WEBVTT
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in this lecture we will go back to some of
the issues that we have been discussing earlier
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and have formed the basis of many of the implementation
that we have been doing in the recent times
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while we looked at the problem sets and
the solutions and we looked at what the students
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are saying about the discussion which are
happening on the internet it has become clear
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that a little bit of background material revise
is very important as we go along because we
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have been looking at implementation aspects
where many of these aspects which we are currently
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describing needs to have a relook so in this
lecture we will look back at some of the
00:58.730 --> 01:04.370
problems that we have already given to you
and then while discussing those problems or
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looking at those problems we will be reviewing
the concepts that has laid the foundation
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of the implementation aspect that we are looking
at in the present lectures so in terms of
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reviewing the concepts let us see how do we
go about doing this in some previous lecture
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we have already looked at solution of
problems from a certain week this particular
01:31.960 --> 01:37.409
the first week solutions were already looked
at this particular reviewing session would
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therefore start with looking at the analysis
of problems from week two because we haven't
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done any of that before in that regard let
us look at the problem from the very beginning
01:51.000 --> 01:56.100
of week two and while doing that as i promise
we will be also reviewing the concepts that
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we have either missed or needs a little bit
more clarification
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so the first question in week two was concerning
the uncertainty principle and the observables
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and how they appear as conjugates at that
time so in order to look at the solution
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which is basically saying that the observables
are conjugates to each other than their corresponding
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operators are going to follow the uncertainty
principle this is what the concept was now
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in order to look at the concept let us
see what the physical observables the wave
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we have defined them in quantum mechanics
the physical observables are operators for
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any two operators say a and b one may always
compute the communicator a b which is basically
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the difference of the product in the two different
ways so if they are commuting which is the
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simplest quantity and is zero for classical
physics but its often non zero in quantum
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mechanics so conjugate variables are pairs
of variables that are mathematically defined
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in such a wave that they become furrier transform
duels of one another such duality relations
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lead naturally to an uncertainty relation
that is the quantum mechanics familiar features
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that we have discussed which is the heisenberg's
uncertainty principle so mathematically the
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uncertainty principle is based on commutators
and it can be written as the variance from
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the actual value of the true observables which
will be always following these inequality
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that have written here and as you all know
the variance of these are defined with respect
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to the operators that we are measuring and
in most cases in quantum mechanics we also
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have realized that the operators that we are
using are going to be having real values
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because the observables are always going to
real values so they are often hermitions and
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other kinds of things which we have discussed
earlier in this previous concepts so with
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this we revisit we have re visited the
uncertainty principle part a little
04:27.240 --> 04:34.530
let us now go back and see what happens
when we go to the particle in a box problem
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so this a concept problem from the particle
in a box where it is asked when the walls
04:41.730 --> 04:48.540
of the one d box are removed which of the
following will happen so the concept of these
04:48.540 --> 04:55.350
problem is to realize that how important the
walls of the particle in a box are when we
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are going to answer this question now the
solution has been given( Refer Time: 05:00)
04:59.590 --> 05:04.750
as the particle is going to have continuous
energy spectrum now to understand that let
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us revisit particle in a one dimensional box
