PROFESSOR: Here is x.
And here is a.
Various copies of the x-axis.
For the ground state, what
is the lowest energy state?
It's not zero energy,
because n begins with 1.
So it's this.
The lowest energy
state is a sine.
So the wave function
looks like this.
This corresponds to sine.
1.
Or n equals 1.
The next one corresponds to n
equals 2, and begins as a sine.
And it just goes up like this.
You add half a wave each time.
Remember, we quantize
k a with n pi.
So each time that you increase
n, you're adding pi to k a.
So the phase, you see,
you have sine of kx.
So you have sine that
goes from 0 up to k a.
And k a is equal to n pi.
So you go from 0 to pi, from 0
to 2 pi, then from 0 to 3 pi.
So it would be one up,
one down, like that.
And I could do one more.
This one would be with four
cycles, two, three, four.
So this is psi 1, psi
2, psi 3, and psi 4.
Wave functions do
more and more things.
So what can we learn
from this wave function?
There are several things
that we need to understand.
So one important thing is
that the wave function,
these are all normalizable
wave functions.
The ground state-- this
is the ground state--
has no nodes.
A node, in a wave function,
is called the point where
the wave function vanishes.
But it's not the endpoints
or the points at infinity,
if you could have a range
that goes up to infinity.
It's an interior
point that vanishes.
And the ground
state has no nodes.
So a node, node, so zero
of the wave function,
not at the end of the domain.
And of the domain.
Because if we
included that, I would
have to say that the ground
states has two nodes already,
you'll see, around 0.
But the 0 at the
end of the domain
should not be counted as a node.
Nodes are the zeros inside.
And look.
This has no nodes, and
it's a general fact
about states of potentials.
The next excited
state has one node.
It's here.
The next has two nodes, and
then the next is three nodes.
So the number of
nodes of the wave
function increases in potential.
You have more and
more wave functions
with higher and higher excited
states, and the number of nodes
increases one by one
on each solution.
That's actually
a theorem that is
valid for general potentials
that have bound states.
Bound states are states
that are normalizable.
So the decay at infinity.
You see, a state that
is not normalizable,
like a plane, where it
is not a bound state.
It exists all over.
And it's a general
theorem that this phase,
this one-dimensional
potentials, whenever
you have bound states,
the number of nodes
increases with the
energy of the eigenstate.
We will see a lot
of evidence for this
as we move along the course,
and a little bit of a proof.
Not a very rigorous proof.
The other thing
I want to comment
on this thing that is
extremely important
is the issue of symmetry.
This potential for simplicity,
to write everything nicely,
was written from 0 to a.
So all the wave functions
are sine of n pi x over a.
But in some ways, it perhaps
would have been better
to put the 0 here.
And you say, why?
What difference does it make?
Well, you have a 0 at the middle
of the interval, the potential
and the domain of
the wave function
are symmetric with respect
to x going to minus x.
So actually, when you look at
the wave functions thinking
you can rethink this as an
infinite box from a over 2
to minus a over 2, and the
solutions, you just copy them,
and you see now, this line
that I drew in the middle,
the ground state is symmetric.
The next state is anti-symmetric
with respect to the midpoint.
The next state is now symmetric.
And the following
one, anti-symmetric.
So this is also a true fact.
If you have bound states
of a symmetric potential--
I will prove this one,
probably on Wednesday.
A symmetric potential
is a potential for which
V of minus x is V of x.
Bound states of a
symmetric potential
are either odd or even.
This is not a completely
simple thing to prove.
We will prove it, but you
need, in fact, another result.
It will be in the homework.
Not this week's homework,
but next week's homework.
In fact, homework that is due
this week is due on Friday.
So the bound states of
a symmetric potential,
a potential that satisfies
this, are either odd or even.
And that's exactly
what you see here.
That's not a coincidence.
It's a true fact.
The number of nodes increase.
And the other fact that is very
important, of bound states,
of one-dimensional potentials--
supremely important fact.
No degeneracies.
If you have a bound
state of a potential that
is either localized like
this or goes to infinity,
there are no degenerate
energy eigenstates.
Each energy eigenstate here,
there was no degeneracy.
Now, that is violated by
our particle in a circle.
The particle in
the circle did have
degenerate energy eigenstates.
But as you will see, when you
have a particle in a circle,
you cannot prove that theorem.
This theorem is valid for
particles in infinitely--
not in a circle.
For x's that go from minus
infinity to infinity or x's
with vanishing conditions
at some hard walls.
In those cases, it's true.
So, look, you're
seeing at this moment
the beginning of very
important general results,
of very fundamental
general results that
allow you to understand the
structure of the wave function
in general.
We're illustrating it here, but
they are very much, truly now,
general potential.
So what are they?
For one-dimensional potential
unbound states, no degeneracy,
number of nodes
increasing one by one.
If the potential is
symmetric, the wave functions
are either even or odd.
