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PROFESSOR: Let me begin by
introducing the subject.

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The subject is resonances.

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And we have seen,
actually, a little bit

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of this in the
resonant transmission

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of the Ramsauer-Townsend effect.

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Because of a
resonance phenomenon

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within the square well
obstacle, somehow,

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for some particular frequencies,
for some particular energies,

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the particles were able to
zoom by without experiencing

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any reflection, whatsoever.

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So let's begin the
subject of resonances

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by asking a question.

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If you have the usual potential,
the short range potential,

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which means, that for some
distance, R, greater than 0,

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the potential is 0.

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Here we put a barrier,
and over there

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could be anything,
some potential.

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We've computed some-- this
concept of time delay,

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there's a formula
for the time delay.

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In fact, it was given by 2 h
bar d delta dE, the time delay,

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2 h bar d delta dE.

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And we discussed that this
time delay can be positive

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or it can be negative.

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If it's positive, it
really means a time delay.

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You send in a wave packet.

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And it takes time to come
back, more time than it

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would have taken if there
had been no potential.

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You see, the time delay,
you have a packet coming

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in from time minus infinity.

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And then it bounces back
a time equal infinity.

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But nevertheless, you
compare that with a situation

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in which there's no potential.

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And you see that there
is some time delay.

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If you time the wave packet
to reach at time equals 0,

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here, it will not reach back
to where you were by time--

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by whatever time-- suppose
you have the wave packet here

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at t equal minus 10, then
it goes here, and it delays,

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and at t equal 10, the
packet has not reached,

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there is a time delay,
a positive time delay.

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A negative time delay
is the opposite.

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The packet arrives
a little earlier.

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And the question
I want to ask you,

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if you have a
negative time delay,

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can it be arbitrarily large.

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Well, if you send
in a wave packet,

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it may find an infinite wall
here, and then may bounce,

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and then yes, it
comes back earlier

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than you expected, because
the free packet would

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have gone here and back.

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But you wouldn't expect it
to be able to come earlier

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than if there was an
infinite wall here,

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because there is no
infinite wall here,

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nor an infinite wall here.

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So it's just not going to
bounce before it reaches here.

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The best it can do is
bounce when it reaches here.

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So you should not expect,
and this, sometimes,

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will [INAUDIBLE],
there is nothing

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that can make it bounce
until you reach here.

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So you cannot expect that the
time advance is larger as if it

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would have bounced before
reaching the obstacle,

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whatsoever.

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So this is important.

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We cannot have a negative time
delay that this infinitely

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large.

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So, in fact, the time delay
as, we're right in here,

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should be greater than
the total travel distance

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that you may save.

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If you bounce here, you
would save 2R over v.

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And you must be greater than
that negative number, which

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is the total travel
time that it would take

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to go back and forth, here.

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So we can do a little
arithmetic, here.

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This is equal to 2 h bar d
delta dk, and here, dE dk.

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This is still greater than or
equal than minus 2R over v.

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And I want to put a sim, because
our argument is not completely

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rigorous as to what's
happening when it reaches here.

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It seems very
plausible classically,

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but there's a bit
of a correction

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if you do it exactly.

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So it's not an exact
inequality we're deriving.

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And what is the E dk is
h bar times the velocity.

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Remember, dE dk, you
are differentiating

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h squared k squared over 2m.

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And you get h bar
times hk over m.

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So therefore, this is
h bar and the velocity.

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And the h bars cancel.

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The velocities cancel.

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Between these two
sides, the 2s cancel.

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And you'll get that d delta dk
must be greater than or equal,

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approximately, to R.

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And that's sometimes
called Wigner's condition

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on scattering.

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And it basically is the
idea that the time delay,

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the time advance
cannot be too large.

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OK, so now we can ask
the second question.

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How about the time
delay, a true time delay,

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can it be very large?

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Can it be arbitrarily large?

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Suppose we have a
barrier of this form.

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And now you send a particle
with a little bit higher energy

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here.

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Now, this particle
is going to have

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very little kinetic energy.

