 So, let me continue today showing a little easier method one of you already pointed out that we do not need to write those matrices to find which irreps the molecular vibrations belong to ok. So, I thought let me just brief you on that before we get on to the groups ok. So, whatever we did I hope you remember we went through an elaborate exercise for a non-linear triatomic molecule where two of the masses were identical. One of the mass was capital M and the other two was small m it is exactly like the water molecule and to find the reducible representation for all molecular motions including translation, rotations everything you put in 3 degrees of freedom for every atom in that molecule clear. So, all these atoms will have three Cartesian coordinates we need to work out the reducible representation or characters for the reducible representation which can account for all the molecular motions ok. So, to do that what we can do is we did this tensor product of states and so on we do not need to really do that what we have to remember is if you do an identity operation on this molecule each atom remains in its own place right. So, each atom remains in its own place. So, you will have character for each atom you know if you add up all the 9 degrees of freedom you will get the character associated with the identity element to be 9 ok. So, that is what you will get, but if you wanted to rotation by an angle theta I am taking theta to be arbitrary which will be different for different groups, but discrete value ok. So, if you take rotation by theta you know the x, y and z coordinate for every atom will undergo this transformation. So, if you take this and I try to find the trace if you write this matrix which does this transformation the trace will give you 2 cos theta right and there will be a 1 for the z a. So, it is a exactly 1 plus 2 cos theta for an atom in the molecule if there are n c atoms which are not displaced ok. So, in this particular example when you do a C 2 rotation what happens these 2 atoms get displaced the number of atoms which do not get displaced is just 1. So, the n c will be 1 in that case ok. So, if n c atoms are not displaced under the rotation then the contribution to the character will be n c multiplied by the character which is the trace of the 3 by 3 matrix which is 1 plus 2 cos theta. Is this clear to you? So, 1 oxygen atom is unmoved under C 2 operation and you are doing C 2, C 2 means theta will be pi. So, cos pi is minus 1 minus 2 plus 1 will be minus 1 and n c is 1 is that clear. So, that will be the character identity operation of course, is identity matrix for each of the atoms. So, the trace of that matrix will be 9. C 2 operation for a general theta operation C n where theta is 2 pi by n this is the formula where n c is the number of atoms which are not displaced under such a operation. So, using that because n c is 1 here and C 2 is only a 180 degree rotation we get the character associated with the C 2 operation for this non-linear molecule is minus 1 ok. I am not writing matrix representation I am just directly writing the molecular motion of this molecule what should be the character for a reducible representation I am not even writing the matrix formula is that clear. So, let us do the same thing for the other two elements also. If n sigma atoms are in the reflection plane then character is just n sigma. So, suppose you take this to be the x z plane which we were discussing in the lecture all the three atoms are in the x z plane. So, if you do a reflection in the x z plane three atoms are there. So, n sigma is 3. So, the character will be 3 ok. You can prove it rigorously, but I am giving you a final answer. So, therefore, the character for the molecular motion reducible representation is going to be 3 for reflection in the x z plane. What will happen for the y z plane can somebody tell me 1 oxygen is in the y z plane. So, the character for reflection in the y z plane is again 1 ok. So, you write this out once you do this the rest of the steps follows. Given a reducible representation of a molecule with 3 n atoms you know how to do the decomposition into irreducible representation when you find that there are totally 9 modes of motion right 3 one dimensional irrep 1 one dimensional irrep 3 plus 1 is 4 and the everything is one dimensional why it is a C 2 v symmetry the molecule is a C 2 v symmetry it is a non it is an abelian group. So, all the irreps have to be one dimensional and all the motions should add up to give you the total degree of freedom. So, 3 into 3 is 9 degrees of freedom and this 9 breaks up into linear combination of this breaking I have not worked it out here, but we have done enough exercise problems that given a reducible representation with this character you should be able to work out the irreps into which it can decompose ok. So, this part I am not doing, but you need to do this ok. Any questions 9 4. Because see finally, I am going to take these as the degrees of freedom right x y z for atom A x y z for atom B x y z for atom C. So, when you do an identity operation x y z will go to x y z ok. So, 3 times 3 which will be 9 is that right. So, you can put even here theta to be 0 if you want in this expression which I have written you