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