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PROFESSOR: So we'll look at
the Schrodinger equation.

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We're going to work
for bound states.

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We could work with
scattering states

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in spherical coordinates,
but it's usually

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done in advanced courses.

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The most important
thing is the calculation

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of the bound states.

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This is kind of a neat
part of quantum mechanics

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because, in a sense, you get
accustomed in quantum mechanics

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about uncertainties.

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You cannot predict the
probability that the photon

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will go through this branch
of the interferometer or this

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other probability.

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You can't be sure and all this.

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But here you get
the energy levels.

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And you get the
exact energy level.

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So it's a nice thing that,
in quantum mechanics,

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energy levels are things
that you calculate exactly.

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Of course, when
you try to measure

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energy levels experimentally,
the uncertainties arise again.

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You get the photon
of some energy

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and then there's some energy
time uncertainty or something

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like that, that can bother you.

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But the end result
is that these systems

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have beautifully the
term in fixed numbers

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called energy levels.

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And that's what we're aiming at.

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It's a nice thing because it's
the most physical example.

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It has many applications.

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The energy levels
of hydrogen atom.

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The more you study,
the more complicated

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they are because you can
include fine effects,

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like the effects of the
spins of the particles.

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What does it do?

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The specs of relativity.

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What does it do?

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All kinds of things
you can put in and do

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more and more accurate results.

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So it's really
unbelievable how much

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you can learn with the fine
spectrum of the hydrogen atom.

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Our equation is minus h squared
over 2m, d second dr squared--

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the radial equation--
plus h squared l times l

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plus 1 over 2mr squared
minus ze squared over r,

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u is equal to Eu.

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So this is the radial equation.

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Remember it was
like the Schrodinger

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equation for a variable u.

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And the wave function was u
over r times psi lm or Ylm,

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actually, a spherical harmonic.

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Here I could have
labeled u with E and l

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because could
certainly u depends--

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u is u of r.

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And u depends on the energy
that we're going to get.

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And it will depend
on l that is there,

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in the differential equation.

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This is the effective potential
that we discussed before.

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It was the original
potential to which you

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add the centrifugal barrier.

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And what are you
supposed to solve here?

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You're supposed to
solve for l equals 0.

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To find some states
for a, go one, two,

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three, four, infinity.

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You should find all the
energy levels of this thing.

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So the wave function is
psi of r theta and phi

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will be this u of r over r
times Ylm of theta and phi.

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And, in fact, plugging that
into the Schrodinger equation

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was what gave us
this radial equation.

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So this was called
the radial equation,

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which we talked about,
but never quite solved it

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for any particular example.

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OK.

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As you know, we
like, in this course,

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to get rid of units
and constants,

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so our first step is to
replace r by a unit free x.

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And we have the right quantity.

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a0.

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a0 would be the perfect thing.

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So you-- if you
do that, you will

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be able to clean up the units.

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But if you do that,
you might not quite

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be able to clean up the
z from all the places

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that you would like
to clean it up.

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So we can improve that--

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maybe it's not too obvious.

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--by putting a 2 over z Here.

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z has no units, of course,
so that you would probably

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do by trial and error.

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You would say, well I want
to get rid of the z as well.

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And I have it in this form.

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And you will see
how it works out.

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So at this moment, I have
to plug r into this equation

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and just clean it up.

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And what should happen is that
everything, all the units,

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should give you a factor
with units of energy.

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Because at the end of the--

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whatever is left, it's
not going to have units.

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Everything out must
have units of energy.

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So this takes about
one line to do.

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I'll skip some algebra
on these things.

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I'll try to post
notes soon on this.

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So if you leave a
line you could do

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that calculation for yourselves
and it might be worth it.

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The claim is that
we get 2z squared e

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squared over a0,
multiplying minus d second,

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dx squared plus l times l
plus 1 over x squared minus 1

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over x, times u, equals Eu.

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That will be the
common factor that

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will come out of everything
by the time you solve it.

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And it has the units of
energy, as you can imagine.

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e squared over a0
has a unit of energy.

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So might as well move
that to the other side

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to get the final form
of the equation, which

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would be minus d second,
dx squared plus l times l

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plus 1 over x squared,
minus 1 over x, u

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is equal to minus
kappa squared u.

