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PROFESSOR: OK,
so, local picture.

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It's all about getting
insight into how

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the way function looks.

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That's what we'll need to get.

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These comments now
will be pretty useful.

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For this equation you
have one over psi,

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d second psi, d x
squared is minus 2 m

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over h squared, E minus v of x.

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Look how I wrote it, I
put the psi back here,

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and that's useful.

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Now, there's a whole
lot of discussion--

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many textbooks-- about
how the way function

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looks, and they say concave
or convex, but it depends.

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Let's try to make it very clear
how the wave function looks.

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For this we need two regions.

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So, the first case, A, is
when the energy minus v of x

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is less than 0.

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The energy is less than v of
x, that's a forbidden region--

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as you can see there--

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so it's a classically forbidden.

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Not quantum
mechanically forbidden,

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but classically forbidden.

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What is the main thing about
this classically forbidden

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region is that the right
hand side of this equation

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is positive.

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Now, this gives you
two possibilities.

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It may be that psi at some
point is positive, in which case

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the second psi must
also be positive,

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because psi and the
second psi appear here.

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If both are positive,
this is positive.

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Or, it may be case two,
that psi is negative,

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and the second psi--

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the x squared--
it's also negative.

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Well, how do we plot this?

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Well, you're at some
point x, and here it

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is, a positive
wave function seems

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to be one type of convexity,
another type of convexity

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for a negative, that's why
people get a little confused

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about this.

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There's a way to see in a way
that there's is no confusion.

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Look at this, it's positive,
second derivative positive.

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When you think of a second
derivative positive,

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I think personally of
a parabola going up.

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So, that's how it could look.

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The wave function is
positive, up, it's all real.

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We're using the thing we
proved at the beginning

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of this lecture: you can work
with real things, all real.

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So, the wave from here
is x, and here negative.

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And the negative
opening parabola,

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that's something they got.

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So nice.

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So, the wave
function at any point

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could look like this
if it's positive,

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or, it could look like
this if it's negative.

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So, it doesn't look like
both, it's not double value.

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So, either one or the other.

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But, this is easy
to say in words,

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it is a shape that is
convex towards the axis.

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From the axis it's convex
here and convex there.

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So, convex towards the axis.

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Now, there's another
possibility I want to just

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make sure you visualize this.

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Sometimes this looks funny--
doesn't mean actually the way

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function can look like that--

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but, it's funny because
of the following reason.

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It's funny because if you
imagine it going forever,

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it doesn't make
sense because you're

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in a classically the
forbidden region.

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And the way function's
becoming bigger and bigger

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is going to blow up.

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So, eventually
something has to happen.

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But, it can look like this.

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So, actually what happens
is that when you're

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going to minus infinity--

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here is x and we
use minus infinity--

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it can look like this.

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This is an example of this
piece that is asymptotic,

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and it's positive, and the
second derivative is positive.

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Or, negative and the second
derivative is negative.

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So that's a left asymptote.

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Or, you could have
a right asymptote,

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and it looks like this.

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Again, second derivative
positive, positive wave

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function.

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Second derivative negative,
negative wave function.

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So, you may find this at
the middle of the potential,

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but then eventually
something has to take over.

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Or, you may find this behavior,
or this behavior, at plus minus

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infinity.

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But, in any case you are
in a classically forbidden,

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you're convex towards the axis.

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That's the thing
you should remember.

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On the other hand, we can be on
the classically allowed region.

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So, let's think of that.

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Any questions about the
classically forbidden?

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Classically allowed, B. E
minus v of x greater than 0,

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classically allowed.

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On the right hand side of
the equation is negative.

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So, you can have, one,
a psi that is positive,

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and a second derivative
that is negative.

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Or, two, a psi that is negative,
and a second derivative

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that this positive.

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So, how does that look?

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Well, positive and second
derivative negative,

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I think of some wave function
as positive, and negative

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is parabolic like that.

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And then, negative and
second derivative positive,

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it's possible to have this.

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The wave function
there it's negative,

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but the second
derivative is positive.

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These things are not very good--

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they're not very
usable asymptotically,

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because eventually
if you are like this,

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you will cross these points.

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And then, if you're still
in the allowed region

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you have to shift.

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But, this is done nicely in a
sense if you put it together

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you can have this.

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Suppose all of this is
classically allowed.

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Then you can have the wave
function being positive,

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the second derivative
being negative,

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matching nicely
with the other half.

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The second derivative positive,
the wave function negative,

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and that's what the
psi function is.

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It just goes one after another.

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So, that's what
typically things look

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in the classically
allowed region.

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So, in this case, we say that
it's concave towards the axis.

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That's probably
worth remembering.

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So, one more case.

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The case C, when E is equal to E
minus v of x not is equal to 0.

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So, we have the negative,
the positive, 0.

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How about when you have
the situation where

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the potential at some point
is equal to the energy?

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Well, that's the
turning points there--

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those were our turning points.

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So, this is how x 0
is a turning point.

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And, something else happens,
see, the right hand side is 0.

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We have that one over
psi, the second psi,

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the x squared is equal to 0.

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And, if psi is different
from 0, then you

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have the second derivative
must be 0 at x not.

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And, the second derivative
being 0 is an inflection point.

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So, if you have a wave function
that has an inflection point,

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you have a sign that you've
reached a turning point.

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An inflection point
in a wave function

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could be anything like that.

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Second derivative
is positive here--

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I'm sorry-- is negative here,
second derivative is positive,

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this is an inflection point.

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It's a point where the
second derivative vanishes.

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So, that's an inflection point.

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And, it should be remarked
that from that differential

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equation, you also get that
the second psi, the x squared,

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is equal to E minus v times
psi, which is constant.

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And, therefore,
when psi vanishes,

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you also get inflection
points automatically

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because the second
derivative vanishes.

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So, inflection points
also at the nodes.

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Turning point is
an inflection point

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where you have this situation.

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Look here, you have negative
second derivative, positive

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second derivative, the
point where the wave

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function vanishes and joins them
is an inflection point as well.

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Is not the turning point--

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turning point are
more interesting--

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but inflection points
are more generic.