M2L6e.txt

#
# File:   content-mit-8-421-2x-subtitles/M2L6e.txt
#
# Captions for 8.421x module
#
# This file has 400 caption lines.
#
# Do not add or delete any lines.
#
#----------------------------------------

Let's now proceed and discuss the helium atom.
So we want to understand now what
are the new effects when we have not only one electron, but two
electrons.

And don't worry.
We're not proceeding to three, four, five, electrons.
I think to go from one to two, we actually
capture the most important step, namely the interaction
between the two electrons, and what the results of that is.
For the helium atom, there are some excellent treatments
in standard textbooks of quantum mechanics.
One is the famous quantum mechanic text
by Cohen-Tannoudji, but also the text of--
The reason why I have edit the helium atom for the curriculum
is because it is a simple system where we can discuss
singlet and triplet states.
And singlet and triplet configuration
is important for population of [? inoptic ?] lattices
for quantum magnetism and such.
Actually you can say, if you have two electrons,
and they align in a triplet state or singlet state,
one you can see is ferromagnetic,
the other one is paired in antiferromagnetic.
It's a simple example where we can discuss magnetism.
So that's my motivation why I want you to know something
about the helium atom.
Let's now discuss energy levels of helium.

Let's just start with the most basic model, where
helium atom has two charges.
So if you regard it as a hydrogen problem,
and we put two electrons into the 1s state,
we would expect that, based on hydrogenic model,
that the binding energy of that is, per electron,
is the [? wood ?] [? that ?] energy, as in hydrogen,
but now we have to scale it with z squared, the nuclear charge,
and this gives us a factor of four.
So we would expect that per electron, the binding energy,
in the most simple hydrogenic model is 54 electronvolt.
So that would mean that the binding energy of the ground
state is minus 108 electronvolt, however,
the experimental result is that it's only 79 electronvolt.
So we find that there is a big discrepancy or 29 electronvolt,
which is really huge.
So, what is responsible for this big discrepancy?
Well, what we have neglected, of course,
is the interaction between the electrons.

So we can fix that in the simplest way,
by keeping the wave function from the hydrogenic model,
but now, calculating the electronic energies,
the electronic, the electron-electron energy,
by using the electron-electron interaction as a perturbation
operator.
So, we still use, a wavefunction for the ground state, electron
one in the 1s state, so 1 0 0 is the designation for N L M,
for the hydrogenic quantum numbers.
And we assume that the ground state is simply
the product of two electrons in the 1s state.
So, if I calculate for this perturbation operator,
the expectation value with this ground state,
we find that there is an energy correction which
is 34 electronvolt. So this removes
most of the discrepancy.
You can improve on it by a variation wave function,
if you use hydrogenic eigenfunctions as your trial
wavefunction, but you are now calculating
those hydrogenic wavefunctions, not for nuclear charge
two, by in nuclear charge C star, which
you keep as a variation parameter,
you find that you find even better wavefunctions,
and you can remove two-thirds of this remaining discrepancy
or five electronvolts.
This variation wavefunction is left to you
as a homework assignment.

So as z is z equals 2, is replaced
by a variation parameter.
Anyway, that's all I want to tell you about the ground
state of helium.
It's pretty much finding a wavefunction which correctly
captures the Coulomb energies, the interaction
between the twin electrons.
But what is much more interesting,
and this is what I want to focus on
for the rest of this lecture, is what
happens in the excited state.
Before I do that, let me just tell you
what we just discussed.
So we have this hydrogenic estimate,
and eventually, the Coulomb energy, this
is the energy level to what we have just discussed.
But, you know, as the ground state has no degeneracy,
and so we all talk about quantitative shifts.
However, when we go to the excited state,
we will find degeneracies, and degeneracies are much more
interesting, because something which was degenerate can split,
you have two different terms.
So suddenly, there is richer physics.
So therefore, we want to discuss now the excited state.
So starting again with the hydrogenic model.
In hydrogen, the 2s and 2p state are degenerate.
So we have two configuration contributing
to the same energy.
One is 2s, and one is 2p.
The binding energy in a hydrogenic model is a quarter
[? of root back ?], [? root ?] [? back ?] over n squared and n
is 2.
We have to scale it by z squared,
and we find 13.6 electronvolt.
But now what happens is that we have to introduce the Coulomb
energy between the electrons.
And if you do that, it shifts up the levels in different ways.
So this is 1s 2s, this is 1s 2p.
So why is this different?

Well, you can see the following--
you have two electrons, and you have the helium nucleus.
And you first put in the 1s electron, and now
the second electron, when it is in a p state, in a 2p state,
it's further out, and it pretty much experiences out there,
the charge off the helium nucleus
shielded by the 1s electron.
And therefore, it sees, in effect,
a smaller nuclear charge.
Whereas the 2s electron penetrates deeper,
gets closer to the nucleus, and will still
realize that the nucleus has a charge of two, and not
a shielded charge.
So therefore you would expect that the shielding effect,
due to the innermost electron, is
more important, has a bigger effect for the 2p electron
then for the 2s electron.

