WEBVTT

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hello and welcome so we are on the day
first of week six so in the last class

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we studied learning about symmetry adapted
linear combination so what is this essentially

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symmetric adapted linear combination we
started in giving some fifth example but

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let's have a look at it in bit detail
so whenever in a molecule we have certain

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atoms and each atom will have their own way
function and this way functions are you know

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also looked at as orbitals so different
you know way functions of different atoms

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will combined whenever they are energetically
preferable and also symmetrically allowed

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so what does that mean
so let's have a look at it so for you know

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various metal complexes the d orbitals
they come into the you know in the play

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so is not that you know two atoms have
been any two d orbitals can combined so they

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are certain symmetry restrictions so for example
if we have a look at the pictures that

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are shown on the screen
so two digits square orbitals they can

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combine which is symmetrically allowed
and they can give rise to two molecular orbitals

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like this ok depending on a which way they
are combining so that can give bonding and

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n type bonding kind of orbitals more with
your orbitals so this the z of sigma

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and sigma star that are formed out of the
combination two digits square orbitals

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they form through a linear combination
of those two digits squares atomic orbitals

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so here we have another example where either
you can take a d x z to d x z orbitals or

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two d y z orbitals and so if you take two
d x z orbitals and this is my d x z plane

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and this is also another d x z orbitals in
the same plane so they can combine ok they

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can combining two different way one is in
this fashion another is in this fashion so

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depending on that you get either a phi
bonding or phi n type bonding molecular

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orbitals so again d x square minus y square
can combine with d x z orbitals because they

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are in the parallel plane so they can have
an overlap d can form like delta bonding

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ok
so delta bonding and delta and n type bonding

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of kind of orbitals where as if you look at
the right hand side panel of this screen you

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can see that d y z and d z square they cannot
be combine thus symmetrically they completely

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different and so is d y z and d x z and
d x square minus y square and d x y so because

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of the different symmetry of this orbitals
there will be no interaction no interaction

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means no bond formation right so no bond formation
means no result and way function or no result

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in orbital formation
now we need to find out at which are the

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orbitals in a given molecules that can form
a linear combination to give rise to molecular

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orbitals meaning the orbitals that are formed
they should complying with the symmetry

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molecular point group right so how to find
that out that we started with in the the

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last class we said that there is truth
called projection operator projection operator

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can give us as the information about which
are the orbitals that can linearly combine

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and give symmetrically allowed you know
molecular orbitals so we studied in

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the last class a complete projection operator
is a capable of generating complete set of

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s a l c
so what we did we we formed this

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projection operator and we derived a sincerely
and this is the form of that projection operator

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that we formed using the general principle
and from their we also said that

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this is the special case when in one talks
about the purely diagonal element of the you

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know matrixes of a representation and there
by this particular special case of projection

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operator will find out whether this phi
t prime belonging to a particular j th

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ire illusive representation occur within the
orthonormal set say phi you know i s

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one of which is here as phi t so if you apply
the projection operator on to any arbitrary

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function belong to a orthonormal set then
either this function itself will be this one

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or it will contain this function in that case
in this particular function will be projected

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out from that set by this projection operator
otherwise this function will be abolished

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or if this function contains this phi t prime
and many other functions also then all

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the other functions will be abolished and
this one only projected that's why it is a

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projection operator
so you start with when you know l i

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dimensional set having orthonormal you
know function and take them as basis and

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you can construct the l i number of projection
operator and this complete set of projection

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operator can give the total information about
you know what are the particular function

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that you know can act as the basis of the
you know whatever the ire illusive representation

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one is concerned about and in in in the
molecular of orbital picture what you can

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think of such a projection operator constructed
form an exhausted you know in list of

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orbitals function list of functions and
if you form all the projection operator

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then the total you know site of projection
operator will give you the overall symmetry

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adapted linear combination that are possible
now this is about complete projection operator

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this this term will be when more clear as
we move ahead and since we are talking

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about complete projection operator there is
also another something called incomplete projection

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operator which which is less efficient
than this complete projection operator while

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we deal with complete orthonormal set and
forming the you know position operators

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involving the each and every element of
the you know of any given representation

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and so you you get the total information
about the s a l c s whereas the incomplete

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projection operator that is you know that
deals with the you know the trace of the the

