Even the past 2, 3 lectures, we have been
discussing the solution of the one dimensional
Schrodinger equation for the particle in a
box of dimension a. And we also considered
before that, the free particle problem. Those
were two important problems in quantum mechanics.
Today, we will be starting our discussion
on the Linear Harmonic Oscillator problem;
this is one of the most important problems
in quantum mechanics and allows one to understand
very clearly the relationship between classical
mechanics and quantum mechanics. The plan
is that, we will first give the solutions
of the Schrodinger equation before doing the
detailed mathematics associated with it. And
then, we will discuss the physics of the solutions
and obtain results, which correspond to the
classical oscillator.
Once we have understood the consequences of
the solution we will then, do a little bit
of algebra and solve a differential equation;
and obtain the exact solution of the Schrodinger
equation of the one dimensional Schrodinger
equation corresponding to the linear harmonic
oscillator problem.
So, let us start with the with our familiar
one dimensional Schrodinger equation, which
corresponds which is given by i h cross as
we all know delta psi by delta t is equal
to H psi, where H is the Hamiltonian a operator
corresponding to the total energy and this
in the one dimensional case is given by minus
h cross square by 2 mu delta 2 by delta x
square plus V of x psi, which is a function
of x and t. Now, we have solved this equation
for the free particle, where V of x is 0.
We have solved this equation for a particle
in a box problem as I just now mentioned,
confined in a one dimensional infinitely deep
potential.
Today, we will give the solutions corresponding
to the harmonic oscillator problem, which
is equal which for which this is V of x is
given by half mu omega square x square. We
know the mu is the mass of the particle, omega
is the classical frequency and x is of course,
the x coordinate, this is the classical expression
for the potential energy corresponding to
the harmonic oscillator. And we will obtain
we will write down first the solution of this
equation corresponding to this potential.
Once again since psi since V is a function
of x only does not depend on time. So, we
can use the separation of variables to solve
this equation and if we carry out this separation
of variables; then, you will have psi of x
of t is small psi of x time's a function
of time, which can be shown to be equal to
minus e i E t by h cross, where psi of x satisfies
the the following eigen value equation minus
h cross square by 2 mu d 2 by d x square.
Now, it is the total differential plus V of
x psi of x, this we had done even in our last
lecture, psi of x is equal to E times psi
of x, this is the time independent Schrodinger
equation, time independent Schrodinger equation.
Actually this is an Eigen value equation as
I had mentioned last time that, H psi of x
is equal to E psi of x, E is a number and
H is the operator. And for a given potential
we have certain discreet values of E, for
the particle in a box problem we had seen
that, that the energy Eigen values were something
like this, E is equal to E n is equal to n
square of E 1 we got discreet energies in
a box. For the free particle problem all values
of E greater than 0 were allowed.
So, we will solve this equation when V of
x is equal to half mu omega square x square.
So, we rewrite this equation for V of x is
equal to half m omega square x square and
if we do that, let me align this properly.
So, we will have d 2 psi by d x square plus
2 mu by h cross square E minus half mu omega
square x square psi of x is equal to 0, this
is the you just rearrange the top equation
and you will get this particular equation.
So, let me put it in a little convenient form,
I introduce a dimensionless variable Xi which
is proportional to x, gamma is the proportionality
constant and I will decide on its value little
later.
So, d psi by d x, if I introduce this independent
variable, so d psi by d Xi into d Xi by d
x, which is equal to gamma I differentiate
it again I will get d 2 psi by d x square
is equal to d of d Xi of this, multiplied
by d Xi by d x; so, this will become d 2 psi
by d Xi square multiplied by gamma square.
So, what I do is, I substitute this in this
equation, in this equation and obtain please
see this d 2 psi by d Xi square multiplied
by gamma square I take 2 mu inside, so I will
get 2 mu E by h cross square minus half mu
omega square and then, I multiply by 2 mu
by h cross square, so 2 mu by h cross square;
and x here is Xi by gamma, so Xi square by
gamma square. Now, psi is a function of Xi
now.
So, I divide by gamma square, so I will get
if I divide the whole equation by gamma square,
so d 2 psi by d Xi square plus let us suppose
I write this as lambda, where lambda is defined
when I have 3 equal to sign here, it implies
define to be equal to. So, this will be equal
to 2 mu E by h cross square gamma square.
Then, if you see this carefully then 2 2 cancelled
out, mu square omega square by h cross square.
