We have discussed how to determine bond by
microwave spectroscopy.
Today we move on and we talk about how to
determine bond strength from IR spectroscopy.
But before that I cannot go away from the
discussion of microwave spectroscopy unless
I show you at least one real microwave spectrum.
So here it is.
This is the rotational spectrum of carbon
monoxide.
Diatomic molecule, dipolar molecule, so it
is microwave active and this is what the spectrum
looks like.
Now generally we are used to seeing spectra
pointing upwards.
Whatever I had drawn on the board had the
peaks pointing upwards.
In this case you see it looks like a negative
spectrum, pointing downwards, why is that?
Yeah, why is that?
Yes, please read the y-axis, it is transmission.
So essentially, the way it has been plotted
here is that you don't plot absorbance but
you look at transmitted light, essentially
y-axis is I-T, I-transmitted about which we
had some questions yesterday after class.
So as you see the spectrum does go through
a maximum, well a minimum in this case because
it's a transmission.
Lines are approximately equispaced and so
you can understand from the spacing of the
lines, you can figure out what is b and consequently
you can figure out what is the bond length.
However, if you look very carefully, do you
think this space is equal to this space?
If you look at the two ends of the spectrum,
the gap between two successive lines are they
the same at the two ends of the spectrum?
Of course, you have to look very, very carefully,
I am talking about this one and this one.
They are not exactly same; they are little
different nad you see the difference when
you compare the two ends of the spectrum.
Why are they different?
They are different because what you are supposed
to do by yourself, I did not complete the
discussion yesterday.
If you consider a non-rigid rotor, in a non-rigid
rotor the spacing does not remain equal anymore.
It keeps decreasing as you go to higher rotational
numbers.
That is why the spacing is not same on the
two sides.
We will have occasion to come back to this
when we talk about high resolution infrared
spectrum a little later today.
In Banvil it is mentioned.
Those are polyatomic molecules, right?
So you are talking about symmetric tops.
Manthan is asking what about symmetric tops
and all.
Right now, we are not going to discuss in
the same sequence as it is discussed in Banvil
or any other textbook.
So the way we have organized this course is
that first we are going to talk about diatomic
molecules only.
We'll complete the discussion of diatomic
molecules, then go on to understand where
your transition moment comes from.
Once we are done, then it is very easy to
work out the transition moment integrals for
the diatomic molecules.
It's like studying hydrogen atom and multielectron
atom.
You want to do hydrogen atom first.
So once we are done with diatomic molecules
then only we'll go on to polyatomic molecules.
That is why we are not talking about symmetric
top and asymmetric top and all that right
now.
We'll come to that later.
But the name says symmetric top.
So the moment you go to polyatomic molecules,
symmetry becomes important.
Using symmetry you can simplify a more complicated,
more complex problem.
So we will have to discuss symmetry as well
a little bit and we are going to focus on
IR and electronic spectroscopy from the point
of your symmetry.
But let that be the story for another day.
Today we want to stick to our simple model,
simple molecule, diatomic molecule.
The only difference is we want to talk about
vibration instead of rotation.
So you'll see instead of rotation is perhaps
not a very fair comment.
We'll elaborate upon that a little later in
these two modules.
So we go on to our model of diatomic molecules
and the fist model that we want to use is
simple harmonic oscillator.
Well, but we are talking about molecules.
So when we talk about simple harmonic oscillator,
it will not be enough, if we talk about a
classical oscillator, we have to discuss a
quantum harmonic oscillator and this is something
that all of you have studied in some quantum
mechanic scores or the other earlier.
So what we'll do is to start with, we are
going to recapitulate the results.
We are not going to do any derivation in that
part.
We'll recapitulate the results, we'll remind
ourselves, what are the energies of simple
harmonic oscillator.
We are going to remind ourselves what do the
wave functions of simple harmonic oscillator
look like.
Then we are going to talk about using these
wave functions, we are going to talk about
what the transition moment integral would
look like and when the transition moment integral
vanishes, when it does not.
From there, we'll come to a discussion of
which transitions are allowed and therefore,
what do the spectra look like for simple harmonic
oscillator.
