 continue ok. You can find tertiary basis by taking a 1 b 1 a 1 and so on. So, many that is the way you get your tensors which have which are you know 3 indexed object moment of inertia has 2 indexed object, quadrupole moment has 2 indexed object. So, if you want to get a 2 indexed object it has to belong to a binary basis. If you want to get 3 indexed object it has to belong to a tertiary basis and how to construct the tertiary basis is to take tensor product of primary basis number of times apply a projection operator and find out the basis ok. In this case it was trivial because they are all 1 dimensions, but we will do a non-trivial C 3 wheel. So, this was with the matrix representations which you can use, but you can also replace this by a formal operator G operator ok. So, you can write P a 1 the projection for the irreducible representation as the character for the group elements in the representation a 1 and then you can also put the G operator and not worry about the matrix representations for it because the matrix representations depends on the dimensionality of your vector space this is also another way of doing it ok. So, this is where you can apply P a 1 if you remember in the harmonic oscillator I said that this Hamiltonian is invariant and a x going to minus x, it has a group symmetry which is isomorphic to permutations of two objects or the C 2 group this is something which I said and you can see that if I operate it on psi of x the projection operator for an irrep a 1 on psi of x using it this is a operator form of it which I am writing ok. So, operators are typically denoted by. So, if you do this then you see that it will give you the plus sign for the C 2 operation this one anyway is plus 1 identity operator and so, this will be your projection operator or any wave function it will project you to a subspace which is an even function. Similarly if you do projection operator of b 1 on psi of x what will happen this has to be replaced by b 1, b 1 character table is 1 and minus 1. So, what will the thing be first element of course, is for the identity element is plus 1 ok the G operator is an identity operator. So, that will remain a psi of x the second element character for the b 1 irrep is minus 1 which wave functions ground state, second excited state, fourth excited state and all belongs to which irrep in the harmonic oscillator. This projector projects it on to a even wave function. So, even wave functions are a 1 is contains ground state, second excited state or even and similarly this means b 1 contains first excited state, third excited state, all odd excited states. So, you do see that the just the group symmetry which is a simplest group symmetry C 2 has helped you to say that the wave functions which you are going to find or the solutions which you are going to find has to be 2 one dimensional representations by one dimensions you mean that it is either going to be the a 1 is a one dimensional irrep, the projector will give you only a one dimensional state. So, it is either even if it is projecting on to an irrep a 1, if it is projecting on to an irrep b 1 it should be an odd wave function ok. So, now, let me come to one non trivial example I did this simple 1 cross 1 matrix which is not that important, but it is a warm up. So, let us take this we have been doing on the C 3 v the 2 cross 2 matrix representations which are nothing, but your rotation matrices right, rotation matrices sigma v we have written it many times. I also told you that you can do a tensor products. So, this tensor thing is not coming, but this will give you a reducible representation ok. So, I think I have a matrix example here. So, e of the this e refers to the 2 dimensional irreducible representation of C 3 v for the element C 3 ok. So, that is what that is just a rotation by 120 degrees which you have gone through many times. What is the corresponding vector space? It is just 2 dimensional vector space and we can take the basis to be the x y basis is that clear? Are you all with me? So, now, you do the tensor product of e cross e. Let us do it gamma e I am going to use the short hand notation as e 4 g I am going to work it out for C 3 in a surplus. It will act on the basis the corresponding basis let me take it as x 1 y 1 take a tensor product. So, I want to do a tensor product. So, let me write it here which I am going to denote it as e cross e this is 2 cross 2 this is 2 cross 2 this will be multiplied is going to give me a 4 cross 4 which is reducible and I am confining myself to C 3. You can do it on every element of C 3 v just for an exercise I am doing it for you here. So, what do you have to do? You have to take this 2 cross 2 matrix take the first element minus half and multiply with all the 4 elements right. So, let us write that here that is the first block. Second block will be minus root 3 by 2 with minus half minus root 3 by 2 you understand what I am doing right you all with me. So, I have written that 2 by 2 block write this this is one block this is another block I need to do the same thing here what will be the multiplication it will be this element multiplying all the 4 elements let me write this as the e ok you understand what I am saying. The last one will be a minus half with the same this is what I have simplified here and put it on the screen for you. So, that you can check it out these are the 4 cross 4 matrix representation