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PROFESSOR: Today, we continue
with our discussion of WKB.

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So a few matters
regarding the WKB

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were explained in the
last few segments.

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We discussed there would
be useful to define

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a position-dependent momentum
for a particle that's

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moving in a potential.

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That was a completely
classical notion,

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but helped our terminology
in solving the Schrodinger

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equation and set up the
stage for some definitions.

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For example, a
position-dependent de Broglie

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wavelength that would
be given in terms

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of the position-dependent
momentum by the classic formula

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that de Broglie used
for particles that move

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with constant momentum.

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Then in this language, the
time independent Schrodinger

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equation took the
form p squared on psi

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is equal to p squared
of x times psi.

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This is psi of x.

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And we mentioned that it was
kind of nice that the momentum

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operator ended up sort of in
the style of an eigenvalue.

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The eigenvalue p of x.

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We then spoke about
wave functions

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that in the WKB
approximation would

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take a form of an exponential
with a phase and a magnitude--

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so the usual notation we
have for complex numbers.

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So the psi of x and p would
be written as rho of x and t

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e to the i over h
bar s of x and t.

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That's a typical form
of a WKB wave function

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that you will see soon.

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And for such wave
functions, it's

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kind of manifest that rho
here is the charge density.

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Because if you take the
norm squared of psi,

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that gives you exactly rho.

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The phase cancels.

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On the other hand, the
computation of the current

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was a little more interesting.

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And it gave you rho times
gradient of s over m--

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the mass of the particle.

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So we identified the
current as perpendicular

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to the surfaces of constant s or
constant phase in the exponent

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of the WKB equation.

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Our last comment had
to do with lambda.

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And we've said that we
suspect that the semiclassical

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approximation is
valid in some way

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when lambda is small compared
to a physical length of a system

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or when lambda changes slowly
as a function of position.

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And those things we
have not quite yet

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determined precisely
how they go.

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And some of what
you have to do today

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is understand more
concretely the nature

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of the approximation.

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So the semiclassical
approximation

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has something to do with lambda
slowly varying and with lambda

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small, in some sense.

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And since you have
an h bar in here

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that would make it
small if h bar is small,

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we also mentioned
we would end up

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considering the limit as a
sort of imaginary or fictitious

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limit in which h bar goes to 0.

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So it's time to try to
really do the approximation.

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Let let's try to write something
and approximate the solution.

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Now, we had a nice instructive
form of the wave function

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there, but I will
take a simpler form

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in which the wave function will
be just a pure exponential.

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So setting the
approximation scheme--

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so approximation-- scheme.

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So before I had
the wave function

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that had a norm and a phase.

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Now, I want the
wave function that

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looks like it just has a phase.

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You would say, of course,
that's impossible.

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So we will for all the time
independent Schrodinger

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equation.

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So we will use an s of x.

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And we will write the
psi of x in the form

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e to the i h bar s of x.

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And you say, no,
that's not true.

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My usual wave functions
are more than just phases.

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They're not just faces.

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We've seen wave functions
have different magnitudes.

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Here that wave function will
have always density equal to 1.

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But that is voided--

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that criticism is voided--

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if you just simply say that s
now may be a complex number.

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So if s is itself a complex
number, the imaginary part of s

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provides the norm of
the wave function.

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So it's possible to do that.

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Any rho up here can be
written as the exponential

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of the log of something and then
can be brought in the exponent.

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And there's no loss of
generality in writing something

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like that if s is complex.

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Now, we have the
Schrodinger equation.

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And the Schrodinger
equation was written there.

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So it's minus h squared d
second d x squared of psi of x.

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It's equal to p squared of
x e to the i over h bar s.

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Now, when we differentiate
an exponential,

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we differentiate it two times.

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We will have a couple of things.

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We can differentiate the first
time-- brings an s prime down.

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And the second time you can
differentiate the exponent

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or you can differentiate
what is down already.

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So it's two terms.

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The first one would be--

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imagine differentiating
the first one.

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The s goes down and then the
derivative acts on the s.

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So you get i over h
bar s double prime.

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Let's use prime notation.

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And the other one is when you
differentiate the ones here.

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It brings the factor
down, again, there.

