 So, just to recall for you that for this particular molecule with 3 atoms, I have shown a big mass with a bigger circle and 2 identical mass with a smaller circle. These are like the bond length changing, this is you know it is an asymmetric kind of it is going here in an asymmetric way this one is going in and this one is coming out ok. These are bond length changing, this is bond angle changing and these are symmetric stretching. So, this you would have seen in the video also. So, we would like to see how to get these pictures from group theory. You know it physically for this 3 atomic you know non-linear molecule with 3 atoms very nicely have given a simple example and is it true if I give you a complex if I give it in your exam some 10 atoms will you be able to do it is the question you will all be little worried, but I am trying to simplify it and tell you now this methodology of what I am going to show in the next few slides will show you how simple and powerful the tool of character table tensor products actually plays a lot of role. It is just manipulating only matrices and you can see that you can get these diagrams drawn from those matrices ok. Still it is a suspense, but you will see it today by the end of today's lecture. I hope I can drive in the fact that normal modes of at least the non-linear 3 triatomic molecule you can understand from this picture ok. So, long method I have gone through right now. So, far whatever I have done in the last half an hour is a really a long method and can we use group theory to determine that ok. Now comes the matrix representation I have tried to use the basis. So, I am going to use the basis q 1. Since I did the long method first let me confine to this long method and I will use the basis. So, this was x 1 plus x 2 right. These things were derived by the long method. Let me for completeness write the basis for these and then worry about how to get these basis also in the next step ok. First since we went through this long method we know that q 1 is one independent degree of freedom, q 2 is another independent degree of freedom and q 3 is the third independent degree of freedom ok, q 2 is actually x 1 minus x 2 and this I think is z 1 plus z 2 ok. This was known the right hand side was known because of the long method ok. Now, in this basis I want to write the matrix representations of all the elements of the C 2 v symmetry of that molecule ok. So, C 2 v is a symmetry of the molecule of the tri-linear non-linear tri-atomic molecule. So, remember the picture I had a bigger mass here, two smaller mass here and I have used a notation z axis, x axis I think this one was x 1, z 1, this one is z 2, x 2 ok. So, this is what I am using as the degrees of freedom in the x z plane for every atom in that molecule yeah. This is capital M, this is identical that side that is the way I am looking at it I am taking it is you assume that the molecule is quite heavy and you are doing putting the axis through that molecule and do the rotation C 2 correct, correct I am assuming all those things ok. So, another way of seeing is that if I put an axis through this, if I do a 180 degree rotation these two will exchange and if I put a mirror here then these two atoms which are identical will go into each other and if I put a mirror on this plane this is like you know it remains it goes into itself, this atom goes into itself, this atom goes into itself that is also symmetry. So, C 2 v is a symmetry of such a molecule under consideration ok. So, now how do I write in this basis in terms of the q 1, q 2, q 3 basis I want to write the matrix representations of all the elements of C 2 v. So, identity element is trivial, but if you want to write the C 2 element what happens? So, what does C 2 do? C 2 takes x 1 to minus x 2, x 2 to minus x 1 in the directions which I have taken. So, what will happen to q 1 which is x 1 plus x 2 under C 2 operation q 1 will go to minus q 1, what about q 2 that is the difference that will go to somebody x 1 will go to minus x 2, x 2 will go to minus x 1. So, q 2 will go to q 2 no change, what about q 3? So, the corresponding matrix representation for this which is a 3 cross 3 matrix it better be reducible because C 2 v is an abelian group with each irrep being one dimensional you cannot get a 3 cross 3 matrix. 3 cross 3 has to be reducible for the C 2 element in this basis q 1, q 2, q 3 will be such that when it acts on q 1 it should go to minus q 1, q 2 should remain this, q 3 should remain this. Tell me what will be the matrix minus 1, 0, 0, 0, 0, 1, is that clear? So, write out for the sigma v treated to be the x z plane, sigma v is also like the identity matrix as I shown in the screen. So, identity element is this, C 2 element is this which I worked it out elaborately. Now, sigma v is the x z plane, x z plane will not change x or z coordinate. So, q 1, q 2, q 3 will remain invariant, but if you take the y z plane x coordinate is going to change sign right and you can show how the elements in q 1, q 2, q 3 are modified ok. So, this element I have not worked it out take it to be the y z plane and do the same exercise I did for C 2 and fix the in the q 1, q 2, q 3 basis write down this matrix ok. So, what have I done by the long method I had found out that q 1, q 2, q 3 essentially will contain the normal modes. Now, I have written a reducible representation for those three coordinates to find which linear combination of q 1, q 2, q 3 will actually give me the normal modes ok. Instead of doing that diagonalization of matrices I am going to now play around with matrix representations for the C 2 v group ok. So, we need to find which linear combination. In fact, we should find that q 1 is untouched right we know that also q 1 equation was completely decoupled from q 2 and q 3 right is that clear ok. So, character table yeah which matrix this matrix. So, I have to do the gamma of C 2 when x on q 1, q 2, q 3 should give me minus q 1, q 2 and q 3. This is the meaning of this q 1 goes to minus q 1, q 2, x 1 goes to minus x 2, x 2 goes to minus x 1. So, x 1 minus x 2 remains same. So, I need a matrix which when acts on q 1, q 2, q 3 should give me this. So, what is that matrix? So, you have to make this diagonal because they are all diagonal and the first element has to be minus 1, the 2, 2 element has to be plus 1 and 3, 3 element has to be plus 1. This means this is the matrix is that clear ok yeah. Any other question? Should we go ahead ok. So, back to our character table basis states and so on ok. This representation reducible I am calling it as the notation I am using on the slide is V, it is a reducible representation and you can find the characters for the identity element of course, trace of the 3 cross 3 matrix is identity and y x z plane will also be an identity matrix. So, that trace is also 3. This one trace of course, you can see that the trace turns out to be 1 right for the C 2 trace is 1 and similarly for sigma V prime which is the y z plane you can find out what is a trace ok. So, once I have this what do I have to do? I have to find out what is the number of times any of these irreps occurs in the reducible representation. Then it will tell me what are the irrep language for the vibrational degree of freedom or the normal inputs ok. So, do that exercise break that gamma V, A 1 appears twice, A 1 appears twice and you have B 1 which appears once. So, totally rank 3 is broken up into 3 rank 1 irreducible representation. So, this is the first step. The next step is what? If I want to find the actual degrees of freedom, what is what do you have to do? You can write a projector in the space in these matrices which I have write a projector and do the projection to get the basis states in the Q 1, Q 2, Q 3 ok. I leave it you to check it out and see what you get you understood. So, given this you can write a projector for the A 1 representation. This projector will involve character of A 1 times the gamma V 3 cross 3, check the rank of this matrix because V is broken up into A 1 twice I would think that the rank is 2 and find the basis states ok. Will you do that? So, P A 1 on an arbitrary vector should give me a piece which is 1 0 0, what does that mean? This is Q 1 ok. One of the vibrational modes is exactly Q 1 ok. So, Q 1 is x 1 plus x 2 clear. One of the vibrational mode is exactly Q 1 if you find this, but I am going to leave it you to check this out and if it is a rank 2 matrix should be able to find two independent basis linearly independent basis. The other basis should involve not this one, but the Q 2 and the Q 3, is that clear? So, this part I am not doing it, but please check it out and then do that a long winded diagonalization to see whether you get the same linear combination of Q 2 and Q 3.