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BARTON ZWIEBACH:
Finite square well.
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So this brings us also
to a little common aside.
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So far, we could
find every solution.
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Now we're going to write the
equations for the finite square
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well, and we're not going to
be able to find the solution.
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But we're going to
understand the solution.
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So you're going to
enjoy a little--
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mathematicians usually say
it's the most important thing,
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understanding the solution.
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Finding it, it's no big deal.
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But we're physicists as well.
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So we sometimes have
to find the solutions.
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Even if we don't
understand them very well,
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it's nice to find them.
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And then you're going to
use numerical methods,
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and this is the part
of the course where
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you're going to be using
numerical methods a lot.
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Here is the finite
square well, and now we
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draw it symmetrically.
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Here it is.
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Here is x.
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We're drawing the
potential V of x.
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It extends from a to minus a.
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It's 0.
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This is the 0 of the potential.
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It's here.
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And it goes down.
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The potential is negative here.
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Its value is minus
V0 with V0 positive.
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And then a possible energy
for a bound state-- we're
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going to look for bound states.
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Bound states are normalizable
states, normalizable solutions
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of the Schrodinger equation.
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So we're going to look for them.
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Look for bound states.
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And they have
energy less than 0.
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So something that has
energy less than 0 is bound.
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You would have to give it some
energy to put it at 0 energy,
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and the particle could escape.
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The other thing
about the bound state
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is that it will be some
probability to find it here.
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Very little probability to find
it in the forbidden region.
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And this energy, which is
negative, it's somewhere here.
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We don't know what are
the possible energies,
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but we will assume there's
some energy that is negative,
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that is there.
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And that's the energy
of the solution.
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Let me write the equation again.
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The equation is psi double prime
is minus 2m over h bar squared,
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E minus V of x times psi.
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This is the
Schrodinger equation.
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You recognize it if you multiply
h squared here, 2m there.
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And we've been solving
already two examples where
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there was no potential, but
finally, there is a potential.
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So this is the energy
of the particle.
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The energy of the
particle can be
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interpreted as the potential
energy plus the kinetic energy.
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Think intuitively here.
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If you have some
energy over here--
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this is the potential energy.
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Well, all this much
is kinetic energy.
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On the other hand,
in this region,
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the energy is smaller
than the potential energy.
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So it has negative kinetic
energy, which is classically
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not understandable.
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And in quantum
mechanics, it just
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will have some
probability of being here,
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but that probability will
go down and eventually
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go to 0 in such a way that the
wave function is normalizable.
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So this is a very mysterious
thing that happens here,
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that the wave function
will not be just over here,
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but it will leak.
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And it leaks because there's
a finite discontinuity.
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If you take the barrier
to be infinitely high,
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it would leak so little that
eventually it would not leak.
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The wave function
would vanish there.
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And it will be the
end of the story,
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and you're back to the
infinite square well.
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So the infinite square
well is a limit of this
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as V0 goes to minus infinity.
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00:05:06,550 --> 00:05:11,600
OK, so a few numbers we
can put in this graph.
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00:05:11,600 --> 00:05:16,030
E, and this-- what is the
energy difference here?
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00:05:16,030 --> 00:05:25,920
It's E minus minus V0
, which is V0 plus E.
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And many times
because E is negative,
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we'll write it as V0
minus absolute value of E.
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A negative number is equal
to minus its absolute value.
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So how about this
whole constant here?
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Well, my tongue slipped,
and I said this constant,
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but it's V of x.
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So what do you mean a constant?
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Well, the potential
is piecewise constant.
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So actually, we are not in
such difficult situation
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because in this region, the
potential is a constant.
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In this region, the
potential is a constant.
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00:06:12,520 --> 00:06:16,830
So this is the
constant here, and we
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00:06:16,830 --> 00:06:19,470
wish to understand
what it is and what
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00:06:19,470 --> 00:06:21,640
are the sines of this constant.
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This constant, that
we call alpha--
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see-- is going to be--
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let's see what it is, the
different circumstances.
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Suppose you are here.
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The energy is bigger
than the potential.
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So energy minus the
potential is positive,
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and the constant is negative.
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So alpha is negative
for x less than a.
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If it's negative
and it's a constant,
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you're going to have
trigonometric solutions
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for x, absolute value
less than a, so trig.
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On the other region,
alpha will be
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positive for x greater
than a, and you will
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have real exponentials of--
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I'll just write exponentials.
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So E to the minus 3x, E to
the 5x, things like that.
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This is the difference
between trigonometric
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and real exponential
solutions depend on the sines.