So, this will be lambda minus mu square omega
square by h cross square, then gamma square
comes here, so it will be Xi square by gamma
to the power of 4. So, there is already a
gamma square here, so I multiply this by gamma
square, so it becomes gamma to the power of
4. So, this multiplied by psi of Xi this is
equal to 0.
I still do not know the value of gamma, I
have not yet defined yet, I now choose gamma
such that, this quantity is equal to 1. So,
that the coefficient of Xi square is unity.
So, I choose gamma to the power of 4 is equal
to mu square omega square by h cross square.
So, if I take the square root then, I will
get gamma square is equal to mu omega by h
cross. And so, therefore, you will have gamma
is equal to I define this equal to mu omega
by h cross, this is the definition of gamma.
So, if I choose gamma equal to under root
of mu omega by h cross then, this quantity
which I have encircled with blue becomes 1.
And so the Schrodinger equation becomes, if
you see this carefully that, the Schrodinger
equation then becomes d 2 psi by d Xi square
plus lambda and this quantity is 1, so this
is Xi square psi of Xi is equal to 0.
So, I have the two definitions, the one definition
is that of gamma, what Xi is equal to gamma
times x, where gamma is equal to square root
of mu omega by h cross, this is a dimensionless
quantity, x has the dimensions of meters.
So, gamma has the dimension of meter inverse
and lambda is defined to be equal to 2 mu
E by h cross square then gamma square; as
you can see here, lambda is equal to 2 mu
E divided by h cross square gamma square,
so this becomes mu omega by h cross, so mu
mu cancel out; so you will get 2 E by h cross
omega.
So, therefore, we can reduce the the transform
the Schrodinger equation we can transform
the Schrodinger equation to this particular
form. Now, we will later solve this equation
we will find that, if I assume that psi of
x or psi of Xi is made to go to 0 as x or
Xi tends to plus infinity or minus infinity,
this is known as the boundary condition these
are known as boundary condition. So, if I
make if I impose the condition that, the wave
function has to go to 0 as x goes to plus
infinity or minus infinity only then it will
be a square integrable function, only then
it will be a well behaved function.
So, if I impose this boundary condition we
will find that, lambda must take only an odd
integer 2 n plus 1, where n is equal to 0,
1, 2 etcetera, these are the Eigen values
of this equation, this concept must be clear.
These are the Eigen values of this problem
that is only for lambda equal to 1, 3, 5,
7, 9 and so on, will the solution of the Schrodinger
equation be well behaved at x equal to plus
infinity and x is equal to minus infinity,
for all other values of E there will exist
a solution, but which will blow up at infinity.
So, therefore, these are the Eigen values
of the problem and since lambda is equal to
2 E by h cross omega, so this means that,
this quantity must be equal to 2 n plus 1
and therefore, this gives us the important
result that, the energy Eigen values are quantized
we will have E is equal to E n n plus half
h cross omega. I must repeat I have not solved
this equation yet, but I plan to do that later
in one of the later lectures.
Today I will just give you the solution and
try to make you understand the importance,
the physical significance concept behind the
solution. For these values of E, the Eigen
functions are given by psi n of x is equal
to some normalization constant N and these
are H n of Xi e to the power of minus half
Xi square; these are the normalized Eigen
functions corresponding to the linear harmonic
oscillator problem.
This is of course, a Gaussian function and
H n of Xi are known as the Hermite polynomials
after the French mathematician Hermite, these
are known as Hermite polynomials. In fact,
H 0 of Xi is equal to 1, H 1 of Xi is equal
to 2 Xi, may be you have read this in your
mathematics class H 2 of Xi is 4 Xi square
minus 2 and H 3 Xi is 8 Xi cube minus 12 Xi
I wrote four of them to show that, they are
alternatively even and odd, this is even,
this is even, this is odd and this is odd,
these are known as Hermite polynomials.
And the normalization constant I think you
must remember this, gamma divide divided by
2 to the power of n n factorial square root
of pi. There is one important property that
may help you to remember the Hermite polynomials
that, H n of Xi the maximum power of Xi will
be Xi to the power of n and the coefficient
of that will be 2 to the power of n. So, please
see this, for n equal to 0, 2 to the power
of 0 is 1.
For n equal to 1, this is 2. For n equal to
2, this becomes 2 to the power of 2, so 4
Xi square; n equal to 3 this becomes 8. For
H 4 of Xi for example, you can write down
immediately that the first term will be 16
Xi to the power of 4 and so on. So, the coefficient
of Xi to the power of n in H n of Xi is always
2 to the power of n. So, therefore, these
are the discreet Eigen values of the problem,
and these are the normalized wave functions
of the problem.