and see how this can be looked at roughly
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speaking we can define three regions all though
region one and region three are the same where
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our potentials are infinity its the region
two were we have zero potential and thats
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where the particle is suppose to reside so
according to classical physics the particle
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can exist any where in the box and follows
a path in accordance to newtons laws however
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as we have noted in quantum physics or quantum
mechanics as we have been studying the particle
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is expressed by wave function and there are
certain areas more likely to contain the particle
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within the box
05:50.460 --> 05:57.120
so we can start by looking at the time dependent
schrodinger equation which is a which contains
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the kinetic energy part the potential energy
part which gives raise to a total energy some
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of the two giving raise a total energy and
the wave function is dependent on time and
06:06.990 --> 06:18.440
position function however for when we integrate
the time out then the time part is going
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to come to is not going to be important we
have only going to look at the position distribution
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and the time independent schrodinger equation
therefore forms the part which we are looking
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at in this case and in this case its only
one variable so we have this time independent
06:37.860 --> 06:43.520
schordinger equation on which we can apply
the boundary conditions and what we will find
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that region one and three has infinity in
terms of its potential so this will basically
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have no wave function no probability of finding
the particle at all so the particle is essentially
06:58.800 --> 07:06.120
a constraint in region two where the potential
is absent is zero
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so we can use this to find the wave function
of this solution and this is similar to the
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general deferential equation or the wave
equation and we have a generalize solution
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which is a sin and cosine function of which
will give the solution and we can then apply
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the boundary condition which is that at both
x is equal to zero and x is equal to l our
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wave function has to vanish so psi has to
goes to zero so we can apply that for both
07:37.460 --> 07:43.460
the cases and what we find is there are two
conditions that we can apply and then we can
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find the energy levels and we often use
h over two pi as our h cross and so we can
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finally get to our energy levels which is
e equal to n square h square over eight m
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l square l being the length of the box and
m being the mass of the particle so our new
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wave function will have the sin function part
because the b there cosine part has gone to
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zero but what is the part which is a which
we can normalize and if we normalize we get
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a equal to root two over l so our normalize
wave function is given here so in this wave
08:30.090 --> 08:36.620
the particle in a box solution is done
previous styles when we will looked at it
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we basically assumed that you were already
knowing these solutions but with time of evolution
08:43.019 --> 08:47.949
of these course we have come to realize its
better that we also sometimes give you the
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background materials so that it can become
clear as we go along
08:52.149 --> 08:59.940
so what we realize is that there are these
wave functions which are discrete at discrete
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energy states for the and that exist only
within the box and this square of the wave
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function gives raise to the probability of
finding the particle so this is the interpretation
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which we talked about earlier which is known
as the born interpretation and for the particle
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in the box the discreetness or the quantum
nature arises due to the confinement and boundary
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conditions thats what we found out but when
it is removed it is a classical particle with
09:28.120 --> 09:34.810
continuous energy which means we have a free
particle where our energy is potentially zero
09:34.810 --> 09:46.050
so we just get this solution which is a free
wave function which can have any direction
09:46.050 --> 09:51.069
its so the particle can go go anywhere it
can be found anywhere with equal probability
09:51.069 --> 09:55.699
as well as the energy values can take any
values continuous values from zero to infinity
09:55.699 --> 10:01.510
so that is what the solution to the problem
was that when we remove the constrain you
10:01.510 --> 10:07.709
know of the particle box then the particle
becomes a free particle and so it can be found
10:07.709 --> 10:14.750
anywhere with equal probability as well as