So let me just write that down.
So the 2p, it sees a shielded nucleus.

In other words, what is experience more
as an inner core, the Coulomb potential of helium plus,
and not so much of helium 2 plus,
and this is due to the 1s electron,
and therefore the 2p electron has a smaller binding energy.
It's actually comparable to the binding
energy of the 2p state in hydrogen,
which is on the order of four electronvolt.
This effect is smaller for the 2s state.
So let's go back to the energy diagram.
So we have now the situation which are just described.
We have two degenerate configurations
for in the hydrogen model, and in the Coulomb energy
between the electrons, there is quite a bit
splitting of several electronvolts.

But now, each level undergoes further splitting,
and this is what I want to discuss.
So we're just still sorting out just the preliminaries,
what I really want to discuss with you
is the singlet and triplet thing,
but now we are there where we can do it.
So if you have a configuration with 1s 2s,
there are two possibilities for the total angular momentum.
Two electrons, in an s state, there
is no orbital angular momentum, but there
are two spins one-half, and they can add up to one,
or can add up to zero.
So therefore we will have two different terms.
One is singlet is zero, and one is triplet is one.
And the splitting is 0.8 electronvolt,
and this is what you want to discuss in the following.
For completeness, but the physics is similar,
let me mention that the 1s 2p state is also
gives rise to two terms.
We have now the total orbital angular momentum
p, orbital angular momentum of one.
The spin angular momentum is zero or one,
so we have singlet and triplet.
And the total angular momentum is one, in this case,
or in this case, two, one, or zero,
and the splitting in this situation
is also on the order of a fraction of an electronvolt.
So what we want to understand now is,
why do we observe a splitting between these two levels, which
seems to depend on the spin?
How can the spin cause the splitting?
Because the spin, so far, has not appeared in our Hamiltonian
We really have a Hamiltonian, which
is on only the Coulomb energy, and the spin,
is not part of it.
So we don't have a magnetic field
to which the magnetic moment of the spin would couple.
And also, we have not yet input new spin orbit coupling.
But if this is on your mind, take it off your mind.
Spin orbit coupling is a much, much smaller effect.
One energy on the order of one electronvolt,
you just cannot get from spin orbit coupling.
Spin orbit coupling is smaller than electronic energies-- as I
will explain to you on Friday-- is smaller
by the fine-structure constant so the typical scale for spin
orbit coupling is maybe 10 or 100 million electronvolt.
It's much smaller.
So therefore, we want to understand now
why do we have spin-dependent energy.
Also, we haven't coupled, at this point,
the spin to any field.
We are focusing now on the splitting.
So we have the 1s 2s configuration.
We get two terms, as I just discussed.
One is singlet, and one is triplet.
And still using the model, hydrogenic model
of non-interacting electrons, we want
to write down the wavefunction.
So, the wavefunction of the two electrons
is we have one electron in the 1s state,
we have one electron in the 2s state.
But now, since we have fermionic atoms,
we have to correctly symmetrize it.
So whether we want the symmetric or anti-symmetric combination,
we exchange the two electrons, so now we
have two and our one in reverse order.

Of course, the total wavefunction for two fermions,
has to be anti-symmetric, but the total wavefunction,
is the product of the spatial wavefunction,
which I just wrote down, times the spine wavefunction.
The spin wavefunction can be anti-symmetric and symmetric,
and the anti-symmetric spin wavefunction
has to combine the symmetric spatial wave function, and vice
versa, to make sure that the total wavefunction is
anti-symmetric, and to correct description for fermions.
The designation here, symmetric and ant-symmetric
for the total wavefunction, reflects the spatial part.
The total wavefunction, of course,
including the spin wavefunction is always anti-symmetric.
OK.
So we have two wavefunctions.
One has symmetric spatial wavefunction, the other one
an anti-symmetric.
The symmetric spatial wavefunction
has an anti-symmetric spin wavefunction,
and that means you have s equals zero.
That's up down minus down up, it's anti-symmetric.
Whereas the anti-symmetric spatial wavefunction
goes together with the symmetric spin wavefunction s equals one.
So this is the situation which gives rise to the triplet
s 1 term, and this here is the singlet s 0 term.
OK.
As long as we have non-interacting electrons,
the two wavefunctions are degenerate.
But now we want to bring in the Coulomb energy between the two
electrons, which we had already discussed before,
and if you calculate, if you now calculate, the energy using
this as a perturbation operator, well, remember,
the wavefunction had two parts.
It was 1 0 0 r 1, 2 0 0 r 2, and the part where r 1 and r 2
were flipped.
And now, if you have the wavefunction, the perturbation
operator, we get a total of four terms.
2 times 2, we have to sort of exit out.
And we will then have this sort of diagonal parts.
And we have the parts which are off-diagonal,
and for the off-diagonal parts, it
matters whether we hit the plus or minus sign.
You know minus, if you have plus plus and minus
minus, they give a positive contribution,
but if you connect the wavefunction
before and after the operator, plus with minus,
we get minus signs.
So in other words, we have one contribution
where it doesn't matter whether we
have the symmetrical and anti-symmetrical wavefunction,
but then we have another term where
it matters whether we have symmetric and anti-symmetric
wavefunction.
That's where the plus or minus sign
from the symmetrized wavefunction appear.
So we have two contributions now to the energy correction.
One is independent of the spin wavefunction,
whether we have the symmetric or anti-symmetric configuration,
the other one is not.
The first term is called the Coulomb energy.
The second term is called the exchange energy.
So, what we find is-- we have to trace back the sign--
but you find that the triplet state, the triplet state
is symmetric in spin, and anti-symmetric
in the spatial wavefunction has the lower energy,
is more strongly bound, because the anti-symmetric spatial
wavefunction that uses the repulsive interaction
between the two electrons.

