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PROFESSOR: There is an
effective potential,

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as it's described here.

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Yes, there is a potential,
the central force,

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but if there is
angular momentum,

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it's like a centrifugal barrier.

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With angular momentum
it becomes very

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difficult to reach the origin.

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Because if you want
to reach the origin,

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you have to spin
faster and faster.

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So it becomes very hard
to reach the origin.

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And this is like a
repulsive potential here.

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So you have an
effective potential.

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You started with
just a potential,

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the central potential,
but that will

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become an effective potential.

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This whole thing in
brackets over there

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is sometimes called the
effective radial potential.

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And it's this V of r plus h
squared over 2mr squared l

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times l plus 1.

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So if you have a
Coulomb potential,

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that would try to
make the electrons go

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all the way to the proton.

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This is attractive.

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Well, even that is not
problematic for l equals 0.

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But for l different from
0, you have to add here

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a potential that diverges.

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And it's positive, say,
for some l positive

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and falls off very fast because
forms of like r squared, this,

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the Coulomb potential
falls off at 1/r.

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So when you combine the
two, so this is just

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the centrifugal barrier,
but you combine the two,

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this diverges faster as
well, so the total potential

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is something like
that in between.

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And then you can
have bound states.

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And all the theorems we've
learned about bound states

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and eigenstates of
one-dimensional potential, now

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you discover that they're very
useful in three dimensions.

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Many things just carry through.

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So this is effective potential.

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And it's going to
recap and remember

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that the solution that
you've written is Re.

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But Re is u of r over r
Y lm of theta and phi.

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And one more thing that
I want you to realize,

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does the function u depend on l?

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Yes.

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The differential
equation depends on l.

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Does the function u depend on m?

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Yes or no?

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STUDENT: [INAUDIBLE]

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PROFESSOR: No.

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There's no m in the
differential equation,

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so that should be good enough.

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So what's happened
is the m dependence

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is kind of very simple,
always very simple.

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It is the e to the im
phi and nothing else.

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The u doesn't know about it.

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On the other hand,
the u depends on l

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because it shows in the
differential equation.

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So I could write here of e
and l because it depends on l,

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and it is the solutions of
the Schrodinger equation

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in one potential.

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So there will be
quantization of energy,

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or there might be
stationary states

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that depend on the energy.

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So this is the function
that knows about l,

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knows about the energy.

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And we've been
totally successful

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with the angular dependence.

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Yes.

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STUDENT: That should
be theta and phi, dy?

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PROFESSOR: Yes, theta and phi.

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Thank you.

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Good.

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Normalization, the last thing
that has to work out nicely.

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Let's try to see what
does the normalization say

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about this function.

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Well, we should find
that the integral of psi

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squared d cubed x is equal to 1.

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But that integral,
as you now know,

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it's the integral of r
squared dr from 0 to infinity.

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And how do you
integrate over volume?

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You integrate over r.

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That has the right units.

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And you integrate over
solid angle, d omega.

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Of psi squared, so let's
do the arithmetic here.

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We have a Y lm star of theta and
phi, a Y lm of theta and phi.

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We have a u squared
and then r squared.

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Poor graph.

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But anyway, you can read it.

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Now, what happens
is just good stuff.

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r squared cancels.

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And this solid
angle angle integral

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is a perfect integral
for our normalization.

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So this gives you 1.

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And therefore the end
result for all this integral

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is just the integral from 0 to
infinity of dr u of r squared.

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And that must be equal to 1.

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So not only is little
u a nice variable

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that satisfies a one-dimensional
Schrodinger equation,

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but you can remember that
your more complicated wave

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function will be normalized
if u is normalized

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in the one-dimensional sense.

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If u squared integrated
over x is equal to 1, yes,

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you're normalized.

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So the set-up to convert the
three-dimensional differential

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equation into a one-dimensional
differential equation

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has been very successful.