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Where kappa squared is
going to be minus E over 2z

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squared little e
squared over a0.

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OK.

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The trick on all
these things is to not

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lose track of our variables.

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And that-- it's a
little challenging.

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So let me just re-emphasize
what's going on here.

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We've passed from r
variables to u variables,

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that we will still
do more things.

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And we have a0 in
there, z in there,

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and then the energy
is really encapsulated

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by kappa squared here.

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And this kappa is unit free.

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So if you know kappa,
you know the energies.

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That's what you
should be looking for.

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Kappa is the thing you
want to figure out.

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Knowing kappa is the same
as knowing their origins

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because this is just a
constant that gives you

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the scale of the energy.

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So this is what
we want to solve.

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And the equation doesn't
look all that complicated,

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but it's actually not yet
that simple, unfortunately.

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So we have to keep
working with it a bit.

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So one reason you can see
that an equation like that

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would not be too
simple, is to look at it

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and see what kind of recursion
relation it would give you.

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And that's a good thing
to look at the beginning.

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And you say, OK here I'm going
to lower the number of powers

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by two.

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Here I'm going to lower the
number of powers by one.

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And here I'm not going to
lower the number of powers.

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You're gonna have three terms.

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So it's not that simple
recursion relation,

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which the next term is
determined by the previous one.

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So that suggests you
better do some work still

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with this equation to
simplify the situation.

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And several things that
you can do-- one thing

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is to look at the behavior of
the equation, near infinity,

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near zero, and see if you
see patterns going on.

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One thing we're
going to do is to--

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which is not urgent,
but it's usually done.

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And people do it
in different ways.

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--is to look at x goes
to infinity and see what

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the solutions may look like.

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So as x goes to infinity, the
differential equation probably

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can be approximated to keep this
storm to see how things vary.

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And x goes to infinity.

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Throw this, throw
this, and keep that.

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So you would have d
second u dx squared is

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equal to kappa squared.

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So this suggests
that u goes like e

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to the plus minus kappa x.

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So exponential behavior.

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e to the plus minus kappa x.

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Ideally, of course,
for our solutions,

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we would like the
minus one, but we

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will see what the equations do.

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Now this suggests, yet another
transformation that people do,

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which is--

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look, kappa is
dimensionless and we

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had x that is unit free also.

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So kappa has no
units, x has no units.

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So let's move to yet
another variable.

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We started with r being
proportionate to x.

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And now we can put factors that
may help us without units here.

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So I'm not suggesting
that this is something

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that would occur to me,
if I'm doing this problem,

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but it's certainly a
possible thing to do.

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To say, OK, I'm going to
define now rho as kappa x.

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And with x being
given by this, rho

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would be 2 kappa z over a0 r.

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So rho is going to
be my new coordinate.

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I'm sorry, we've gone to x.

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And now we've gone to rho,
a new variable over here.

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So what happens to the
differential equation?

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Well, it's going to
be a little better,

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but in particular the solutions
may be a little better.

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But here it is.

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If you look at the
differential equation.

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The differential equation here.

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Think of moving the
kappa squared here below.

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And then you see, immediately,
it fits perfectly well

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00:13:26,330 --> 00:13:27,660
in the first two terms.

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So you get minus d second, d rho
squared plus l times l plus 1,

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00:13:37,150 --> 00:13:38,690
rho squared.

201
00:13:38,690 --> 00:13:42,290
And here it doesn't
fit all that well.

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00:13:42,290 --> 00:13:51,800
You would have 1 kappa
leftover, so 1 over kappa rho, u

203
00:13:51,800 --> 00:13:53,230
equals minus u.

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Yes it's kind of suggestive.

205
00:14:06,440 --> 00:14:09,940
The kappa has almost
disappeared from everywhere,

206
00:14:09,940 --> 00:14:13,410
but it better not
disappear from everywhere.

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00:14:13,410 --> 00:14:15,385
If it would have
disappeared from everywhere

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00:14:15,385 --> 00:14:18,670
we would have been in problems
because the equation would've

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00:14:18,670 --> 00:14:20,500
not fixed the energy.