what is the basis vector this one is x 1 y 1 let us take for this one x 1 y 1 this one as x 2 y 2 ok. So, the basis vector for this will be x 1 x 2 x 1 y 2 then y 1 x 2 y 1 y 2 is that right am I right this is a reducible vector space I want to use the projector to find out what is the irreducible subspace is that fine. Here mechanically we took b 1 times b 1 and said it is a 1 by looking at it, but technically you should be able to take the reducible representation and break it up and see how many components are there. So, you can do that here which same no no it is a same representation, but one is acting on particle one another one is acting on particle 2 or one is acting on the position space one is acting on the momentum space. So, you can take it like that you can also do it on the same particle twice that is a different thing then you will have quadratic powers that is what I wrote here like you can do c 1 c 2 c 1 squared is like on the same basis, but this is on a two different spaces that is why there. So, all those finer details can be work depending upon whether both are same space or different space right. Now, I am taking it as like let us take a this is something which you have been doing mechanically I am sure you have all been doing two spin half particles you combine and then you find the irreducible subspace. Now, I am trying to tell you that this is purely from this projection purely from multiplicity when you take the tensor product what are the irreps how many times it occurs and then the projection operator will tell you what are the subspaces this one which I have written is a reducible space. What I am going to show is that that S matrix well break it up into a piece which is x 1 x 2 plus y 1 y 2 that is one piece then you will have a another piece which belongs to A 2 and then another piece which belongs to E which are all binary basis. The binary basis is nothing, but the dot product of two vectors belongs to your trivial representation. This is like taking dot product of x 1 y 1 with x 2 y 2 and you will get it naturally. How will you get this? If you do a projection operator A 1 it has to give you this if you do a projection operator A 2 it has to give me this which is also 1 cross 1. If you do a projection operator E if you do a projection operator E the projection operator should be what should be the rank of this projection operator. It is a two dimensional projection operator or it will give you two independent basis which means it is a rank is it is a rank two projection operator and the rank two projection operator will give you these two basis. And why have I said A 1, A 2 and E? I have not really done it here, but if you try to work this out you will find it is a direct sum of A 1 plus A 2 plus E ok. I have not really worked it out, but this breaking is using that multiplicity argument you can check this out I am not done it here. So, basically I am trying to say that E cross E in the short hand notation which I am using will turn out to be A 1 plus A 2 plus E. So, this is what? This is two dimension, this is two dimension it is four dimension, this one is one dimension, this one is one dimension, this is two dimension. It breaks up into irreducible subspaces which is one cross one, one cross one under two cross one. The corresponding matrices will be block diagonal ok. The corresponding matrices will be block diagonal with first block diagonal as one cross one, the next one is one cross one, but it belongs to the irrep which is different. If it is one cross one, you can look at the character table and see that it is A 2 characters entries. And then the last one will be the conventional two dimensional irrep ok. So, this is what I am trying to show on the screen here. So, I have explicitly tried to P A 1 in the case of C 3 V group all the characters are plus one using this formula ok, but now you replace it by all the elements of C 3 V and do not worry about this. So, this is what I have written it of the screen here. Order of the group is six, the dimension of the matrix, dimension of the irrep A 1 is one cross one, one dimensionally irrep. And then the all the characters of this A 1 which is the unit representation is all one. So, you add up all the matrices I wrote only one matrix you have to write the matrices for all the elements and add it up. What you get is what you call it is a prediction operator for A 1. So, I have shown that here P A 1 turns out to be this sorry this is not right is that right yeah it is right. So, this is a 4 cross 4 matrix write the 4 cross 4 matrix sum it up and what you get are the prediction operator for A 1 again do it for using the characters of the irrep E please check this and the last one. And you can check that the rank of these matrices are these two are one and one and this one will have a rank two ok. The immediate step after this is you take an arbitrary vector just like I took x, y, z or a, b, c take a arbitrary four dimensional vector you know what is the basis state right. I am using here the i cap, j cap notation, but it is like x, y coordinate if you want to take it. So, i 1 cap, i 2 cap, j 1 cap, j 2 cap ok. So, what is happening is that if you do the prediction operator for each of the irreps on this arbitrary vector you get this answer especially for P E I get this answer ok. What does it mean? I get a linear combination for arbitrary alpha, beta, gamma, delta. So, this