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So it's going to travel quite
slowly here, and go back.

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And this time, it's going to
delay quite a bit, probably.

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But the problem is, if you
create-- there's nothing

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very peculiar about this,
if you go a little lower,

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than you're advanced, and then
suddenly, it gets delayed.

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It's not that evident, but
the phenomenon of resonance

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is precisely what we get when
we, sort of, trap the particle.

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And then we make it be, as far
as it seems, arbitrarily large,

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if you design a well properly.

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But the thing that
we have to design,

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the example of what
we're going to design,

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is different from all the
things I've drawn so far.

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It's the following way,
this is just an example.

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I have this zero
line of the energy.

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This is v of x.

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This is x.

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And then I put an
attractive potential here.

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And here is minus v0.

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And then I put a
barrier here with a v1.

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So what I'm going to
aim at is, you see,

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if v1 will be extremely
large, there will be--

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well, if v0 is
extremely large, then

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begin there would be
bound states here,

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but these are not
scattering states.

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On the other hand, if v1
will also be infinite,

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you would have
bound states here,

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but they could not escape.

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So certainly, if I
combine these two,

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I put a v0 and
maybe a larger v1,

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I can almost create
bound states here.

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But they're not
really bound states,

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because they can leak out and
produce scattering states.

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But these are going
to be resonances.

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This part and this,
this being a attractive,

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trying to keep the particle
in, and this being a barrier,

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can combine to produce a
state that gets trapped here,

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and stays a very long, time,
will have a very long time

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delay.

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And that's the
phenomenon of resonances.

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We need to trap that
particle, somehow.

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And we're going to see now
the details of how this works,

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and what the properties are.

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Now, it's very interesting
that actually, these resonances

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occur at some
particular energies.

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And they have
different properties.

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But we can identify
energies of resonances.

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And these are not bound states.

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They're just resonances.

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They eventually escape.

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And they're not normalizable,
really, but in some ways

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they behave as bound
states for awhile.

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They stay there for a
while and do nice things.

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So let's set this off.

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Now we're going to
spare you a little bit

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of these calculations,
because the important thing is

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that you know how to set it
up, and if you get an answer,

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you know how to plot it,
how to get the units out,

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how to try to understand it.

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So that's what
we're going to do.

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I'm going to put an
energy here, an energy, E.

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And I'm going to receive
E to be less than v1

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and greater than 0.

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I don't expect true
resonances beyond,

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because the particle
just bounces out.

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It doesn't get trapped.

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The phenomenon of resonance is a
little more intricate than just

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having a long time delay.

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There's more that has to happen.

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Another thing that
will happen, is

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if the particle spends
a lot of time here,

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you would find, in this
spirit of resonance,

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that the amplitude
of the wave function

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here is going to be very big.

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So you will scan the
energy and the amplitude.

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It will be normal,
normal, normal.

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And suddenly for some
energy it becomes very big.

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And we're going to do that.

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The way I'm going to develop
that, we're going to calculate

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this, plot these things.

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And then we are going
to ask whether there

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is a mathematical condition
that picks resonances.

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Well, how do I, if I want
to explain to somebody in 30

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seconds where are
the resonances,

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how do you calculate them, you
cannot tell that somebody, OK,

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calculate it for all
energies, do all the plots,

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and see some peak in some
thing, and this is a resonance.

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This is what we're going
to do to begin with,

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but then we'll get
more sophisticated.

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So let's put k in.

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So let's call this k prime,
the wave number in this area.

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Kappa here, because it's
a forbidden region, and k

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over here, as usual.

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So k squared is 2mE
over h bar squared.

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K prime squared is equal to
2m, the total kinetic energy

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is E plus v0, over h squared.

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And kappa squared is again,
similar formula, but this time

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is the energy differential
between v1 and E,

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so 2m v1 minus E over h squared.

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All of these three
numbers are positive.

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And they are the
relevant constants

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to write wave functions.

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So we have to write
a wave function.

