In the study of vibrational spectroscopy,
one of the most important ideas that you will
need is that of the harmonic oscillator. Now,
the harmonic oscillator system is that of
a particle moving in a harmonic potential,
which I will tell you about and we want to
study how this particle behaves by using the
laws of quantum mechanics. So, let us start
at the beginning let us first understand what
is a harmonic oscillator, in other words,
what is this harmonic potential?
So, for that, consider a system like this
where you have a small mass which is free
to move on a frictionless surface that is
this surface here and it is attached to a
wall via spring like this. Now, if you take
this mass and make displace it by x, about
its equilibrium position. So by equilibrium
position, I mean this position when the spring
is not stretched or compressed and then if
I stretch the mass by a distance x and leave
it, then the mass begins to oscillate back
and forth.
And this oscillation is what is called harmonic
motion now if you plot the motion of this
particle as a function of time, so, I am going
to plot time in this direction and the amplitude
of the motion, then the particle initially
has an amplitude like that, then it comes
down it has an amplitude in this direction
it goes up and it oscillates back and forth
like this, about this equilibrium position
this motion is what is called harmonic motion.
And the force that the spring exerts on the
mass F = - k times x that is the force is
proportional to the displacement that the
particle is has with respect to the un displaced
position of the spring. The corresponding
potential energy is v x is = 1 / 2 k x square
and the classical motion of this particle
is given by x as a function of t is = x 0
cosine of omega t, where omega which is the
frequency of the oscillation is related to
the spring constant.
So, k is called the spring constant 
and omega is related to the spring constant
as square root of k over m. So, the classical
motion is given by this x as a function of
time and the graph of that is what you see
here, the total classical energy of the spring,
total energy is half k x 0 squared, where
x 0 is the maximum displacement of the spring
the total energy is a constant and is a sum
of kinetic energy and potential energy.
So, when the particle is moving fast the energy
is primarily kinetic energy and when the particle
is turning around and it is slowing down,
then the energy is primarily potential energy.
So this total energy can be written as a sum
of kinetic energy and potential energy and
classically, this is k x 0 square by 2 sine
squared omega t. This is the kinetic energy
plus cosine squared omega by 2, which is the
potential energy.
If we graph the potential energy as a function
of time the graph looks something like this.
Initially it is all potential energy and then
it decreases again its potential energy and
decreases. So, this is potential energy and
the kinetic energy is initially zero and then
that increases when the potential energy becomes
zero and then it oscillates like this. So,
this is the kinetic energy and the graphs
are here for the kinetic energy and here for
the potential energy.
And you see that the total energy which is
the sum of the potential and kinetic energy,
that is constant, this is the classical picture
of the motion of the particle in a harmonic
potential. So what we have in this page is
the classical mechanical picture. Our goal
is to understand the quantum mechanical description
of the harmonic oscillator and that is what
we are going to look at in very great detail.
Now, our goal is to obtain a quantum mechanical
description of a particle moving in a harmonic
potential. More precisely, we want to find
the wave function of a particle so, psi of
x, t which is moving in the harmonic potential
that is moving with potential V of x is = half
k x square. So, for this, we have to write
the Hamiltonian for this system and then solve
its Schrodinger equation the Hamiltonian for
the system H is the kinetic energy operator
plus the potential energy operator the kinetic
energy is simply the one dimensional kinetic
energy operator.
Which is - h bar square by 2m d square by
dx square and the potential energy is half
k x square. Now, our goal is to find a psi
x, t which satisfies the Schrodinger equation
ih bar del psi by del t is equal H of psi
this is the Schrodinger equation. We have
seen in the lectures on the basics of quantum
mechanics, that if the Hamiltonian does not
depend on time which is the case here, then
solving the Schrodinger equation is equivalent
to solving the eigen value equation of the
Hamiltonian H psi is = E psi since, h does
not depend on time.
So, let us look at the solution of this equation
H psi is = E psi where h is this Hamiltonian
H bar squared by 2m d square by dx square
+ half k squared and let us go and derive
what are the actual eigen functions and eigen
values of this Schrodinger equation.
Our goal is to solve the following Schrodinger
equation -h bar square by 2m d square by dx
square + half kx square psi of x is = E times
psi of x. We note that k is = m omega square
from the classical solution of the harmonic
oscillator. We also see that this is consistent
with k having dimensions of mass by time square.
Because half kx squared has dimension of energy
which is mass length square divided by time
squared.
