WEBVTT
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PROFESSOR: OK.
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That is the first step.
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But to make things
really clear, we
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have to do the second step that
it involves the Bi solution.
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So let me say something
about Bi as well.
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So Bi has similar
asymptotic expansions.
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A Bi solution, if psi, you
have an exact Bi solution.
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When psi on the left would
behave as minus 1 over pi,
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this is for u negative.
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1 over u to the 1/4 sine of 2/3
u to the 3/2 minus pi over 4.
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And psi on the right
of x would behave as 1
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over square root of pi,
1 over u to the 1/4, e
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to the 2/3, u over 3/2.
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OK.
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This is the asymptotic
behavior of B. So
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what's noteworthy about it?
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First, that it oscillates
for u negative.
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That's to the left.
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It oscillates.
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Instead of a cosine, a
sine, the same phase shift.
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And then that on the other side
for u positive, it blows up.
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So it uses the
other exponential.
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And this all is reasonable.
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And in fact, it allows
you to connect as well.
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So in this case, we would have
this term looks like this one.
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And the left one looks like--
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so indeed, if you
had the Bi solution,
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the matching requires
that the D term over there
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is matched to the B term
over here because that's how
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the Bi solution is connected.
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So we now need to
relate the B to the D.
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So I will write down
the relation here.
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So in this case,
there's no factor of 2,
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but there is a minus sign.
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So we have B must
be equal to minus D.
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So you can take B equals to
minus 1 and D equals to 1.
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And we get 1 over square root
of kfx sine x to A. Kfx prime dx
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prime minus pi over 4 connects
in principle to minus 1
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over square root of kappa of x
e to the integral from A to x,
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kappa of x prime dx prime.
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I think I have it all right.
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B is there.
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B is minus 1.
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It's over there.
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D is equal to 1.
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That's the D term.
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That's the other
connection condition.
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OK.
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So the technicalities are gone.
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But the concept now requires
a serious discussion.
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I'll give you a little
time to take it in.
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We've connected every functions.
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That was the first work
we did at the beginning
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of this lecture.
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We did, in the top blackboard
justify, this expansions.
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We discussed how the Bi function
would originate from a counter
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integral as well.
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And we didn't derive the
Bi asymptotic expansions,
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but they're similarly derived.
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And then we used those
to connect things.
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And now the question is whether
these equations can be used.
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If I know the solution
on the left is like that,
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can I say the solution on
the right is like that?
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Or if I know the solution
on the right is like that,
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can I say the equation
is on the left?
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And the same thing here.
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If I know the
solution on the left,
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do I know the
solution on the right?
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And if I know the
solution on the right,
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do I know the
solution on the left?
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Now it looks like yes,
that's what you derive.
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But remember, it's
not quite simple
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because our whole discussion
assume that OK, the potential
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is strictly linear.
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But in reality, the
potential starts to deviate.
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So these things
that were right here
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are not exact
asymptotic expansions.
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That is the exact dominant term
of the asymptotic expansions.
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But here there
may be extra terms
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because the solution
does not correspond
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to a potential is
exactly linear.
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And anyway, this is
not an exact solution
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of the differential equation.
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So there is a possibility
of ambiguities here
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that actually indicate
the following.
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I claim this connection can
only be done in this way.
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You can only take this equality
from the right to the left.
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That is, if you
know that you just
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have a decaying
exponential, you know
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it connects to this function.
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Why?
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Because if you know you
have a decaying exponential,
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there's 0 chance
there is a growing
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exponential on the right.
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You may have a barrier
that extends forever,
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and there's definitely only
a decay in exponential,
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however the barrier looks.
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If there's only a
decaying exponential,
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you know that this growing
exponential is just not there.
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The coefficient here is 0.
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So you don't have the
ambiguity that maybe you
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have a little bit of a
growing exponential that
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connects to this.
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So you can go from
here to there.
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On the other hand, you
cannot quite go from the left
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to the right.
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Why is that?
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The reason is the following.
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If you just know
this is on the left,
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you would say, OK, now I predict
a decaying exponential there.
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But your calculation
is not totally exact,
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so there could be an
infinitesimal wave
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of the other type on the
left, an infinitesimal wave
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of this kind.
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But that infinitesimal
wave of this kind
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goes into a growing exponential,
and a growing exponential
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eventually overtakes this
one and ruins your solution.
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So you cannot really use this
equation from right to left.
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The small possibility of
error on the left side,
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because you don't
have exact solutions,
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translates into an
ambiguity concerning
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a positively
growing exponential,
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and it would overtake this.
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So if you say oh, I
have this, therefore I
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have this decaying
exponential, this
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may be a very inaccurate
thing because maybe you
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have a little bit
of the sine that you
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didn't see because it was
much smaller than this term.
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But that would give you
a growing exponential.
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So you cannot use that
equation in that direction.
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Similarly here, you
have another direction.
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The direction is this one.
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It's crucial.
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You can go that way, but you
cannot go the other direction
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as well.
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And the reason is
kind of similar.
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Suppose you have
here, you say, OK,
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I have this growing exponential.
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And say somebody comes in, no,
there's a little bit of cosine
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here.
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And you say well, but
a little bit of cosine
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gives me a decaying
exponential that is irrelevant
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compared to this one.
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So you are safe in going
from this side to that side.
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You're going into
growing exponential.
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Any error on the left
will produce a small error
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on the right.
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Of the other hand, you
cannot go, honestly,
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from right to left because
if you had a decaying
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exponential-- this is
a growing exponential.
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If you had the
decaying exponential,
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you wouldn't see it here.
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This one takes over.
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But the decaying exponential
produces a nice cosine
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that is quite
comparable to this one.
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So a decaying exponential
that you don't see here
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produces on this side a term
that is comparable to this one.
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So you also cannot go from this
direction to that direction.
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So these are the
connection formulas
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and these are the directions.
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We've done this for this
configuration in which we have
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a turning point of this form
at the point A, in which you're
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allowed to the left and
forbidden to the right.
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For convenience, it's
useful to have a formula--
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let's go here-- where you have
the opposite direction in which
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you have a B here, and you
have allowed to the right
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and forbidden to the left.
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I will write those formulas
because they are many times
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used and you will
need them in general.
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And Griffiths and other
books don't have the patience
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to give you all the formulas.
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This is kappa of
x prime, vx prime.
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This is growing or
decaying exponential?
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This should be called
the growing exponential
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on the left.
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This is the forbidden region.
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As x becomes more and more
left, the interval is bigger.
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So this is a growing
exponential on the left.
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1 over k of x sine b to x,
k of x prime, vx prime minus
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pi over 4.
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And this one, again,
just like that one,
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you go into the
growing exponential.
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So that's that relation.
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And I kind of write these
things following this picture.
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So you're allowed to the right
and forbidden on the left side.
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The allowed way
function the right,
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the forbidden way
function to the left.
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1 over kappa of x,
exponential of minus x
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to B. Kappa of x prime
vx prime goes into--
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a decay in exponential you can
follow into the other region.
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And that's what we
do here, as before.
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The decaying exponential
is safe because if you
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know there's just a
decaying exponential,
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there's no possibility of
having the growing exponential.
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So these are your conditions
for that turning point,
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and that's how you use them.
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So one example that we will
not get to discuss now,
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but will be in the notes, is
tunneling across a barrier.
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And you will use the connection
conditions in this case
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here, and for this case, the
other ones that we've done.
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So you need for tunneling
both connection conditions.
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It's a very nice exercise.
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I recommend that
you play with it,
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and we'll put it in
the notes as well.
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So let's stop here.