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PROFESSOR: Here
is your potential.

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It's going to be a smooth,
nice potential like that.

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V of x.

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x, x.

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And now, suppose you
don't know anything

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about the energy eigenstates.

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Now, this potential will
be assumed to be symmetric.

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So here is one thing you can do.

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You can exploit some things that
you know about this potential.

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And here's the wave
function that we're

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going to try to plot.

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And we could say the following.

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Let's see.

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Whatever energy are
here, for bound states,

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I'm going to eventually be
in the forbidden region.

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So far on the right here, I
will be in the forbidden region.

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And I must meet
the wave function

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that looks like the forbidden
region wave function.

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And the only possibility
is something like that.

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You could say it's
the [INAUDIBLE] of 1,

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but actually, if it's
a [INAUDIBLE] then

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I could multiply by minus 1 and
use this one conventionally.

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That wave function
is always going

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to be like that over there.

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On the other hand,
very important-- the

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on the very left, how will
the wave function look?

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Well it also has to decay,
so it can decay like that,

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or it might be
decaying like this.

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And in fact, you don't
know until you figure out

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what's happening in the middle.

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It may be decaying
like this or like that.

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I fix the sign here,
so whatever this

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does it should either end up
like that or end up like that.

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So these are the guidelines
that you have to solve this.

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Should begin like
that, and we'll see.

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And it should be either
symmetric or anti-symmetric.

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This would be the
case anti-symmetric,

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this would be the
case symmetric.

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OK.

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So let's draw one, two,
three, four lines there.

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One, two, three, and four.

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And I don't know where
the energies lie.

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I don't know what is
the ground state energy.

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And I want to give you
an insight into how

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you can figure out why you
get this energy one decision

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when this happens.

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So let's plot the wave
function for the first case.

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I don't know if I have a
label, but let's assume

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this is E0, E1, E2, E3.

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Three energies [INAUDIBLE],
and here's the one for e0.

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OK.

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So I begin here,
that's how it goes.

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And then I go through
my Schrodinger equation,

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integrate it.

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You see?

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Numerical, you can always
integrate the Schrodinger

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equation.

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And this should be always
in this region, let me--

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like this.

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And it's growing.

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And, oops, there should
be turning points here.

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There should be turning points--

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suppose this is--

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I'm not going to get this so
well, but it goes like this.

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And now this should
be a turning point.

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So I should change to the
other type of curvature,

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curvature down.

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But what probably will happen
with E0 is that it will switch

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and it will go like and
start to curve, maybe.

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Well, if it looks
like that, it must

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match to the development of the
odd piece or the even piece.

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Now, it's never going to
match with the odd one,

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so it might be with the even.

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And yes, it would match
turning point here,

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but look what has happened here.

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You got the corner there.

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You know, this was
turning slowly,

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and this is starting
to turn slowly,

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but here there is
a discontinuity

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in the derivative.

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So this is not the solution.

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You try, but you fail.

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But that's-- right.

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Not every energy
gives a solution.

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So they should have
matched continuously

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and derivative
continuously at that point,

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but it didn't have
enough time to do that.

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On the other hand, if we
try the next one, maybe.

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The turning points will be here.

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Let's see what happens.

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Well, now the forbidden
energies are over here,

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and now you have a
turning point here that--

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in here, the
curvature is negative,

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the second
derivative's curvature.

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And it's larger
than it was here.

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Here it was small,
here it's larger.

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So it's going to curve faster.

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Maybe if you get
the E1 right, it

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will curve enough so this
flat here, in which case

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the other side will match nicely
and you've got the solution.

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So you probably have
to go little by little

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until this becomes
flattened and, boom,

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you've got the solution.

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Energy eigenstate.

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Let's go a little further.

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This graph continues there.

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Now I want to go to E2.

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How am I going to do that?

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I'm going to do it this way.

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So this was here,
this was there.

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There is the vertical line here.

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And for E2, the turning
points are even further out.

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And here is the wave function.

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And let's look at this
thing that I have.

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Now, the turning point
in this one corresponds

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to the E2 turning point.

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This is E1.

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And now this will go
in here, we'll turn,

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and we will go curve and
maybe do something like that.

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Because it's curving more
and more, and faster.

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So by the time you
reach here, this

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is no good, because this
one will be symmetric.

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You know, you would have
an anti-symmetric one

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that is no use.

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But now you don't have
a solution, again.

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So as you increase the energy,
this is starting to do this,

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and that is not quite so good.

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And then when you go to E3, you
have a turning point over here.

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So maybe in this case
it will go up here

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and it will start turning, and
it will turn enough to just--

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this dip go to the origin.

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OK.

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You're saying no good either,
because this is terrible, this

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continuous.

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But, ah, you were supposed to
draw the other one as well.

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The old one is actually
perfect for it.

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So this is dash,
it doesn't exist,

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and this one matches here.

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So by the time the dip--

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this is not a solution, but
the dip goes down and down,

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and eventually goes to zero,
it matches with this one.

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That's why I said sometimes you
don't know whether this matches

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with the one that
comes from here

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or the one that comes
from the bottom.

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So there you go.

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This is an energy
eigenstate again.

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It's odd and it has one node.

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And that gives you
the intuition how,

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as you sort of come from the
end and you reach the middle,

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you sometimes match things
or sometimes don't match.

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And explains why you
get energy quantization.

