in this lecture we will go back to some of the issues that we have been discussing earlier
and have formed the basis of many of the implementation that we have been doing in the recent times
while we looked at the problem sets and the solutions and we looked at what the students
are saying about the discussion which are happening on the internet it has become clear
that a little bit of background material revise is very important as we go along because we
have been looking at implementation aspects where many of these aspects which we are currently
describing needs to have a relook so in this lecture we will look back at some of the
problems that we have already given to you and then while discussing those problems or
looking at those problems we will be reviewing the concepts that has laid the foundation
of the implementation aspect that we are looking at in the present lectures so in terms of
reviewing the concepts let us see how do we go about doing this in some previous lecture
we have already looked at solution of problems from a certain week this particular
the first week solutions were already looked at this particular reviewing session would
therefore start with looking at the analysis of problems from week two because we haven't
done any of that before in that regard let us look at the problem from the very beginning
of week two and while doing that as i promise we will be also reviewing the concepts that
we have either missed or needs a little bit more clarification
so the first question in week two was concerning the uncertainty principle and the observables
and how they appear as conjugates at that time so in order to look at the solution
which is basically saying that the observables are conjugates to each other than their corresponding
operators are going to follow the uncertainty principle this is what the concept was now
in order to look at the concept let us see what the physical observables the wave
we have defined them in quantum mechanics the physical observables are operators for
any two operators say a and b one may always compute the communicator a b which is basically
the difference of the product in the two different ways so if they are commuting which is the
simplest quantity and is zero for classical physics but its often non zero in quantum
mechanics so conjugate variables are pairs of variables that are mathematically defined
in such a wave that they become furrier transform duels of one another such duality relations
lead naturally to an uncertainty relation that is the quantum mechanics familiar features
that we have discussed which is the heisenberg's uncertainty principle so mathematically the
uncertainty principle is based on commutators and it can be written as the variance from
the actual value of the true observables which will be always following these inequality
that have written here and as you all know the variance of these are defined with respect
to the operators that we are measuring and in most cases in quantum mechanics we also
have realized that the operators that we are using are going to be having real values
because the observables are always going to real values so they are often hermitions and
other kinds of things which we have discussed earlier in this previous concepts so with
this we revisit we have re visited the uncertainty principle part a little
let us now go back and see what happens when we go to the particle in a box problem
so this a concept problem from the particle in a box where it is asked when the walls
of the one d box are removed which of the following will happen so the concept of these
problem is to realize that how important the walls of the particle in a box are when we
are going to answer this question now the solution has been given( Refer Time: 05:00)
as the particle is going to have continuous energy spectrum now to understand that let
us revisit particle in a one dimensional box and see how this can be looked at roughly
speaking we can define three regions all though region one and region three are the same where
our potentials are infinity its the region two were we have zero potential and thats
where the particle is suppose to reside so according to classical physics the particle
can exist any where in the box and follows a path in accordance to newtons laws however
as we have noted in quantum physics or quantum mechanics as we have been studying the particle
is expressed by wave function and there are certain areas more likely to contain the particle
within the box
so we can start by looking at the time dependent schrodinger equation which is a which contains
the kinetic energy part the potential energy part which gives raise to a total energy some
of the two giving raise a total energy and the wave function is dependent on time and
position function however for when we integrate the time out then the time part is going
to come to is not going to be important we have only going to look at the position distribution
and the time independent schrodinger equation therefore forms the part which we are looking
at in this case and in this case its only one variable so we have this time independent
schordinger equation on which we can apply the boundary conditions and what we will find
that region one and three has infinity in terms of its potential so this will basically
have no wave function no probability of finding the particle at all so the particle is essentially
a constraint in region two where the potential is absent is zero
so we can use this to find the wave function of this solution and this is similar to the
general deferential equation or the wave equation and we have a generalize solution
which is a sin and cosine function of which will give the solution and we can then apply
the boundary condition which is that at both x is equal to zero and x is equal to l our
wave function has to vanish so psi has to goes to zero so we can apply that for both
the cases and what we find is there are two conditions that we can apply and then we can
find the energy levels and we often use h over two pi as our h cross and so we can
finally get to our energy levels which is e equal to n square h square over eight m
l square l being the length of the box and m being the mass of the particle so our new
wave function will have the sin function part because the b there cosine part has gone to
zero but what is the part which is a which we can normalize and if we normalize we get
a equal to root two over l so our normalize wave function is given here so in this wave
the particle in a box solution is done previous styles when we will looked at it
we basically assumed that you were already knowing these solutions but with time of evolution
of these course we have come to realize its better that we also sometimes give you the
background materials so that it can become clear as we go along
so what we realize is that there are these wave functions which are discrete at discrete
energy states for the and that exist only within the box and this square of the wave
function gives raise to the probability of finding the particle so this is the interpretation
which we talked about earlier which is known as the born interpretation and for the particle
in the box the discreetness or the quantum nature arises due to the confinement and boundary