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you know matrixes that of prompt while forming
the representation

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so we will talk about that little while
from now so this is the same thing that

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is written over the board and the we we can
write the explicit ire illusive representation

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while replacing this generalized term j right
so here just for an example we can write this

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as b three u suppose i am concerned about
a particular character table of say d two

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eight ok so an in the function that you
can choose as any of the bond vector or you

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can choose as any of the you know atom
orbital is anything so you know by that

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the projection operator may be applied to
be an arbitrary function say phi t and

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only if that function itself or some term
in happens to be phi t prime the result

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will be you know non zero if phi t prime is
a component of arbitrary function five is

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prime that will be projected out of it and
the rest will be abolished so we have this

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term in this as a special case where we
talk about the projection operator involving

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only the diagonal elements of the representation
so t prime t prime is a diagonal element right

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now so ultimately this p t prime t prime
corresponding to a j th ire illusive representation

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that will project this you know phi t prime
out of an n arbitrary function as phi t now

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if i you know what we started with l i no
of or you know you can call it l j number

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of orthonormal functions ok so if you take
a set of you know l j number of orthonormal

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function then you can actually produce l j
number of projection operator because using

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that l j number of orthonormal functions you
can form and you know the number of

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diagonal elements that will be found with
also so you know we can use that particular

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set of the projection operator to completely
find out you know you can use any general

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arbitrary function phi t correspond to
any particular ire illusive representation

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i and the function that form a basis for the
j th ire illusive representation which you

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termed as phi t prime t
now how does the projection operator work

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actually ok so before way into exact formation
of symmetry adapted linear combination let's

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take simple example and see how the projection
operator helps finding out that which of the

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of any given arbitrary function which part
of a actually form the basis for any given

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ire illusive ok so here we have an example
where we consider the group c three v and

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we take an arbitrary function which is as
x z plus y z plus z square ok so this is any

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arbitrary function now what we will try to
do you will try to you know get from this

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arbitrary function we would like to have a
pair of function which will form the basis

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for the ire illusive representation e that
is that is two dimensional ire illusive representation

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that c three v point group has we would like
to find two functions you know which would

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like to take out from the this arbitrary function
x z plus y z plus z square and that those

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two functions will act as the basis for the
representation e

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now here we have all the matrix representation
of all the symmetry operation that c three

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v point group has ok this one we have already
dealt with earlier so must be familiar with

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you now here talking about complete projection
operator like so when i talk about complete

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projection operator i will be always dealling
with the matrix elements ok so matrix elements

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is specifically as a special case we will
take all those diagonal element with that

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sort sort of purpose now what we have to
do so we to first form the projection operator

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so projection operator has the form that you
can see on your screen on there so you

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have to find out dimension of that ire representation
so here for representation e l j will be

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equals to two h for c three v it will be equals
to six because it has order six then you

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have to get the matrix elements so this is
the representation and if essentially we are

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taking the elements of the representation
so here we have we have found on the screen

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that each and every you know every symmetry
operation has their matrix representation

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so they are two by two matrix correct so now
you have total you two diagonal elements

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in that each and every matrixes so what you
have to do you have to select first the one

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one element because for any given make
fix which is two by two so you have say

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a b c and d so this particular one is essentially
termed as one one ok one two two one and two

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two right so that's all in matrixes algebra
we termed this we use the terms so what

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we have to do we have to find out this t prime
t prime right you know matrix projection

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operator form for this p j t prime t prime
diagonal element so we will be concerned with

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this one and this one
so here is a two diagonal matrix and we will

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be two two dimensional prior ire illusive
representation because we are concerned with

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e so therefore we have two projection operators
because we have two diagonal elements so first

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what we have to do we have to form the projection
operator for this term one one and apply that

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on the function that we choose so here we
have like x z plus y z plus z square that's

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an arbitrary function so we have to operate
this projection operator corresponding to

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this one one element on to the arbitrary
function x z plus y z plus z square and then

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we have to form the projection operator involving
the term two two matrix element two two and

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again do the same thing so we will get two
functions out ok when we operate this projection

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operators this two projection operators on
this arbitrary function all right so what

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we have to do is written here
so first we have dealing with this the p one

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one term correct so if you you know can
correctly identify this you know this symbol

