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PROFESSOR: Today we have to
continue with our discussion

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of the hydrogen atom.

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We had derived or
explained how you

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would derive the corrections
to the original Hamiltonian,

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the Hamiltonian you've studied
already in a couple of courses,

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this h0 Hamiltonian
for the hydrogen atom

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that has a kinetic term
and a potential term.

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This Hamilton we've
studied for a while.

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And what we've said was that the
Dirac equation provided a way

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to get systematically all the
corrections, the first order

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of that Hamiltonian, and
that's what we have here.

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That was the end product of that
discussion in which the better

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Hamiltonian or the perturbed
Hamiltonian including more

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effects was the original one
plus a relativistic correction

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because this is not the
relativistic kinetic energy.

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Then there was a
spin orbit coupling

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in which you couple the
spin of the electron

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to the orbital angular
momentum of the vector.

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And finally, there
is a Darwin term.

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It's a surprising term.

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This is Charles Gaston
Darwin, a British physicist,

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that discovered this term.

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And all these terms were
suppressed with respect to h0,

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in the sense that
h0 had energies.

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The energies
associated to h0 went

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like alpha squared mc squared.

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The fine structure constant,
that's for h0 veera.

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But for h1 or for
delta h, the energies

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go like alpha to the
fourth mc squared,

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where m is the mass
of the electron,

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and that's about 20,000,
19,000 times smaller.

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So this is our
perturbation, and that's

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what you have to understand.

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So we'll begin with
the Darwin term.

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Then we'll turn to
the relativistic term,

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then to the spin orbit
term, and, by the end

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of today's lecture, we'll
put it all together.

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So that's our plan.

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So let's start with
the Darwin term.

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What is this term?

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So Darwin term, I will
try to evaluate it.

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So it depends on the
potential energy.

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It has the Laplacian of
the potential energy.

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So the potential energy
is minus e squared over r.

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So how much is the Laplacian
of the potential energy?

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Well, the Laplacian of v
would be minus e squared

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times the Leplacian of 1 over r.

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And that's minus e squared times
minus 4 pi delta function of r.

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That's something you study in
in E and M. The Laplacian of 1

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over r is related
to a delta function.

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It's the charge density
of a point particle

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that produces that potential.

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So our result here is that
this Laplacian is e square 4

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pi e squared delta of r.

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It's a delta function
contribution.

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So let's write it here.

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The delta h Darwin
is pi over 2 e

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squared h squared over m
squared c squared delta of r.

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All right, so that's
our correction.

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And we want to do
perturbation theory with it,

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how it affects the energy
levels of the various states

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of hydrogen. So how
are they changed?

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What does it do to them?

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Now, there is a simplifying
fact here, the delta function.

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So you remember, first
order corrections

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are always obtained by taking
the interactive Hamiltonian

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and putting states in it.

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And the first thing we notice
is that unless the wave

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function of the states
does not vanish at 0,

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the correction vanishes.

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So this will pick up the
value of the wave functions.

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When you have two
states, arbitrary

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state 1 and state 2 and
delta h Darwin here--

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there's two wave
functions that you're

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going to put here if you're
trying to compute matrix

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elements of the perturbation.

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And if either of
these wave functions

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vanishes at the origin,
that's not possible.

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But all wave functions
vanish at the origin,

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unless the orbital angular
momentum l is equal to 0.

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Remember, all wave
functions go like r

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to the l near the origin.

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So for l equals 0, that
goes like 1, a constant,

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and you get a possibility.

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So this only affects
l equals 0 states.

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That's a great simplicity.

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Not only does that, but
that's another simplification.

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Because at any energy
level, l equals

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0 states is just one of them.

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If you consider spin,
there are two of them.

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Remember in our table
of hydrogen atoms

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states go like that.

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So here are the l
equals 0 states.

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And those are the
only ones that matter.

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So the problem is
really very simple.

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We don't have to consider
the fact that there are

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other degenerate states here.

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We just need to
focus on l equals

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0 states because both
states have to be l equals 0

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and l equals 0.

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So our correction that we can
call e1 Darwin for n 0 0--

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because l is equal to 0, and
therefore m is equal to 0--

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would be equal to psi n zero
zero delta h Darwin psi n 0 0.

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So what is this?

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We have this expression
for delta h Darwin.

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So let's put it here, pi over
2 e squared h squared over m

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squared c squared.

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And this will pick up the
values of the wave functions.

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This overlap means integral
over all of space of this wave

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function complex conjugated,
this wave function and delta

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h Darwin.

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So at the end of the day,
due the delta function,

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it gives us just psi n 0
0 at the origin squared.

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That's all it is.

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Simple enough.

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Now, finding this
number is not that easy.

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Because while the wave
function here for l equals 0

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is simple for the
ground state, it already

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involves more and
more complicated

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polynomials over here.

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And the value of the wave
function at the origin

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requires that you normalize
the wave function correctly.

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So if you have the wave
function and you have not

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normalized it
properly, how are you

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going to get the value of the
wave function at the origin?

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This whole perturbation
theory, we always

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assume we have an
orthonormal basis.

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And indeed, if you change
the normalization--

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if the normalization
was irrelevant here,

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this number would change
with the normalization,

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and the correction would change.

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So you really have to
be normalized for this

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to make sense.

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And finding this
normalization is complicated.

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You could for a few problems
maybe look look up tables

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and see the normalized
wave function, what it is.

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But in general here,
there is a method

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that can be used to find this
normalization analytically.

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And it's something you will
explore in the homework.

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So in the homework
there is a way

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to do this, a very clever way,
in which it actually turns out.

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So in the homework, you will
see that the wave function

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at the origin for l
equals 0 problems is

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proportional to the
expectation value

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of the derivative of the
potential with respect to r.

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So it's something you will do.

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This potential, of
course, is the 1

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over r potential in our case.

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And the derivative
means that you

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need to evaluate the
expectation value of 1

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over r squared in this state.

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But the expectation values of 1
over r squared in the hydrogen

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atom is something you've already
done in the previous homework.

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It's not that difficult.

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So the end result is that
this term is calculable.

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And psi of n 0 0 at
the origin squared

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is actually 1 over
pi n cube a0 cube.

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With that in, one can
substitute this value

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and get the Darwin correction.

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You can see there is no
big calculation to be done.

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But you can rewrite it in
terms of the fine structure

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constant Darwin.

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And it's equal to alpha
to the fourth mc squared 1

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over 2n cubed.

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And it's valid for
l equals 0 states.

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So this number goes in
here, and then write

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things in terms of the
fine structure constant,

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and then out comes
this resolved.