So since x squared has dimension of length
squared, k must have dimension of mass by
time squared. So taking k to be m omega squared,
we write the Schrodinger equation as -h bar
square by 2m d square by dx square + m omega
squared by 2 x squared psi x is = E times
psi of x, we need to find a psi x, which satisfies
this equation to obtain solutions of the Schrodinger
equation, that is to obtain the value of Psi
of x that satisfies this equation.
We will use the method of ladder operators
and for that, let me begin by defining dimensionless
coordinate q 
which is related to x in the following manner
so, q is equal to square root of m omega by
h bar times x. Let us verify that this q is
indeed dimensionless. So, if you take the
dimensions of all the other quantities, m
has dimensions of mass, omega has dimensions
of inverse of time, h has dimensions off so,
this is units of Joule second.
So, dimensions are mass length squared time
inverse so, you see that inside the square
root the mass in the numerator and denominator
cancel and the time inverse in the numerator
and denominator cancel and you are left with
one over L squared square root, which is one
over L. So the dimension of this entire square
root is L inverse and the dimension of the
X is L. So the dimension of this entire Q
is dimensionless.
We will now write the Schrodinger equation
in terms of Q and to do that, we need to write
d squared by dx squared in terms of q. Let
us start by writing d by dx in terms of q
so d by dx is = d by dq and dq by dx and we
know that dq by dx is = square root of m omega
by h bar. So d by dx is = square root of m
omega h bar d by dq d squared by dx squared
is another derivative with respect to x which
gives dq by dx and then a another derivative
with respect to q and again we write the expression
of dq by dx as square root of m omega by h
bar.
So finally, we have d squared x by d squared
by dx squared is equal to m omega h bar d
squared by dq squared using this expression
we will now write the Hamiltonian in terms
of the dimensionless coordinate q and solve
the Schrodinger equation for the dimensionless
coordinate and in the end, you can always
come back to the coordinate x by using the
conversion factor. The reason we use the dimensionless
coordinate is to simplify the math, which
will follow in the ladder operator approach
and we will understand this in detail as we
proceed
Using the relation of q and x, which is m
omega by h bar x and d squared by dx squared,
which is m omega h bar d squared by dq squared
in terms of q the Schrodinger equation becomes
h bar omega by 2 - d squared by dq squared
+ q squared psi of q is equal to E of psi
of q. We notice that in the Schrodinger equation,
this operator has the form minus alpha squared
plus beta squared, which as you know, can
be factorized as minus alpha plus beta and
alpha plus beta.
So, just as an experiment, let us write an
operator, which is h bar omega, this h bar
omega here and whatever is in the brackets
as 1 over square root of 2 - d by dq + q multiplied
by 1 over square root of 2 d by dq + q. Now,
if we expand this out, we get h bar omega
by 2 - d squared by dq squared + q squared
and the additional terms - h bar omega by
2 d by d2 q times q multiplied by q times
d by dq. So, here we have the operator like
in the Hamiltonian and additionally we have
operator which we see here.
So, the question is what is the value of this
operator? Let us try to determine this d by
dq n q operating on f of q - q d by dq operating
on f of q gives and you have to use the chain
rule here f of q + q times f prime of q - q
f prime of q now, these 2 will cancel and
you get this to be simply f of q. So, the
operator here is nothing but just one and
this entire operator he had that we have written
here is effectively h the Hamiltonian - h
bar omega by 2.
Now, if we give some special names to the
operators here and here so, we call the first
one as the b dagger and this other operator
as b, then we can see that the Hamiltonian
which is the operator A + h bar omega by 2
is = h bar omega b dagger b + h bar omega
by 2 that is h bar omega b dagger b + half.
So, effectively, we have written the Hamiltonian
operator in terms of 2 new operators which
we have defined, which are b dagger and b
and we will see how this will help us actually
solve the Schrodinger equation H psi = E psi.
Our goal now is to solve H psi = E psi, where
h is written in terms of the two new operators
which we have defined b dagger and b. So,
this is h bar omega b dagger b plus half do
this, it is helpful to derive a commutation
relation between these 2 new operators’
b dagger and b. So, we want a commutation
relation between b and b dagger. By commutation
relation, the mean what is the value of the
commutator b b dagger
So, this is the symbol of a commutator the
square brackets and the commentator simply
means, b b dagger - b dagger b. In general
the commutation operator between 2 operators
A and B is AB - BA. So, in our case, the commutation
between BB dagger is written out here and
if you write this explicitly in terms of the
q this becomes b is d by dq + q and this one
over square root of 2 has been taken out as
half outside and d dagger is - d by dq + q
and we have minus half b dagger is minus d
by dq + q multiplied by d by dq + q.
