WEBVTT
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PROFESSOR: Now,
what should happen?
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Somehow, this
equation probably has
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solutions for all
values of the energy,
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but those solutions diverge
and are not normalizeable.
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It's the kind of thing you will
find with a shooting method
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that you're doing
with your computer.
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Count the solution and
it suddenly diverges up
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or diverges down and
cannot be normalized.
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But for some specific
values, it can be normalized.
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So what we need is an intuition
why this differential equation
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has normalizeable solutions,
only if the energies, curly E,
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take specific values.
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That's intuition that we need.
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For that, we'll look at the
equation a little closer,
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and try to understand what
happens at the place where
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it can get in trouble, which
is large X. That's where you
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expect it to get in trouble.
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So what does this
equation become as u
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goes to plus minus infinity?
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Well this equation, at that
stage, becomes like this.
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The second phi, the u squared is
roughly equal to u squared phi.
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Because e is a constant,
it doesn't blow up,
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so the differential
equation, the terms
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that are supposed to be largest
in this right hand side,
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is the u squared.
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So how does this
look as a solution?
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And you can see that
that's definitely not
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like a power solution.
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If you have a power of u,
say phi equals u to the n,
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after u differentiates,
the power goes down.
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But here it goes up.
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You have a u to the end.
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The relative is supposed to
give you u to the n plus 2,
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but that doesn't work.
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So that's not a power solution.
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So it has to be different.
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So what it is,
you can try a phi.
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And what it should be is of
the form e to the u squared.
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It's kind of like that,
because this is the function
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that when you're
differentiate, you
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bring down the derivative of
this quantity, which is a u.
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When you differentiate
again, you
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can bring another u
to get the u squared,
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or you can
differentiate this one.
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But if you differentiate the
thing that is in the bottom,
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you get something
that diverges less.
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So morally speaking, this
function is about right.
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So I'll put here, for example,
e to the alpha over 2.
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And I'll even put here a uk.
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Can this work?
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A u to the k times
alpha u squared.
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Well if we take two derivatives,
if we take one derivative,
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if I differentiate the u to the
k, I get u to the k minus 1.
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I lose powers of u.
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If I differentiate
the exponential,
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I can get alpha times u
times the same function.
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See, that's, roughly
speaking, what's happening.
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You differentiate the thing
that diverges the most.
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So if you differentiate
twice, each time
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you differentiate you get
a factor of alpha times u
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squared phi, roughly.
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This is plus subleading.
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When you differentiate
a function
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and you're wanting to show
the most divergent thing, then
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you--
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because we're looking at
the most divergent part,
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you will always differentiate
this, and this u to the k
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is really a spectator,
it doesn't do anything,
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because when you differentiate
that, you get something much
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smaller, that doesn't matter.
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So yes, with these exponentials,
we get something like this.
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Beside double pronged
should have been here.
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And therefore, you see
that alpha is plus minus 1.
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Alpha is plus minus
1, and those are
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likely to be
approximate solutions
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as x goes to infinity.
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So we could expect solutions
for alpha equals 1,
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and I will write that it,
and all this, I should say
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is always as u goes to infinity.
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So always as u go to infinity,
all of this in this blackboard.
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So also as u go to infinity,
we would expect maybe solutions
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of the form A u to
the k, e to the minus
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u squared over 2 plus B u to
the k e to the u squared over 2.
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The two values of alpha
equal plus minus 1
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are the possibilities for
these two equations to match.
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So you would expect things
like this to be solutions.
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And here you are seeing the
beginning of the danger.
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Well a minus u
squared over 2 times
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a power sounds pretty good,
but a u squared over 2 times
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a power sounds pretty bad.
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So maybe this is what
we want to happen.
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This is not an exact
solution of anything yet.
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We're just looking at u going
to plus and minus infinity,
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and maybe we'll have such a
behavior or such behavior.
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But we want this one,
otherwise we will never
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be able to normalize it.
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So here it is.
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Without any loss of generality
and inspired by this,
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this analysis is absolutely
crucial, you see,
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we're following a very
logical procedure.
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Cleaning the
equation then looking
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where the divergence
would happen
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and learning something about
the form of the solution.
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Now without any loss of
generality, I can write,
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I will write 5x
is going to be a--
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not of x anymore.
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U, h of u times e to the
minus u squared over 2.
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So it's an an sat, but it's
without any loss of generality,
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because you can always write
any function as another function
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times e to the minus u over 2,
because you take the function,
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you multiply by e
to the plus u over 2
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and e to the minus u over 2
and it's written like that.
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But this should be nice,
because it has sort
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of the right behavior already.
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And here is the hope.
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The hope is that this function
now is a proxy for phi.
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If you know h, you know
phi, which is what you want.
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And this function,
hopefully, won't diverge.
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This will be into [INAUDIBLE].
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So if this function
doesn't diverge,
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it will be a great thing.
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In particular, we could hope
that h of u is a polynomial.
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You see, if somebody
with have come and said,
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look at that equation.
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Could that be a polynomial?
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A polynomial is something
that ends up to some power
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x to the 20 or x to the 30,
but it doesn't go up forever.
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An exponential has all powers.
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This equation doesn't have
a polynomial solution.
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No polynomial will
ever solve this.
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But now that you've
isolated the divergence,
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there is a hope that a
polynomial will work.
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So for doing that,
exploring that hope,
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I now have to substitute--
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this is no assumption--
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the differential
equation for phi
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implies a differential
equation for h,
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you just substitute
this and look at it.
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That's a one line computation
or two line computation.
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I'll give the answer.
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So what is the differential
equation for h?
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So back in star, you'll
get the second h,
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the u squared minus 2u dh du
plus e minus 1h equals zero.
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So you substitute that
into this equation
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and you get this
differential equation.
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And now this is a
differential equation for h.
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We hope it has a
polynomial solution.
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You will see that it wants to
have a polynomial solution,
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but it doesn't quite make it.
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And then you will discover
quantization helps,
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and suddenly you get the other
normal solution and everything
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works out.