WEBVTT
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so in the last lecture we looked at the levels
of implementation that you can do with optical
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approaches to quantum computing and there
may be several more of it but what we will
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do now is that we will look at those kinds
of implementations with other aspects as we
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mentioned in the last class also that there
are other aspects like sprintronics were optics
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and other kinds of optical implementations
as well as teleportation we have looked at
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it before all those become very interesting
and important those we will deal with it as
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we look at those particular kinds of implementations
from todays lecture onwards for this week
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for the rest of this week we will be solving
the assignment problems that we have been
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given giving you for the last six weeks and
maybe i will not be able to do it in the
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last week but generally as much as possible
we will try to do because while doing these
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assignment problems we will also revisit and
refresh some of the problems and some of the
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aspects at we were been looking into in this
area of implementation of quantum computing
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we will also link you at every time while
we do the problems to the key concepts that
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we are looking at while we are doing this
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so let us now look at this principle of solving
the problems so the first problem set is the
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one which we had given to you in the very
first week and the first question in that
01:40.080 --> 01:46.530
problem is as we mentioned here and the solution
had been given to you but the basic concept
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that we were discussing and looking for in
that problem was the principle of the particle
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nature of light
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so with that let us now look at all the different
aspects that we can do for this particular
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problem started with in the first week where
we had these multiple choice questions and
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their solutions were given so the first one
was based on the photoelectric effect um which
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was questioning the concept of photoelectric
effect that we discussed and so the question
02:19.470 --> 02:27.629
basically was given to find out which of
these different choices was important
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so it sort of was looking for the particle
nature of light to be understood and in this
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case most important parameter or understanding
gain from the discretization of the energy
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that was first proposed by max planck in terms
of the max in terms of the plancks constant
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h um and this is a visual representation of
how the continuous energy picture change to
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the discretized energy picture discrete or
quantize energy picture after the principle
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in given by a max planck
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so this quanta or the discreteness of the
energy in terms of the plancks constant was
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the key to bringing the photoelectric effect
explanation by einstein where he was able
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to show that the radiation field is quantized
because no matter how much of a photons you
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provide beyond a certain energy level they
will be no electron ejection for the photoelectric
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effect essentially showing that the light
below a frequency of say in this particular
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example of certain hertz or wavelength longer
than six eighty three nanometers would not
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eject electrons the fact that this plot was
not dependent on the intensity of the incident
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light implied that the interaction was like
a particle which gave all its energy to the
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electron and and ejected it with that energy
minus that which took to escape the surface
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so this extra energy that was necessary to
make this happen it is often known as the
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work function of the um surface so on so forth
but generally speaking this was the basic
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concept on the particle nature of light which
was been asked in this first question
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the second question goes into the uncertainty
principle which states that if you can measure
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position and time simultaneously then you
cannot measure so here um one of the important
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things to remember that it actually questions
both position and time and then gives you
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choices with respect to that and so the correct
choices essentially the correlation between
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these two which we now discuss here so this
was first realized by heisenberg where he
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was able to understand in the world of very
small particles one cannot measure any property
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of a particle without interacting with it
