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BARTON ZWIEBACH: --that
has served, also,
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our first example of solving
the Schrodinger equation.
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Last time, I showed you
a particle in a circle.
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And we wrote the wave function.
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And we said, OK, let's see
what is the momentum of it.
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But now, let's solve,
completely, this problem.
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So we have the
particle in the circle.
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Which means particle
moving here.
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And this is the coordinate x.
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And x goes from 0 to L.
And we think of this point
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and that point, identify.
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We actually write this as,
x is the same as x plus L.
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This is a strange
way of saying things,
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but it's actually
very practical.
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Here is 2L, 3L.
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We say that any point is the
same as the point at which you
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add L. So the circle is
the whole, infinite line
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with this identification,
because every point here,
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for example, is the
same as this point.
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And this point is the
same as that point.
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So at the end of
everything, it's
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equivalent to this piece,
where L is equivalent to 0.
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It's almost like if I was
walking here in this room,
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I begin here.
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I go there.
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And when I reach
those control panels,
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somehow, it looks like a door.
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And I walk in.
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And there's another classroom
there with lots of people
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sitting.
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And it continues,
and goes on forever.
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And then I would conclude
that I live in a circle,
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because I have just
begun here and returned
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to the same point that is there.
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And it just continues.
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So here it is.
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You are all sitting here.
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But you are all sitting there.
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And you are all sitting there,
and just live on a circle.
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So this implies that in
order to solve wave functions
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in a circle, we'll have to
put that psi of x plus L
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is equal to psi of x,
which are the same points.
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And we'll have 0 potential.
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V of x equals 0.
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It will make life simple.
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So the Hamiltonian is
just minus h squared
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over 2m d second dx squared.
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We want to find the
energy eigenstate.
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So we want to find
minus h squared over 2m
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d second psi dx squared
is equal to E psi.
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We want to find those solutions.
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Now it's simple, or
relatively simple
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to show that all the
energies that you can find
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are either zero or positive.
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It's impossible to find
solutions of this equation
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with a negative energies.
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And we do it as follows.
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We multiply by dx and psi star
and integrate from 0 to L.
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So we do that on this equation.
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And what will we get?
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Minus h squared over
2m integral psi star
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of x d dx of d dx psi
of x is equal to E times
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the integral psi star psi x dx.
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And we will assume, of
course, that you have things
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that are well normalized.
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So if this is well
normalized, this is 1.
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So this is the energy is
equal to this quantity.
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And look at this quantity.
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This is minus h squared over 2m.
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I could integrate by parts.
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If I do this quickly,
I would say, just
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integrate by parts over here.
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And if we integrate by
parts, d dx of psi of x,
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we will get a minus sign.
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We'll cancel this minus
sign, and will be over.
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But let's do it a
little bit more slowly.
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You can put dx, this is
equal to d dx of psi star
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d psi dx minus d psi
star dx d psi dx.
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I will do it like this,
with a nice big bracket.
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Look what I wrote.
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I rewrote the psi
star d second of psi
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as d dx of this
quantity, which gives me
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this term when the derivative
acts on the second factor.
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But then I used an extra
term, where the derivative
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acts on the first factor that is
not present in the above line.
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Therefore, it must
be subtracted out.
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So this bracket has
replaced this thing.
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Now d dx of something, if
you integrate over x from 0
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to L, the derivative
of something,
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this will be minus h bar
squared over 2m psi star d psi
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dx integrated at L and at 0.
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And then minus cancels.
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So you get plus h squared
over 2m integral from 0
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to L dx d psi dx
squared equal E.
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And therefore,
this quantity is 0.
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The point L is the same
point as the point 0.
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This is not the
point at infinity.
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I cannot say that the wave
function goes to 0 at L,
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or goes to 0, because
you're going to infinity.
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No, they have a better
argument in this case.
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Whatever it is,
the wave function,
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the derivative, everything,
is periodic with L.
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So whatever values it
has at L equal 0 it has--
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at x equals 0, it has at
x equals L. So this is 0.
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And this equation shows
that E is the integral
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of a positive quantity.
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So it's showing that E is
greater than 0, as claimed.
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So E is greater than 0.
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So let's just try a couple
of solutions, and solve.
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We'll comment on
them more in time.
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But let's get the
solutions, because,
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after all, that's what
we're supposed to do.
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The differential equation
is d second psi dx
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squared is equal to minus
2mE over h squared psi.
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And here comes the thing.
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We always like to define
quantities, numbers.
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If this is a number,
and E is positive,
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this I can call
minus k squared psi.
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Where k is a real number.
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Because k real, the
square is positive.
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And we've shown that
the energy is positive.
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And in fact, this
is nice notation.
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Because if you were setting
k squared equal to 2mE
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over h squared,
you're saying that E
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is equal to h squared
k squared over 2m.
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So, in fact, the
momentum is equal to hk.
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Which is very nice notation.
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So this number, k,
actually has the meaning
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that we usually associate,
that hk is the momentum.
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And now you just have to solve
this. d second psi dx squared
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is equal to minus k squared psi.
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Well, those are solved by
sines or cosines of kx.
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So you could choose sine of
kx, cosine of kx, e to the ikx.
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And this is, kind of
better, or easier,
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because you don't
have to deal with two
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types of different functions.
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And when you take k and minus
k, you have to use this, too.
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So let's try this.
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And these are your
solutions, indeed.
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psi is equal to e to the ikx.
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So we leave for
next time to analyze
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the [INAUDIBLE] details.
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What values of k are
necessary for periodicity
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and how we normalize
this wave function.