PROFESSOR: Last time we talked
about particle on a circle.
Today the whole lecture is
going to be developed to solving
Schrodinger's equation.
This is very important,
has lots of applications,
and begins to give you
the insight that you
need to the solutions.
So we're going to be
solving this equation
all through this lecture.
And let me remind
you what with had
with a particle on a circle.
The circle is segment 0 to
L, with L and 0 identified.
More properly we actually
think of the whole x-axis
with the identification
that two points related
in this way that differ by
L, or therefore for 2L, 3L,
are the same point.
As a result, we
want wave functions
that have this periodicity.
And that implies the same
periodicity for the derivatives
as well.
We looked at the
Schrodinger equation
and we proved that the
energy of any solution
has to be positive or 0.
And therefore the differential
equation, the Schrodinger
differential equation,
can be read then
as minus k squared psi, where
k squared is this quantity
and it's positive, so
k is a real number.
That makes sense.
And finally, once you
have this equation,
you know that if the second
derivative of a function
is proportional to
minus the function,
the solution are trigonometric
functions or exponentials.
And we decided to
go for exponentials,
that they are perhaps a
little more understandable,
though we will go back to them.
Now that's where we
stopped last time.
And now we apply the
periodicity condition.
So we must have e to the ik x
plus L equal to e to the ikx.
If you cancel the e to
the ikx on both sides,
you get to e to the ikL must
be equal to 1, which forces kL
to be a multiple of pi.
That is, kL equal 2 pi n--
of 2 pi, I'm sorry--
2 pi n, where n is an integer.
So those are the values of k.
We'll write them
slightly differently.
We'll write kn with
the subscript n
to represent the k
determined by the integer n.
So it will be 2 pi n over
L. Now from that equation,
for k squared equal
2mE over h squared,
you get e is equal to h
squared k squared over 2m.
And k in these solutions
represents therefore
the momentum.
That is, the momentum
Pn is h bar kn,
and it's 2 pi h bar n over L.
And the energies associated
with solutions
with kn value is En
would be h squared
kn squared, so
4 pi squared n squared
over L squared over 2m.
So this is equal to 2
pi squared h squared
n squared over m L squared.
Those are numbers.
It's good to have them.
Our solution is psi n of
x is equal to e to the i,
or is proportional
to, e to the i knx.
So far so good.
But we can now
normalize this thing.
This is the beauty
of the problem
of a particle on a circle.
If you have a particle
in free space,
psi squared is equal to 1
and the integral is infinite.
On the other hand, these
ones are normalizable.
That is, we can demand that the
infinite over the circle of psi
n squared be equal to 1.
So how do we do that?
Well I'll write it a little
more explicitly here.
Psi n of x will be some
constant times e to the iknx.
And therefore this thing
is the integral from 0
to L dx of the constant squared.
The constant can be
chosen to be real.
N squared times psi
n squared, which is--
this exponential
squared is just 1.
This is 1.
So it just gives you L times
N squared is equal to 1.
So N is equal to 1
over square root of L.
And finally, our psi n's of
x are 1 over square root of L
e to the iknx or 1 over square
root of L e to the 2 pi inx
over L. Oops.
All right.
So these are our wave functions.
These are our
energy eigenstates.
Our full stationary
states, where we're
finding stationary states--
stationary states have a psi
of x times a time dependence.
The time dependence is
e to the minus i e--
so you could say that
psi n of x and t,
the full stationary
state, is psi n of x times
e to the minus i
e n t over h bar.
And that solves the full
Schrodinger equation.
That's our stationary state.
So one thing that should
be emphasized here
is the range of the integers.
n is an integer and
we better realize
if there are some exceptions.
Maybe just the positive?
Is 0 included?
Is 0 not included?
And here, it's really
as stated here.
It's all the integers,
n from minus infinity
to plus infinity.
All of them must be included.