in some way this interaction introduces an
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unavoidable uncertainty into the result and
one can never measure all the properties exactly
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now this was the realization of heisenberg
that gave rise to this uncertainty principle
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its illustrated here in some detail so if
you shine light on electron and detected reflected
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light it just using a microscope then assume
this kind of a way cartoonist picture where
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the electron photon collision happens before
and minimum uncertainty in position is given
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by the wavelength of light so to determine
the position accurately it is necessary to
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use light with short wavelength however
a after the electron photon collision by plancks
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law we know that there is a photon with short
wavelength has a large energy which would
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mean thus it would impart a large kick to
the electron but to determine its momentum
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accurately electron must only give a small
kick
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so the moment you want to actually determine
the position very accurately you were almost
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lost in a terms of its momentum and thats
the basic idea here because of the re coil
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of this so this means using light of long
wavelength so the measurement disturbs the
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system and that is the consequence of this
entire problem so because the particles are
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so small the interaction is large enough to
do these issues so the uncertainty principle
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essentially means that there is a fundamental
limit to the accuracy of a measurement determined
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by the heinsenbergs uncertainty principle
and this is different from the classical uncertainty
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where the measurement of uncertainty is due
to limitations of the measurement apparatus
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there is nothing to do with the fundamental
limit which is there in terms of quantum
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mechanics and in this particular case for
example in classical there is really no limit
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in principle to how a accurately measurement
can be made it all depends on the instrumental
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apparatus that we are talking about
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in quantum mechanics we have these two very
important aspects one related to the position
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which is the position accuracy and the
simultaneous measurement of the linear momentum
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its with precision delta p then the product
of the two can never be less than this this
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is one part of the uncertainty heisenberg
also recognized the time energy uncertainty
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relationship which is related to the energy
of the system as well as the time that is
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be used and so and there are two different
uncertainties and the problem probe both of
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them thats why the answer was for both the
cases when position and velocity are the once
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which cannot be measured simultaneously and
similarly time and energy cannot be mentions
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simultaneously
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when all these four parameters are there thats
the reason why the answer was given in that
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particular wave this particular problem is
perhaps easier which basically introduces
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and looks at the concept of the wave function
which is just a mathematical description of
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a quantum entity and in a little bit more
detail manner here is the description which
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we have already looked at before but in very
precised sense this is the schrodinger's a
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proposed equation that accurately describes
the motion of wave that accompanies the electron
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this wave represented by the greek letter
psi gave precise predictions for the behavior
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of atoms
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the wave function is a complex function a
function containing quantities that are complex
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numbers and is constructed in such a way that
performing an operation to take the modulus
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of the complex function for a given state
results in a real value that represents the
09:22.340 --> 09:30.199
probability density of a result that corresponds
to that state this is how wave function is
09:30.199 --> 09:38.310
transformed into a meaningful physical prediction
so thats the importance of the wave function
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that it actually embeds the all the physical
um a important information that are necessary
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so the wave function represented by greek
letter psi is a mathematical description of
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a quantum entity and thats what the choice
answer would have appealed
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similarly the next question is on the combined
condition of atomic orbitals which is a molecular
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orbital and this is defined in many different