So, therefore, if I if I plot the plot the
potential. So, it will be the parabolic potential,
so this is my V of x and this is x, so the
discreet energy states are half H cross omega,
3 halves H cross omega, 5 halves H cross omega
and so on. And the first one is just a Gaussian
function, the second one is a Gaussian function
multiplied by Xi, so it will be sorry it will
be something like this, and then this will
be something like, these are alternately symmetric,
anti symmetric, symmetric I will prove that
theorem little later.
So, this is my V of x is equal to half mu
omega square x square. So, we have we have
now proved that, we have the expressions for
H psi n is equal to E n psi n, n takes values
0 1 2 3 to infinity, so there are infinite
number of wave functions; and E's of n is
equal to n plus half H cross omega. And psi
n of x as I told you, these are N's of n
H n of Xi e to the power of minus half Xi
square, where Xi is equal to gamma x.
As I had discussed in my last lecture, these
wave functions form a set of orthonormal functions;
and the coefficient N's of n that is orthonormal
function means psi m star of x psi n of x
d x is equal to 0, if m is not equal to n,
all limits as before are from minus infinity
to plus infinity. And when m is equal to n,
mod psi n square d x I always I can always
choose the value of N's of n such that,
this integral is 1.
And in fact, the N's of n are chosen in
such a way that, you have this condition that
minus infinity to plus infinity psi m star
of x psi n of x d x is equal to the Kronecker
delta function. So, this is equal to 0, if
m is not equal to n; and is equal to 1, if
m is equal to n. So, this condition is known
as the Orthonormality conditions this condition
and these wave functions will satisfy satisfy
this Orthonormality condition.
There was yet another condition that we had
said and that was, that was the completeness
condition that summation psi n star of x prime
psi n x summed over n, this is equal to delta
of x minus x prime, this is the completeness
condition. This tells us that, any well behaved
function of x any well behaved function of
x like phi of x can always be expanded in
terms of this complete set of something like
the Fourier series, that in a particular interval
any well behaved function can be expanded
as you may have already read in terms of the
Fourier series.
Similarly, in this space defined by from minus
infinity to plus infinity, the Hermite gauss
function that I had written down the Hermite
gauss functions are, these these are known
as the Hermite gauss functions. So, the Hermite
gauss functions form a complete set of function
that, any arbitrary function any arbitrary
square integrable single valued function can
be expanded in terms of this function. And
therefore, the general solution the most general
solution most general solution of the one
dimensional Schrodinger equation that is i
h cross delta psi by delta t is equal to H
psi, where H corresponds to the linear harmonic
oscillator problem will therefore, be given
by psi when I write capital psi; that means,
it is a function of x and time.
When I write small psi then, that is a function
of x only. So, psi of x comma t is equal to
a superposition that is C n just as we had
done for the particle in a box problem e to
the power of minus i E n t by h cross. But
in this case and the summation is over n equal
to 0, 1, 2, 3 to infinity in this case, I
know the values of E n. E n is equal to n
plus half h cross omega, so this becomes this
becomes equal to summation C n psi n of x
e to the power of minus i n plus half omega
t, this is the most general solution for the
harmonic oscillated problem. That is any state
of the oscillator can be described by this
sum, and this is how it will evolve with time.
So, let me give you an example. So, let us
suppose I tell you and I will give you an
example with which all of you will be familiar.
Let us suppose I know the wave function at
time t equal to 0, so this we know I know
the state of the oscillator at time t equal
to 0 and I want to find out that, how will
the wave function sorry how will the wave
function evolve with time, that is my question.
So, I substitute t equal to 0 here and I will
get C n psi n of x I know this function. So,
using this equation I will find out C of n,
I know psi of x comma 0, I know psi n of x
these are the Hermite gauss functions. So,
I will now use the Orthonormality condition
to determine C of n. And then, I will substitute
it in this equation and carry out the summation.
In most cases, it is not possible to obtain
an analytical form, but in some situations,
it is possible to sum it analytically and
give a physical description of this system.
So, I come back to this equation and and I
am sure all of you know this, let me let me
write this psi of x comma 0 is so much. So,
how do I find this, I am sure you know this
we have done this before I multiply both sides
by psi m star of x. So, I multiply both sides
multiply both sides by psi m star of x d x
and carry out the integration from minus infinity
to plus infinity.