The reason we can
understand that
is that the momentum of each
of these states, the momentum
is 2 pi h n over
L. And therefore
these are all states
of different momentum.
There's no question that
these are different states.
It cannot be that one is
just the same as another one.
They have different momentum.
They represent
the particle going
with some momentum
around the circle,
and that momentum
is quantified by n
and it could be in
the positive direction
or negative direction.
Now you could be suspicious
about n equals 0.
But there's actually nothing
to be suspicious about it.
It's surprising.
But psi 0 is 1 over square
root of L, has no x dependence.
And therefore it has 0 energy.
And that's-- I'm sorry, here.
There's some psi missing.
The second derivative
of a constant is 0.
And if e is equal to 0,
that's a consistent solution.
The constant is important.
And now you also
realize that psi,
you have a nice phenomenon,
that psi minus 1 and psi 1,
for example, they correspond
to n equals 1 and minus 1,
have the same energy.
Because energy
depends on n squared,
so these are degenerate
states with energy E1
equal to E minus 1.
And so are psi 2
and psi minus 2.
And of course just
psi minus k and psi k.
They are degenerate states.
And now this hits
into a property
that is going to be
important in the future about
degenerate states.
Whenever somebody gives you a
couple of degenerate states,
you know they have
the same energy.
But you must not stop there.
If they are degenerate states
and there are two states,
it means that they
are not the same.
So there must be something
physical about them
that distinguishes them.
Whenever you have
degenerate states,
you have to work
until you figure out
what is different about
one state and the other.
And here we got the answer.
The answer is simply that
they are degenerate states
with a different momentum.
So the momentum is an
observable that distinguishes
those degenerate states.
In fact, as we've
written here, p on psi n
of x is equal to Pn psi n.
And P n given by this quantity.
OK.
Our eigenstates are orthonormal.
They're eigenstates.
So why are they orthonormal?
They are eigenstates
of a Hermitian operator
with different eigenvalues.
They're eigenstates of p
with different eigenvalues.
So they're orthonormal.
The argument with the
energy would have not
worked out so well because there
you have degenerate states.
So these two states are
degenerate with respect
to energy.
So you could wonder, how do
you know they are orthogonal?
But in this case it's simple.
They have different momentum.
Momentum is a
Hermitian operator,
and it should be orthogonal.
So the states are orthogonal.
They are complete.
You could write any wave
function of the circle
as a superposition
of those psi n's.
So any psi of x periodic can be
written as psi of x the sum a n
psi n's over all the integers .
And one last remark.
We could have worked
with sines and cosines.
And therefore we
could have worked
with psi k plus psi minus k.
This psi k and psi minus
k have the same energy.
Therefore this sum is an energy
eigenstate of that same energy.
The Hamiltonian acting
on psi k gives you
the energy times psi k.
Here, the same energy
times psi minus k,
so this is an energy eigenstate.
And this is proportional
to cosine of kx.
And this is an energy
eigenstate you know,
because two
derivatives of a cosine
will give you back that cosine.
Similarly, psi k
minus psi of minus k
is proportional to sine of kx.
And that's also an
energy eigenstate.
Both are energy eigenstates.
So this is kind of the
way you can reformulate
Fourier's theorem here.
You could say anything
can be written
as a superposition of
all the exponentials,
including the
exponential with n equals
0, which is just a constant.
Or alternatively,
everything could
be written in terms
of sines and cosines,
which is another way of
doing the Fourier theorem.
These are energy eigenstates,
but they're not P eigenstates
anymore.
This, when you take a
derivative, becomes a sine.
When this, you take a
derivative, becomes a cosine.
They're not energy.
They're not momentum
eigenstates.
So you can work with
momentum eigenstates,
you can work with
energy eigenstates.
It's your choice.
It's probably easier to work
just with momentum, I can say.
So that's it for the
particle in a circle.
We have three problems
to solve today.
Particle in a circle, particle
in a box, and particle
in a finite well.