base but this is a particular definition which
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is been looked at here and there is a there
is a little right up on if that i have put
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up for you now
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so that it becomes clear a molecular orbital
is a mathematical function that describes
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the wave like behavior of an electron in a
molecule the term orbital was introduced by
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mulliken in nineteen thirty two as an abbreviation
for one electron orbital wave function and
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thats the reason why the most appropriate
um connection between the molecular orbital
10:46.839 --> 10:53.839
these to the one electron orbital wave function
at an elementary level molecular orbital is
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used to describe the region of space in which
the function has a significant amplitude they
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are usually constructed by combining atomic
orbitals or hybrid orbitals from each atom
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of the molecule or other molecular orbitals
from groups of atoms molecular orbitals can
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be calculated quantitatively thats why molecular
orbitals have a lot of important positions
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and intec interactions to take here and so
i thought a little bit more detail on what
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we have done before is necessary and here
the most important part to keep in mind is
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that the is a mathematical entity which has
to do with one electron orbital wave function
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and it sort of like connecting to the molecular
orbital in that sense
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there are two questions on de broglie so the
first one is just trying to see that you
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connect the de broglie to the concept that
he essentially introduced this is a a matter
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wave characteristic and so anything related
to the wave like nature of matter is now considered
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to be related to de broglie so that was the
point of this particular question and this
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is the concept that he introduced through
his thesis these thesis in nineteen twenty
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four where he proposed that just like light
both wave like and particle like properties
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electrons with momentum p also have wave like
properties like wavelength wave like properties
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like sorry with wavelength lambda such
that lambda is related to the plancks constant
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over the momentum
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now this relationship is now known to hold
for all types of matter and not just for electrons
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although it was initially proposed by him
in his thesis for the electrons but it was
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for everything and in the next question
taking on to the fact that this matter wave
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is universal fact the question is raised on
how the ratio of that goes between them
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and so the ratio of de broglie wavelength
for a cricket ball of mass of a given mass
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to that of helium atom both of which are
traveling at the same speed it just for argument
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sake is being looked at and for to do this
problem its important that you use the de
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broglie relation with that relates matter
with momentum and its wavelength and take
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a ratio of the two cases and realize that
the wavelength is inversely proportional to
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mass and the velocity
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now since velocity has been kept same so it
essentially means that you have to only consider
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the mass of the system for both of them and
once you do that you just get a ratio of the
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two masses in this particular case point four
and four coming from the helium is canceled
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and you are left with a order of magnitude
value which is simply relating the conversion
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hm between the two
so the next question looks at a degeneracy
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of a particular state of so for example
this is a thirteen point six over n square
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electron volt is the energy this is the as
you know for the hydrogen atom and it is under
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the constraints that the state has all these
other characteristics and you are suppose
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to find the solution and this involves realizing
that each quantum level n for an atom ok thats
14:51.209 --> 14:59.360
the principle quantum number there are zero
to n minus one angular momentum states each
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of which have a degeneracy of again zero
plus minus one all the way to plus minus l
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which means that each l state will have two
l plus one degeneracy
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so the total degeneracy given that each n
state can have any volume from zero to n minus
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one would then mean a summation of this entire
two l plus one over zero to n plus one so
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once you do the summation will find that the
answer it's a its series where the answer
15:33.920 --> 15:42.019