Then we'll do something a little more complicated.
But let's start with this.
Simple harmonic oscillator, first thing that
we know is if it's simple harmonic, what is
the potential energy for the simple harmonic
oscillator 1/2 kx2.
So potential energy I can write is 1/2 kx2and
1/2 kx2is the equation of a circle.
Is that right?
No?
Then what is right?
It's the equation of parabola.
Do not say yes to anything and everything
I say.
Sometimes I can make a mistake to see if you
are away.
Sometimes I can make a mistake because I made
a mistake.
So please don't take everything for granted.
This then is the potential energy surface,
this is x.
While drawing this it makes sense to remind
ourselves what the meaning of x is.
What is the meaning of x?
You are talking about an oscillator and we
are going to use this model for a diatomic
molecule.
So once again, like a rigid rotor it is a
two-body problem.
We have two atoms connected by a bond and
they are vibrating.
As you know as we have discussed several times
in the previous classes you don't want to
handle two-body problems.
Wherever possible, you want to reduce two-body
problems to ta one-body problem.
So what does this reduce to?
Say let's say this is m1 mass, this is m2
mass.
It reduces to one mass of reduced mass ĩ,
I'll just write ĩr here because again I am
going to use ĩ for dipolar moment.
I don't want you to get confused.
So let us say ĩr reduced mass and it is attached
to a massless centre.
It's vibrating with respect to that.
And then let us say this is the displacement.
So this is the mean position.
x is this, displacement from mean position
that is x. of course, x can be positive, x
can be negative.
So that's what we are doing here.
It's important not to forget something that
we might think is obvious.
This is x.
If this is v, and since we have defined x,
what would the time-independent Schrodinger's
equation be?
(-h^2)/(2ĩr) (d^2 ?)/(dx^2 )+1/2 kx^2 ?=E?.
Once again it is important to remember, x
is not cartesian x. x is displacement from
mean position here.
And you know that if you solve the Schrodinger's
equation using the boundary conditions, what
are the boundary conditions used in case of
simple harmonic oscillator?
See, for that f part I remember the boundary
condition very well, periodic boundary condition.
What is the boundary condition used for a
simple harmonic oscillator?
Yes, so what she is saying is the way function
must vanish at -8 and 8.
So when you use appropriate boundary condition
and all, you get this energy expression, I'll
write it in the form that I find to be very
convenient.
Ev = hvvib(V+1/2) where V is the vibrational
quantum number and it varies from what to
what?
What is the value of V?
0, 1, 2, so on and so forth.
We'll come to the wave function shortly.
But even before we go there we can see that
the lowest energy that I have is E0 which
is 1/2hvvib.
By the way what is vvib?
vvibis the frequency of vibration.
We are not talking about light yet.
vvibis frequency of vibration and how is it
related to this k and ĩr?
1/2p(k/ĩr)1/2.
So if we can determine this frequency of vibration,
knowing the reduced mass, we can find out
what is k, the spring constant and if we know
what the spring constant, k, is then we essentially
know how strong the bond is.
That's the basic idea of this whole exercise.
Now coming back to E0.
This is the minimum energy.
So what we see is a very major difference
between a classical harmonic oscillator and
quantum harmonic oscillator is that energy
can never be equal to 0.
Minimum energy that we have is for V=0 which
is 1/2hvvib.
Since you have studied this problem in some
other course already, can you tell me why?
Why is it that for a quantum harmonic oscillator,
energy can never become 0.
Tell us Vishnu.
Rajdeep?
Why is it that for a quantum harmonic oscillator,
I cannot have 0 energy?
0 energy means what?
The oscillator is not oscillating anymore.
The vibration has stopped, right?
Stopped completely.
x is 0 mean it's at equilibrium position.
Well, you are going one step forward perhaps.
Easier.
Well, that's a right answer but perhaps to
a little different question.
Yes, Sushnatu.
Sushnatu is saying then it violates the answer,
that's the answer I was looking for.
