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PROFESSOR: I'll begin
by reviewing quickly

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what we did last time.

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We considered what are called
finite range potentials,

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in which over a distance
R, in the x-axis,

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there's a non-zero potential.

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So the potential is some v of x
for x between capital R and 0,

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is equal to 0 for x
larger than capital R,

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and it's infinity
for x negative.

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So there's a wall at x equals 0.

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And there can be some
potential, but this

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is called a finite range
potential, because nothing

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happens after distance R.

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As usual, we considered
scattering solutions, solutions

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that are unnormalizable
with energies,

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h squared k squared over 2m,
for a particle with mass m.

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And if we had no potential,
we wrote the solution phi

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of x, the wave function,
which was sine of kx.

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And we also wrote it
as a superposition

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of an incoming wave.

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Now, an incoming
wave in this set up

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is a wave that propagates
from plus infinity towards 0.

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And a reflected wave is
a wave that bounces back

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and propagates towards
more positive x.

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So here we'll
write this as minus

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e to the minus ikx over 2i,
plus e to the ikx over 2i.

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This is the sine
function rewritten

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in terms of exponential
in such a way that

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here is the incoming wave.

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Remember the time dependence
is minus iet over h bar.

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So this wave
combined with a time

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is a wave that is moving
towards the origin.

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This wave is moving outwards.

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Then we said that there would
be, in general, with potential.

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With a potential, you would
have a solution psi effects,

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which we wrote after some
tinkering in the farm

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i delta sine of kx plus delta.

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And if you look at the
part of the phase that

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has the minus ikx would have a
minus delta and a delta here.

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So they would cancel.

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So this solution has
the same incoming wave

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as the no potential solution.

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On the other hand here, you
would have e to the 2i delta,

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e to the ikx over
2i, and this solution

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is only valid for x
greater than R. You see,

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this is just a plane
wave after all.

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There's nothing more than a
plane wave and a phase shift.

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The phase shift,
of course, doesn't

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make the solution any more
complicated or subtle,

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but what it does is, by
depending on the energy,

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this phase shift delta depends
on the energy, and we're on k.

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Then, it produces
interesting phenomena

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when you send in wave packets.

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So if we write psi,
we usually write

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psi is equal to
the phi plus psi s,

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where psi s is called
the scattered wave.

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You see, the full wave that
you get, for x greater than R,

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we would have to solve and work
very carefully to figure out

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what is the wave function
in the region 0 to R.

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But for x greater
than R is simple,

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and for x greater than
R the wave function psi

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is the free wave function,
in the case of no potential,

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plus the scattered wave.

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Quick calculation with this,
things [? give to ?] you

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the scatter wave is e to the i
delta sine delta e to the ikx

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is an outgoing wave.

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And this coefficient is called
the scattering amplitude.

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It's the amplitude of
the scattered wave.

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This is a wave that is going
out, and this is its amplitude.

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So it has something to do with
the strength of the scattering,

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because if there
was no scattering,

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the wave function would
just behave like the no

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potential wave function.

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But due to the potential,
there is an extra piece,

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and that represents
an outgoing wave

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beyond what you get outgoing
with a free no potential wave

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function.

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So it's the
scattering amplitude,

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and therefore sometimes we
are interested in as squared,

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which is just sine
squared delta.

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Anyway, those are the
things we did last time.

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And we can connect to some
ideas that we were talking about

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in the past, having to
do with time delays,

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by constructing a wave packet.

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That's what's usually done.

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Consider the process
of time delay, which

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is a phenomenon
that we've observed

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happens in several
circumstances.

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If you have an incident
wave, how do you

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construct an incident wave?

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Well, it has to be
a superposition of e

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to the minus ikx, for sure.

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So we'll put the function in
front, we'll integrate over k,

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and we'll go from 0 to infinity.

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I will actually add the
time dependence as well.

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So let's do phi of x and t.

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Then, we would have e to the
minus i, e of k, t over h bar,

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and this would be valid
for x greater than R.

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Again, as a solution of
the Schrodinger equation.

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You see, it's a free wave.

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There's nothing extra from what
you know from the de Broglie

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waves we started
a long time ago.

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So if this is your
incident wave,

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you have to now realize that
you have this equation over here

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telling you about the general
solution of the Schrodinger

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equation.

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The general solution of
the Schrodinger equation,

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in this simple region,
the outside region,

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is of this form, and it
depends on this delta

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that must be calculated.

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This is the incoming wave,
this is the reflected wave,

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and this is a solution.

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So by superposition, I construct
the reflected wave of x and t.

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So for each e to
the minus ikx wave,

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I must put down
one e to the ikx,

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but I must also put an e to
the 2i delta of the energy,

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or delta of k.

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And I must put an
extra minus sign,

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because these two
have opposite signs,

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so I should put a minus
0 to infinity dk f of k.

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And we'll have the e to the
minus i, e of k, e over h bar.

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And just for reference, f of
k is some real function that

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picks at some value k naught.

