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PROFESSOR: We're going to
complete our study of WKB

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today.

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There's a lot of important
things we haven't said yet.

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And in particular,
we have not explained

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how it all works, really, in
the sense of the connection

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formulas.

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That's the most non-trivial
part of the WKB story.

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And in some ways it's some of
the most interesting things

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that you have to
learn from this.

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It's an opportunity to learn
something about differential

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equation, Fourier
transforms, and the limits

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of the quantities you
can really compute.

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So we will begin our discussion
with some reconsideration

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of these Airy functions.

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You've started to
hear about, probably

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some in the homework,
some in recitation.

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So this relates to this
person, George Biddle

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Airy from the Airy
functions, who

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lived in England for 90 years.

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He was the seventh
Royal Astronomer.

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And he was an astronomer
from 1835 to 1881.

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Was located, his work
area, in Greenwich.

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It's a little bit
outside of London.

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And that's where astronomers
were working hard

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to establish all the data
associated with the Prime

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Meridian.

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France also tried to make the
their own meridian the most

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important one.

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But at the end of
the day, it all

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depended of which
astronomers did

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the most work in measuring
precise, accurate determination

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of times, motion of the
moon, Jupiter's satellites,

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all kinds of things that were
pretty important at that time,

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especially also
the manufacturing

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of good telescopes
and good clocks.

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So Mr. Airy also
worked in optics.

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And that's where he discovered
these functions we're studying.

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And they're very relevant
to the WKB approximation.

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So let me remind you a
little of what you've

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been considering in the homework
and extend some of those ideas

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as well.

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So you considered a differential
equation, d 2nd PSI du

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squared equal u PSI.

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Pretty innocent looking
differential equation.

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I'm always surprised how a
differential equation that

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looks so simple can
be so intricate.

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But anyway, it is.

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And that's the Airy equation.

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And it shows up for this problem
that you have, for example,

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an infinite wall and a
potential goes up like that.

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The bound states are controlled
by solutions of this equation

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at the end of the day.

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This is after you clean up all
the units and do all the work.

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But it shows in
other places as well.

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And you probably found already
that this PSI function, PSI

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of u, is given as a constant
times an integral that

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came from Fourier
transformations of e

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to the ik cubed
over 3 e to the iku.

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And so this is our integral.

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This is our solution
of this equation.

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And the nice thing is that
this integral tells you

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a lot about the function.

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It's a perfectly nice
definition of the function.

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You may have
noticed, and I think

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it was discussed to some
degree in recitation

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as well, that the Fourier
transform converts

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this equation to first
order differential equation.

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Because this becomes
p squared times PSI.

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Well, you know from
Schrodinger equations

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this part of the differential
equation comes from p squared.

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On the other hand,
multiplication by x

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is the same as dvp
in Fourier space.

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Therefore, you get
the dvp of PSI equal p

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squared PSI equation.

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And that equation is a first
order differential equation.

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First order differential
equations have one solution.

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So you would say, OK, I tried
to do this by Fourier transform,

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and I'm getting just
half of the solutions.

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And that's true.

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The reason is that the other
solution of this differential

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equation doesn't quite
have a Fourier transform,

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so it's a little more
difficult to get at.

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But once you think
of this integral,

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it is kind of useful to try
to be a little more general

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and think of it in terms of the
complex plane, the complex k

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plane.

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So here is the k plane.

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And this integral is over k
from minus infinity to infinity.

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So this is the standard
countour of integration.

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We can call it the countour c1.

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When you're doing this
integral, and you're

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doing this in the
complex plane you

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may want to consider that
if you go off the real line,

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maybe things are even better.

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In particular, as
you're integrating

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u is a fixed number,
it's a fixed coordinate.

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We usually think of u
as real, but we could

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think of u as complex as well.

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And this integral
has these factories.

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It's an oscillatory factor
that becomes faster and faster

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oscillating in time.

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But the magnitude of the
integrand is 1 at every place.

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It oscillates very fast,
but the magnitude is 1.

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Not the best situation possible.

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You know?

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When you're doing
an integral, you

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have the feeling
that eventually you

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get 0 because it's
oscillating so fast.

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But it would be nicer
if you had a decay.

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So for that, consider
that this function

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would be k if k, if the
imaginary part of k cube,

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is positive, the ky.

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If the imaginary part of k cube
is positive, you have a k cube

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would be i times a positive
number. i items i is minus 1,

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so you get the
suppression factor,

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if the imaginary part
of k cube is positive.

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So if you're integrating
over a region

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where the imaginary part
of k cube is positive,

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you will be in good shape.

