WEBVTT

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PROFESSOR: Let's do
E less than V not.

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So we're back here.

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And now of the energy e
here is v not is x equal 0.

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X-axis.

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And that's the situation.

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Now you could solve this again.

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And do your
calculations once more.

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But we can do this
in an easier way

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by trusting the principle
of analytic continuation.

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In this case, it's very
clear and very unambiguous.

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So the big words,
analytic continuation,

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don't carry all the
mathematical depth.

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But it's a nice, simple thing.

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We first say that the solution
is the same for x less than 0.

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So for x less than 0, we
write the same solution.

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Because the energy
is greater than 0,

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or all what we said
here, the value of k

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squared, a into the ikhd
e to the minus i k x.

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It's all good.

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And k squared is still
2 m e over h squared.

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The problem is the region
where x is greater than 0.

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Because there you
have an exponential.

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But now you must have
a decaying exponential.

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But we know how
that should work.

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It should really be an e
to the minus some kappa x.

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So how could I achieve that?

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If I let k bar replace--

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everywhere you see k bar,
replace it by i kappa.

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Park then one thing
that happens is that--

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you learn from here,
from this k bar squared,

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would be minus kappa
squared equal to that.

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So kappa squared would be 2 m,
v not minus e over h squared.

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The sign is just the
opposite from this equation.

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That's what that equation
becomes upon that substitution.

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Now that substitution would
not make any difference, how

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they put a plus i or minus
i, I wouldn't have gotten

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my sine change, and this.

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But if I look at
the solution there,

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the solution psi becomes--

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on the region x greater than
0, turns into c, e to the i.

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kappa bar is a i kappa x.

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Therefore, it's equal to
c, e to the minus kappa

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x, which is the right thing.

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And that sine that I
chose, of letting k equal i

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kappa proves necessary
to get the right thing.

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So it's clear that to get the
right thing, you have that.

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And now you know
that, of course,

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if you would have written the
equation from the beginning,

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you would have said,
yes, in this region,

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there is a decaying thing.

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And looking at the
Schrodinger equation,

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you have concluded that
kappa is given by [? that. ?]

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But the place where
you now save the time

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is that, since I just must do
this change in the equations,

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I can do that change in
the solutions as well.

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And I don't have to write the
continuity equations again,

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nor solve them.

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I can take the solutions
and let everywhere

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that was a kappa bar replaced by
i, that was a k bar, replace it

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by i kappa bar.

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So what do we get?

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It should go here and
believe those circuits.

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OK, so b over a,
that used to be.

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Top blackboard there.

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Middle, k minus k bar
becomes k plus i--

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no, minus I k bar,
minus i kappa.

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And k plus i kappa.

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so it has changed.

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Suddenly this ratio
has become complex.

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It's kind of interesting.

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Well, let's make it clearer
by factoring a minus i here.

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So this becomes kappa.

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And you need plus k, so this
must be plus i k over i.

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This would be kappa minus i k.

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So this is just minus kappa
plus i k over kappa minus i k.

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But when you see
that ratio, you're

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seeing the ratio of two complex
numbers of equal length.

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And therefore, that
ratio is just a phase.

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It's not any magnitude.

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So this is just a phase,
and it deserves a new name.

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There is a phase shift
between the b coefficient

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and the a coefficient.

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And we'll write it as e minus--

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the minus I'll keep.

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e to the 2 i delta.

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That depends on the energy.

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I'll put delta of the energy,
because after all, kappa, k,

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everybody depends on the energy.

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So let's call it 2 i delta of e.

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And what is delta of e?

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Well, think of the
number kappa plus i k.

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This is the k, i k here.

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The angle-- this complex
number has an angle

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that, in fact, is delta.

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e to i delta is that phase.

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And delta is the arc
tangent, k over kappa.

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So I'll write it like that.

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Delta.

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Now you get a delta from the
numerator, a minus delta,

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as you can imagine,
from the denominator.

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And that's why you get
a total 2 i delta here.

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So delta of e is 10
minus 1, k over kappa.

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And if you look at what k
and kappa were, k over kappa

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is like the ratio of the
square root of the energy

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over v not minus the energy.

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So delta of e is equal to 10
minus 1 square root of energy

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over e v not minus and energy.

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Now I got the question
about current conservation.

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What happens to current
conservation this time?

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Well, you have all
these waves here.

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But on the region x greater
than 0, the solution is real.

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If the solution is real, there
is no probability current

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on the right.

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There's really no probability
that you get this thing,

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and you get current
flowing there.

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And you get this
pulse, or whatever

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you send to keep moving and
moving and moving to the right.

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Indeed, the solution decays.

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And it looks like the one of
the bound state in this region.

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So eventually there's
no current here,

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because there's no current here.

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No current there.

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No current there.

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Because the solution drops down.

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But it's a real solution anyway.

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So there's no current there.

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So Ja-- Jc is equal to 0.

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0.

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Solution is real for
x greater than 0,

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and any way goes
to 0 at infinity.

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So the fact that it's real
is a mathematical nicety

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that help us realize
that it must be 0.

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But the fact that there's no
current far away essentially

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telling you better be 0.