tells me that there are. So, this one will be according to that it will be x 1, x 2 minus y 1, y 2 and this one is x 1, y 2 minus or plus y 1, x 2. This is the basis for the E projection and for the A 1 anyway I have written and A 2 will be this is minus I have used I 1, I 2, but it is the same as the I component and the j component. The thing which I want you to do is I will put this on Moodle please check this algebra it is it is just to I have given you the concept today, but unless you work it out all the 4 cross 4 matrices and substitute and check it out you will not understand what I am saying. So, I have given you this data this also you should check. You should also check that the projection operator P A 1 when operates on an arbitrary alpha, beta, gamma, delta it gives you only this subspace whose basis is going to be given by x 1, x 2 plus y 1, y 2 or I call it as I 1, I 2 plus j 1, j 2 and you can check P A 2 will give you this and P E will give you this ok. So, what does that tell me that I can start writing binary basis, z squared is a trivial binary basis which will come from A 1 cross A 1. If you do A 1 times A 1 you will get a z squared. How are you getting x squared plus y squared? It is this one when x 1 and x 2 are taken to be the same x you get x squared plus y squared that comes from the tensor product of E cross E and it belongs to the A 1 A revenue. Similarly, you can work out all the tensor product. So, we will anyway this I have already discussed with you this also I was telling you that typically a system will have a bigger group and then you break it up into a smaller group ok. This you can try it out and there was one more assignment problem which we are supposed to do it and take a look at this. Essentially, you will have a system sudden perturbation will break it up into a lower symmetry system. Once it does whatever irrep in which the system was there it becomes a reducible representation and it will break it up into irreps of the lower group represent ok. So, this is what will happen in nature ok. So, to sum up what I am trying to tell you I just wanted to flash this for you since I have done so much. For any system with symmetry group G union H will commute with the elements of G that is the definition of saying that if suppose the Hamiltonian has a symmetry Hamiltonian will commute with every element of this C 2 group the non-trivial element is only the C 2 element Hamiltonian will commute with that. Once it commutes whenever there is a commutation relation like this in quantum mechanics you allow for you allow for state psi of x and G psi of x to have the same energy you all know that right. If you have operator A and operator B commuting you can have a simultaneous Eigen states of both operator the two operators which commute here the first operator is Hamiltonian. So, I am saying the energy Eigen states have to be degenerate in general. When will it not be degenerate if G on psi of x is proportional to psi of x that is what happens in this harmonic oscillator G on psi of x is proportional to psi of x then you will get non-degenerate Eigen state. How do I see it in this irrep language? How do I see it in this irrep language? If it was degenerate then you will have G of psi of x will be a linear combination of the set of degenerate Eigen function this is known in quantum mechanics. Now, in the irrep language look at the character table. So, when you take the character table C 3 V has two one dimensional irrep, one two dimensional irrep, phi 2 has two one dimensional irrep. Once you look at this character table the corresponding matrix representation will have to act on a vector space which will have basis states. So, xi i is what I am calling it as a basis states where I will run from one to L alpha and what is the meaning whenever there is a group symmetry in the system when I operate this on xi i it will be a linear combination of xi i. If it is a one dimensional vector space if L alpha is 1 what is the linear combination available? You do not have any other linear combination. So, gamma G of xi i and xi i will have to have same energy if the Hamiltonian had the group symmetry and gamma G of xi i will give you a linear combination of basis states of that irrep. If there is only a one dimensional irrep what is the linear combination? So, it is non degenerate. If you are looking at one dimensionally irrep xi i when you do gamma on xi i it better be only proportional to xi i you cannot get anything linear combination. The linear combination will happen only when you go to C 3 V or other groups with two dimensionally reps where the basis will be two dimensional basis or three dimensional basis depending on what group symmetry we are looking at. Is this clear? So, now with this information you take a relook at the particle in a two dimensional box that is why you start getting degeneracies when you go to two dimensional box is because of the C 4 V symmetry because C 4 V symmetry will allow for a two dimensionally irrep and then gamma G of xi i will be a linear combination that is why you start getting degenerate states for 2D harmonic oscillator or 2D particle in a box, but not for a 1D harmonic oscillator. 1D harmonic oscillator has to be non-degenerate because the irreps of the character table are only one dimensional you cannot get degenerate energy eigenstates. Is this clear? So, let me stop here.