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Now, if you think of k
as a complex number, e

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to the i theta k, k
cube would be this.

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An imaginary part of k
cube is positive requires

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that this angle 3i theta
k be between pi and 0.

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Why?

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Because if you have a complex
number, any complex number,

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its imaginary part is positive
if the argument of that angle

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goes between 0 and pi.

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Moreover, if you get
an imaginary part

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positive for some angle
theta k, if m of k cube

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is positive for some
theta k, it is also

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positive for theta
k plus 2pi over 3.

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Because if you add 2
pi over 3 to theta k,

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you change this by 2pi.

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And therefore, the
exponential doesn't change.

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From this inequality, I'm sorry,
I have an i too many here.

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The imaginary part
of this requires

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that the argument of
this be between 0 and pi,

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or theta k between
pi over 3 and 0.

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So actually, here, I
should draw 0 to pi over 3.

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This is at 60 degrees.

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And that's the region
in k space where

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the integral is suppressed.

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Now, we also said that
this region is unchanged

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if you add 2pi over 3.

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So if I add 2pi over 3, that's
precisely 1/3 of a turn.

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This is 30 degrees.

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It goes this area into this.

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And finally, if I add,
again, 2pi over 3,

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it goes into 30, 30,
30, and to this region.

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I guess the sign of nuclear
material, hazardous materials

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or something that.

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It's not hazardous
to our health.

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We'll survive this lecture.

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So what's happening
here actually

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is something quite interesting.

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For example, you're
doing this integral here.

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If you started
doing this integral,

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once you are far, far
away, you start going up,

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the integral over here
is going to give nothing,

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because you're very far away.

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And therefore,
you're in the region

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of absolute exponential
suppression.

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We said this thing has a
positive imaginary part.

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And therefore, you
have a suppression.

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It's k cubed, so
it's becoming like e

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to the minus radius cubed.

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It's killed.

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So it's no problem to go up
here, and you get nothing.

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So what it means is
that actually there's

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no singularities in
the integrand either.

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This integral that
we've done here

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could have been done over
this contour like that.

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And then you could
have gone like that.

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You could do it in Mathematica.

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The answer will be the same.

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Or you could have
gone like that.

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Or you could have
gone here like this,

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also you're
suppressed over here.

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And you're suppressed over here.

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And actually, by contouring
the formation, you could

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instead of doing
this in theory, you

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could have come here
and gone up like that.

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And done that.

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That also would have been the
same result. There's nothing,

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the integrand is analytic here.

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So this you can push it
down and nothing goes wrong.

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Or you could have done even
this integral like that.

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Because the integral here is
0 and the integral here is 0,

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so you can bring them down.

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In fact, if you
did this, I think,

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with Mathematica,
Mathematica when

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you try to do this
integral, kind of does it.

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But it complains a little.

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It's worried that,
you know, it's

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not getting the answer right
because the integrand is still

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big.

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It's oscillating very
fast, but it's big.

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If you go here, I think
Mathematica will complain less

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and will give you
the same answer.

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So this is the power
of complex analysis

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that allows you to do this
integral in all kinds of ways.

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And there is something even
more important about this

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that I will explain
in the notes,

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but I can convey the idea here.

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When you prove that
this is a solution,

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and when you went
to Fourier space,

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you had to do an
integration by parts.

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And an integration by
parts is always dangerous.

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And if you're trying to find a
true solution of your problem,

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you have to make sure it works.

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P, the operator P, is
supposed to be her mission.

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And here what happens is
that you turn this into dvp.

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And you have to fall
back this factor.

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And here is where this
discussion is very relevant.

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00:16:08,770 --> 00:16:14,050
Because when you fold out this
factor, you have to integrate.

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You couldn't get contributions
from the boundary.

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And you have to make sure
those contributions are 0.

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Otherwise, you're not solving
the differential equation

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really.

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And the happy thing
about these contours,

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and now that you
understand these regions,

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is that, yes, the
boundary contributions

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are going to be 0 from
the two sides because

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of this suppression factor.

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So this analysis actually turns
it into a rigorous analysis.

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And the part that has
to do with the boundary

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conditions, the fact that the
operations that you're doing

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are illegal when you try
to solve the differential

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equation, require--

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this is the details that
I will not derive now--

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require precisely that e to the
i k cubed over 3 e to the iku

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vanish at the ends of
the countour gamma that

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you're using to
solve the problem.

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So if you're in the
wave from here to here,

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it should vanish at this
end and at that end.

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Rigorously speaking,
this thing only

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vanishes if you're
a little bit up.

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But this is what you really
need for the equation

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to be a solution.