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PROFESSOR: Here is
the answer, answer.

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It's easier apparently to
write 1 over T. And 1 over T

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is equal to 1 plus 1 over 4 V0
squared over E times E plus V0

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times sine squared of 2k2a.

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So the one thing to
notice in this formula,

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it's a little complicated,
is that the second term

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is positive.

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Because V0 squared is positive,
the energy is positive,

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and sine squared is positive.

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So if this is positive,
the right-hand side

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is greater than 1.

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And therefore, the
T is less than 1.

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So this implies T less
than or equal to 1.

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And there seems to
be a possibility

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of T being equal to 1
exactly if the sine squared

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of this quantity, or the sine
of this quantity, vanishes.

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So there is a possibility of
very interesting saturation,

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in which the transmission
is really equal to 1.

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So we'll see it.

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The other thing you can notice
is that, as E goes to 0,

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this is infinite.

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And therefore, T is going to 0.

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No transmission as
the energy goes to 0.

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As the energy goes to infinity,
well, this term goes to 0.

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And you get transmission,
T equals to one.

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So these are interesting limits.

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Now, to appreciate
this better, we

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can write it with
unit-free language.

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So for that, I'll
do the following.

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It's a little rewriting,
but it helps a bit.

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So think of 2k2
times a, this factor,

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as the argument of
the sine function.

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Well, it's 2.

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k2 was defined up there, so it's
2m a squared E plus V0 over h

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squared.

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And I put the a inside
the square root.

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So what do we have here?

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2 times the square
root of 2m a squared.

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Let's factor a V0, so that
you have 1 plus E over V0.

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And you have h squared here.

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So this is OK.

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There's clearly two
things you can do.

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First, define a
unit-free energy.

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So the energy is now
described by this little E.

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Without units, that compares
the energy of your energy

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eigenstate to the
depth of the potential.

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So it should be over V0.

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So this is nice.

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You don't have to talk
about EVs or some quantity.

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Just a pure number.

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And here, there is another
number that is famous.

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This is the number Z0
squared of a potential well.

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This is the unit-free
number that tells you

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how deep or profound
is your potential,

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and controls the
number of zeros.

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So at this moment,
this is simply 2 Z0,

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because the square root is there
and takes the Z0 squared out

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as Z0.

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Square root of 1 plus
e, which is nice.

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So here, you can divide
by V0 squared, numerator

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and denominator.

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So you have an E over V0,
and a 1 plus an E over V0.

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So the end result is that 1
over T is now 1 plus 1 over 4e 1

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plus e sine squared of 2
Z0 square root of 1 plus e.

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So it's ready for
numerical calculation,

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for plotting, and doing all
kinds of things with it.

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But what we want to
understand is this phenomenon

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that you would expect, in
general, some reflection

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and some transmission.

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But there is a possibility
when T is equal

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to 1, and in particular, when
this sine squared function

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is equal to 0, and that
will make T equal to 1,

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then you have a
perfect transmission.

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So let's see why
it is happening,

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or under what
circumstances it happens.

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So for what the
energies will we have?

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For what energies?

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Energies is T equal to 1.

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It's perfect transmission.

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No reflection whatsoever.

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So we need, then, that the
argument of this sine function

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be equal to multiples of pi,
2 Z0 square root of 1 plus e

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is equal to a multiple of pi.

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Now, we would say what
the multiple of pi?

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Well, it could be 0, 1, 2, 3.

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Not obvious, because the only
thing you have here to adjust

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is the energy.

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The energy is positive.

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And that's that
little e in here.

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So this number n must
exceed some number,

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because this left-hand side
never becomes very small.

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The smallest it can be is 2 Z0.

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So n must be greater than
or equal to 2 Z0 over pi.

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This is because e, since
e is greater than 0.

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So the left hand
side is a number

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that is greater than 2 Z0,
and the right-hand side

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must therefore be that way.

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All right, so this
is a possibility.

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But then, let's calculate
those values of the energies.

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Calculate those en's.

