PROFESSOR: E less than v0.
So you have an incident wave, e to the ikx, incident,
and a reflected wave that you have, e to the minus ikx--
remember minus the other face-- and e to the 2i delta of E.
So this is the incident wave, and this is the reflected wave.
They correspond to your Ae to the ikx
plus Be to the minus ikx.
Remember, when the energy was less than v0,
the range of B over A was minus e to the 2i delta.
And since I take A equals 1, you get this thing.
So suppose I construct a psi incident of x less than 0
and t as the sum 0 to infinity dk f of k e to the ikx
e to the minus iEt over h bar.
Whew.
So I superimpose the incident thing here.
Then the reflected one should be superimposed too
and would be 0 to infinity, a minus in front
because there's a minus, dk f of k e to the minus ikx
e to the 2i delta of E e to the minus iEt over h bar.
Whew.
That's the reflected wave superimposed.
So now you've constructed everything.
Here the reflected wave is more interesting
than the transmitted wave, because there's
no real big transmitted wave.
It just whistles out.
But the reflective thing is interesting.
If you're doing the experiment, you send in a particle,
you want to see what you get back.
That's going to tell you what kind of potential
you can expect it encountered.
So let's do the stationary phase for this one,
for the reflected.
Let's see how it moves.
We know how the incident moves.
The incident moves with x equals--
we've done it there-- hk0 over m t.
But how about this one?
Well, this one you would have to do d dk of minus kx
plus 2 delta of E minus Et over h bar, all that at k0 equals 0.
And you'll probably remember that this thing
was in your midterm and your first test.
You had this wave, and you had to analyze,
what did stationary phase do?
And it does that.
So what do you get?
Well, when you take the derivative,
you have to take the derivative of delta with respect
to energy, that's delta prime, and then derivative of energy
with respect to t.
Let me save you a little time.
The answer is minus h bar k0 m t minus 2h bar delta
prime of E. OK.
That's what you get.
That's how this packet moves.
And what does it do really?
Remember, this is defined for x less than 0.
So this is valid--
forget about this little term here--
this is valid for t positive.
For t positive, you're going to get this to satisfy.
So this is a big wave packet for t positive.
It's the reflected wave.
That's what you would expect.
This is the reflected.
Now, if this factor was not here,
it is as if, well, the incoming packet hit the origin at t
equals 0.
And this will be perfect bouncing in which
the packet gets reflected.
And at t equals 0, it starts to move to the left.
And as t increases, it's moved more and more to the left.
You see it there, because x must be negative.
But if there is this term, it really
doesn't start to move to the left
until t is bigger than that so that x is negative.
So only at t equal to this amount of time
the packet reflects.
So there's a delay, and the delay
is 2h bar delta prime of E.
So this is a technology people use
in scattering theory to figure out what kind of potential
you have, figure out how much things get
delayed from the bouncing.
Now, this derivative-- we've plotted it there-- delta
prime of E. You get a big delay for low energy,
for energies near v0, and in the middle it's not so big.