PROFESSOR: OK, degenerate
perturbation theory.
OK, we've done nicely
our non-degenerate case.
So we'll get degenerate done
in a very clear way, I think.
One more blackboard.
So the first thing I want
to say about degenerate
perturbation theories, when
it fails and why it fails
and what goes wrong.
So degenerate
perturbation theory.
Trivial example to begin
with, why not, h of lambda,
again, is equal to h0
plus lambda delta h.
And for h0, we'll take original
Hamiltonian with energies 1
and 1, and for
delta h, I'll take
an off-diagonal
Hamiltonian, a Pauli matrix,
and that's our delta h.
And then we say OK, blindly,
not worry about this degeneracy,
and use the formulas that we
have for non-degenerate states.
And we would say well,
for the first state
and the second state
what do we have?
E, I'll call them 1
and 2, not 0 and 1.
So let's call them
1 of lambda will
be equal to the first value, 1,
plus, if the formula is right,
delta h times lambda, 1, 1.
The matrix element of the
perturbation in the 1, 1 state.
Similarly, E2, the
other eigenvalue,
should be 1, there's the
second eigenvalue, plus lambda,
delta h 2, 2.
That's what the
formula would say.
But this is
completely incorrect.
Why?
Because delta h 1, 1 is the 1, 1
element of delta h, and it's 0.
And the 2, 2 element
of delta h is also 0,
and therefore this says 1 and
1, that the eigenvalues haven't
changed.
On the other hand, we know that
the matrix, the total matrix,
which is 1 lambda, lambda
1 has eigenvectors,
1 over square root of 2, 1, 1,
and 1 over square root of 2,
1 minus 1, with eigenvalues--
this is our eigenvectors
and eigenvalues--
1 plus lambda and
1 minus lambda.
So the eigenvalues of the
matrix are 1 plus lambda
and 1 minus lambda.
And this formula, to first order
in lambda, gave us nothing.
It failed badly.
It's a disaster.
So you can say,
well, what happened?
And maybe a clue is on the
fact of the eigenvectors.
These are the eigenvectors
of the total matrix.
These are in fact,
also the eigenvectors
of the perturbation.
On the other hand,
when we said OK, this
is the first state,
the first eigenvector,
we thought of it as 1,
0 on the Hamiltonian,
on that eigenvector
gave you the energy 1.
The second energy corresponds
to the eigenvector 0, 1.
That's what we usually
use for our matrices.
So here is the strange thing
that seems to have happened.
For the original
Hamiltonian, [INAUDIBLE]
we had this eigenvector
and this eigenvector,
and then suddenly you
turn on the perturbation
and lambda can be 10
to the minus 1,000,
and already the eigenvectors
jump and become this one.
They go from this
one to that one.
You just add a perturbation
that you even cannot measure,
and the eigenvectors change.
That's crazy.
The explanation is this
thing that we usually,
sometimes forget to say in a
mathematically precise way.
These are not the eigenvectors
of the original matrix.
Since the original
matrix is degenerate,
this is one possible
choice of eigenvectors.
Any linear combination of
them is an eigenvector.
So this space of
eigenvectors of this matrix
is the span of these things.
So there was no reason
to say, oh, these
were the eigenvectors before,
and now they've changed.
They're this.
No.
Before, you don't know
what are the eigenvectors.
They're ambiguous.
There's no way to
decide who they are.
This, this, or any
linear combination.
Remember, when you have
a degenerate eigenstate,
any superposition of them
is a degenerate eigenstate.
So the explanation is
that, in some sense,
these eigenvectors of
the perturbed Hamiltonian
is one of the possible choices
of a basis of eigenvectors
for the original space,
and what the perturbation
does is break the degeneracy.
All those equivalent
eigenvectors suddenly
are not all equivalent.
There are some preferred ones.
So the fact is that the
formulas are not working right,
and we have to do this again.
So here is what
we're going to do.
We're going to set this up
for a systematic analysis
to get this right.
So systematic analysis.
So again, we say h0 is known
and it will have an E of, say,
1, 0, an E 2, 0, like that,
until you encounter an E n, 0
that happens to be equal to
the next one, E, n plus 1, 0,
and happens to be
equal to the next one,
all the way to an
E n plus capital
N minus 1, 0, which
is then smaller
than the next one, which is
E n plus N and then again.
So I'm saying in
funny ways the fact
that, yes, you had some
states and suddenly you
hit a collection of N energy
eigenvalues that are identical.
They're all equal to En0.
All n states have energy en0.
So it is as if you
have this matrix h0
and there are all
kinds of numbers.
And then there's a whole block
where all the entries of size
n by n where all the
entries are the same.
So n, n plus 1, n plus 2, up
to N minus 1 is n entries,
and they all have
the same energy.
And the corresponding
eigenstates
are going to be called n--
we're going to use the
label n for all of them,
even though the energies
are all the same.
So there's just an E n energy.
So we have N state,
so there will
be an n1, an n2, up to an nN.
All those states are going
to form an orthonormal basis
for a space that we're
going to call vn.
So the space spanned by this
state, we're going to go vn,
with the n reminding you
it has to do with the n
states, states with energy
described by the label n.
So this is the
degenerate subspace.
This is the space of all those
states that have energy En0.
So I'll state that here.
So it's quite important to
get our notation right here.
Otherwise we don't
understand what's going on.
So h0, n0k, they all
have energy En0, n0k.
And this is for k equals
1 up to N. Now OK,
this is your
degenerate subspace.
These are all the states.
And we now want to know
what happens to this space,
to all these states,
when the energy turns on.
This is not just a
theoretical construct.