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So we're going to have to
impose boundary conditions as
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well because we're going
to solve the equation here
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inside with one value of
alpha and then outside
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with another value of alpha.
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And then we're
going to match them.
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So that's how this will go.
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Now in this process,
somehow the energy
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will be fixed to some
value, some allowed values.
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There's a counting we could
do to understand that,
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and we'll probably
do it next time.
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At this moment,
we'll just proceed.
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But we imagine there must
be a quantization because
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in the limit as V0
goes to infinity
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and the potential well
becomes infinitely deep,
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you're back to an
infinite square well
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that has quantized energies
that we calculated.
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So that should be quantized,
and it should be no problem
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OK, ready to do some work?
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Let's-- Yes?
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AUDIENCE: So when alpha is 0 and
you get polynomial solutions,
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does that case not matter?
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BARTON ZWIEBACH: When
alpha is 0-- well,
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alpha is going to be never 0.
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You see, either E is
less than the potential
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or is better than the potential.
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So you're not going
to get alpha equals 0.
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You could say, do we have some
energy maybe here or here?
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Well, you could have those
energies, in which case
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you will see what happens.
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It's not quite
polynomial solutions.
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There's one property about
potentials that actually is
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in the homework due this week,
which is that there cannot be
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solutions of the Schrodinger
equation with less energy than
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the minimum of the potential.
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You have to have more
energy than the minimum
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of the potential.
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So alpha equal to 0 is not going
to show up in our analysis.
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OK, so let's begin.
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Moreover, we have
to use this result.
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The potential is symmetric,
completely symmetric,
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so there are going to be even
solutions and odd solutions.
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So let's consider
the even solutions.
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Solutions.
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So those are solutions.
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Psi of minus x is
equal to psi of x.
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Now, you will see how
I solve this thing now,
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and the lesson of all
of this is the relevance
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of unit-free numbers.
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Unit-free numbers are going
to be your best friends
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in solving these equations.
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So it will look like
I'm not solving anything
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except making more and
more definitions, and all
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of the sudden, the
solution will be there.
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And it's a power of
notation as well.
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This problem can be
very messy if you
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don't have the right notation.
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If you were solving it alone,
if we would stop the lecture now
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and you would go home,
you would probably
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find something very
messy for quite a bit.
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And then maybe you clean
it up little by little,
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and eventually,
something nice shows up.
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So here's what
we're going to do.
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We're solving for x in between
a and minus a, even solutions.
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So what does the equation
look like? d second psi d
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x squared is minus 2m over
h squared E minus V of x--
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in that region, the
potential is minus V0.
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So this is minus 2m
over h squared, V0
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plus E, which I write as minus
absolute value of E, psi.
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And I forgot a psi here.
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OK.
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V0 is positive.
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Minus V0 is going down, and V0
minus the absolute value of E
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is positive.
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So this whole quantity
over here is positive.
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And here it comes,
first definition.
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I will define k squared
to be that quantity--
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2m over h squared V0
minus absolute value of E.
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And that's greater
than 0 in this region,
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as we're trying to solve, and
I'll call this equation one.
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And the differential
equation has become psi
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double prime is minus k psi.
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00:13:24,270 --> 00:13:27,570
So the solutions are simple.
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00:13:27,570 --> 00:13:32,360
It's trigonometrics, as we
said, and the only solution-- k
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00:13:32,360 --> 00:13:34,070
squared, I'm sorry.
200
00:13:34,070 --> 00:13:35,970
k squared-- and
the only solution
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00:13:35,970 --> 00:13:39,120
that is possible, because
it's a symmetric thing,
202
00:13:39,120 --> 00:13:41,520
is cosine of kx.
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00:13:41,520 --> 00:13:48,080
So psi of x is going
to be cosine of kx,
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00:13:48,080 --> 00:13:51,570
and that will hold
from a to minus a.
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End of story, actually, for
that part of the potential.
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00:14:01,290 --> 00:14:04,200
You could ask-- one second.
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00:14:04,200 --> 00:14:07,320
You could say, you're going
to normalize these things,
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aren't you?
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00:14:09,090 --> 00:14:13,830
Well, the fact that I
won't normalize them,
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00:14:13,830 --> 00:14:18,750
it's just a lot of work,
and it's a little messy.
211
00:14:18,750 --> 00:14:22,140
But that's no problem.
212
00:14:22,140 --> 00:14:25,610
You see, you're finding
energy eigenstates,
213
00:14:25,610 --> 00:14:28,560
and by definition,
solving the differential
214
00:14:28,560 --> 00:14:31,870
equation is not going to
give you a normalization.