210
00:14:20,500 --> 00:14:24,520
You see we're hoping that the
differential equation will have

211
00:14:24,520 --> 00:14:27,760
additional solutions for
some values of the energy

212
00:14:27,760 --> 00:14:29,470
and for the others no.

213
00:14:29,470 --> 00:14:31,780
So the energy better
not disappear.

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00:14:31,780 --> 00:14:34,840
It came close to
disappearing, the kappa,

215
00:14:34,840 --> 00:14:39,850
but it's still here,
so we're still OK.

216
00:14:39,850 --> 00:14:46,410
And now, you will
say, OK, this is nice.

217
00:14:46,410 --> 00:14:52,050
If you look at rho
going to infinity again.

218
00:14:52,050 --> 00:14:53,550
It works as we wanted.

219
00:14:53,550 --> 00:15:00,960
You get d second u, d rho
squared is equal to u.

220
00:15:00,960 --> 00:15:10,030
And that means u goes to e
to the plus minus rho, which

221
00:15:10,030 --> 00:15:13,030
is what inspired this.

222
00:15:13,030 --> 00:15:15,280
And the other part
of the solution

223
00:15:15,280 --> 00:15:20,740
is the solution at
near r going to 0.

224
00:15:20,740 --> 00:15:24,580
Or your x going to 0
or near rho equal--

225
00:15:24,580 --> 00:15:26,810
going to 0.

226
00:15:26,810 --> 00:15:30,530
So near rho going to 0,
you have these two terms.

227
00:15:33,060 --> 00:15:36,120
And we actually
did it last time.

228
00:15:36,120 --> 00:15:41,950
We analyzed what was going on
with this equation last time.

229
00:15:41,950 --> 00:15:43,850
Near rho equal to 0.

230
00:15:43,850 --> 00:15:50,970
And we found out that u must
behave like rho to the l plus 1

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00:15:50,970 --> 00:15:52,830
for rho going to 0.

232
00:15:57,240 --> 00:16:02,380
So it was from this two terms
the differential equation.

233
00:16:02,380 --> 00:16:04,650
And for rho going to 0.

234
00:16:04,650 --> 00:16:08,070
Remember the wave
function must vanish.

235
00:16:08,070 --> 00:16:09,990
And how fast it should vanish?

236
00:16:09,990 --> 00:16:14,971
Should vanish to this
power to the l plus 1.

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00:16:14,971 --> 00:16:15,470
OK.

238
00:16:15,470 --> 00:16:18,630
So you know lots of things
about this function.

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00:16:18,630 --> 00:16:21,560
So first of all, that
this thing doesn't

240
00:16:21,560 --> 00:16:26,630
have necessarily polynomial
solutions because it

241
00:16:26,630 --> 00:16:29,000
behaves exponentially.

242
00:16:29,000 --> 00:16:35,180
Moreover, you know it doesn't
start with constant plus rho

243
00:16:35,180 --> 00:16:36,070
plus rho squared.

244
00:16:36,070 --> 00:16:39,130
It starts with rho
to the l plus 1.

245
00:16:39,130 --> 00:16:42,440
So it is pretty important
to see all these things

246
00:16:42,440 --> 00:16:46,180
before you try to do a recursion
relation because recursion

247
00:16:46,180 --> 00:16:50,780
relation might lead
you to funny things.

248
00:16:50,780 --> 00:16:53,580
So here we go.

249
00:16:53,580 --> 00:16:55,670
What do we do based on all this?

250
00:16:55,670 --> 00:16:57,360
We try something better.

251
00:16:57,360 --> 00:17:03,690
Which is-- we said u of rho, the
solution-- we're writing ansets

252
00:17:03,690 --> 00:17:06,930
--is going to be rho
to the l plus 1, that

253
00:17:06,930 --> 00:17:12,359
will have the right
behavior for rho going to 0.

254
00:17:12,359 --> 00:17:16,715
An unknown function
w of rho times

255
00:17:16,715 --> 00:17:23,010
e to the minus rho, which is
the right behavior we want.

256
00:17:23,010 --> 00:17:26,700
Now there is no
assumption, whatsoever,

257
00:17:26,700 --> 00:17:29,700
when you write in
ansets of this form.

258
00:17:29,700 --> 00:17:32,080
You're just expressing
your knowledge.