with continuous energy so this is our revisit
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to the particle in a box you know next
problem enable us to revisit the harmonic
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oscillator and the rigid rotor part so the
question essentially asked as to the form
10:29.810 --> 10:34.459
of the solution for the eigen functions of
the one d simple harmonic oscillator and the
10:34.459 --> 10:41.080
rigid rotor would involve what kind of polynomials
now this is possible to be known if you
10:41.080 --> 10:46.999
have an idea would how the solutions for these
eigen functions of simple harmonic oscillator
10:46.999 --> 10:53.410
and rigid rotor are so the solution is given
here because thats what we had told you before
10:53.410 --> 11:00.459
but i dont think we had really gone through
again the details of these solution of
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the harmonic oscillator or the rigid rotor
which we now show you for clarity so for
11:07.350 --> 11:13.660
a simple harmonic oscillator is simply a particle
which is held which has a spring constant
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k and is undergoing a displacement in terms
of the applied force f so the force is the
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restoring force so the force is the restoring
force which is trying to bring back the
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particle to its position and so we get this
potential with respect to position which is
11:33.680 --> 11:40.629
quadratic in nature and we can do a tailor
expansion of the potential function to
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redefine the minima of the potential and the
zero potential we find that the potential
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can be written in terms of half v two x
minus x zero square and when we substitute
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these to the wave equation we are able to
get this wave equation solution that we are
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used to so this sort of brings you back to
the original problem that we have talked about
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and here we have therefore given you the values
of the beta as well as the alpha which are
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often used in this contexts this is a very
good depiction of a diatomic molecule in many
12:18.079 --> 12:24.180
ways the bottom of the potential for a
diatomic molecule roughly follows the simple
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harmonic motion which we show in this particular
graph for you so this parabolic potential
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well enables us to bound the system and if
the lowest energy level of the system is set
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to zero it would violate the uncertainty principle
because is not possible to have a system
12:45.430 --> 12:51.649
completely well defined because in this
particular case when it is only under vibrational
12:51.649 --> 13:00.439
mode all the other motion processes can
be set to go to zero and so it will violate
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the principle that it can be defined in
a specific specific position so thats why
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it has to have a certain value and the wave
functions solutions are in terms of the hermitt
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polynomial which is psi n h n x e to the
power minus alpha x square over two which
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is were it is shown where the h n x is the
hermitt polynomial of order n
13:27.059 --> 13:32.600
in contrast to the particle in a box problem
where the oscillator wave function is sinusoidal
13:32.600 --> 13:39.120
curve in this case the oscillatory behavior
is due to the polynomial which dominates at
13:39.120 --> 13:46.680
small x the exponential tail is provided by
the gaussian function which dominates at large
13:46.680 --> 13:53.680
x and so there is a competition between the
two which makes this potential this wave
13:53.680 --> 14:00.209
function look the way it is in order to understand
it a little better let us look up the quantum
14:00.209 --> 14:05.839
harmonic oscillator in slightly more detail
here the schrodinger equation for a one dimensional
14:05.839 --> 14:13.360
harmonic oscillator can be written in the
simplest possible form in one dimensions
14:13.360 --> 14:20.279
just this a x is the dimension here one
displacement and we can write the schrodinger
14:20.279 --> 14:27.970
equation which is which can be than re written
to solve for the problem and the v as we know
14:27.970 --> 14:35.269
is already we have seen in the earlier
cases is half x square because my force restoring
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force is minus k x so the energy
for this can be found out to be h cross omega
14:45.439 --> 14:54.019
n plus half with n going from zero to all
the way up but because half h cross omega
14:54.019 --> 15:00.059
is always going to be there so there is a
zero point energy the frequency of oscillation
15:00.059 --> 15:09.370
is one over two pi root force constant
k over mu mu is the reduce mass of the
15:09.370 --> 15:17.221
system the angle of frequency of the oscillation
omega is basically two part omega and when
15:17.221 --> 15:23.370
we look at the zero point energy its as
i mention half h cross omega and this sort
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of provides the fact that we can have the
zero point energy in this case to be non zero
15:28.520 --> 15:33.369
because of the consequence that you have to
follow the uncertainty principle you cannot
15:33.369 --> 15:36.569
determinately say where the particle is at
any point of time
15:36.569 --> 15:41.481
so the wave functions corresponding to the
eigan values for the harmonic oscillator are
15:41.481 --> 15:45.240
non degenerate and they have the form as we