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So it's plus i over h
bar s prime squared.

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And then the phase
is still there,

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and it can cancel between
the left and the right.

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So this is equal
to p squared of x.

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So I took the derivative and
cancelled the exponentials.

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So cleaning this
up a little bit,

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we'll have this term over here.

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The h's cancel,
the sine cancels,

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and you get s prime of x squared
minus i h bar s double prime.

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It's equal to p squared of x.

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OK.

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Simple enough.

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We have a derivation
of that equation.

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And the first thing
that you say is,

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it looks like we've
gone backwards.

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We've gone from a
reasonably simple equation,

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the Schrodinger
equation-- second order,

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linear differential equation--

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to a nonlinear equation.

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Here is the function squared,
the derivative squared, second

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derivative, no--

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there's nothing linear
about this equation in s.

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If you have one solution for
an s and another solution,

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you cannot add them.

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So this happens because we put
everything in the exponential.

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When you take double
derivatives of an exponential

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you get a term with
double derivatives

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and a term with a
derivative squared.

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There's nothing you can do.

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And this still represents
progress in some way,

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even though it has
become an equation that

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looks more difficult. It can
be tracked in some other way.

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So the first and
most important thing

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we want to say
about this equation

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is that it's a nonlinear
differential equation.

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The h bar term appears in
just one position here.

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And let's consider
a following claim--

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I will claim that i h
bar s double prime--

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this term-- is small when
v of x is slowly varying.

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You see, we're having
in mind the situation

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with which a particle is moving
in a potential-- a quantum

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particle.

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So there it s, b of
x is well-defined.

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That's a term that goes
into this equation.

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So this is partly known.

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You may not know the energy,
but the potential you know.

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And my claim-- and perhaps a
very important claim about this

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equation that sets
you going clearly--

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is that when v of s
is slowly varying,

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this term is almost irrelevant.

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That's the first thing we want
to make sure we understand.

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So let's take v of x is
equal to v0 at constant.

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So this is the extreme case
of a slowly varying potential.

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It just doesn't vary at all.

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In that case, p of x is
going to be a constant.

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And that constant is the
square root of 2m e minus V0.

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And what do we have here?

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We have a free particle.

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This v of x is a constant.

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So the solution of the
Schrodinger equation,

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that you know in
general, is that psi

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of x is e to the
i p0 x over h bar.

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That solves the
Schrodinger equation.

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Now, we're talking
about this equation.

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So to connect to that
equation in this situation

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of constant potential
constant, momentum

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in the classical sense,
and a free particle

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with that constant momentum,
remember that s is a term here

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in the exponential.

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So for this solution here,
s of x is equal to p0 x.

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That's all it is.

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It's whatever is left when you
single out the i over h bar.

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And let's look at that thing.

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That should be a solution.

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We've constructed the
solution of the momentum.

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equation for constant potential.

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We've read what s of x is.

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That should solve this equation.

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And how does it
manage to solve it?

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It manages to solve it because
s prime of x is equal to p0.

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s double prime of
x is equal to 0.

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And the equation works out
with the first term squared--

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p0 squared-- equal
to p of x squared--

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which is p0 squared.

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So this term, first term
in the left-hand side,

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is equal to the right-hand side
This term is identically 0.

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So when the potential
is constant,

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that term, i h bar s double
prime, plays no role, It's 0.

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So the term i h bar s
double prime is equal to 0.

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So the claim now follows from
a fairly intuitive result.

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If the potential is constant,
that term in the solution is 0.

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If the potential
will be extremely,

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slowly varying, that term
should be very small.

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You cannot expect that
the constant potential has

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a solution.

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And you now do infinitesimal
variation of your potential,

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and suddenly this
term becomes very big.

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So for constant v, i h bar
is double prime equal 0.

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So for slowly varying v of
x i h bar is double prime,

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it should be small in the
sense that this solution

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is approximately correct.

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So if we do say that,
we've identified the term

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in the equation that is small
when potentials are slowly

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varying.

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Therefore, we will
take that term

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as being the small
term in that equation.

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And this will be nicely
implemented by considering,

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as we said, h bar going to
0, or h as a small parameter.

