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PROFESSOR: Find
our final solution,
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we just have to
match the equations.
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So Psi continues at x equals a.
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And what do we have?
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Well from two, you
have cosine of ka.
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And from four you would have
a equals e to the minus ka.
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This is the value of this--
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so the interior
solution at x equals
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a must match the value of
the exterior solution of k
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equals a.
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Psi prime must be continuous
at x equals a as well.
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Well what is the derivative
of this function?
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It's minus the sine of this.
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So it's minus k sine
of kx, that becomes ka,
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is equal to the derivative
of that one, which
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is minus Kappa A e to
that minus Kappa little a.
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Two equations, and
how many unknowns?
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Well there's A and some
information about Kappa and k.
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And the easiest way to eliminate
that is to divide them.
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So you divide the bottom
equation by this equation.
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So what do we get?
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Divide the bottom by the top.
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Minus k and and
the minuses cancel,
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we can cancel those
minus signs and you
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get k tan ka is equal to Kappa.
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But you already are
convinced, I hope, on the idea
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that we should not use
equations that have units.
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So I will multiply by
little a and a little a
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to get a [INAUDIBLE] size, and
therefore, the right hand side
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becomes Xi equals, and the
left side become Eta tan Eta.
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OK, I want to make a little
comment about these quantities
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already.
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So all the problem has turned
out into the following.
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You were given a potential and
that determines a number z0.
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If you know the width and
everything, you know z0.
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Now you have to
calculate Eta and Xi.
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If you know either Eta or
Xi, you know Kappa or k.
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And if you know either k or
Kappa, since you know v0,
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you will know the energy.
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So it's kind of neat to
express this more clearly,
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and I think it's maybe
easier if one uses Xi.
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And look at Xi squared is
Kappa squared times a squared.
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And what is Kappa
squared, it's over there.
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2m absolute value of e, a
squared over h bar squared.
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Now you want to find e, you're
going to get in some units.
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Even e is nice to
have it without units.
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So I will multiply
and divide by v0.
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2m v0 a squared over h squared,
absolute value of e over v0.
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After all, you probably
prefer to know e
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over v0, which tells you how
proportional the energy is
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to the depth of the potential.
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And this is your
famous constant z0.
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So e over v0 is actually
equal to Xi over z0 squared.
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And this is something
just to keep in mind.
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If you know Xi, you
certainly must know z0,
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because that's not
in your potential,
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and then you know how
much is the energy.
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All very convenient things.
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So punchline for solutions.
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So what do we have?
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We have two equations.
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This equation maybe
should be given a number.
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Xi equals 8 at tan Eta and Eta
and Xi squared giving you z
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squared.
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So how do we solve it?
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We solve it graphically.
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We have Psi, Eta,
and then we say,
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oh, let's try to plot
the two equations.
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Well this is a circle.
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Eta squared plus Psi squared.
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Now Xi and Eta must
be positive, so we
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look at solutions
just in this quadrant.
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Let's put here pi over 2.
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Pi 3 pi over 2, 2 pi, and
here is Eta and there is Xi.
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Well this is a circle, as we
said, but let's look at this.
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Xi equals Eta tan Eta.
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That vanished as Eta goes
to 0 and will diverge
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at Eta equals pi over 2.
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So this part, at
least, looks like this.
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And then it will go
negative, which don't care,
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from this region, and
then reach here at pi.
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And after pi, it will
go positive again
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and it will reach
another infinity here.
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And then at 3 pi, at 2 pi,
it will go again and reach
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not another infinity like that.
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So these are these curves.
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And the other curve, the
circle, is just a circle here.
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So, for example, I could
have a circle like this.
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So the radius of this
circle is radius z0.
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And there you go, you've
solved the problem.
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At least intuitively
you know the answer.
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And there's a lot of things that
come out of this calculation.
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If the radius z0 is 3
pi over 2, for example,
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and the radius z0 represents
some potential of some depth
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and width, there will
be just two solutions.
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These are these solutions.
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These points represent values
of Xi and values of Eta,
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from which you could
read the energy.
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In fact, you can look
at that state and say,
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that's the state of
largest Xi, and therefore
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it's the state of the largest
absolute value of the energy.
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It's the most
deeply bound state.
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Then this is next
deeply bound state.
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So there's two bounce
states in this case.
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Interestingly, however shallow
this potential might be,
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however small, z0, the
circle, will always
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have one intersection,
so there will always
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be at least one solution.
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That's the end of that story.
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Let me say that for the
odd case, odd solutions,
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I will not solve it.
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It's a good thing to do
in recitation or as part
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of the home work as well.
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The answer for the odd case is
that Psi is equal to minus Eta
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[? Cot ?] Eta.
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And in that case, I'll give
you a little preview of how
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the this [? Cot ?] looks.
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It looks like this.
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And then there are more
branches of this thing.
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So for the odd solution,
you have these curves.
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And if you have a
circle, sometimes you
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don't have a solution.
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It doesn't intersect this.
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So these odd solutions, you
will see and try to understand,
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they don't always exist.
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You meet a potential
that is sufficiently deep
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to get an odd solution.
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And then the odds
and even solutions
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will interweave
each other and there
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will be a nice
story that you will
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explore in a lot of detail.
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But the is, you've reviewed
the problem to unit free
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calculation, in which you
can get the intuition of when
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solutions exist and
and when they don't.
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But solving for the
particular numbers
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are transcendental
equations, and you
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need a computer to solve.