comes out to be n squared so thats how this
little maths is suppose to take you there
15:42.019 --> 15:51.790
the next couple of questions on this are
essentially looking at the properties of these
15:51.790 --> 15:58.490
molecules to some level but based on very
simplistic arguments and so considering a
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harmonic oscillator model for hydrogen molecule
and an hd molecule and assuming that the same
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force constants exist for both molecules the
lowest energy state for h two is you have
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to find out that to basically you would need
to know the how many oscillator model form
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of energy and the harmonic for a harmonic
oscillator energy states are given by this
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where b is the quantum number where w can
go from anywhere from zero to whatever number
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and then here the way the energy essentially
exist makes sure that there is a zero point
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energy and a because at v equal to zero the
energy exits its half h cross omega
16:46.800 --> 16:54.399
h cross by the way as we are a known is h
over to two pi ok now the frequency of vibration
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is given by the spring constant over that
or the force constant over the a mass or reduce
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mass as the system may be and assuming thats
this has been given in the problem that the
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same force constant exists for the both the
molecules the lowest energy state for both
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will have an inverse relationship with respect
to their masses with respect to their their
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respective mass which means that you will
have the e zero being proportional to one
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over root m and since hydrogen mass is a lower
mass as compare to the hd molecule the lowest
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energy for h two will then therefore be higher
than that of hd because a its inversely proportional
17:38.480 --> 17:45.060
so you will be slightly lower in a energy
then a h two so that is the level of identification
17:45.060 --> 17:49.980
that we are suppose to make for these kinds
of the problems
17:49.980 --> 17:55.930
now similarly the other simple question was
based on the ideas to how to looks for wave
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functions and realize how many variables are
necessary for describing a wave function so
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for a schrodinger wave function have of this
is helium atom am he was asked how many coordinates
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or how many variables this depend on so in
order to do that the best thing is to draw
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a simplistic picture of say the helium atom
which has one nucleus and two electrons now
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the potential part of this will have the electron
electron reflection part where there is a
18:28.210 --> 18:37.370
r one two which is the relative distance
between the two electrons and the two columbic
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attractions would be between the nuclei and
the nucleus
18:40.800 --> 18:46.170
so the potential energy part will have an
three components one of them is repulsive
18:46.170 --> 18:51.550
and the other two being attractive and therefore
the wave function will depend on each of these
18:51.550 --> 19:01.450
three parameters and each of these parameters
are dependent on entire set of three possibilities
19:01.450 --> 19:07.440
so what we are essentially saying is if
we consider all possibilities we get an nine
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dimensional problem more clearly helium
atom has one nuclei as we mentioned and two
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electrons each of which would depend on
three variables implying that its a wave
19:21.991 --> 19:27.410
function would therefore depend on all possible
nine variables
19:27.410 --> 19:34.690
so this is how this problem has been looked
at and this is the sort of the way this problems
19:34.690 --> 19:41.920
have been designed and done in this kind of
problem set the there is a one issue in the
19:41.920 --> 19:48.380
last problem of this week one which while
i was going through a solving it i realize
19:48.380 --> 19:55.330
that this question were the lowest rotational
energy gap of the linear carbon dioxide molecule
19:55.330 --> 20:01.220
is being questioned under the assumption that
it's a three d rigid rotor this is a completely
20:01.220 --> 20:08.850
um thought problem because its thought of
as a linear molecule and assume to be a three
20:08.850 --> 20:24.780
d rigid rotor where the carbon is set to be
at the center of the rotating axis and
20:24.780 --> 20:35.960
um and two oxygen atoms exactly at bohr radius
so its taken as fifty three um pecometer on
20:35.960 --> 20:41.420
both sides from the central carbon atoms
so the carbon is consider to be the center
20:41.420 --> 20:47.450
and this has been asked with the idea that
whether you can calculate this to be a number
20:47.450 --> 20:54.490
which comes out to be a exactly one of these
unfortunately am perhaps i didn't do it very
20:54.490 --> 21:00.270
with a lot of effort but have when i was doing
it quickly i could not get exact match to
21:00.270 --> 21:05.390
one of these problems and there one of these
answers which i given here and so thats actually
21:05.390 --> 21:08.200
little bit a way a difficulties
21:08.200 --> 21:13.500
so what have decided is that i will open the
problem for you and then we will look in
21:13.500 --> 21:18.390
to this problem further into the next lecture
because i dont think will have much time in
21:18.390 --> 21:23.430
this lecture to finish the problem but what
we will do is will open up this problem
21:23.430 --> 21:31.310
for this now and if we are not able to
get any answer then we will have to see why
21:31.310 --> 21:37.830
the must have been and printing typing error
on my side on while making the question and
21:37.830 --> 21:38.830
let us see
21:38.830 --> 21:44.660
so the question is based on the idea that