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p superscript e and subscript one one
and you can correlate with the general form

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that we have earlier use the set p j t prime
t prime ok then you are find if you have any

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difficulties go back and again check the previous
part that we have talked about and then come

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back to this portion again so now we are going
to operate this projection operator on this

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arbitrary function x z plus y z plus z square
and see what its gives out ok

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so in order to do that what do you have to
do you have to you know get the form of this

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projection operator that i just said so
l i is to h is equal to six and then if you

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look at the form of the projection operator
over there you get the matrix elements

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corresponding to particular symmetry operation
right so you start with say e then you go

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for c three then you go for c three square
then you go for sigma v x z sigma v prime

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sigma beta whole prime that way all right
so you see how we have done this one how

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did we operated this projection operator on
this function so l i by h is common for all

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and then you have selected the matrix element
here you see this is for the first term is

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for identity operation
so here we have this term one if you quickly

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go back you will see the this one one element
of the operation e is one all right and then

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you need to know how this identity operation
changes the function right so e doesnt do

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anything so x remains x y remains y z remains
z and then overall the function x z plus y

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z plus z square remain as such all right so
that therefore you have you know multiplication

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by the matrix element term that's a one
and then whatever is happening to the function

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due to the application of symmetry operation
on that function so it returns the same one

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now you move on because you have a some over
all r so you have keep doing each and every

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r the symmetry operation and then all together
you have to add

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so i have just told you about the e so next
step you have to go to the next operation

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that is say c three so here on this table
you have all the transformation property this

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particular table you have all the transformation
properties of x y and z axis upon this

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symmetry operations ok for c three v so these
are the transformation terms and at the end

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you have actually the overall transformation
of the function x z plus y z plus z square

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so what do you can do you can verify this
table yourself ok and i will suggest you to

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do so so here coming back so what do we
have to do first we have to find out a one

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one element for the c three v operation so
let's quickly go back to this one and c three

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the one one element this minus half all right
and let's look here so we are first multiplying

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by minus half and then you to be multiplied
by the you know resultant of the operation

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of c three on to the function x z plus
y z plus z square and this particular term

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that is given here is it can be found to
over here so you can see that is you know

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this particular part right
so in that way you keep going for c three

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square and sigma v and sigma prime and sigma
v double prime so when you complete this sum

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you have got your result for the you know
operation projection operator on this arbitrary

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function and this projection operatory is
specifically for the each ire illusive representation

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which is two dimensional ire illusive representation
corresponding to the c three v point group

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so after you get the somewhat you will get
you you you will get you know some you

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know terms with containing x z some from
containing y z some containing z square so

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ultimately if you can rearrange them properly
so you get the co efficient corresponding

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to x z co efficient corresponding to y z co
efficient corresponding to z square because

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this three terms are they are in my arbitrary
function

21:40.950 --> 21:46.761
so my intension will be to a you know rearrange
the in result of this projection operator

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on this function to also express as co efficient
of this three functions you know individually

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components individually x z y z and z square
so that's what it shown here when we you know

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we collect all these terms and separate
the co efficient then i get the co efficient

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of x z to be one while co efficient of y z
zero and that of z square to be zero so my

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projection operator p one one corresponding
to the ire illusive representation e for c

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three v is projecting out x z as the basis
for e representation out of the arbitrary

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function x z plus y z plus z square all right
so we have to get the complete result by using

22:40.889 --> 22:46.169
the you know all the projection operator that
can be formed explicitly because we are dealing

22:46.169 --> 22:51.359
with the complete projection operator
so the next projection operator will be

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taking the other diagonal element of this
two by two representation so that will be

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corresponding to two by two diagonal element
of each and every symmetry operations so if

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we just quickly look at the representation
for e the two by two element is also again

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one so it will be no different and for c three
v c three again it is minus half for c three

23:15.729 --> 23:23.140
square minus half and so on so you get those
particular element out ok and then multiply

23:23.140 --> 23:35.019
that with the resultant of the symmetry
operation on the particular arbitrary function

23:35.019 --> 23:45.350
so very similar way when we do it and ultimately
you collect all the terms and find out the

23:45.350 --> 23:56.879
co efficients we get the co efficient for
y z to be equal to zero s is equal to n and

