PROFESSOR: The molecules and Born-Oppenheimer approximation.
OK, we all know that molecules are
a lot harder to solve than atoms,
and even atoms are not that easy once you
have more than one electron because
of the electrostatic repulsion, but molecules
are significantly different in that one of the greatest
simplicities that we had with atoms
is that the atom is such that the potential created
by the nucleus is spherically symmetric.
When you have a molecule you have separate nuclei
and therefore your spherical symmetry is gone,
and whether you have one electron
or more than one electron, there is no spherical symmetry.
All our tools of angular momentum don't help us much.
We have to start the problem anew.
So the difficulty with molecules is basically
that the potential for the electrons, where they move
is not spherically symmetric.
There is another thing that helps
us, however, is that there's a nice separation of mass scales.
You have the mass, little m, of the electron
and the nuclear mass, this could be the mass of one nucleus
or a mass of a few nuclei or doesn't matter,
but this is small, and typically it is 10 to the minus 4.
You know, an electron is about 2,000 times lighter
than a single proton.
If you already have a nice nucleus with a couple of them,
you're several thousandths already there,
so this is a nice number.
So we want to understand the scales of the physics of this,
so let's assume we have a molecule,
maybe a has few centers and then there's electronic cloud,
and it has a size of the order of a.
So we can estimate at first sight what
are the energies involved here?
What is the energy of the electron, for example?
And you can say the momentum of the electron
is of the order of h bar times the size of the cloud.
It's an estimate, and the energies, the electron
energies, are basically determined
by the size in which there are confined,
which is a, h bar, and the mass of the electron.
Those are the electronic energy, so h squared over
m a squared, those are the right units,
is p squared over m, the units, so it's natural like this.
So this is the electronic configuration
and typical electronic energy, so what
is the approximate picture of the physics of a molecule?
Basically you have nuclei, and the nuclei repel each other,
they want to be away.
Now the electrons come in and then they
fly around the nuclei, and the electrons attract the nuclei
and attract the nuclei, so the electrons, reasonably placed,
compensate the repulsion of the nuclei
by creating some new attraction.
At the end you get some kind of equilibrium.
It's not classical because you cannot create stable
equilibrium between classical particles.
This is one of the amazing things about quantum mechanics.
You know very well an atom would not
exist without quantum mechanics, you
would have stable classical orbits, circular orbits,
but they radiate energy so they decay.
If you have a molecule you have two repelling positive
charged particles.
In [? E&M ?] you saw the case when
you put the negative charge in the middle
and you create equilibrium, but that's absolutely unstable.
You move that electric charge a micron and then off it goes,
the whole thing is unstable.
But quantum mechanics solves those problems
and our physical picture is that of repelling nuclei,
an attracting cloud of electrons creating
an equilibrium in which the nuclei are reasonably
well-localized, they're quantum particles
but they're reasonably well-localized,
while the electrons are completely delocalized.
That is our picture, they are moving in that way.
So our picture will be-- we have that
and we have slow vibrations of the nuclei that
are mostly localized.
And you can imagine calculating the electronic configuration
for any particular separation of the nuclei
and that's the cloud for this separation.
As the nuclei vibrate, the electronic configuration
probably adjusts adiabatically to this thing
in the instantaneous eigenstates,
and they work that way.
We're going to do that in trying to understand things.
I also want to emphasize this kind
of remarkable property of an electron
to create the bound state.
So we are not amazed on the fact that if we have a proton
and we can have an electron here, we create a bound state,
and that's a hydrogen atom.
But suppose I bring in another proton from infinity.
Well, I-- if I bring it, I'm going
to start bringing it here, this kind of is neutral
but eventually this proton is gonna polarize this thing.
OK, I'm going to bring it and then if I come too close
it repels it so it will be a force repelling it
and we have to figure out what it does.
Suppose I try to bring it in.
Do you think that an electron is capable of stabilizing
two protons?
Can they create a cloud in which an electron stabilizes the two
protons?
Is that possible or not possible?
There's a lot of protons for one single electron.
Is that possible or not?
All right, you say yes, yes, any other opinions?
Can that happen?
Can the electric produce a bound state of two protons like that?