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So if the current Jc is
0, Ja must be equal to Jb.

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And therefore that means a
squared is equal to b squared.

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And happily that's what
happened because b over a

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is a complex number
of magnitude one.

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So the fact that b and a
differ by just the phase

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was required by
current conservation.

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A over b is a number
that has norm equal to 1.

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So that's a consistent picture.

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This phase is very important.

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So what happens for
resolution for x less than 0?

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Well, psi of x would
be a into the i k x,

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plus b all the way
to the left there.

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Your solution is a plus
b equal minus i k x.

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Of course, we now know what
the b is, so this is minus a

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from the ratio over there.

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e to the 2 i delta of e.

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E to the minus i k x.

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For x less than 0.

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And for x greater than 0,
psi of x is going to c e

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to the minus kappa x.

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And I'm not bothering
to write the coefficient

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c in Terms of a.

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Now this expression
for x less than 0

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can be simplified a little.

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You can factor an a.

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But it's very nice,
and you should

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have an eye for those
kind of simplifications.

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It's very nice to
factor more than an a

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and to factor one
phase like an i delta.

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I delta of e,
because in that way,

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you get e to the i k x minus
delta of e from the first term,

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where the two delta
appearances cancel each other.

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Because the first term
didn't have a delta.

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But then the second
term will have

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the same argument here,
of the exponential

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but with a minus sign.

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Minus i k x.

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And I claim also a
minus delta of e.

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And this time minus and
minus gives you plus.

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And the other e to the i
delta gives you back the 2.

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But now you've created the
trigonometric function,

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which is simpler to work with.

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So psi effects is equal to
2 i a e to i delta of e,

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sine of k x minus delta of e.

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And if you wish psi squared,
the probability density

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is proportional to 4 a
squared, sine squared

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of k x minus delta of e.

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So this can be plotted.

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Sine squared is like that.

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And where is x equal 0.

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OK, I'll say x equal
0, say is here.

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So this is not
really true anymore.

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But this point here is x 0.

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It vanishes.

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Would be the point at which
k x 0 is equal to delta of e.

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And the sine squared vanishes.

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So this is not a
solution either.

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Solution is like that.

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And then this is
for psi squared.

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And then it must couple to the
k exponential on this side.

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So that the true solution must
somehow be like this and well,

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whatever.

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I don't know how it looks.

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That is the e to the minus 2
kappa x decay and exponential.

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It must decay.

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There's continuity of the
derivative and continuity

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of the wave function.

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So that's how this should look.

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A couple more things we can
say about this solution that

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will play a role later.

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I want to get just a little
intuition about this phase,

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delta of e, this phase shift.

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So we have it there.

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Delta of e, I'll write it
here, so you won't have to--

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10 minus 1, square root
of e over v not minus e.

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So this is interesting.

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This phase shift just applies
for energies up to v not.

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And that corresponds to the
fact that we've been solving

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for energies under the barrier.

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And if we solve for
energies under the barrier,

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well, the solutions
as we're writing

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with these complex
numbers, apply up

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to energies equal to the
barrier, but no more.

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So we shouldn't plot
beyond this place.

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And here is delta of e.

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The phase shift.

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And when the energy is
0, when your particle

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you're sending in, or
the packet eventually

00:16:45.580 --> 00:16:48.340
is very low energy here.

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Then the phase shift is the
arctangent of 0, which is 0.

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As the energy goes
to the value v not,

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then the denominator goes to 0.

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The ratio goes to infinity.

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And the arctangent is pi over 2.

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So it's a curve that
goes from here to here.

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And it's not quite
like a straight line.

00:17:18.910 --> 00:17:22.109
But because of the
square roots, it sort of

00:17:22.109 --> 00:17:30.140
begins kind of vertical,
then goes like this.

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It's not flat either,
in the middle.

00:17:33.150 --> 00:17:35.820
So maybe my curve
doesn't look too good.

00:17:40.230 --> 00:17:42.665
Those more vertical here.

00:17:42.665 --> 00:17:46.570
Wow, I'm having a
hard time with this.

00:17:46.570 --> 00:17:47.530
Something like this.

00:17:51.050 --> 00:17:53.620
In fact, it's kind of
interesting to plug

00:17:53.620 --> 00:17:59.810
the derivative, d
delta, d energy.

00:17:59.810 --> 00:18:03.430
A little calculation will
give you this expression.

00:18:03.430 --> 00:18:09.180
You can do this with
mathematica or v not minus e.

00:18:09.180 --> 00:18:15.470
And shows, in fact, that here
is v not, and here is v delta,

00:18:15.470 --> 00:18:16.240
v e.

00:18:16.240 --> 00:18:18.810
We could call it
delta prime of e,

00:18:18.810 --> 00:18:24.420
because we wrote the phase shift
as a function of the energy.

00:18:24.420 --> 00:18:30.960
So that the delta, d e is
really delta prime of e.

00:18:30.960 --> 00:18:34.690
And it sort of infinite--

00:18:34.690 --> 00:18:38.300
goes to a minimum and infinite
again, in that direction.

00:18:38.300 --> 00:18:40.390
That's how it behaves.