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And I'm going to
write a wave function

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because it takes
a little tinkering

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to do it in an efficient way.

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There is one that you
don't have to think,

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you just have to remember.

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It's the one outside.

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It's the universal
formula, e to the i delta

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sine kx plus delta is valid
for for x greater than R.

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This one we derived at the
beginning of our analysis

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of scattering.

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How about the other region.

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Oops, I should have
put letters here.

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These are a and 2a
they are positions.

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And therefore,
it's not R in here.

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Well, it's R, it's the range of
the potential, but here is 2a.

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How about the other one?

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In this region, it's
kind of simple again.

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The wave function has to
vanish here, has to be

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sines or cosines of k prime.

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So it has to be a sine
function of k prime.

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And since we don't put an
extra constant in here,

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we kind of put an
extra constant in here,

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there must be a constant
here, A sine of k prime x.

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And that must be for
x between 0 and a.

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We used k prime, the
wave from over there.

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And there is A. And what we
were saying about resonances,

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is that, well, A
may depend on k.

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And when you have
a resonance, A is

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going to [INAUDIBLE], presumably
because the particle spends

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a long time inside the well.

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And now I have to
write this one in here.

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And this is the
one that, you can

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do it, do a little bit more
work, or do it kind of,

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efficiently.

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In that region we
have exponentials,

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like we have e to the kappa
x and e to the minus kappa x.

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Or I may want to have sinh
of kappa x and cosh of kappa

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x to write my solutions.

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But I actually don't
want either of them

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too much, because I
would like to write

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00:16:16,920 --> 00:16:23,070
an answer that almost imposes
continuity in a nice way.

247
00:16:23,070 --> 00:16:37,190
So I could use sinh of kappa
x minus a and cosh of kappa x

248
00:16:37,190 --> 00:16:38,095
minus a.

249
00:16:38,095 --> 00:16:40,610
These are all solutions.

250
00:16:40,610 --> 00:16:43,130
You can choose
whichever pair you want.

251
00:16:46,490 --> 00:16:51,500
So for example, if I want
to implement continuity

252
00:16:51,500 --> 00:16:54,770
with this thing,
this wave function,

253
00:16:54,770 --> 00:16:56,615
I want to write
something that I don't

254
00:16:56,615 --> 00:16:59,290
have to write another
equation for continuity.

255
00:16:59,290 --> 00:17:08,210
So I will write A
sine of k prime a--

256
00:17:08,210 --> 00:17:11,839
so far, this wave
function, if x equal a,

257
00:17:11,839 --> 00:17:13,910
coincides with this one.

258
00:17:13,910 --> 00:17:17,369
But this is no wave function
yet, not with an x dependence,

259
00:17:17,369 --> 00:17:19,349
so I have to put more.

260
00:17:19,349 --> 00:17:24,589
But then, I know that cosh
is 1 for x equals zero.

261
00:17:24,589 --> 00:17:32,830
So I put here a cosh
kappa x minus a.

262
00:17:32,830 --> 00:17:39,470
And now this is a solution that
matches that one at x equals a.

263
00:17:39,470 --> 00:17:44,050
At x equal a, the cosh
becomes 1 and matches.

264
00:17:44,050 --> 00:17:46,400
But this kind of need
a complete solution.

265
00:17:46,400 --> 00:17:47,600
It's not general enough.

266
00:17:47,600 --> 00:17:56,280
So you have to put a B
sinh of kappa x minus a.

267
00:17:56,280 --> 00:17:59,690
And this won't ruin the
matching, because at x equal a,

268
00:17:59,690 --> 00:18:01,670
that second term vanishes.

269
00:18:01,670 --> 00:18:05,780
So we're still
matching well there.

270
00:18:05,780 --> 00:18:09,060
And Well matching
here is non-trivial

271
00:18:09,060 --> 00:18:11,810
when I impose some conditions.

272
00:18:11,810 --> 00:18:16,010
So you still have to match
derivatives and do a little bit

273
00:18:16,010 --> 00:18:19,060
of work but not too much work.