AUDIENCE: Isn't that just hydrogen?
PROFESSOR: Sorry?
AUDIENCE: Isn't that just hydrogen?
PROFESSOR: It is what?
AUDIENCE: Isn't that just hydrogen?
Or [INAUDIBLE] hydrogen [INAUDIBLE]??
PROFESSOR: OK, yes, it can produce it,
and this would be called, with an electron here,
the hydrogen molecule ion, so H2+.
It's a hydrogen molecule that had these two things
and it lost one electron, and yes, this can be done
and we'll analyze it.
And now you can get greedy and say,
can you get an electron to stabilize three protons?
Yes or no?
AUDIENCE: [INAUDIBLE]
PROFESSOR: No, cannot get it, it's just too much work
for an electron.
It cannot do it.
There are theorems like that, Elliott
Lieb, a mathematical physicist, has proven
all kinds of these things, and we
might discuss some of that in the homework,
it's kind of nice stuff.
So we will be looking at the physics of those things
with a little detail.
So now of let's go back to our molecule
here and estimate the vibrational nuclear energy.
So we have this picture, the electrons are there,
the nuclei are kind of semi-classical,
they are roughly localized and they're varying,
and how would we estimate the nuclear vibration frequency?
So estimate nuclear vibration.
So we think of a harmonic oscillator,
b squared over 2 mass of the nucleons,
plus a restoring force, H subnuclear oscillation.
So this is an estimate so it's not very rigorous,
but I think it will be clear enough.
Now this restoring force has to do
with the cloud of the electrons that adjusts and does things.
As you separate the nuclei you have
to restore the restoring energies
due to the electrons and the Coulomb forces,
but it does not depend on the mass, M, the restoring force.
So this k depends on the electrons and Coulomb forces
but definitely not on the mass of this quantity.
So k has units of E over L squared,
and for energy we have this energy
of the electron and length, the length,
the scale of the molecule, so k is proportional to h
squared over ma to the fourth.
It just cannot depend on the mass of the nuclei,
that's definitely not the origin of the restoring force,
so if the only mass you can use is the mass of the electron it
can only be that.
But then the frequency, omega, is square root
of k over M, that is the mass of the nuclei,
so the nuclear frequency is this,
and it's therefore equal to h squared
over ma fourth one over M, that was k,
and this is square root of little m over capital M times
square root of h squared over m squared a to the fourth,
I borrowed a little m up and a little m down.
So from here we get that h bar omega
nuclear is square root of little m over capital M,
h squared over ma squared.
If you take the square root here you get h bar over ma squared,
but there was another h bar I put in.
And now we get a sensible equation
in which we see that the nuclear energies for oscillation
are much smaller, significantly smaller,
so E nuclear oscillations go like square root of m over M
is 10 to the minus 4, we say square root is 10
to the minus 2, times E electronic.
So it's considerably less which is this, so if you think--
I want to make another observation here on timescales,
so omega n, we have it here, omega n,
or we have it here as well, is equal to this times
this quantity, so let me say it this way.
The omega electronic for vibrations
is the electronic energy divided by h bar, I need h bar over ma
squared, but here we see that that's precisely
that square root, so the omega nuclear is
equal to little m over M omega electronic.
Now that is nice because it says that the frequency associated
to the nuclear thing is much smaller
than the timescales associated with electronic changes.
So this is a good reason for our idiomatic thinking
that as the nuclei slowly oscillate,
the electronic cloud remains in that particular eigenstate
and thus and jump.
It's our use of the adiabatic approximation.
A last comment is that the rotational energy
of the molecule goes like the angular momentum squared
over twice the moment of inertia,
L squared is h squared times some numbers,
as usual, L times L plus 1, and here
you get the mass of the molecule times the size squared
of the molecule.
These energies are like little m over capital M times,
this is roughly h squared over little ma squared,
and therefore the rotational energies
are even smaller than the electronic energies.
The rotational energies go like little m
over capital M times the electronic energies.
So you have the electronic energies, the nuclear energies
of oscillation, which go like square root,
and then finally the smallest of them
are the rotation, in which the molecule rotates
like a solid object.
We'll continue this next time and calculate a few more
things and molecules.