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00:14:31,870 --> 00:14:36,300
So this is a good solution
of the differential equation,
216
00:14:36,300 --> 00:14:38,550
and let's leave it at that.
217
00:14:38,550 --> 00:14:41,060
This we'll call solution two.
218
00:14:43,980 --> 00:14:47,820
How about the region
x greater than a?
219
00:14:47,820 --> 00:14:50,800
Well, what does the
differential equation look like?
220
00:14:50,800 --> 00:14:55,230
Well, it looks like psi double
prime is equal to minus 2,
221
00:14:55,230 --> 00:15:08,070
m over h squared, E psi
because the potential is 0.
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00:15:08,070 --> 00:15:12,170
V0 outside.
223
00:15:12,170 --> 00:15:15,122
x equals a.
224
00:15:15,122 --> 00:15:21,310
And here again, you want
to look at the equation
225
00:15:21,310 --> 00:15:22,610
and know the sine.
226
00:15:22,610 --> 00:15:26,770
So maybe better to put
the absolute value here.
227
00:15:26,770 --> 00:15:34,230
So this is 2m absolute value
of E over h squared psi.
228
00:15:34,230 --> 00:15:36,435
And one more definition--
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00:15:36,435 --> 00:15:45,740
kappa squared is going
to be 2mE over h squared.
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That's equation number three.
231
00:15:52,730 --> 00:15:56,930
The equation has become
psi double prime equals
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kappa squared psi, and
the solutions for that
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00:16:03,190 --> 00:16:05,990
are exponentials.
234
00:16:05,990 --> 00:16:12,650
Solutions are psi-- goes like
E to the plus minus kappa x.
235
00:16:12,650 --> 00:16:15,885
You see the solution we're
constructing is symmetric.
236
00:16:15,885 --> 00:16:17,240
It's even.
237
00:16:17,240 --> 00:16:19,940
So let's worry just
about one side.
238
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If one side works, the other
side will work as well.
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00:16:24,990 --> 00:16:31,550
So I will just write
that for psi of x
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00:16:31,550 --> 00:16:34,760
is equal to the minus kappa x.
241
00:16:34,760 --> 00:16:39,170
That's a solution
for x greater than a.
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00:16:39,170 --> 00:16:44,690
If I just did that, I
would be making a mistake.
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00:16:44,690 --> 00:16:47,810
And the reason is
that yes, I don't
244
00:16:47,810 --> 00:16:51,510
care about the normalization
of the wave function,
245
00:16:51,510 --> 00:16:55,100
but by not putting
a number here,
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00:16:55,100 --> 00:16:57,800
I'm selecting some
particular normalization.
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00:16:57,800 --> 00:17:01,910
And The wave function must
be continuous and satisfy
248
00:17:01,910 --> 00:17:02,990
all these nice things.
249
00:17:02,990 --> 00:17:07,069
So yes, here I can maybe
not put a constant,
250
00:17:07,069 --> 00:17:10,090
but here, already, I
must put a constant.
251
00:17:10,090 --> 00:17:13,880
It may be needed to match
the boundary conditions.
252
00:17:13,880 --> 00:17:15,920
I cannot ignore it here.
253
00:17:15,920 --> 00:17:22,430
So I must put the constant
a that I don't know,
254
00:17:22,430 --> 00:17:25,140
and that's going to be
equation number four.
255
00:17:30,330 --> 00:17:32,226
Now, look at your definitions.
256
00:17:36,250 --> 00:17:40,910
k squared for a
trigonometric, kappa--
257
00:17:40,910 --> 00:17:45,130
many people use kappa for things
that go with exponentials.
258
00:17:45,130 --> 00:17:48,930
But look, k squared
and kappa squared
259
00:17:48,930 --> 00:17:51,250
satisfy a very nice relation.
260
00:17:51,250 --> 00:17:54,710
If you add them up--
261
00:17:54,710 --> 00:17:59,080
k squared plus kappa squared--
262
00:17:59,080 --> 00:18:04,840
the energy part cancels,
and you get 2mV0, which
263
00:18:04,840 --> 00:18:09,250
is positive, over h squared.
264
00:18:09,250 --> 00:18:18,210
Well, that's not so bad, but we
want to keep defining things.
265
00:18:22,040 --> 00:18:24,590
How can I make this really nice?
266
00:18:24,590 --> 00:18:28,730
If it didn't have units,
it would be much nicer.
267
00:18:28,730 --> 00:18:31,190
This is full of units.
268
00:18:31,190 --> 00:18:36,855
k times a length has no units,
and kappa times a length
269
00:18:36,855 --> 00:18:38,420
has no unit.