So, the left hand side will become all limits
are from minus infinity to plus infinity,
psi m star of x psi of x comma 0 d x is equal
to summation integral psi m star of x psi
n of x d x, this we know from the Orthonormality
condition. This is equal to delta m n. Therefore,
this equation becomes say, right hand side
becomes C n delta m n, m is a fixed number,
n is a running index, n goes from 0 to infinity.
So, in this series only the n equal to m term
will survive and all the other terms will
be 0. So, this will be equal to C's of m,
so C of m will be given by this. And so, therefore,
once I know C of m, so I can so I know that,
C of n is equal to minus infinity to plus
infinity psi m star sorry psi n star now of
x psi of x comma 0 d x. If I know C of n then,
in principle not in principle actually I can
substitute it psi of x t is equal to C n psi
n of x e to the power of minus i n plus half
omega t. There is one more thing that, I I
would like so, therefore, if I know psi of
x comma 0 I can find out C of n. Once I know
C of n I substitute it here and carry out
the summation.
Before I proceed further let me mention one
more thing that, psi of x comma 0 we wrote
as C n psi n of x. The complex conjugate of
that is, psi star x comma 0 is equal to C
n star psi n star x. So, if I multiply this
by this, when I have to multiply this sum
with this sum, so that this end does not get
both summations are over n, this does not
get confused with this I replace n here by
m.
So that, what I am trying to tell you is,
psi star x comma 0 multiplied by psi of x
comma 0, if I multiply the two equations and
integrate. So, I will have two sums, one over
n and one over this n, but this n should not
get confused with this n. So, I will write
here as m C m star C n psi m star x psi n
of x, if I multiply by d x and carry out the
integration then, these are the only two quantities,
which depend on x, so I can carry this.
But the limits are all of course, from minus
infinity to plus infinity, this from the Orthonormality
relation we know that, this is equal to delta
m n. So, only the m equal to n term will survive
and therefore, this will be modulus of C n
square. So, therefore, if initially I choose
a wave function, which is normalized then
mod C n square is 1, summation 1 2 3 and therefore,
we can interpret C n square as the probability
of finding the oscillator in the n-th state.
So, C n square will be the probability of
finding the oscillator in the n-th state 
and this does not change with time. So, when
I write psi of x, this is a characteristic
of quantum mechanics, that is system can be
in a superpose state. So, it is a superposition
of various states. So, let us suppose C 1
is equal to 1 over root 2, C 2 is equal to
1 over root 2, so only two states are superposed.
The probability of finding in the n equal
to 1 state is half; probability of finding
in the n equal to 2 state is half. So, the
Eigen value corresponding to this is 1 plus
half that is 3 by 2 h cross omega and the
eigen value corresponding to 5 by 2 h cross
omega.
Then you ask yourself the question that, does
the oscillator have a given fixed energy?
The answer is no. There is a half probability
of finding 3 by 2 h cross omega, that is if
you make a measurement of energy, then there
is a half probability of finding 3 by 2 h
cross omega; and there is a half probability
of finding 5 by 2 h cross omega, this is the
concept of superposition, which is one of
the most important parts of quantum mechanics,
one of the important aspects of quantum mechanics,
it can be in a superposed state.
Let me take another very simple example that
C 1 is equal to let us suppose 1 over root
3, C 2 is equal to say 1 over root 3 and C
3 is 1 over root 3, I have taken a very simple.
So, there is a one-third probability of finding
it in E 1, one-third probability of finding
in E 2 and one-third probability of finding
in E 3 and these probabilities will not change
with time. And therefore, these states are
known as stationary states, the probability
of finding them in that state will not change
with time.
However when I wrote down psi of x comma 0
is equal to C n psi n of x and psi of x comma
t will be equal to how will it evolve with
time, psi n of x e to the power of minus i
n plus half omega t, the phase factor will
change, the coefficient of psi C n the sorry
the coefficient of C n the modulus square
of that does not change with time. So, C n
times this C n times the exponential factor
let us suppose this I write as phi. So, minus
i ph the modulus square of that is mod C n
square. So, that does not phi depends on time
that does not change with time, but different
states will superpose with different phases
at different times. And I will tell you a
consequence of that.
Now, let us suppose that, I choose a system
I have my initial oscillator in this state
that psi of x comma 0 is equal to say, this
is a normalized function gamma under root
of pi e to the power of minus half Xi minus
Xi 0 whole square. If Xi 0 is 0, then this
is just the ground state wave function. But,
I assume that, Xi 0 is finite and this is
the initial state of the system at time t
equal to 0, the wave function describing the
harmonic oscillator is given by this equation.
