that
this Laplacian is e square 4
pi e squared delta of r.
It's a delta function
contribution.
So let's write it here.
The delta h Darwin
is pi over 2 e
squared h squared over m
squared c squared delta of r.
All right, so that's
our correction.
And we want to do
perturbation theory with it,
how it affects the energy
levels of the various states
of hydrogen. So how
are they changed?
What does it do to them?
Now, there is a simplifying
fact here, the delta function.
So you remember, first
order corrections
are always obtained by taking
the interactive Hamiltonian
and putting states in it.
And the first thing we notice
is that unless the wave
function of the states
does not vanish at 0,
the correction vanishes.
So this will pick up the
value of the wave functions.
When you have two
states, arbitrary
state 1 and state 2 and
delta h Darwin here--
there's two wave
functions that you're
going to put here if you're
trying to compute matrix
elements of the perturbation.
And if either of
these wave functions
vanishes at the origin,
that's not possible.
But all wave functions
vanish at the origin,
unless the orbital angular
momentum l is equal to 0.
Remember, all wave
functions go like r
to the l near the origin.
So for l equals 0, that
goes like 1, a constant,
and you get a possibility.
So this only affects
l equals 0 states.
That's a great simplicity.
Not only does that, but
that's another simplification.
Because at any energy
level, l equals
0 states is just one of them.
If you consider spin,
there are two of them.
Remember in our table
of hydrogen atoms
states go like that.
So here are the l
equals 0 states.
And those are the
only ones that matter.
So the problem is
really very simple.
We don't have to consider
the fact that there are
other degenerate states here.
We just need to
focus on l equals
0 states because both
states have to be l equals 0
and l equals 0.
So our correction that we can
call e1 Darwin for n 0 0--
because l is equal to 0, and
therefore m is equal to 0--
would be equal to psi n zero
zero delta h Darwin psi n 0 0.
So what is this?
We have this expression
for delta h Darwin.
So let's put it here, pi over
2 e squared h squared over m
squared c squared.
And this will pick up the
values of the wave functions.
This overlap means integral
over all of space of this wave
function complex conjugated,
this wave function and delta
h Darwin.
So at the end of the day,
due the delta function,
it gives us just psi n 0
0 at the origin squared.
That's all it is.
Simple enough.
Now, finding this
number is not that easy.
Because while the wave
function here for l equals 0
is simple for the
ground state, it already
involves more and
more complicated
polynomials over here.
And the value of the wave
function at the origin
requires that you normalize
the wave function correctly.
So if you have the wave
function and you have not
normalized it
properly, how are you
going to get the value of the
wave function at the origin?
This whole perturbation
theory, we always
assume we have an
orthonormal basis.
And indeed, if you change
the normalization--
if the normalization
was irrelevant here,
this number would change
with the normalization,
and the correction would change.
So you really have to
be normalized for this
to make sense.
And finding this
normalization is complicated.
You could for a few problems
maybe look look up tables
and see the normalized
wave function, what it is.
But in general here,
there is a method
that can be used to find this
normalization analytically.
And it's something you will
explore in the homework.
So in the homework
there is a way
to do this, a very clever way,
in which it actually turns out.
So in the homework, you will
see that the wave function
at the origin for l
equals 0 problems is
proportional to the
expectation value
of the derivative of the
potential with respect to r.
So it's something you will do.
This potential, of
course, is the 1
over r potential in our case.
And the derivative
means that you
need to evaluate the
expectation value of 1
over r squared in this state.
But the expectation values of 1
over r squared in the hydrogen
atom is something you've already
done in the previous homework.
It's not that difficult.
So the end result is that
this term is calculable.
And psi of n 0 0 at
the origin squared
is actually 1 over
pi n cube a0 cube.
With that in, one can
substitute this value
and get the Darwin correction.
You can see there is no
big calculation to be done.
But you can rewrite it in
terms of the fine structure
constant Darwin.
And it's equal to alpha
to the fourth mc squared 1
over 2n cubed.
And it's valid for
l equals 0 states.
So this number goes in
here, and then write
things in terms of the
fine structure constant,
and then out comes
this resolved.
