PROFESSOR: Good morning.
We're going to continue doing perturbation theory today.
We've finished doing non-degenerate perturbation
theory, and then turned to the degenerate perturbation theory
as well.
We had computed already the first order corrections
to the state, first order corrections to the energy,
and second order correction to the energy.
So here in this formulas that I've written
are all the results we have so far.
And a useful thing is to ask yourself
whether you know what every symbol here means.
This is the state who change the deformation
of the original state n0 of the Hamiltonian H0.
We have the Hamiltonian H of lambda,
which was equal to H0 plus lambda delta H.
And n0 was an eigenstate of H0 with energy En0.
And we found that as you turn on lambda,
lambda becomes non-zero.
The energies change.
The eigenstate changes.
And here is lambda, first order in lambda, second order
in lambda.
We didn't compute the state to second order in lambda.
Moreover, we have the symbol delta Hkn
that was defined to be the matrix element
on the unperturbed eigenstate of the operator delta H. Well,
that's another symbol in here.
We are summing over all states in the spectrum,
except for the state n here because the denominator here
would give you a 0.
And this would be problematic in general.
So that was what we derived.
And these are the formulas we have.
It's worth making a few remarks as to what
we see from these formulas because they're
a little complicated.
And you may not have immediate intuition about it.
So a few remarks.
And the first one--
in fact, both remarks that I'm going
to make concern the behavior of the energy as represented
by this formula.
First remark is that if you were to look at the ground state
energy of the system, the first two terms here, so for n
equals 0, if we call the ground state by the label n
equals 0 for this state n equals 0,
these first two terms on the energy overestimate
the ground state energy.
They always give you more than what the true ground state
energy is for any value of lambda.
That is kind of plausible, given that you see here
that for the ground state energy, when n is equal to 0,
and it's the lowest energy state,
these differences are all positive.
All these states have more energy.
Therefore, this term is negative.
And the next correction to order lambda square
tends to lower the energy of the ground state.
So the claim is that the first v lambda ground state
energy overestimate the true ground state energy.
And so how do we see that?
So we consider E0, because we're doing n equals 0.
This is the ground state.
We start labeling with n equals 0 sometimes.
Plus lambda H00.
That is the order lambda estimate for the ground state
energy.
The first term can be viewed as the expectation
value of the original Hamiltonian on the ground
state.
That is E0 because this H0 and 0 is E00.
So this is E00.
Plus lambda.
I'm sorry here.
I made a little mistake.
That should be delta H. Delta H00
plus lambda, the expectation value of delta H
on the same state.
So these two things together are nothing
but the expectation value on the ground state of H0 plus lambda
delta H.
That's kind of nice.
Isn't it?
That these two terms really are nothing else
but the expectation value on the ground state of this thing
that we called H of lambda.
And now comes the variational principal
that says that if you evaluate the Hamiltonian on an arbitrary
state, the expectation value of the Hamilton in an arbitrary
state, you get more energy than the ground state
energy of the system.
You always get more.
And when you hit the ground state,
you get the lowest value.
So look at that.
This is the Hamiltonian.
And this is the unperturbed ground state.
This is not the real ground state.
The real ground state is what we're trying to find.
So this is like saying, OK, you're
evaluating the exact Hamiltonian on some arbitrary state.
Therefore, this is greater than or equal to the true ground
state energy that we would call E0 lambda.
True ground state energy.
Which in that notation, indeed, is E0 lambda.
So that's a nice result. And it matches with the idea
that for the ground state the order lambda squared
correction is minus lambda squared, the sum over k
different from 0 delta Hk0 squared over Ek minus E00.
And this thing, as we mentioned, the numerator is positive.
The denominator is positive because this
was the ground state.
And this is a non-degenerate ground state.
And K, therefore, has more energy here, Ek0.
So everything is positive.
So this is negative.
So actually here, there is a generalization
of this observation.
And you can imagine that you have now a particular state,
n here.
Then you have all these states that are above
and all these states that are below.
So let's look at the second order correction
to the energy of the state En, the second order correction.
The first order correction shifts.
The energy is proportional to delta Hmn.
But the second order correction is
minus lambda squared, the sum over k different from n.
So let's split it.
k greater than n.
And we'll write the same thing Hkn squared over Ek0 minus En0.
And then I want to write the other states, the states where
k is less than n.
But let me do one thing here.
On those states, I'll switch the order of the sum.
I'll put En0, the order of the sign in the denominator, Ek0,
and change the sign here.
So these two changes of sign are correlated.
I changed the sign in the denominator.
And I changed the sign in front.
And now, we see the following.
All these states that are above n, k is greater than n.
This difference is positive.
This is manifestly positive.
This is negative.
So these states are kind of pushing your state down.
The fact of this states on top is a negative correction
are pushing the states down.
On the other hand, the lower states,
the states with k less than n, again, I've
ordered now convenient.
The other states k now have less energy.
So this is positive.
This is positive.
So these states over here are actually pushing that state up.
So the upper states are pushing it down.