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The other way in which you're
going to gain intuition

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is with the so-called
shooting method,

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which is the last thing I
want to discuss for a minute.

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So the shooting method
in differential equations

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is quite nice.

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Shooting method.

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Suppose you have a potential
that this symmetric may

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be something like this--

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it doesn't look very symmetric.

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It looks a little better now.

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And you want to find
energy eigenstates.

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You do the following.

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You say, well, the normalization
of the energy eigenstates

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is not so important.

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Let's look for even states.

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Now, you can look for
even or odd states

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if the potential is symmetric.

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Sometimes the potential will
have a wall, in which case

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you have to require a
symmetric potential.

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It's easy to solve, as well.

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But let's consider the case
when the potential is symmetric

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and you look for even states.

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So what you do is just,
say, you pick an energy.

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Pick some energy E0.

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And then you put some
boundary condition.

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You say that the wave
function at x equals 0 is 1.

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And then you say that the
derivative of the wave function

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at x equals zero is how much?

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Any suggestion?

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How much should it be?

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You see, you have a second
order differential equation.

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The second psi is equal to E
minus V. That's the Schrodinger

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equation.

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You need-- the boundary
conditions are the value of psi

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and the derivative at one
point, and then the computer

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will integrate for you.

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Mathematica will do it.

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But you have to give
me the derivative, so

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what should I put?

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A number there, 1, 2, 3?

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Is that an unknown?

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What should I pick?

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We must put in the 0, because
if you had a wave function whose

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derivative is not 0, and
it's an even wave function,

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it would look like this.

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And there would be a
discontinuity in psi prime--

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00:12:40,410 --> 00:12:41,400
discontinuous.

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And that's not possible
unless you have a hard wall

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or you have a delta function.

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So you must put this.

202
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And then you
integrate numerically.

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Numerically.

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And what will happen?

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Well, if you
integrate numerically,

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00:13:03,920 --> 00:13:11,050
the computer is just going to
integrate and see no problem.

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00:13:11,050 --> 00:13:13,110
Basically, it's just
going to do the interview.

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00:13:13,110 --> 00:13:15,860
Ask the computer to calculate
the wave function out

209
00:13:15,860 --> 00:13:19,920
to x equal 5, it
will calculate it.

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00:13:19,920 --> 00:13:22,610
So the problem is that--

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you can see visually,
if you pick some energy,

212
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the wave function will
do like something,

213
00:13:30,960 --> 00:13:34,135
and then will start blowing up.

214
00:13:38,740 --> 00:13:40,810
And then you say,
oh, that energy

215
00:13:40,810 --> 00:13:46,820
is no good because the wave
function won't be normalizable.

216
00:13:46,820 --> 00:13:48,790
And then you go
back to the computer

217
00:13:48,790 --> 00:13:52,030
and change the
energy a little bit,

218
00:13:52,030 --> 00:13:56,766
and then you will
find, well, maybe this.

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Now it blows up in
the other direction.

220
00:14:00,210 --> 00:14:01,930
No good either.

221
00:14:01,930 --> 00:14:06,700
But in some energy in
between, as you change,

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00:14:06,700 --> 00:14:09,700
there must be one
in which it does

223
00:14:09,700 --> 00:14:14,020
the right thing, which is BK.

224
00:14:14,020 --> 00:14:15,620
Somewhere in between.

225
00:14:15,620 --> 00:14:18,852
And numerically, you change
the value of the energy,

226
00:14:18,852 --> 00:14:20,810
you go-- in the shooting
method, when you shoot

227
00:14:20,810 --> 00:14:23,110
it either goes up or down.

228
00:14:23,110 --> 00:14:26,980
And you start working
within those two numbers

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00:14:26,980 --> 00:14:29,350
to restrict it
until you get here.

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00:14:29,350 --> 00:14:33,400
If you have a solution
with five-digit accuracy,

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00:14:33,400 --> 00:14:37,180
it will do this, this,
this, and then blow up.

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00:14:37,180 --> 00:14:39,530
If you have a solution
with 10 digits after it,

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00:14:39,530 --> 00:14:43,460
it will do this and go up
to here and blow up again.

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00:14:43,460 --> 00:14:48,100
You need 500-digits
accuracy to get that wall.

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00:14:48,100 --> 00:14:51,460
But it's a fun thing that you
can do numerically and play

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00:14:51,460 --> 00:14:52,450
with it.

237
00:14:52,450 --> 00:14:56,830
You can calculate five digits
accuracy, ten digits accuracy

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00:14:56,830 --> 00:15:00,280
within a matter of minutes.

239
00:15:00,280 --> 00:15:03,430
It's very practical,
and it's very nice,

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00:15:03,430 --> 00:15:07,550
but one thing you have to
do is clean up your equation

241
00:15:07,550 --> 00:15:08,830
before you start.

242
00:15:08,830 --> 00:15:13,120
You cannot have an equation in
Mathematica with h-bar and m

243
00:15:13,120 --> 00:15:14,380
and all that.

244
00:15:14,380 --> 00:15:17,830
So you have to clean up the
units, is the first step,

245
00:15:17,830 --> 00:15:20,060
and write it as an equation
question without units.

246
00:15:20,060 --> 00:15:23,840
Your And this plots very
nicely in Mathematica,

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00:15:23,840 --> 00:15:26,430
and you will have
lots of practice.