PROFESSOR: So here comes the
point that this quite fabulous
about Hermitian operators.
Here is the thing that it
really should impress you.
It's the fact that any,
all Hermitian operators
have as many eigenfunctions and
eigenvalues as you can possibly
need, whatever that means.
But they're rich.
It's a lot of those states.
What it really means is that the
set of eigenfunctions for any
Hermitian operator--
whatever Hermitian operator,
it's not just for some
especially nice ones--
for all of them you
get eigenfunctions.
And these eigenfunctions,
because it has vectors,
they are enough to span
the space of states.
That is any state can be
written as a superposition
of those eigenvectors.
There's enough.
If you're thinking finite
dimensional vector spaces,
if you're looking at
the Hermitian matrix,
the eigenvectors
will provide you
a basis for the vector space.
You can understand anything
in terms of eigenvectors.
It is such an important theorem.
It's called the spectral
theorem in mathematics.
And it's discussed in
lots of detail in 805.
Because there's
a minor subtlety.
We can get the whole
idea about it here.
But there are a couple
of complications
that mathematicians
have to iron out.
So basically let's
state we really
need, which is the following.
Consider the collection of
eigenfunctions and eigenvalues
of the Hermitian operator q.
And then I go and say, well,
q psi 1 equal q 1 psi 1 q psi
2 equal q2 psi 2.
And I actually don't
specify if it's
a finite set or an infinite set.
The infinite set, of course,
is a tiny bit more complicated.
But the result is true as well.
And we can work with it.
So that is the set up.
And here comes the claim.
Claim 3, the
eigenfunctions can be
organized to satisfy the
following relation, integral dx
psi i of x psi j of x
is equal to delta ij.
And this is called
orthonormality.
Let's see what this all means.
We have a collection
of eigenfunctions.
And here it says
something quite nice.
These functions are like
orthonormal functions,
which is to say each
function has unit norm.
You see, if you
take i equal to j,
suppose you take psi 1 psi 1,
you get delta 1 1, which is 1.
Remember the [INAUDIBLE] for
delta is 1 from the [INAUDIBLE]
are the same.
And it's 0 otherwise.
psi 1 the norm of psi 1
is 1 and [INAUDIBLE]
squared [INAUDIBLE]
psi 1, psi 2, psi3, all of
them are well normalized.
So they satisfied this thing
we wanted them to satisfy.
Those are good states.
psi 1, psi 2, psi 3,
those are good states.
They are all normalized.
But even more, any two
different ones are orthonormal.
This is like the 3
basis vectors of r3.
The x basic unit vector, the y
unit vector, the z unit vector,
each one has length 1, and
they're all orthonormal.
And when are two
functions orthonormal?
You say, well, when vectors are
orthonormal I know what I mean.
But orthonormality for functions
means doing this integral.
This measures how different one
function is from another one.
Because if you have
the same function,
this integral and this
positive, and this all adds up.
But for different
functions, this
is a measure of the inner
product between two functions.
You see, you have the dot
product between two vectors.
The dot product of two functions
is an integral like that.
It's the only thing
that makes sense
So I want to prove
one part of this,
which is a part that is doable
with elementary methods.
And the other part is a
little more complicated.
So let's do this.
And consider the case if
qi is different from qj,
I claim i can prove
this property.
We can prove this
orthonormality.
So start with the integral
dx of psi i star q psi j.
Well, q out here at psi j is qj.
So this is integral dx
psi i star qj psi j.
And therefore, it's equal to qj
times integral psi i star psi
j.
I simplified this by
just enervating it.
Because psi i and psi
j are eigenstates of q.
Now, the other thing I
can do is use the property
that q is Hermitant and move
the q to act on this function.
So this is equal to integral
dx q i psi i star psi j.
And now I can keep
simplifying as well.
And I have dx.
And then I have the complex
conjugate of qi psi i psi i,
like this, psi j.
And now, remember q is an
eigenvalue for Hermitian
operator.
We already know it's real.
So q goes out of the
integral as a number.
Because it's real,
and it's not changed.
Integral dx psi i star psi j.
The end result is that we've
shown that this quantity is
equal to this second quantity.
And therefore moving this--
since the integral is the same
in both quantities, this shows
that q i minus qj, subtracting
these two equations,
or just moving one to one side,
integral psi i star psi j dx
is equal to 0.
So look what you've proven
by using Hermiticity,
that the difference
between the eigenvalues
times the overlap between
psi i and psi j must be 0.
But we started
with the assumption
that the eigenvalues
are different.
And if the eigenvalues are
different, this is non-zero.
And the only possibility
is that this integral is 0.
So this implies
since we've assumed
that qi is different than qj.
We've proven that psi i
star psi j dx is equal to 0.
And that's part of
this little theorem.
That the eigenfunctions
can be organized
to have orthonormality
and orthonormality
between the different points.
My proof is good.
But it's not perfect.
Because it ignores one
possible complication,
which is that here we wrote the
list of all the eigenfunctions.
But sometimes something
very interesting
happens in quantum mechanics.
It's called degeneracy.
And degeneracy
means that there may
be several eigenfunctions
that are different but have
the same eigenvalue.
We're going to find that soon--
we're going to find,
for example, states
of a particle that move in
a circle that are different
and have the same energy.
For example, a particle moving
in a circle with this velocity
and a particle moving in a
circle with the same magnitude
of the velocity in
the other direction
are two states that
are different but have
the same energy eigenvalue.
So it's possible that this
list not all are different.
So suppose you have like three
or four degenerate states,
say three degenerate states.
They all have the
same eigenvalue.
But they are different.
Are they orthonormal or not?
The answer is-- actually
the clue is there.
The eigenfunctions can
be organized to satisfy.
It would be wrong if you say
the eigenfunctions satisfy.
They can be
organized to satisfy.
It means that, yes, those ones
that have different eigenvalues
are automatically orthonormal.
But those that have
the same eigenvalues,
you may have three
of them maybe,
they may not necessarily
be orthonormal.
But you can do linear
transformations of them
and form linear combinations
such that they are orthonormal.
So the interesting
part of this theorem,
which is the more difficult
part mathematically,
is to show that when
you have degeneracies
this still can be done.
And there's still enough
eigenvectors to span the space.
