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