And my question is that if this is so, then
how will this wave function evolve with time?
And the answer is simple I first calculate
C's of n that is psi n star of x psi x comma
0 d x, the algebra is bit complicated involved,
it is given in many books including my book
with professor Lokanathan; and the final result
is that, slightly complicated square root
of n factorial half Xi square raised to the
power of n by 2 e to the power of minus quarter
Xi naught square. So, this is if psi of x
comma 0 at time t equal to 0 is given by this
expression then, the coefficient C of n if
you substitute this expression in this and
use for psi and f x the Hermite gauss function
carry out the integration from minus infinity
to plus infinity, it is little cumbersome,
but very straightforward.
And it is given in many books including the
the in my book that is myself and Professor
Lokanathan S Lokanathan quantum mechanics
and published by Macmillan, so all the details
details steps are given there, this is the
5 th edition. Now, the next step will be to
substitute this, so I want to now find out
how it will evolve with time psi of x t. So,
what I will do is, I will substitute in this
equation, this has been the general recipe
and sum it. So, I substitute this expression
here, and carry out this summation. And it
is one of those rare occasions that, you can
actually sum the series and the final result
is very straight very very beautiful.
So, you will get if I sum that and take the
complex conjugate of that and then, multiply
that is mod psi x t square is equal to gamma
by root pi e to the power of minus Xi comma
Xi 0 cos omega t whole square. So, this is
the probability distribution, p of x comma
t, d x represents the probability of finding
the particle between x and x plus d x. So,
that probability distribution function is
actually oscillating with time and I will
show this to you in a moment graphically.
If I take if I calculate the expectation value
of x, the average value of x then that as
we all know was psi square x d x I multiply
this by x you may remember that, Xi is equal
to gamma x you can carry out this integration
very easily just a few steps and you will
find that, this will be equal to x naught
cos omega t. And similarly, if you find calculate
x square, so you will find that, mod psi square
x square d x you will find that, this is equal
to x naught square cos square omega t plus
1 over 2 gamma square.
I would request all of you to work out these
integrals, this is very straightforward and
you only have to use the integral that, e
to the power of minus alpha x square plus
beta x d x the same integral that I had mentioned
couple of times before pi by alpha e to the
power of beta square by 4 alpha, all limits
are from minus infinity to plus infinity.
So, then you can calculate delta x is equal
to x square minus x average square under the
root. So, this is this minus x 0 square cos
square omega t, so this term will cancel out
with this. So, I will get 1 over root 2 gamma
and so this will be under root of h cross
by 2 mu gamma. Let me show you the temporal
evolution and then, we will come back to this
slightly later.
So, we discussed we were we have been discussing
the harmonic oscillator problem. And as I
told you this is just briefly going through
what I have already done. So, this is my Hamiltonian,
this is p square by 2 m and this is the potential
energy function half mu omega square x square,
this is the potential energy function corresponding
to the oscillator.
So, you have, so you solve this Eigen value
equation and you find that, for the wave function
to be well behaved we will do this in detail
that, the energy Eigen values are n plus half
h cross omega; and the corresponding wave
functions are the Hermite gauss functions,
these are known as the Hermite gauss functions.
So, therefore, of the time dependent Schrodinger
equation, this is the one dimensional time
dependent Schrodinger equation , the most
general solution will be given by this, these
are psi n of x multiplied by this time or
something like the modes of the system, the
normal modes of the system. As you may have
seen that, if you have a stretched string,
it has certain normal modes of the system,
these I have worked out in my book on optics
and then, you displace the string in a particular
way and then, you want to study it's time
evolution.
So, you have to express this as a superposition
of the normal modes of the system 
and therefore, these are the the psi n of
x and these are the corresponding frequencies
of normal modes, so this is how it will evolve
with time. So, we this is the most general
solution of the time dependent Schrodinger
equation. Let me come back to this in a moment,
but tell you we as I had told you earlier
we have developed this software and these
are the wave functions, so this is the ground
state wave function and the energy eigen value
is 1 times E 0, E 0 is half h cross omega.
Let me plot the first Eigen function, so this
an anti symmetric function. So, the Eigen
value is 1 plus half h cross omega 3 by 2
h cross omega and you can see, this is an
anti symmetric function. The second state,
which corresponds to n equal to 2, so the
Eigen value is 5 by 2 h cross omega, so this
is the second function second Eigen function.