23:56.879 --> 24:04.749
the rest of the two that is x y and z square
they give us zero that means when we apply

24:04.749 --> 24:14.529
p two two it projects out this y z you know
that can for the basis for the representation

24:14.529 --> 24:19.190
e ok
so now if we compile the total result that

24:19.190 --> 24:27.019
we have complete set of projection operator
meaning p one one and p two two for the representation

24:27.019 --> 24:41.859
e all together it is giving us x z and y z
so this means that x z and y z out that arbitrary

24:41.859 --> 24:50.509
function x z plus y z plus z square can formed
the basis for e which is not surprised right

24:50.509 --> 25:02.559
because e is a two dimensional representation
so we will have two different functions acting

25:02.559 --> 25:14.890
as a basis ok so this one will be x z so
you know x z and y z will formed the basis

25:14.890 --> 25:22.820
for representation e and also you see that
z square is completely abolished by this

25:22.820 --> 25:27.649
projection operator so it will tell you exactly
the which are the functions that can from

25:27.649 --> 25:33.399
the basis of any particular ire illusive representation
and whichever is unnecessary that whichever

25:33.399 --> 25:38.440
do not form the you basis for the particular
representation they will be automatically

25:38.440 --> 25:42.799
abolished so you will get zero result when
you operate projection operator on the function

25:42.799 --> 25:48.700
for that particular component all the time
so here z square give you know zero contribution

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in both the cases for p one one as well as
p two two ok all right

25:57.840 --> 26:04.059
so that is about complete projection operator
so now now this complete projection operator

26:04.059 --> 26:14.619
it deals with the you know diagonal element
so for an one dimensional representation

26:14.619 --> 26:18.979
things are easy because for one dimensional
representation the matrix is also one dimension

26:18.979 --> 26:25.700
so its its kind of trivial but the moment
you go for higher dimensional ire illusive

26:25.700 --> 26:31.009
representation say it two dimensional three
dimensional thing become bit complicated because

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getting the matrix representation finding
the particular you know elements and then

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doing it it if this white cumbersome and it
cannot be you know finding this completely

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set of projection operator its not
really can be done by a machine alone so it

26:52.779 --> 27:01.330
needs so called human intervention all
right so you know it will be a you know

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like someone dealing with this problem
has to construct the you know completed

27:08.759 --> 27:16.299
to compute the total position of patterns
if it on a particular function then again

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you know go for the next one so every time
someone has to instruct it continuously and

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all together its its its really you know cumbersome
but you know getting the character of

27:29.759 --> 27:34.099
the representation is always easy because
you have the character table quality so just

27:34.099 --> 27:39.099
have at look at the character table you get
it and ultimately what you are doing if

27:39.099 --> 27:44.710
you think about so you are taking any representation
whatever the dimension be it and you are just

27:44.710 --> 27:51.669
taking the diagonal element to form the projection
operator right and then collecting all

27:51.669 --> 27:56.019
the diagonal you know position operators
corresponding to all the diagonal element

27:56.019 --> 27:59.509
you are forming the total set or complete
set of projection operator which is known

27:59.509 --> 28:03.889
as complete set of projection operator now
if you look at this one if you look at the

28:03.889 --> 28:09.210
tries of the representation that actually
contains the information that you know that

28:09.210 --> 28:13.779
can be found from each individual diagonal
element that we have already talked about

28:13.779 --> 28:17.289
and that's why this tries is also called the
character

28:17.289 --> 28:27.159
so it should be pretty fine if we take the
character and try to form you know a projection

28:27.159 --> 28:33.940
operator out of that so in some sense it will
be incomplete but it will be extremely useful

28:33.940 --> 28:39.169
now it will be incompete when you are particularly
go for extremely you know complicate

28:39.169 --> 28:45.529
ire illusive representations but otherwise
for actual practical purpose is this incomplete

28:45.529 --> 28:52.779
projection operator which involves the
characters only should work absolutely fine

28:52.779 --> 28:59.999
and in the next class we will talk about
this incomplete projection operator and we

28:59.999 --> 29:07.929
will try to derive this incomplete projection
operator and use it and you know we try to

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vulgate the points that i just discussed ok
so we will stop here today and in our following

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class we discuss more about this thing till
then have a good day