I am not talking about one simple harmonic
oscillator; I am not talking about a collection
of -- so when you talk about a collection
of operators, then that residual entropy business
comes in.
But here, the answer I am looking for is answer
-- see if it not vibrating at all, that means
what?
x=0.
Position is completely well-defined.
No uncertainty in position.
Also, since it's not moving, momentum is also
equal to 0, 0ą0.
So no uncertainty in position, no uncertainty
in momentum.
That is what would violate uncertainty principle.
An uncertainty principle is something, remember,
which cannot be violated.
It is nature's threshold beyond which you
cannot go.
It's not a question of making a better instrument
or anything.
So if quantum harmonic oscillator stopped
completely, then it would have violated uncertainty
principle and here in lies a major difference
between a quantum harmonic oscillator and
a quantum rigid rotor that we discussed yesterday.
Remember, rigid rotor could have stopped.
Why?
Because when it stops it can take up any value
of ?, it can take up any value of f.
So uncertainty in position is 8.
Here, if it stops uncertainty in position
also would become 0 and thereby, uncertainty
principle would be violated.
That is why quantum harmonic oscillator cannot
have 0 energy.
Minimum energy, even if you go down to 0 kelvin
and 0 kelvin is what prompted Manthan to give
that answer, residual entropy, then also you
get this hvvib energy.
So this is called, I know you think, what
is it called zero point energy.
Where is the next one?
Where is E1.
If you put V=1, 3/2 hvvib.
Where is E2?5/2 hvvib.
In fact, if you just write the general expression
for Ev+1-Ev, what do you get?hvvib.
So we learn that allowing for my poor artistic
skills, the energy spacings are all equal
for a quantum harmonic oscillator and the
spacing between two successive levels is hvvib.
Now that 
we have talked about energy, let us talk about
populations quickly and then we move on to
the wave function.
Again, we use a concept that we have discussed
for your rigid rotor.
It is your Boltzmann distribution.
Using Boltzmann distribution, you understand
that Pv+1 or maybe I'll even like, well it
doesn't matter.
Pv what would that be equal to?
Now degeneracy of each of these levels is
actually 1.
In case of rigid rotor you have degeneracy
of 2J+1 because the same angular momentum
vector could have been pointed in different
directions.
Here, there is no such thing.
So degeneracy of every vibration level is
1 and only 1.
So degeneracy factor is 1 and then you have
Boltzmann factor, what is that?
e-?E/kT.
What is ?E for two successive levels?
hvvib/kT.
Now, of course, I am jumping a few steps at
this moment but let me just tell you that
if you use, say, ?vibthis is what I remember,
?vib,or vvib.
How is ?vibrelated to vvib?
By factor of c.
So if it is 1000 centimetre inverse, well
for typical vibrational levels, let us say,
typical quantum harmonic oscillator, this
ratio trans out to be something like 0.008.
Well, I am using an energy gap, the wave number
equivalent of which is 1000 centimetre inverse
and I am using a temperature of 300 kelvin.
If I use those numbers the ratio becomes 0.008.
What does that mean?
Start with the 0th level.
If there are 1000 molecules in the 0th level,
how many molecules should be at v=1 level?
8.
And how many would be in v=2 level?
Well, 0 because you cannot really have factional
molecules.
So for all practical purposes, for that kind
of energy gap the only populated level is
v=0.
So all upward transition will actually begin
here.
Is that point made?
This becomes more important when we talk about
an un-harmonic oscillator because in un-harmonic
oscillator the energy gaps are not one and
the same.
Here see, if there is a transition between
these two, and if there is a transition between
these two, would you be able to differentiate?
No, because energy gaps are all the same.
If there is a transition from here to here,
of course, it will be different.
But as we'll see, that is not allowed.
The only allowed transitions are ?V=ą1.
So for ?V=ą1 all energy gaps are same.
So even if a transition originated in one
of the higher energy levels, you would not
be able to tell, but the issue is it doesn't.
The only populated level for all practical
purposes, for 1000 centimetre inverse kind
of energy gaps is V=0.
Are we clear about this?
Then we move on to the wave function.