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We've reduced it to a
one-dimensional problem.

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We have to solve those.

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Each time somebody gives
you a spherical potential,

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you look at that equation,
the radial equation,

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and try to solve for u's.

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If you solve for
u's, then you can

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append the angular
dependents that

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correspond to angular
momentum eigenstates.

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We call this angular
moment eigenstates.

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They are the most you can
ask from angular momentum.

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And then you have
many solutions.

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So I want to conclude
that part of the analysis

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by mentioning something
about solutions

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with the appropriate
boundary conditions.

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We have x, in the
one-dimensional problem

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we call it x.

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And you'd run from minus
infinity to infinity.

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The one difference here
is that you have r,

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and r runs from 0 to infinity.

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So you may wonder if you have
some issues with r going to 0.

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What should the wave
function do when r goes to 0?

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OK.

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Then the way we think about
it is not completely general

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but is good enough.

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We think of the
differential equation

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as we have here and
imagine a potential

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that when r goes to 0, the
centrifugal barrier dominates.

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So our potential of the
form 1 over r to the fourth

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would be even more
singular than the barrier.

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And I don't know what
happens in that case.

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Most likely it's no
good, not interesting.

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You cannot find solutions.

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But if the potential, like
the Coulomb potential,

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is weaker than the centrifugal
barrier as r goes to 0,

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the centrifugal barrier
dominates when r goes to 0.

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And this differential
equation, as r goes to 0,

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has a potential infinite
term which corresponds

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to the centrifugal barrier.

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So we think of this
as r goes to 0,

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the differential
equation roughly

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becomes minus h squared over
2m d second u dr squared

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plus h squared l times l plus 1
over r squared of u, roughly 0.

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At least the leading
behavior of these things

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should work out correctly.

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So the h squareds--

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here is a 2m, as
well, I'm sorry.

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The h squared over 2m's
cancel, and you get

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d second u dr squared plus--

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no, is equal to l times l
plus 1 over r squared u.

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And we're only interested
for this as r goes to 0.

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And it's not an exact statement.

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It's a discovery.

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We're trying to discover
what's happening with the wave

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function near r equals 0.

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Well, this has two
kinds of solutions.

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You can try a polynomial.

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So this could go like
r to the l plus 1.

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If you take two
derivatives, that works out.

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You get the l plus 1 and l.

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And it solves the equation.

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Or r to the minus l also
solves the equation.

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As you can imagine,
this is going

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to be problematic in general.

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It's too divergent.

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It will not be possible to
normalize it, in general,

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for arbitrary values of l.

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So this is a very
brief analysis.

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I'm not going into
all the detail

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that probably this
deserves at this moment.

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But this is ruled out,
and this is ruled in.

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The only thing-- so it's true
that this one is ruled out,

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and it has problems
for normalization.

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It is too divergent as l.

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But for l equals 0,
it's not divergent.

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But for l equals 0,
there's another reason

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why this is not good.

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It turns out that for l
equals 0 this doesn't quite

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solve the Schrodinger equation,
the exact Schrodinger equation.

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So the bottom line
of this analysis

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is that we will have u of r
behave like r to the l plus 1

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as r goes to 0.

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And in particular, when
l is equal to 0, u of r

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will behave like r.

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So it will vanish
as r goes to 0.

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So the only question is
how fast it vanishes.

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It vanishes as r goes
to 0 for l equals 0.

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It vanishes even
faster for higher l.

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So it always vanishes, the
wave function at r equals 0.

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And that's why we can
usually think of it

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as having an infinite barrier.

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The wave function could not
exist for r less than 0.

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That physically doesn't exist.

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And the boundary conditions are
such that that cannot happen.

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All right.

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We've finished the discussion
of the radial equation.

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Now, our main interest with
the radial equation, of course,

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is, at this moment,
the hydrogen atom.

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00:14:26,530 --> 00:14:29,270
So we're going to turn to that.