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So you see, just
like what we did

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in the case of the step
potential, in which we

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had an incident wave,
a reflected wave

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packet, a transmitted wave
packet, the wave packets go

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along with the basic solution.

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The basic solution had
coefficients A, B, and C,

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and you knew what B was
in terms of A and C.

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Therefore, you constructed the
incoming wave with A e to ikx,

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and then the reflected wave
with B e to the minus ikx.

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The same thing we're doing
here inspired by this solution,

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the psi affects we
superpose many of those,

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and that's what we've done here.

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Now of course, we can do the
stationary phase calculations

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that we've done several
times to figure out

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how the peak of the
wave packet moves.

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So a stationary phase
at k equal k naught.

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As you remember, the
only contribution

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can really come when k is near
k naught, and at that point,

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you want the phase to be
stationary as a function of k.

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I will not do here
the computation again

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for psi incident.

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You've done this computation
a few times already.

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For psi incident, you find
the relation between x and t,

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and I will just write it.

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It's simple.

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You find that x is
equal to minus h

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bar k naught over mt, or
minus some v velocity, group

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velocity, times t.

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That is the condition
for a peak to exist.

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The peak satisfies
that equation,

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and this makes sense
when t is negative.

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This solution for psi incident
only makes sense for x positive

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if in fact x greater than R. So
this solution needs x positive.

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So it needs t
negative [? indeed. ?]

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This is a wave that is
coming from plus infinity,

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x equal plus infinity,
at time minus infinity,

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and it's going in
with this velocity.

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For psi reflected,
the derivative

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now has to take the derivative
of delta, with respect to e,

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and then the derivative of e
with respect to the energy.

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And the answer, in this case--

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you've done this before--

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it's v group times t minus
2 h bar delta prime of E.

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So yes, in the reflected
wave, x grows as t grows

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and it's positive.

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t must now be
positive, but in fact,

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if you would have a
just x equal v group t,

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this would correspond
to a particle that

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seems to start at the origin
at time equals 0 and goes out.

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But this actually there is
an extra term subtracted.

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So only for t greater
than this number

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the particle begins to appear.

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So this is a delay, t
minus some t naught,

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the packet gets delayed
by this potential.

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Now, this delay can
really get delayed.

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Sometimes it might
even accelerated,

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but in general, the delay
is given by this quantity

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So I'll write it here.

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The delay, delta t, is 2
h bar delta prime of E.

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And let's write it
in a way that you

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can see maybe the units better
and get a little intuition

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about what this
computation gives.

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For that, let's differentiate
this with respect to k,

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and then k with
respect to energy.

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So v delta with respect to k,
and dk with respect to energy.

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This is 2 over 1 over h
bar dE with respect to k.

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I do a little rearrangement
of this derivative

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is one function
of one variable k

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and neither is a
single relation.

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So you can just invert it.

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This is more dangerous when
you have partial derivatives.

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This is not necessarily true but
for this ordinary derivatives

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is true, and then you have
this 2 to the left here.

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The h bar went all the way
down, and I have d delta dk.

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And here, we recognize
that this is 2,

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and this is nothing else
than the group velocity

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we were talking before.

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The E, the energy, is h
squared k squared over 2m.

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You differentiate,
divide by h bar,

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and it gives you the group
velocity hk naught over m.

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Because these
derivatives all have

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to be evaluated at k naught.

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So this derivative is really
evaluated at k naught.

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00:16:15,930 --> 00:16:18,490
This is also
evaluated at k naught.

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So this is the group velocity,
d delta, dk, and finally,

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let me rewrite it in a
slightly different way.

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I multiply by 1 over R. Why?

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Because d delta dk, k has
units of 1 over length.

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00:16:46,360 --> 00:16:50,860
So if I multiply by 1 over
R, this will have no units.

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00:16:50,860 --> 00:17:04,869
So I claim that one over R d
delta dk is equal to delta t,

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00:17:04,869 --> 00:17:16,250
and you'll have 2 over vg
and R. So I did a few steps.

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00:17:16,250 --> 00:17:22,190
I moved the 2 over
vg down to the left,

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00:17:22,190 --> 00:17:25,880
and I multiplied by
1 over R, and now we

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have a nice expression.

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This is the delay.

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Delta t is the
delay, but you now

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have divided it by 2R divided
by the velocity, which

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is the time it takes the
particle with the group

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velocity to travel back and
forth in the finite range

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potential.

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So that gives you an idea.

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So if you compute
the time delay,

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again, it will have
units of microseconds,

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and you may not know if
that's little or much.

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But here, by computing
this quantity,

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not exactly delta prime
of v but this quantity.

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You get an [? insight,
?] because this is

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the delay divided by
the free transit time.

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It's kind of a nice quantity.

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You're dividing your delay
and comparing it with the time

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that it takes a particle,
with a velocity that

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00:18:45,410 --> 00:18:49,940
is coming in, to do the
bouncing across the finite range

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00:18:49,940 --> 00:18:51,790
potential.