conditions thats what we found out but when it is removed it is a classical particle with
continuous energy which means we have a free particle where our energy is potentially zero
so we just get this solution which is a free wave function which can have any direction
its so the particle can go go anywhere it can be found anywhere with equal probability
as well as the energy values can take any values continuous values from zero to infinity
so that is what the solution to the problem was that when we remove the constrain you
know of the particle box then the particle becomes a free particle and so it can be found
anywhere with equal probability as well as with continuous energy so this is our revisit
to the particle in a box you know next problem enable us to revisit the harmonic
oscillator and the rigid rotor part so the question essentially asked as to the form
of the solution for the eigen functions of the one d simple harmonic oscillator and the
rigid rotor would involve what kind of polynomials now this is possible to be known if you
have an idea would how the solutions for these eigen functions of simple harmonic oscillator
and rigid rotor are so the solution is given here because thats what we had told you before
but i dont think we had really gone through again the details of these solution of
the harmonic oscillator or the rigid rotor which we now show you for clarity so for
a simple harmonic oscillator is simply a particle which is held which has a spring constant
k and is undergoing a displacement in terms of the applied force f so the force is the
restoring force so the force is the restoring force which is trying to bring back the
particle to its position and so we get this potential with respect to position which is
quadratic in nature and we can do a tailor expansion of the potential function to
redefine the minima of the potential and the zero potential we find that the potential
can be written in terms of half v two x minus x zero square and when we substitute
these to the wave equation we are able to get this wave equation solution that we are
used to so this sort of brings you back to the original problem that we have talked about
and here we have therefore given you the values of the beta as well as the alpha which are
often used in this contexts this is a very good depiction of a diatomic molecule in many
ways the bottom of the potential for a diatomic molecule roughly follows the simple
harmonic motion which we show in this particular graph for you so this parabolic potential
well enables us to bound the system and if the lowest energy level of the system is set
to zero it would violate the uncertainty principle because is not possible to have a system
completely well defined because in this particular case when it is only under vibrational
mode all the other motion processes can be set to go to zero and so it will violate
the principle that it can be defined in a specific specific position so thats why
it has to have a certain value and the wave functions solutions are in terms of the hermitt
polynomial which is psi n h n x e to the power minus alpha x square over two which
is were it is shown where the h n x is the hermitt polynomial of order n
in contrast to the particle in a box problem where the oscillator wave function is sinusoidal
curve in this case the oscillatory behavior is due to the polynomial which dominates at
small x the exponential tail is provided by the gaussian function which dominates at large
x and so there is a competition between the two which makes this potential this wave
function look the way it is in order to understand it a little better let us look up the quantum
harmonic oscillator in slightly more detail here the schrodinger equation for a one dimensional
harmonic oscillator can be written in the simplest possible form in one dimensions
just this a x is the dimension here one displacement and we can write the schrodinger
equation which is which can be than re written to solve for the problem and the v as we know
is already we have seen in the earlier cases is half x square because my force restoring
force is minus k x so the energy for this can be found out to be h cross omega
n plus half with n going from zero to all the way up but because half h cross omega
is always going to be there so there is a zero point energy the frequency of oscillation
is one over two pi root force constant k over mu mu is the reduce mass of the
system the angle of frequency of the oscillation omega is basically two part omega and when
we look at the zero point energy its as i mention half h cross omega and this sort
of provides the fact that we can have the zero point energy in this case to be non zero
because of the consequence that you have to follow the uncertainty principle you cannot
determinately say where the particle is at any point of time
so the wave functions corresponding to the eigan values for the harmonic oscillator are
non degenerate and they have the form as we have been saying earlier also of this type
with a polynomial as well as a gaussian function exponential which happens to be a gaussian
and these polynomials are known as the hermit polynomial there is a normalization factor
also a n which is been shown here and we find that each of these wave functions go between
even odd even odd kind of a feature so all the even ones are even functions whereas all
the odd n values are odd functions so thats how these wave functions are and we can see
that it here in terms of the picture that we have shown here so once we know that the
harmonic oscillator problem is a hermit polynomial now let us look into the rotational motion
problem and as we know the smallest i mean the first possible root rotational rigid
root of problem would be a two dimensional root rotation because whenever we rotate we
involve two dimensions so here is the picture which shows that two particle which are
rotating it has a center of mass and now this is a center of mass problem in contrast to
the reduce mass problem that we saw in the last problem of the vibrating particle
vibrating system which was simple harmonic oscillator and here in this case the center
of mass problem would be represented in terms of the two dimensional rotation with the easiest
to deal with this in terms of polar coordinates because the for rigid rotor the r which is
the radial distance is going to be constant the radius is not going to change so one
of the one of the coordinates can be easily handle compare to the other
so the two dimensional rotation motion can be written in terms of the radial coordinates
and the hamiltonian can therefore be written in terms of the radial coordinates and
the particle is constraint it is root rotation and there is no external potential working
on it so its just their kinetic energy of the system so its h cross two mu central
mass minus h cross two mu l square which is going to be written in terms of the
angular coordinates which is what is written and since the since the r is going to be rigid
that its the rigid rotor so its going to be constant so [vocalized noise] we are only
going to have the variation coming due to the phi angle the angular motion and