have been saying earlier also of this type
15:45.240 --> 15:51.019
with a polynomial as well as a gaussian function
exponential which happens to be a gaussian
15:51.019 --> 15:57.050
and these polynomials are known as the hermit
polynomial there is a normalization factor
15:57.050 --> 16:06.199
also a n which is been shown here and we find
that each of these wave functions go between
16:06.199 --> 16:12.910
even odd even odd kind of a feature so all
the even ones are even functions whereas all
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the odd n values are odd functions so thats
how these wave functions are and we can see
16:19.480 --> 16:27.100
that it here in terms of the picture that
we have shown here so once we know that the
16:27.100 --> 16:32.639
harmonic oscillator problem is a hermit polynomial
now let us look into the rotational motion
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problem and as we know the smallest i mean
the first possible root rotational rigid
16:40.029 --> 16:45.220
root of problem would be a two dimensional
root rotation because whenever we rotate we
16:45.220 --> 16:50.970
involve two dimensions so here is the picture
which shows that two particle which are
16:50.970 --> 16:56.009
rotating it has a center of mass and now this
is a center of mass problem in contrast to
16:56.009 --> 17:01.309
the reduce mass problem that we saw in the
last problem of the vibrating particle
17:01.309 --> 17:07.100
vibrating system which was simple harmonic
oscillator and here in this case the center
17:07.100 --> 17:15.000
of mass problem would be represented in terms
of the two dimensional rotation with the easiest
17:15.000 --> 17:23.110
to deal with this in terms of polar coordinates
because the for rigid rotor the r which is
17:23.110 --> 17:28.070
the radial distance is going to be constant
the radius is not going to change so one
17:28.070 --> 17:33.110
of the one of the coordinates can be easily
handle compare to the other
17:33.110 --> 17:37.570
so the two dimensional rotation motion can
be written in terms of the radial coordinates
17:37.570 --> 17:44.110
and the hamiltonian can therefore be written
in terms of the radial coordinates and
17:44.110 --> 17:52.160
the particle is constraint it is root rotation
and there is no external potential working
17:52.160 --> 17:59.020
on it so its just their kinetic energy of
the system so its h cross two mu central
17:59.020 --> 18:06.950
mass minus h cross two mu l square which
is going to be written in terms of the
18:06.950 --> 18:12.310
angular coordinates which is what is written
and since the since the r is going to be rigid
18:12.310 --> 18:16.930
that its the rigid rotor so its going to be
constant so [vocalized noise] we are only
18:16.930 --> 18:24.500
going to have the variation coming due
to the phi angle the angular motion and
18:24.500 --> 18:32.370
therefore although its a two dimensional problem
it reduces to a single dimensional nature
18:32.370 --> 18:37.550
so then we can solve the single dimensional
problem by using the schrodinger equation
18:37.550 --> 18:44.200
and once we apply the s i equal to e s i form
here and in this case the variable is on
18:44.200 --> 18:49.460
the phi which is the rotating angle then we
can find how this constrain is going to work
18:49.460 --> 18:55.550
because at every two pi the particle is going
to retrace it person thats its constrain so
18:55.550 --> 19:02.370
we can put the boundary conditions and we
can find out how the energy and the
19:02.370 --> 19:08.750
quantum numbers vary for these kinds of problems
so the quantum number is n and this is the
19:08.750 --> 19:16.570
rotational rigid rotor problem so the solution
can be taken in terms of these form
19:16.570 --> 19:25.430
and the solution of this is phi this is
the angular part of the motion which is been
19:25.430 --> 19:32.080
shown the wave function angular wave function
and the corresponding energy and momentum
19:32.080 --> 19:38.990
can be found out by this process where
the energy is m squad h squad over two i i
19:38.990 --> 19:46.290
is the momentum of inertia of this system
and the and since it is rotating about the
19:46.290 --> 19:55.810
z axis then its the l z square is the good
angular momentum so it it it is l z is is
19:55.810 --> 20:01.980
the one which is conserved so its m m h cross
is the l z operator
20:01.980 --> 20:06.910
for the two dimensional rigid rotor we have
this solution that we are already familiar
20:06.910 --> 20:12.760
with that here we have try to show you have
you get to it and what we find is that it
20:12.760 --> 20:19.380
is dependent on the angular momentum and the
operator for that is l z m where we are getting
20:19.380 --> 20:24.090
all these values from only one quantum number
is required in this case to determine the
20:24.090 --> 20:29.510
state because [vocalized noise] the radial
part is constant and so we have kept it as
20:29.510 --> 20:37.000
a constant so this is the solution that we
have used and when instead of using two dimensional
20:37.000 --> 20:43.410
motion if it is allowed to move in all the
three dimensions then it a three d rotor and
20:43.410 --> 20:48.120
we have to use spherical coordinates and in
terms us spherical coordinates is curious
20:48.120 --> 20:55.180
to note that while two d two pi is the motion
that we need to do for the rotation constrain
20:55.180 --> 21:00.480
to be put in along the other dimension all
you need to do is to move ( Refer Time:
21:00.480 --> 21:07.330
21:00) which is the theta angle that is that
corresponds to just a pi rotation which can
21:07.330 --> 21:12.810
then generate all the rest of the motions
which is what is being shown here in these
21:12.810 --> 21:19.310
three dimensional cartoons and the theta
angle therefore is constrain to go between
21:19.310 --> 21:26.050
zero and pi were is the five angle is going
to go between two zero two to pi and the [vocalized
21:26.050 --> 21:32.240
noise] radial part the radius r is going to
be given by the three coordinates root of
21:32.240 --> 21:39.750
that and we can use this to get to our coordinate