When we expand this out, this becomes half
- d squared by dq squared + dy by dq q - q
d by dq + q squared and + d squared by dq
squared and + d by dq of q - q d by the dq
and - q squared. Several terms here cancel
so for example, this first term cancels with
this term, and the q squared term cancels
the - q squared term and furthermore, this
term dq d by dq times q, and d by d q times
q appears twice and similarly this term here
appears twice.
So, we can write this b b dagger commutator
as the by dq q - q d by dq and this operator
we have seen is simply equal to what we have
just derived this in the previous slide. So,
the final result we get is the commutator
of B B dagger is equal to one which we will
use in our derivation going ahead.
Let is again write the Eigen value equation
for the Hamiltonian H psi is = E psi so, h
bar omega b dagger b + half times psi is = E
psi and we want to find what Psi satisfies
this equation. So, let us pre multiply or
multiply from the left by b dagger on both
sides of the equation 
that gives h bar omega b dagger b dagger b
+ b dagger by 2 psi is = E times b dagger
of psi . b dagger is a linear operator and
so, I could write b dagger E of psi as E times
b dagger of psi, which is what I have here.
Now, we notice in this equation that we have
a b dagger operating on psi here and a b dagger
operating on psi here, but in this term, we
have b be operating on Psi. So, let us try
to interchange the b dagger and b and for
this we can use the commutation relation which
we have just derived which is b b dagger is
= 1 or in other words b b dagger - b dagger
b is = 1 or b dagger b is equal to b b dagger
-1.
So, if I substitute this b dagger b here then
I will get h bar omega b dagger b b dagger
minus one which is simply b dagger + b dagger
by 2 psi is = E times b dagger of psi and
this gives h bar omega b dagger b b dagger
- b dagger by 2 times psi is equal to E b
dagger of psi and this gives h bar omega b
dagger b - half B dagger of psi is equal to
E times b dagger of psi. If we want to make
this operator on the left hand side look like
the Hamiltonian operator.
Then we had from here you add h bar omega
b dagger psi on both sides and this gives
h bar omega b dagger b + half b dagger psi
is equal to E + h bar omega b dagger psi.
Now, this operator on the left hand side here
is nothing but the Hamiltonian operator. So,
we have Hamiltonian operating on b dagger
psi gives E plus h bar omega b dagger of psi.
This implies that if h psi is = E psi then
H of b dagger psi is = E + h bar omega b dagger
psi so, if psi is an eigen function 
then b dagger psi is also an eigen function.
And if psi has eigen value E then b dagger
psi has Eigen value E + h bar omega so, if
you have an eigen function of the harmonic
oscillator Hamiltonian, then operating with
b dagger on that eigen function gives another
eigen function, but with an eigen value which
is increased by h bar omega. So, this operator
b dagger is raising the energy of the Eigen
function and giving another Eigen function
and this b dagger operator is sometimes called
the ladder up operator.
To summer rise up to now we have seen that
the Hamiltonian of the harmonic oscillator
is h bar omega b dagger b + half and we have
seen that if psi is an eigen function with
eigen value E, then b dagger psi is also an
eigen function with Eigen value e + h bar
omega. Now, if we consider this b dagger psi
to be another eigen function, let us say phi
then h phi is = E prime of phi and this implies
that b dagger of phi is also an eigen function
with eigen value.
E prime + h bar omega that is b dagger of
b dagger psi phi is just b dagger psi is also
an eigen function with eigen value E prime
is E + h bar omega so, the total Eigen value
is E + h bar omega + h bar omega. So, in summary,
b dagger b dagger psi E is an Eigen function
of the Hamiltonian with Eigen values E + 2
h bar omega. We see that the operator b dagger
operates on an Eigen function and gives another
Eigen function with Eigen value 
increased by h bar omega.
So, this suggests that the Eigen values of
the harmonic oscillator Hamiltonian are spaced
by equal values and the spacing in each case
is h bar omega. So, all of these are Eigen
values corresponding to different Eigen functions
and these are all obtained by operating the
b dagger operator on one of these Eigen functions
the question now is what is the lowest Eigen
value and what are the functional forms of
these different Eigen functions.
