WEBVTT
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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.
00:14:24.580 --> 00:14:28.120
And let's write it
in a way that you
00:14:28.120 --> 00:14:31.420
can see maybe the units better
and get a little intuition
00:14:31.420 --> 00:14:34.420
about what this
computation gives.
00:14:34.420 --> 00:14:39.910
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.
00:14:56.120 --> 00:15:08.460
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.
00:15:18.150 --> 00:15:21.110
So you can just invert it.
00:15:21.110 --> 00:15:23.745
This is more dangerous when
you have partial derivatives.
00:15:23.745 --> 00:15:28.950
This is not necessarily true but
for this ordinary derivatives
00:15:28.950 --> 00:15:34.200
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.
00:15:42.570 --> 00:15:46.680
And here, we recognize
that this is 2,
00:15:46.680 --> 00:15:50.280
and this is nothing else
than the group velocity
00:15:50.280 --> 00:15:51.740
we were talking before.
00:15:51.740 --> 00:15:57.510
The E, the energy, is h
squared k squared over 2m.
00:15:57.510 --> 00:16:00.300
You differentiate,
divide by h bar,
00:16:00.300 --> 00:16:05.220
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.
00:16:11.460 --> 00:16:15.930
So this derivative is really
evaluated at k naught.
00:16:15.930 --> 00:16:18.490
This is also
evaluated at k naught.
00:16:21.640 --> 00:16:30.380
So this is the group velocity,
d delta, dk, and finally,
00:16:30.380 --> 00:16:35.100
let me rewrite it in a
slightly different way.
00:16:35.100 --> 00:16:41.320
I multiply by 1 over R. Why?
00:16:41.320 --> 00:16:46.360
Because d delta dk, k has
units of 1 over length.
00:16:46.360 --> 00:16:50.860
So if I multiply by 1 over
R, this will have no units.
00:16:50.860 --> 00:17:04.869
So I claim that one over R d
delta dk is equal to delta t,
00:17:04.869 --> 00:17:16.250
and you'll have 2 over vg
and R. So I did a few steps.
00:17:16.250 --> 00:17:22.190
I moved the 2 over
vg down to the left,
00:17:22.190 --> 00:17:25.880
and I multiplied by
1 over R, and now we
00:17:25.880 --> 00:17:29.430
have a nice expression.
00:17:29.430 --> 00:17:31.220
This is the delay.
00:17:31.220 --> 00:17:34.820
Delta t is the
delay, but you now
00:17:34.820 --> 00:17:41.690
have divided it by 2R divided
by the velocity, which
00:17:41.690 --> 00:17:46.010
is the time it takes the
particle with the group
00:17:46.010 --> 00:17:55.190
velocity to travel back and
forth in the finite range
00:17:55.190 --> 00:17:55.980
potential.
00:17:55.980 --> 00:17:57.920
So that gives you an idea.
00:17:57.920 --> 00:18:00.680
So if you compute
the time delay,
00:18:00.680 --> 00:18:03.530
again, it will have
units of microseconds,
00:18:03.530 --> 00:18:08.240
and you may not know if
that's little or much.
00:18:08.240 --> 00:18:11.210
But here, by computing
this quantity,
00:18:11.210 --> 00:18:15.800
not exactly delta prime
of v but this quantity.
00:18:15.800 --> 00:18:17.720
You get an [? insight,
?] because this is
00:18:17.720 --> 00:18:24.510
the delay divided by
the free transit time.
00:18:34.680 --> 00:18:37.330
It's kind of a nice quantity.
00:18:37.330 --> 00:18:43.060
You're dividing your delay
and comparing it with the time
00:18:43.060 --> 00:18:45.410
that it takes a particle,
with a velocity that
00:18:45.410 --> 00:18:49.940
is coming in, to do the
bouncing across the finite range
00:18:49.940 --> 00:18:51.790
potential.