270
00:18:38,420 --> 00:18:43,250
So multiply by a
squared, and you have k
271
00:18:43,250 --> 00:18:46,400
squared a squared
plus kappa squared
272
00:18:46,400 --> 00:18:52,780
a squared is equal to 2mV0
a squared over h squared.
273
00:19:03,690 --> 00:19:07,770
Now, say, well yes,
this looks nice.
274
00:19:07,770 --> 00:19:10,670
So let's make the
new definitions.
275
00:19:13,980 --> 00:19:17,740
So don't lose track
of what we are doing.
276
00:19:17,740 --> 00:19:19,230
We need to find the energy.
277
00:19:19,230 --> 00:19:21,540
That's basically what we want.
278
00:19:21,540 --> 00:19:23,790
What are the possible energies?
279
00:19:23,790 --> 00:19:29,130
And we already included
two constants--
280
00:19:29,130 --> 00:19:33,610
k and kappa- and they have
these properties here.
281
00:19:33,610 --> 00:19:45,370
So let me define
psi to be kappa a,
282
00:19:45,370 --> 00:19:47,830
and it will be defined
to be positive.
283
00:19:47,830 --> 00:19:51,970
Kappa is defined to
be positive, and k
284
00:19:51,970 --> 00:19:53,630
is defined to be positive.
285
00:20:00,280 --> 00:20:08,050
eta, you will define it to
be ka. a It's unit-free.
286
00:20:08,050 --> 00:20:08,580
No units.
287
00:20:11,838 --> 00:20:19,350
And from that equation,
now we have eta
288
00:20:19,350 --> 00:20:25,270
squared plus psi squared is
equal to the right-hand side,
289
00:20:25,270 --> 00:20:28,320
which I will call z0 squared.
290
00:20:28,320 --> 00:20:31,920
So z0 squared-- that's
another definition--
291
00:20:31,920 --> 00:20:38,400
is 2mV0 a squared
over h squared.
292
00:20:38,400 --> 00:20:40,670
This is the list of
your definitions.
293
00:20:44,041 --> 00:20:44,540
OK.
294
00:20:44,540 --> 00:20:45,540
What did we do?
295
00:20:45,540 --> 00:20:51,110
We traded kappa and k
that control the behavior
296
00:20:51,110 --> 00:20:54,200
of the wave function--
k inside the well,
297
00:20:54,200 --> 00:20:56,600
kappa outside the well--
298
00:20:56,600 --> 00:21:02,010
we traded them for eta
and psi, unit-free.
299
00:21:02,010 --> 00:21:06,870
And a new constant came up, z0.
300
00:21:06,870 --> 00:21:07,670
What is z0?
301
00:21:10,200 --> 00:21:13,535
Well, z0 is a very
interesting constant.
302
00:21:13,535 --> 00:21:17,420
It's a number that
you can construct out
303
00:21:17,420 --> 00:21:20,900
of the parameters
of your potential.
304
00:21:20,900 --> 00:21:23,780
It involves V0 and the width.
305
00:21:23,780 --> 00:21:28,760
If V0 is very
large, z0 is large.
306
00:21:28,760 --> 00:21:33,500
If the width is
very big, z0 is big.
307
00:21:33,500 --> 00:21:40,340
If the potential is shallow
or very narrow, z0 is small.
308
00:21:40,340 --> 00:21:43,190
But the most important
thing about z0
309
00:21:43,190 --> 00:21:48,550
is that it will give you
the number of bound states.
310
00:21:48,550 --> 00:21:52,440
If z0 is very big, it's
a very deep potential,
311
00:21:52,440 --> 00:21:55,090
we'll have lots of bound states.
312
00:21:55,090 --> 00:21:59,630
If z0 is very shallow, there
will be one bound state,
313
00:21:59,630 --> 00:22:00,890
no more.
314
00:22:00,890 --> 00:22:02,540
You will see it today.
315
00:22:02,540 --> 00:22:05,990
But z0 controls the
number of bound states.
316
00:22:05,990 --> 00:22:08,760
And this is its role, and
it will be very important--
317
00:22:08,760 --> 00:22:11,900
these dimensionless
quantities and number.
318
00:22:11,900 --> 00:22:16,940
If somebody says, I have a
potential with z0 equals 5,
319
00:22:16,940 --> 00:22:19,010
you can tell immediately
three bound states
320
00:22:19,010 --> 00:22:21,500
or some number of bound states.
321
00:22:21,500 --> 00:22:23,430
That's the nice thing about z0.
322
00:22:23,430 --> 00:22:25,975
And this is a very
nice-looking equation,
323
00:22:25,975 --> 00:22:30,980
this equation of a circle
in the 8x psi plane.