we know how the energy obey three d rigid
21:44.660 --> 21:51.870
rotor looks like and that is sort of given
in terms of a this expression where the energy
21:51.870 --> 21:59.630
of the lth state is can be written in this
term very often you are used to the h cross
21:59.630 --> 22:04.720
terms however in this particular case since
we are going to write in terms wave numbers
22:04.720 --> 22:10.400
i have taken out the h cross and therefore
we have gotten these a pi sitting in there
22:10.400 --> 22:17.950
if you write in terms of h cross these pi
square term vanish and its a perhaps sometimes
22:17.950 --> 22:26.390
easier to remember them a in that forms so
we often write them in the h cross terms with
22:26.390 --> 22:36.290
two i so thats roughly what we do because
four pi squared gets absorbed when as soon
22:36.290 --> 22:42.480
as we say h cross in there by irrespective
whatever we have been number a your energy
22:42.480 --> 22:49.390
is going to look like this and this results
in an energy gap in terms of l to l plus one
22:49.390 --> 22:57.270
let say you will get something of this form
and if you write it in terms of in wave
22:57.270 --> 23:03.120
numbers then you are going to divide it divide
the answer by h c thats the idea and in this
23:03.120 --> 23:10.470
particular case when you divide by h c you
have to remember to use your value of hm c
23:10.470 --> 23:15.820
in terms of a centimeters per second
23:15.820 --> 23:22.180
so generally in these cases what is done is
it is taken as three times roughly three
23:22.180 --> 23:28.380
times ten to the power ten centimeters per
second and this is the velocity of light
23:28.380 --> 23:33.350
which is taken instead of ten to the power
eight three times ten to the power eight meters
23:33.350 --> 23:38.540
per second this is just to make sure that
we can get the wave number unit at the end
23:38.540 --> 23:49.210
so the i is the so we have here the quantum
number l rotational rotation constant
23:49.210 --> 24:03.640
be i now i is related to the reduced mass
in this case so i is mu r squared r is the
24:03.640 --> 24:12.200
distance um so in our in our particular model
when we have the carbon sitting in the center
24:12.200 --> 24:18.680
and the oxygen sitting linearly on the other
sides the r is basically this distance that
24:18.680 --> 24:24.040
we are looking at which has been given to
us the mu is the reduce mass
24:24.040 --> 24:30.630
now the reduce mass can be taken in many different
ways in this particular case the simplest
24:30.630 --> 24:38.610
one is to consider only the carbon and the
oxygen because those are the only two points
24:38.610 --> 24:45.450
that we are looking at the h the rest of the
same however in reality um by just considering
24:45.450 --> 24:53.620
it to be a three body system you can always
utilize the harmonic mean for this system
24:53.620 --> 24:59.160
also in spectroscopy however there is a little
bit more involved manner of doing this thats
24:59.160 --> 25:08.720
why the problem were was simply simplified
by providing the assumption that its an a
25:08.720 --> 25:16.430
linear level by this particular case um in
in in the simplest way of looking a three
25:16.430 --> 25:22.780
body kind of a situation like this reduce
mass will just be the three masses in this
25:22.780 --> 25:36.120
case oxygen um carbon and oxygen this could
also be used but to make it even simpler we
25:36.120 --> 25:43.980
just kept between the two to get rough estimate
and we have to always use the unit for conversion
25:43.980 --> 25:50.980
into the kilogram right was it goes and the
r has been given as we mentioned in the question
25:50.980 --> 26:01.350
so the moment of inertia i can then be calculated
although this particular way of calculating
26:01.350 --> 26:07.790
the moment of inertia has a lot of assumption
in it and any way by going by this root we
26:07.790 --> 26:14.160
can get some form from where we can get the
lowest rotational energy gap which we have
26:14.160 --> 26:19.790
described here by just plugging in the values
it so turns out that once we plug in these
26:19.790 --> 26:26.530
values in order to get the number in terms
of wave numbers and at least i did not find
26:26.530 --> 26:33.410
it to come close enough to the values that
we are used to the best be came to us i think
26:33.410 --> 26:35.780
not what we have given as choice
26:35.780 --> 26:42.380
so as we had been talking to you about this
problem it does not give any of the results
26:42.380 --> 26:51.090
that we have been talking that we have provided
in this problem thats because most of the
26:51.090 --> 26:59.470
so this last problem in that next problem
set had the difficulty that the printed solutions
26:59.470 --> 27:05.410
that were provided as options did not actually
come as any of the answers that we are looking
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at here either when you do it in the approximate
sense that we present here or any of the other
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sense that we do here and therefore um i dont
think you will find any solution which will
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be in connection correct to the option that
are given so in that way what i am going to
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do is we are not going to have that as a grade
numbering for you will take that part of
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but the solution part was important to
show you as how to do this problem because
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there is more important to realize how to
do a problem rather than to just be able to
27:41.830 --> 27:43.140
get some marks for it
27:43.140 --> 27:49.040
so what i have done here is that i given you
an idea as to how to do this problem in case
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something of similar nature comes out it can
be very many very much more complicated as
27:54.220 --> 27:58.730
i have been mentioning here but many times
will be essentially looking at the problem
27:58.730 --> 28:03.220
in a much more simpler way to make sure make
sure that we are getting the main aspects
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of the problem so in the next week we will
be looking at the other problems such that
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we have been giving you and with this i
would like to close todays lecture see you
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next time
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thank you