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We will learn, as we
do the approximation,

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how to quantify what something
that we call slowly varying is.

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But we will take h now
as a small parameter--

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that makes that term small--

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and setup an expansion
to solve this equation.

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That is our goal now.

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So how do we set it up?

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We set it up like
this-- we say s of x,

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as you've learned in
perturbation theory,

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is s0 of x plus h bar s1 of
x-- the first correction--

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plus h bar squared is s2
of x and higher order.

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Now, s already has
units of h bar--

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so s0 will have
units of h bar too.

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s1, for example, has no units.

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So each term has a
different set of units.

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And that's OK, because
h bar has units.

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And this will go like
1 over h bar units

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and so forth and so on.

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So we have an expansion.

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And this we'll call our
semiclassical expansion.

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As we apply now this
expansion to that equation,

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we should treat h as we treated
lambda in perturbation theory.

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That is, we imagine that
this must be hauled order

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by order in h bar,
because we imagine

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we have the flexibility
of changing h bar.

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So let's do this.

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00:19:25,150 --> 00:19:27,220
What do we have in the equation?

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We have s prime.

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So the equation has s0
prime plus h bar s1 prime.

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00:19:40,770 --> 00:19:43,860
Now, I will keep terms
up to order h bar.

248
00:19:43,860 --> 00:19:53,700
So I will stop here-- plus
order h bar squared minus i h

249
00:19:53,700 --> 00:19:56,930
bar s double prime.

250
00:19:56,930 --> 00:20:02,950
So that should be s0
double prime plus order

251
00:20:02,950 --> 00:20:08,800
h bar equal b squared of x.

252
00:20:15,540 --> 00:20:17,365
So let's organize our terms.

253
00:20:21,170 --> 00:20:30,220
We have s0 prime primed
squared on the left-hand side.

254
00:20:30,220 --> 00:20:32,760
Minus p squared of x--

255
00:20:32,760 --> 00:20:36,940
I bring the p to
the left-hand side--

256
00:20:36,940 --> 00:20:39,640
plus h bar.

257
00:20:39,640 --> 00:20:45,100
So these are terms
that have no h bar.

258
00:20:45,100 --> 00:20:50,530
And when we look at the h bar
from the first term there,

259
00:20:50,530 --> 00:20:53,140
we square this thing.

260
00:20:53,140 --> 00:20:56,920
So you get the cross
product between the s0 prime

261
00:20:56,920 --> 00:20:58,310
and the s1 prime.

262
00:20:58,310 --> 00:21:03,990
So you get 2--

263
00:21:03,990 --> 00:21:08,250
the h prime already is there--

264
00:21:08,250 --> 00:21:11,175
s0 prime s1 prime.

265
00:21:14,240 --> 00:21:19,190
And from the second
term, you get

266
00:21:19,190 --> 00:21:32,275
minus i s0 double prime plus
order h squared equals 0.

267
00:21:39,960 --> 00:21:43,020
So I just collected
all the terms there.

268
00:21:46,920 --> 00:21:50,160
So if we're believing
in this expansion,

269
00:21:50,160 --> 00:21:53,720
the first thing we should
say is that each coefficient

270
00:21:53,720 --> 00:21:57,500
in the power series
of h bar is 0.

271
00:21:57,500 --> 00:21:59,210
So we'll get two equations.

272
00:22:03,120 --> 00:22:10,854
First one is s0 prime squared
is equal to p squared of x.

273
00:22:10,854 --> 00:22:13,260
That's one.

274
00:22:13,260 --> 00:22:16,850
That's this term equal to 0.

275
00:22:16,850 --> 00:22:21,500
And here, we get--

276
00:22:21,500 --> 00:22:25,340
let's write it in a way that
we solve for the unknown.

277
00:22:25,340 --> 00:22:30,110
Supposedly from this first
equation we, can solve for s0.

278
00:22:30,110 --> 00:22:34,110
If you know s0, what do you
want now to know is s1 prime.

279
00:22:34,110 --> 00:22:39,950
So let's write this as
s1 prime is equal to i s0

280
00:22:39,950 --> 00:22:47,120
double prime over
2 s0 zero prime.

281
00:22:47,120 --> 00:22:50,200
So these are my two equations.