So in other words, we do not have any spin
term in the Hamiltonian, it's just
that whether the spin wavefunction is
symmetric or anti-symmetric, it requires
the spatial wavefunction to be the opposite.
And now, when we calculate the Coulomb energy,
for the spatial or anti-symmetrical spatial
wavefunction, we find a big difference,
and the big difference is the exchange energy.
So it is a spin-dependent term for the total energy,
but what is behind this energy is simply the Coulomb energy.
So the spin through the symmetry of the wavefunction,
leads to a difference in the Coulomb energy.
And actually, what I'm telling you
is the explanation why we have magnetism at room temperature.
It was Heisenberg's idea when he realized,
for the first time, what can cause ferromagnetism.
It was pretty much the model of the helium atom expended
to many, many electrons in the lattice.
So the result is that the Curie temperature of thousands
of degrees, we have magnetism below 1,000 degrees,
this energy scale is an electronic energy,
it's a Coulomb-energy scale, and not
an energy scale where spin and direction comes into play.
If it were not for the exchange energy,
we would not have to have magnetic materials above one
Curie.
So this is what you see here in the Helium atom.
How spin leads to an energy splitting,
which is an electronic-Coulombic energy splitting.
OK, let's make it maybe even more obvious.
The above equation can be rewritten.

So this energy splitting can be written
as a constant alpha, plus a constant beta,
times the product of S 1 and S 2.
Well, what happens is, the product of S 1
and S 2 can be written as minus S 1 squared minus S 2
squared plus S squared is 1 squared is
one-half times three-halves is three quarters is 2 squared is
three quarters, and S squares is either zero or 2,
depending whether you're in the singlet or triplet state.
In other words, I just want to-- with the sidebar,
which you've seen many, many times-- just to remind you,
that the product of S 1 and S 2 has only two values, one
for the singlet state, one for the triplet state.
So therefore, if I have a singlet level
and a triplet level, I can always parameterize it
like this, and I can make it more obvious
by showing what this formula, on the right hand side,
has two values.
One for S equals zero, and one for S equals 1.
And this can be rewritten as alpha--
you'll find a more explicit calculation in the Wiki,
but it's also just one more line of algebra.
You can write it in the following way.
So therefore, the alpha and beta parameter
are just a way of-- you have two energy levels,
and you, with two constants, you can always
describe two energy levels.
Here I've done with alpha and beta.
But before, I did it with the Coulomb energy and the exchange
energy, which we obtained in this equation, when
we perturbatively calculated the integral.
So by writing it as S 1 dot S 2, I even suggest that the two
spins interact like a dipole-dipole interaction,
but they don't.
It comes from the Coulomb interaction,
but it is equivalent to a gigantic dipole-dipole
interaction.
So therefore, the conclusion of this
is that what I've derived for you for the helium atom,
it looks like the ferromagnetic spin-spin interaction.
And well if it looks like it, is is actually
an effective ferromagnetic spin-spin interaction.
However, the coupling is purely electrostatic here,
and not magnetic.

OK.
Yes.
So, since we're still using the hydrogenic wavefunctions
as a basis, how closely does this
get to the actually measured [? ohm? ?]
The question is, how-- how close do we
get with hydrogenic wavefunctions
to the actual measured energy?

I don't know, I don't know the exact numbers for the excited
state.
I assume that it's similar to the ground state.
You saw that for the ground state we had a big discrepancy.
Most of that was closed by using hydrogenic wavefunctions,
and just calculating the perturbative terms.
So pretty much you get, quantitatively
or semi-quantitatively, you get the picture out of it.
And unless you're really interested in the absolute
values, you can stop there.
But one way to go further and reduce a discrepancy by 75%
is to use this variation wavefunction
values, hydrogenic wave function,
but you use Z, the nuclear charge,
as a variation parameter.
And I mean this is-- it's amazing.
I mean you a two-electron atom and you
use a hydrogenic wavefunction with one [? feet ?] parameter
and you get binding energies which are on the order of 70
electronvolt accurate to within better than 3%, 4%, 5%.
But by adding other terms, we're using little bit more fancy
wavefunctions.