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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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of half-waves in this region.
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So we can do on a
little numerical example
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to show how that works.