PROFESSOR: So let's do an example where
we can calculate from the beginning
to the end everything.
Now, you have to get accustomed to the idea of even
though you can calculate everything,
your formulas that you get sometimes are a little big.
And you look at them and they may not
tell you too much unless you plot them with a computer.
So we push the calculations to some degree,
and then at some point, we decide
to plot things using a computer and get some insight
on what's happening.
So here's the example.
We have a potential up to distance, a, to 0.
The wall is always there, and this number is minus v naught.
So it's a well, a potential well.
And we are producing energy.
Eigenstates are coming here.
And the question now is to really calculate the solution
so that we can really calculate the phase shift.
We know how the solutions should read,
but unless you do a real calculation,
you cannot get the phase shift.
So that's what we want to do.
So for that, we have to solve the Schrodinger equation.
Psi of x is equal to what?
Well, there's a discontinuity.
So we probably have to write an answer in which we'll
have a solution in one piece and a solution in the other piece.
But then we say, oh, we wrote the solution
in the outside piece already.
It is known.
It's always the same.
It's universal.
I don't have to think.
I just write this.
E to the i delta.
I don't know what delta is, but that's
the answer, E to the i delta sine kx plus delta should
be the solution for x greater than a.
You know if you were not using that answer,
it has all the relevant information
for the problem, time delays, everything,
you would simply write some superposition of E
to the i kx and E to the minus i kx with two coefficients.
On the other hand, here, we will have, again, a wave.
Now, it could be maybe an E to the i kx or E
to the minus i kx.
Neither one is very good because the wave function
must vanish at x equals 0.
And in fact, the k that represents the kinetic energy
here, k is always related to E by the standard quantity,
k squared equal to mE over h squared or E equal
the famous formula.
On the other hand, there is a different k here
because you have different kinetic energy.
There must be a k prime here, which is 2m E plus v naught.
That's a total kinetic energy over h squared.
And yes, the solutions could be E to the ik prime
x equal to minus ik prime x minus ik prime x,
but since they must vanish at 0, should be a sine function.
So the only thing we can have here
is a sine of k prime x for x less than a and a coefficient.
We didn't put the additional normalization here.
We don't want to put that, but then we
must put the number here, so I'll put it here.
That's the answer, and that's k and k prime.
Now we have boundary conditions that x equals a.
So psi continues at x equals a.
What does it give you?
It gives you a sine of ka is equal to E to the i delta
sine of ka plus delta.
And psi prime continues at x equals a will give me
ka cosine ka equal--
I have primes missing; I'm sorry, primes--
equals k E to the i delta cosine ka plus delta.
What do we care for?
Basically we care for delta.
That's what we want to find out because delta tells us
all about the physics of the scattering.
It tells us about the scattering amplitude, sine squared delta.
It tells us about the time delay, and let's calculate it.
Well, one way to calculate it is to take
a ratio of these two equations so that you get rid
of the a constant.
So from that side of the equation,
you get k cotangent of ka plus delta
is equal to k prime cotangent of k prime a.
Or cotangent of ka plus delta is k prime over k.
We'll erase this.
And now you can do two things.
You can display some trigonometric wizardry,
or you say, OK, delta is arc cotangent of this minus ka.
That is OK, but it's not ideal.
It's better to do a little bit of trigonometric identities.
And the identity that is relevant
is the identity for cotangent of a plus b is cot a cot b minus 1
over cot a plus cot b.
So from here, you have that this expression
is cot ka cot delta minus 1 over cot ka plus cot delta.
And now, equating left-hand side to this right-hand side,
you can solve for cotangent of delta.
So cotangent of delta can be solved for--
and here is the answer.
Cot delta is equal to tan ka plus k prime over k cot k
prime a over 1 minus k prime over k cot k prime a tan ka.
Now, who would box such a complicated equation?
Well, it can't be simplified any more.
Sorry.
That's the best we can do.