Then the third, this is an anti symmetric.
So, alternately the wave functions are symmetric
and anti symmetric, symmetric and anti symmetric.
The fourth in fact, the fourth wave function
has 4 zeros has 4 zeros and similarly, the
fifth and the sixth, so these are the normalized
Hermite gauss functions, which are the Eigen
functions and these are the corresponding
Eigen values n plus half h cross omega. So,
these are this is for example, the psi psi
6 of x this is a this is a symmetric function
of x, so it is 1 2 3 4 5 6 this. So, we will
come back to this in a moment. So, these are
my Eigen functions.
Now, as I mentioned before let us assume a
particular case, that at time t equal to 0
the wave function is given by this expression,
it is a displaced Gaussian it is a displaced
Gaussian. Let me tell you in advance this
is known as a coherent state of the oscillator.
So, I want to, if this is my question is,
if this is psi of x comma 0, what will be
psi of x comma t? And the answer is simple,
I first find out C's of n, hopefully I can
find an analytical expression and then, substitute
it back in this equation and carry out the
summation.
So, in this particular case in for this particular
form of psi of x comma 0, I know I can determine
analytical expression for C of n I substitute
it here, carry out the summation and and I
will obtain psi of x comma t mod square is
equal to this, which represents the classical
oscillator.
So, if I do this, this is known as a coherent
state. Let me then let me then tell you that,
if I superposed different states, so I am
trying to superpose two states with the following
coefficient. Let us suppose the first state
0.707 is 1 over root 2 and 0.707 is 1 over
root 2, so there is a half probability of
finding in the first state, half probability
of finding in the second state. So, let me
ok this and then evolve. So, this is how the
probability distribution will dance.
So, let me make it slowly. So, you can see
that, this is how the probability distribution
will evolve. So, is the system has a particular
energy? The answer is no, there is a half
probability of finding E 1 and half probability
of finding E 2. Let me do this once again,
but let me put C 1 equal to 1 and C 2 equal
to 0, so that it is in the first state, then
what will happen is, if I evolve it will remain
the same psi of x comma t whole square remains
the same for all times, it does not change
and that is because it is a stationary state,
I hope you understand.
So, let me take say three states. So, let
me have C 2 is equal to 1 and C 3 also equal
to 1, equal probabilities of three states,
so they are not normalized because, there
should be actually 1 over root 3, 1 over root
3, 1 over root 3; so, the coefficients are
not normalized. The software is asking me,
do you want the program to normalize them?
Let say yes and let us evolve, so this is
like this, this is how the state of the oscillator
will evolve with time.
Now, I go back to my coherent state and I
say that, let C of n be given by this, only
thing is I have to give a value of Xi 0 do
you understand this, I then say let it be
choose C n equal to so much, C 1 equal to
so much, C 2 equal to so much, C 3 equal to
so much, C 4, C 0 equal to so much, let C
n be like this, it is a superposition of an
infinite number of states.
But, I have to give the value of Xi 0 let
me do that. So, that is known as a coherent
state and I choose the value Xi 0 equal to
8 and I evolve with time, this is my classical
oscillator. So, let me make Xi 0 large as
I will tell you in a moment let me take it
100 and you will find that, this dances back
and forth. So, let me make it in little more
time steps, see this, so this is my classical
harmonic oscillator.
So, we when you see in your first year laboratory,
a long pendulum oscillating like this and
you ask yourself the question, that in quantum
mechanics we have solved this, to which Eigen
value does it belong? The answer is, it does
not belong to a particular value of E, it
is a superposition of a very large number
of states and these different states superposed
so beautifully with time, with different phases
that the wave packet oscillates back and forth
just as you would see in a classical harmonic
oscillator.
So, if I make it still these values of Xi
0, I will explain that in my next lecture,
so then it will become even sharper. And let
me make it in in larger number of time steps,
this is a software which is available with
my book on basic quantum mechanics, which
is also been published by Macmillan. So, you
will see that, the wave packet oscillates
back and forth.
So, therefore, I end this lecture by mentioning
that, my classical oscillator when it is moving
back and forth to which energy does it correspond
to and I will detail it in my next lecture
that, it does not correspond to a particular
energy, it is a superposition of billions
of states, but but very closely packed energy,
closely spaced energy levels. Does it have
a definite energy? The answer is no, but the
in the energy is extremely small. We will
discuss this in more detail in my next lecture,
thank you.