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So what do we have?

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We squared the left
hand side for Z0

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squared times and 1 plus en is
equal to pi squared n squared.

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And en is equal to minus
1 plus n squared pi

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squared over 4 Z0 squared.

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OK, this is quantitatively nice.

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But probably still doesn't
give us much intuition about

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what's going on.

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So let me go back
to the total energy.

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en, remember, was
energy divided by V0.

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So multiply all terms by V0.

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E equals minus V0 plus n
squared pi squared V0 over 4.

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Z0 squared, I'm going
to go all the way back

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to conventional language.

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And, too, 4 times Z0
squared, which is 2ma

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squared V0 over h bar squared.

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So E is minus V0 plus
n squared pi squared.

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The V0s cancel.

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h squared over 2n
times 2a squared.

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I think I got every term right.

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So what does this say?

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Well, think of the potential.

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In this region,
there's an e here.

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And there's minus V0 there.

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So it says E is minus
V0 plus this quantity.

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So minus V0 plus this
quantity, which is n

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squared pi squared h squared
over 2m times 2a squared.

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So the resonance
happens if the energy

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is a distance above the
bottom of the potential, which

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is equal to this quantity.

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And now, you see
something that we could

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have seen maybe some other way.

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That what's happening here is a
little strange at first sight.

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These are the energy levels
of an infinite square

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well of width, 2a.

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If you remember, the energy
levels of an infinite square

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well are n squared
pi squared h squared

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over 2m times the width squared.

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And those are exactly it.

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So the energies at which
you find the transmission,

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and the name is going to
become obvious in a second,

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it's called the
resonant transition,

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are those in which
the energy coincides

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with some hypothetical
energy of the infinite square

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well that you would put here.

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If it is as if you would have
put an infinite square well

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in the middle and
look at where are

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the energies of bound states
that are bigger than 0,

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that might be bouncing
the energies here,

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but those are not
relevant, because you only

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consider energies positive.

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So if you find an
energy that is positive,

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that corresponds to a would-be
of infinite square well, that's

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it.

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That's an energy for which
you will have transmission.

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And in fact, if we
think about this

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from the viewpoint
of the wave function,

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this factor over here, look
at this property over here.

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So what do we have?

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The condition was
that k2 time 2a,

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the argument of
the sine function

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would be a multiple of pi.

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But k2 is 2 pi over the
wavelength of the wave

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that you have in
this range, over 2a.

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It's equal to n pi.

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So we can cancel
the pis and the 2s

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so that you get 2a over
lambda is equal to n over 2.

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And what does that say?

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It says that the de
Broglie wavelength

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that you have in this region
is such that it fits into 2a.

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Let me write it
yet in another way.

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Let me try this a as--

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I won't write it like that.

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Leave it like that.

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The wavelength lambda fits
into 2a a half-integer number

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of times.

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And that's exactly what you
have in an infinite square well.

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If you have a width, well, you
could have half a wavelength

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there for n equals 1, a
full thing for n equals 2,

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3 halves for n equals 3.

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You always get half and
halves and halves increasing

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and increasing all the time.

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Yeah.

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So the way I think
I wanted to do it,

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this equation can
be written as n is

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equal to 2a over lambda over 2.

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That's the same equation.

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So in this way, you see an
integer a number of times

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is 2a divided by lambda
over 2, which is precisely

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the condition for infinite
square well energy eigenstate.

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So there is no infinite square
well anywhere in this problem.

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But somehow, when the
wavelength of the de Broglie

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representation of the
particle in this region

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is an exact number of
half-waves, there's resonance.

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And this resonance is
such that it allows a wave

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to go completely through.

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It's a pretty
remarkable phenomenon.

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So the infinite
square well appears

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just as a way to think of what
are the energies at which you

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will observe the resonances.

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But the resonance is simply
due to having an exact number

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00:17:37,930 --> 00:17:43,040
of half-waves in this region.

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00:17:43,040 --> 00:17:47,380
So we can do on a
little numerical example

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to show how that works.