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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.