259
00:17:32,080 --> 00:17:34,590
Because at the end
of the day if rho--

260
00:17:34,590 --> 00:17:39,030
if omega-- or w here,
actually, is undetermined.

261
00:17:39,030 --> 00:17:42,810
This w will have an e to the
rho that cancels this factor.

262
00:17:42,810 --> 00:17:47,130
And may be off with some
funny power that cancels this.

263
00:17:47,130 --> 00:17:50,910
This is just a hope
that we're expressing

264
00:17:50,910 --> 00:17:54,600
that by writing this
solution in this form,

265
00:17:54,600 --> 00:17:57,000
this quantity may be simple.

266
00:17:57,000 --> 00:18:02,190
Because we know this is present
in the solution for larger rho,

267
00:18:02,190 --> 00:18:06,270
this is present for small
rho, well in between

268
00:18:06,270 --> 00:18:07,120
we might have that.

269
00:18:10,420 --> 00:18:13,750
And when you write
analysis of this form,

270
00:18:13,750 --> 00:18:31,070
you hope for a simple
differential equation for w.

271
00:18:31,070 --> 00:18:32,310
So what do we get?

272
00:18:35,810 --> 00:18:43,280
We can plug that ansets into
the differential equation

273
00:18:43,280 --> 00:18:49,470
that we already have
and see what happens.

274
00:18:49,470 --> 00:18:53,000
And indeed that's what we'll do.

275
00:18:53,000 --> 00:18:55,520
I'll skip the calculation.

276
00:18:55,520 --> 00:18:58,990
In my notes, it
took me half a page

277
00:18:58,990 --> 00:19:04,690
and I write big, so
it's not too long.

278
00:19:04,690 --> 00:19:05,760
So what do we get?

279
00:19:09,530 --> 00:19:13,130
You get an equation that
doesn't look that simple.

280
00:19:13,130 --> 00:19:14,080
I'm sorry.

281
00:19:14,080 --> 00:19:19,640
It just looks like a
step back, but it's not.

282
00:19:19,640 --> 00:19:28,370
d second w, d rho squared plus
2 times l plus 1 minus rho.

283
00:19:28,370 --> 00:19:29,830
Strange.

284
00:19:29,830 --> 00:19:42,059
dw, d rho plus, 1 over kappa
minus 2 times l plus 1,

285
00:19:42,059 --> 00:19:44,160
w equals 0.

286
00:19:49,130 --> 00:19:49,890
OK.

287
00:19:49,890 --> 00:19:52,100
Aesthetically, it looks worse.

288
00:19:52,100 --> 00:19:57,000
Certainly that equation on
the left board looked nicer,

289
00:19:57,000 --> 00:20:04,830
but actually it's pretty
good because, again, now look

290
00:20:04,830 --> 00:20:06,630
at your recursion relation.

291
00:20:06,630 --> 00:20:08,580
How will it be?

292
00:20:08,580 --> 00:20:11,790
If you take some power--

293
00:20:11,790 --> 00:20:13,140
fixed power.

294
00:20:13,140 --> 00:20:17,100
Here you lose one power.

295
00:20:17,100 --> 00:20:21,150
Here with the l plus 1 of
these, you lose one power.

296
00:20:21,150 --> 00:20:23,790
Here you lose nothing.

297
00:20:23,790 --> 00:20:25,470
And here you lose nothing.

298
00:20:25,470 --> 00:20:29,050
So you have either one
power less or your power.

299
00:20:29,050 --> 00:20:33,250
So it's a one step recursion
relation without the gap.

300
00:20:33,250 --> 00:20:36,120
It's not like the
two steps that we

301
00:20:36,120 --> 00:20:39,960
had for the harmonic oscillator,
for the Legendre polynomials.

302
00:20:39,960 --> 00:20:42,660
Here is one step
recursion relation.

303
00:20:42,660 --> 00:20:46,950
ak plus 1 determined by ak.

304
00:20:46,950 --> 00:20:51,720
So we say, excuse
me to the equation,

305
00:20:51,720 --> 00:20:55,320
you don't look that good,
but you're very solvable.

306
00:20:55,320 --> 00:20:58,940
So we can proceed.