When you start asking what
happens to the first excited
states in the hydrogen
atom, you already
have four states there, l
equals 1, l equals 0 states,
n equals 2.
And you have degenerate
states and immediately you're
stuck in this situation.
So I will also use the
notation that the total space,
a curly h, is the sum
of vn plus a v hat.
This is called the direct sum.
At this point it just
doesn't matter too much.
These are linearly independent
subspaces, vn and v hat,
and basically you think as v
hat as all the other states.
Yes, there were these
degenerate subspace
and all the other
states of our v hat.
So v hat is the span
of all other states.
And we'll call them
p0, or what letter?
Yeah, p0, with p
equals some numbers.
So this is a good notation.
It allows you to distinguish.
The degenerate
states use 2 labels.
Maybe you only needed
1 label, but you
will get confused
with the notation
if you just use 1 label.
So if that's the case,
degenerate state,
but by putting
the n here you now
know that you're talking
about degenerate states.
When you have a
single label you're
talking about the other
states in the space.
So it's very good
when your notation
makes it easy to
think and recognize
the equations that you have.
And v hat in this,
are orthogonal spaces.
v hat is perpendicular to vn.
The inner product of any state
outside of degenerate space,
with a state in the
general space, is 0.
OK.
Very good.
So let's now do the
important thing.
I can do it here, I think.
I'll do it here.
So what do we want to do?
What is the question here?
The question is the following.
We have a state n0k.
We have N of those
states, and we
want to figure out
what they become as you
turn on the perturbation.
So using the old notation,
we'll say they become this.
And what is that going to be?
It is going to be n0k,
the original state, what
it was, plus order lambda,
the first correction,
so n1 for this k state,
plus dot dot dot.
For the energies, we have En0.
That was the energy
of every state
in the degenerate subspace.
Now it's going to
become En lambda.
But if I say En lambda I'm
already making a mistake,
because I'm looking
at the fixed k.
And a fixed k means we
chose one of these states,
and any single one of
the states, the energy
can grow in a different way.
So I should call this
Enk, because we're
talking about this state,
will be equal to En0--
yes, they all have the same
first zeroth order energy.
But then Enk1 plus lambda
squared Enk2 and so on.
So this is the fate
of the kth state.
And the degeneracy will
be broken to first order
if these Enk's become different
numbers, because if they become
different numbers in
lambda they're going
to start splitting the states.
Remember this picture we had
last time of several states
at one point?
If the order lambda
corrections are different,
the states split.
The degeneracy is resolved.
And we can calculate
things more easily.
If it's not resolved to
first order, it's harder.
It will be all of next
lecture to figure out
what happens in that case.
We discussed last time
that the n1 correction
didn't have any component
along the n0 vector.
We could always
arrange that to happen.
Here, we can arrange
something similar to happen.
So we will still
assume, and we can
check that's always
possible, that vnp,
the corrections to the state k
at order p for p equals 1, 2,
3, because it's not 0,
are orthogonal to n0k.
So this state doesn't have
any component along n0k,
and this state doesn't
have any component
along the original one.
It's always possible to do
that, just like we did it
in the non-degenerate case.
Moreover, this is something
I want you to notice.
We're saying that
the state n1 doesn't
have any component along
the state n0 for a given k,
but the state n1 can
have components along n0
for a different k.
So while they've nth order
or a pth order correction
doesn't have
component along this,
it can have component
for an l here,
where l is different from k.
So it means that npk still
can have a component in vn.
Remember, vn is
this whole thing.
So if you're doing
the kth one, well, n1k
doesn't have a component
along that one,
but it may have a component
along all the others.
So it may still have a
component, that correction,
in vn.
In fact, that's what will
make the problem a little hard
to do, but it's a good thing to
try to figure out this thing.
So what is the equation
we want to solve?
The usual equation, h
of lambda on nk lambda
is equal to Enk of
lambda nk of lambda.
So we produce the
general state that's
its energy for the
full Hamiltonian.
That's our Schrodinger equation.
So what do we have to do?
We have to do exactly
what we did before,
plug in that series, separate
the terms with various lambdas,
and see what we get.
So here is what we get.
To order lambda to the 0,
you get h0 minus En0 on n0k
equals 0.
And that equation, as it was
the case for the non-degenerate
case--
sorry for the redundancy--
it's trivially satisfied.
We've stated that those are the
energies in the top equation
on that blackboard.
So this equation we don't
need to worry about.
That's not too difficult.
Lambda 1, h0 minus
En0 and 1k will
be equal to Enk1
minus delta h n0k.
Last equation.
I'm starting to get lazy to
write them out, but we must.
n2k is equal to minus delta
h and 1k plus Enk2 and 0k.
OK and these equations
are equations for k fixed,
but every term in the
equation has the same k,
but k then can run
from 1 up to n.
So three equations times
n times, there we go.
OK, these are our
equations this time.
And we need to understand them.
In fact, we need to solve them.
What we've gained experience
with the non-degenerate case
is going to come
very useful here,
although some things are not
going to be exactly the same.
We're going to try to find
the energy corrections here,
but calculating the state n1
is going to be a little harder.
We won't finish today--
we won't have much time left--
but what's going to
happen is that you
can calculate the state n1,
the part in the space v hat.
But the part in the
degenerate subspace,
where I said that npk
still can have a component
in the degenerate subspace,
cannot be calculated from this
equation.
So even when we're finished
doing this first equation,
you're going to
find this equation,
you still have not calculated
all of the states n1.
You're going to have to go
through the second equation
to find the missing
part of the state n1.
So it's going to be
pretty interesting.