system and the parallel between the spherical
21:39.750 --> 21:45.150
and the Cartesian cartesian coordinate can
be utilize to get to know the volume element
21:45.150 --> 21:50.140
which is being described here and we can make
a parallel between this which is being shown
21:50.140 --> 21:56.950
in this graph where [vocalized noise] we have
made the volume element which is being
21:56.950 --> 22:05.340
created because of this motion and the d theta
d r and d five are the dimensions that we
22:05.340 --> 22:13.980
are looking at when we do the volume element
so the spherical polar coordinates minimum
22:13.980 --> 22:20.260
volume element would therefore be given by
this d b where it can have this elements which
22:20.260 --> 22:29.630
is been looked at r d theta r sine theta
d phi and so as result we get and d five so
22:29.630 --> 22:41.070
we get r square sine theta d five d theta
m d r so the rigid rotor in quantum mechanics
22:41.070 --> 22:47.340
can therefore be written in this form where
the spherical coordinates hamiltonian would
22:47.340 --> 22:52.690
be having the form which is given here the
wave function must contain both theta and
22:52.690 --> 22:59.720
pi now because we are now having two different
coordinates and when you solve for this we
22:59.720 --> 23:04.870
will be getting the solution which is y
theta and five which are known as spherical
23:04.870 --> 23:10.880
harmonics and this is the solution which was
given to this problem when it was been asked
23:10.880 --> 23:16.400
and this background information is something
which i thought is important for you to at
23:16.400 --> 23:21.520
least be able to be familiarized yourself
to go along with this
23:21.520 --> 23:29.010
the next question concept of degeneracy
of the energy levels were been asked and looked
23:29.010 --> 23:35.260
at in order to ensure that you understand
how this is done so all these different cases
23:35.260 --> 23:44.230
that have been studied has been looked at
one d box three d cuboid box third rotational
23:44.230 --> 23:51.780
level in a rigid diatomic rotor first excited
energy level of a hydrogen atom and ground
23:51.780 --> 23:57.840
energy state of a hydrogen molecule so how
do you address this this is to essentially
23:57.840 --> 24:04.040
look at what are the possible degeneracy's
that can arise in each of these problems so
24:04.040 --> 24:08.800
for a one d box of a infinite high there
is only one possibility of getting the answer
24:08.800 --> 24:13.790
because the number of parameters available
are only as gone to be only one which is one
24:13.790 --> 24:23.240
dimensional x and so it is a singles process
it is non degenerate set the three d cuboid
24:23.240 --> 24:32.000
box of infinite length would have three
parameters left three parameters dependence
24:32.000 --> 24:37.990
because its a three d condition it is x y
z which is there and its three fold degenerate
24:37.990 --> 24:46.720
and and so will be having the degeneracy as
three the third rotational level in a rigid
24:46.720 --> 24:54.470
diatomic rotor will have a degeneracy of
two l plus one because each of the l values
24:54.470 --> 25:01.910
which is corresponding to the theta rotation
can take two plus one values because for
25:01.910 --> 25:08.870
each l value which can go from zero to n minus
one there are correspondingly minus l through
25:08.870 --> 25:17.250
zero to plus l number of degeneracy and so
for a third rotational level of a rigid diatomic
25:17.250 --> 25:25.470
rotor l is equal to two zero one two and so
our degeneracy is going to be two times two
25:25.470 --> 25:27.630
plus one which is five
25:27.630 --> 25:33.690
the first excited level of a hydrogen atom
would correspond to a degeneracy because we
25:33.690 --> 25:38.880
know that for the hydrogen atom which is coming
later in this review also were we look
25:38.880 --> 25:43.860
at the development and treatment of the
hydrogen atom the degeneracy of the hydrogen
25:43.860 --> 25:51.450
atom is n squared and so for this particular
case which is the first excited energy level
25:51.450 --> 25:56.250
of hydrogen where n starts from one so the
first case n equal to two and so we have a
25:56.250 --> 26:04.910
degeneracy of four the ground energy state
of a hydrogen molecule would be singly
26:04.910 --> 26:10.610
degenerate when we consider that schrodinger
solution without relativistic inclusion of
26:10.610 --> 26:15.310
spin because in our particular case of
solution of the the schrodinger equation we
26:15.310 --> 26:25.440
only have three quantum number n l and
m l so for this particular case when we
26:25.440 --> 26:32.900
do not include the spin part it will only
be singly degenerate because for a given
26:32.900 --> 26:40.510
n value the ground state of the system
we will only have n equal to one we will
26:40.510 --> 26:47.340
only have one particular degeneracy which
is n squared one squared is one
26:47.340 --> 26:54.790
so based on this what we have done is we have
managed to reflect some of the initial
26:54.790 --> 27:00.911
parts of the discussion that we have been
looking at as we go along because we have
27:00.911 --> 27:08.640
only covered say four of the different
concepts that we tested in week two problem
27:08.640 --> 27:15.140
set we will be looking back and reviewing
the rest of the problems in that in the subsequent
27:15.140 --> 27:20.960
lecture where will be able to see how many
more of those concept that are necessary
27:20.960 --> 27:25.730
to be looked at once again because at the
time of implementation all of these are being
27:25.730 --> 27:32.010
assumed to be known to you when we are going
through the steps so this particular process
27:32.010 --> 27:36.100
right now that we are going through is very
critical for you to be able to understand
27:36.100 --> 27:42.350
and be able to address all the problems
that we are currently doing and be able
27:42.350 --> 27:47.500
to understand the principle of the implementations
that we are doing in terms of quantum complete
27:47.500 --> 27:48.870
so see you in the next lecture