therefore although its a two dimensional problem it reduces to a single dimensional nature
so then we can solve the single dimensional problem by using the schrodinger equation
and once we apply the s i equal to e s i form here and in this case the variable is on
the phi which is the rotating angle then we can find how this constrain is going to work
because at every two pi the particle is going to retrace it person thats its constrain so
we can put the boundary conditions and we can find out how the energy and the
quantum numbers vary for these kinds of problems so the quantum number is n and this is the
rotational rigid rotor problem so the solution can be taken in terms of these form
and the solution of this is phi this is the angular part of the motion which is been
shown the wave function angular wave function and the corresponding energy and momentum
can be found out by this process where the energy is m squad h squad over two i i
is the momentum of inertia of this system and the and since it is rotating about the
z axis then its the l z square is the good angular momentum so it it it is l z is is
the one which is conserved so its m m h cross is the l z operator
for the two dimensional rigid rotor we have this solution that we are already familiar
with that here we have try to show you have you get to it and what we find is that it
is dependent on the angular momentum and the operator for that is l z m where we are getting
all these values from only one quantum number is required in this case to determine the
state because [vocalized noise] the radial part is constant and so we have kept it as
a constant so this is the solution that we have used and when instead of using two dimensional
motion if it is allowed to move in all the three dimensions then it a three d rotor and
we have to use spherical coordinates and in terms us spherical coordinates is curious
to note that while two d two pi is the motion that we need to do for the rotation constrain
to be put in along the other dimension all you need to do is to move ( Refer Time:
21:00) which is the theta angle that is that corresponds to just a pi rotation which can
then generate all the rest of the motions which is what is being shown here in these
three dimensional cartoons and the theta angle therefore is constrain to go between
zero and pi were is the five angle is going to go between two zero two to pi and the [vocalized
noise] radial part the radius r is going to be given by the three coordinates root of
that and we can use this to get to our coordinate system and the parallel between the spherical
and the Cartesian cartesian coordinate can be utilize to get to know the volume element
which is being described here and we can make a parallel between this which is being shown
in this graph where [vocalized noise] we have made the volume element which is being
created because of this motion and the d theta d r and d five are the dimensions that we
are looking at when we do the volume element so the spherical polar coordinates minimum
volume element would therefore be given by this d b where it can have this elements which
is been looked at r d theta r sine theta d phi and so as result we get and d five so
we get r square sine theta d five d theta m d r so the rigid rotor in quantum mechanics
can therefore be written in this form where the spherical coordinates hamiltonian would
be having the form which is given here the wave function must contain both theta and
pi now because we are now having two different coordinates and when you solve for this we
will be getting the solution which is y theta and five which are known as spherical
harmonics and this is the solution which was given to this problem when it was been asked
and this background information is something which i thought is important for you to at
least be able to be familiarized yourself to go along with this
the next question concept of degeneracy of the energy levels were been asked and looked
at in order to ensure that you understand how this is done so all these different cases
that have been studied has been looked at one d box three d cuboid box third rotational
level in a rigid diatomic rotor first excited energy level of a hydrogen atom and ground
energy state of a hydrogen molecule so how do you address this this is to essentially
look at what are the possible degeneracy's that can arise in each of these problems so
for a one d box of a infinite high there is only one possibility of getting the answer
because the number of parameters available are only as gone to be only one which is one
dimensional x and so it is a singles process it is non degenerate set the three d cuboid
box of infinite length would have three parameters left three parameters dependence
because its a three d condition it is x y z which is there and its three fold degenerate
and and so will be having the degeneracy as three the third rotational level in a rigid
diatomic rotor will have a degeneracy of two l plus one because each of the l values
which is corresponding to the theta rotation can take two plus one values because for
each l value which can go from zero to n minus one there are correspondingly minus l through
zero to plus l number of degeneracy and so for a third rotational level of a rigid diatomic
rotor l is equal to two zero one two and so our degeneracy is going to be two times two
plus one which is five
the first excited level of a hydrogen atom would correspond to a degeneracy because we
know that for the hydrogen atom which is coming later in this review also were we look
at the development and treatment of the hydrogen atom the degeneracy of the hydrogen
atom is n squared and so for this particular case which is the first excited energy level
of hydrogen where n starts from one so the first case n equal to two and so we have a
degeneracy of four the ground energy state of a hydrogen molecule would be singly
degenerate when we consider that schrodinger solution without relativistic inclusion of
spin because in our particular case of solution of the the schrodinger equation we
only have three quantum number n l and m l so for this particular case when we
do not include the spin part it will only be singly degenerate because for a given
n value the ground state of the system we will only have n equal to one we will
only have one particular degeneracy which is n squared one squared is one
so based on this what we have done is we have managed to reflect some of the initial
parts of the discussion that we have been looking at as we go along because we have
only covered say four of the different concepts that we tested in week two problem
set we will be looking back and reviewing the rest of the problems in that in the subsequent
lecture where will be able to see how many more of those concept that are necessary
to be looked at once again because at the time of implementation all of these are being
assumed to be known to you when we are going through the steps so this particular process
right now that we are going through is very critical for you to be able to understand
and be able to address all the problems that we are currently doing and be able
to understand the principle of the implementations that we are doing in terms of quantum complete
so see you in the next lecture