The lower states are pushing it up.
This is usually referred to as level repulsion.
The levels repel each other.
The upper states don't want the state n to go close to them.
The lower states don't want them to go close to them either.
So it's a nice dynamic that helps you understand
what's going on here.
Another point we want to make has
to do with the validity of this expansion.
So in general, when you have series expansions
the issues of convergence are delicate.
So we can get a lot of insight just by doing an example.
So let me talk about the validity of the perturbation
series.
This is supposed to give us some insight.
One thing we've said is that we think delta H is
supposed to be smaller than h0.
Now, these are operators.
So the statement that they're smaller
has to be made more precise.
What is the size of an operator?
And you could think, well, we could
say the entries of the matrix elements
should be much smaller.
And that is true, but it's not enough.
For that, let's consider an example.
So this will be a two by two matrix Hamiltonian, in which
H of lambda will be given by some H0 plus lambda v
hat, which will be E1 0 E2 0 plus lambda.
That v hat matrix will be of the form v v star.
And must be her mission.
So v is a number.
v star is the complex conjugate.
These are the two energies, because the Hamiltonian
is known.
Each time we talk about the Hamiltonian that is known,
we write that there's a diagonal matrix.
We're saying, yes, that is the matrix element
of the Hamiltonian in the basis of eigenstates, which we know.
So a known Hamiltonian you can represent by a diagonal matrix.
So here is our whole matrix H of lambda.
And you could say, all right, our perturbation theory,
practically speaking, is these formulas that
allow us to compute the eigenvalues of this matrices,
which are of this matrix, which is
the energies of the system of the eigenstate
and compute the eigenstates.
For H0, these are the eigenvalues.
And the eigenstates are 1 0 and 0 1.
So perturbation theory is really,
you can think in terms of Hamiltonians,
but also you can think of it in terms of matrices.
It's allowing you to find the eigenvalues and eigenvectors
of this matrices.
So this is a two by two matrix with numbers.
You know how to compute the eigenvalues.
So I'll give you the answer.
The eigenvalues E plus minus are E1 plus E2/2
plus minus E1 minus E2/2 times the square root of 1
plus lambda squared absolute value of V2 squared E1
minus E2/2 squared.
Long formula.
That's it.
Those are the exact things.
If you wanted to see this as a result in perturbation theory,
you would say, well, lambda squared v squared
or lambda v is small.
And I'm going to think of this term as small.
And I'm going to expand the square root.
And if I expand this square root,
I'm going to get all kinds of terms
with different powers of lambda.
You can also see there that there's no energy correction
linear in lambda.
Because when you expand 1 plus epsilon square root,
it's roughly 1 plus epsilon over 2 plus dot, dot.
And therefore, when you span the square root,
the first correction is going to be
proportional to lambda squared.
And that actually conforms to that
because the matrix elements of the Hamiltonian delta
Hnn along the diagonals for the perturbation are 0.
So there is no order lambda correction.
So what do you do here then?
You must do this expansion.
And here, I write the relevant series.
f of z, if you define it, as 1 plus z squared
is equal 1 plus z squared over 2 minus z 4th over 8 plus z6/6.
And it's not that simple after that.
Minus 5/128z to the 8 plus order z to the 10.
So that's this perturbative series expansion
that you would use here.
And how good is the convergence of this series?
Well, it's OK.
It's something that when you study complex analysis
you see this function 1 plus square root of z squared
has branch cuts at i and at minus i in the complex plane,
in the z complex plane.
Those are the branch cuts.
These are the places where this square root becomes 0.
And you have to deal with them.
And the function is convergent only up to here.
There is a radius of convergence.
A radius of convergence is 1.
You can try it with Mathematica or with the program
and take 30 terms, 40 terms.
And you will see that as long as you take a point z here,
it converges.
You take a point a little bit out, it blows up the series.
So it has a radius of convergence.
That's not so bad.
Radius of convergence is OK.
But basically, we need z to be small, which corresponds here
to the statement that lambda v z small for a fast convergence
corresponds to lambda v absolute value
being smaller than E1 0 minus E2 0 absolute value over 2.
Roughly smaller.
And now, we see that the thing that matters for a perturbation
series to be good is also that the perturbation be
small compared with the energy differences.
Not just it should be small, the perturbation.
It should be small compared with the energy difference.
So if you have a Hamiltonian with a state of energy 100
and a state of energy 101, you might say, well,
if I take a perturbation of size two
or three, that's very small compared to the energies.
But it's not been small compared to the difference of energies.
And that can cause the perturbation expansion
to go wrong.
So this gives you extra insight that, in fact,
being a small perturbation not only means small compared
to the energies, but also small compared
with the energy differences.
Something that you see here as well.
Energy differences are controlling things.
And if the energy differences are--
if the perturbation is not small compared to energy differences,
then these ratios can be rather large.
And the perturbation terms are very large.
And nothing is very simple.
So that's what I wanted to say about the convergence
of a perturbation expansion.
More rigorous statements can be made,
but we're not going to try to make them here.