can take theta to be 0 if you do that then you will get 3 and 3 are unchanged 3 into 3 is 9 is that clear? Yeah any other question? So, once we have this we have the 9 modes of vibrations sorry non 9 modes includes translation rotations and vibrations ok. So, this splitting which I have done here in order to do this you do not need to write the matrix representation to do this. You can do this argument to actually figure out how many vibrational modes are going to be one dimensional that means, that frequency will occur only once. If suppose you had another group like C 3 B the way I have given it in the assignment if it belongs to a 2 dimensional irrep then you can looking at it you can say that experimentally if they start looking at the spectra they will see degenerate frequency they will see 2 spikes with the same frequency ok. So, these are things which you can guess by just working out with the groups is that clear? So, in this C 2 B case you have each irrep to be one dimensional. So, each frequency has to be distinct did you work it out Sachin was supposed to do the discriminant you know? Frequencies are different or? Very good. So, why is it different now you know it has to be different because it is 1 D irreps. So, none of the frequencies can be degenerate if you find the eigenvectors and 2 eigenvectors shares the same frequency. Eigenvectors. Eigenvectors are very big, not simple I see maybe you can show it to me and I will simplify it to you ok fine ok. So, what I want is this we have already done once we have broken it up into this irreps then you have to subtract out. What I did last time was I treated it as if we are in the 2 D plane. Now I have just taken it someone was saying is it important to work in 2 D. So, now today I am showing that even if you work in 3 D you will get back the same vibrational modes that is all I am trying to justify by using this shortcut way of writing the characters for the reducible representation ok. So, 3 D if you take the motion to be in 3 D you will have 3 into 3 which is 9 degrees of freedom to start with earlier when we were doing we were taking it as a 2 D and then I said 2 into 3 which is 6 degrees of freedom. So, that is the way we started with ok. So, here I am doing it as if it is a 3 3 D situation and then you have to remove the translation and the same procedure there we did only the translation along x and y we subtracted, but now you have to also subtract the translation in the z coordinate because we are doing it like a 3 D problem ok. So, associated with the z associated with translation along z you will have an a 1 which should get removed ok associated with x translation you have to remove 1 b 1 and similarly y translation will remove 1 b 2 out ok. So, essentially it becomes 2 a 1 plus a 2 plus 2 b 1 plus b 2 is that clear? This is after we removed the translation in the x y and z direction. So, now what we are left with is whatever we have got after subtracting the translation is the rotational irreps plus the vibrational irreps ok. So, now I have to remove the rotational irreps out of this then what will remain will be the vibrational modes. So, rotations you can see about z axis is a 2 about y axis is b 1 x axis is b 2 ok. So, I need to subtract them. So, once I do that I get the probably missed out there is a b 1 rotation and what happened a 2 b 1. So, the answer will be there should be 3 rotations know what happened. Louder did I miss something some factor. So, it is 1 a 2 and then the b 2 will also get removed ok b 2 will get removed and 1 b 1 will get removed ok excellent. So, what we have essentially respect of whether we looked at it as a 2 D problem or a 3 D problem the vibrational degrees of freedom should have to be 3 and the irreps which it belongs to will not change ok. We did this in a long process one was using the normal coordinates and the other process where I said let us look at it as a tensor product of 3 atoms with x y coordinates there also we got 2 a 1 plus b 1. Now, I have shown you by doing it as if it is a 3 D problem still we get the vibrational modes to belong to irreps 2 a 1 plus b 1 is that clear ok. So, this anyway I have worked it out for you the projectors and basis states in 2 D you can try to redo it for the 3 D case. Just as an exercise you can write the matrix representations for the reducible one which I have here for this case you can work out the matrix representation find the projectors, subtract out the projectors for translations you can do that elaborately here again you will start getting a similar expressions ok. It is tedious, but it is straightforward ok that is why when I did it in the class last time I just confined to as if it is constrained to move on a plane and then it became a little simpler ok. So, I am not really discussed if you had roto reflection operation ok. What happens to the contribution to the character I leave it you to try it out and see what happens when you do a roto reflection. If n s atoms are not displaced under roto reflection then what is the contribution to the character that is one I will leave it you and if you use those information for rotation which I have already discussed you could actually work out the character for the molecular motion of let us say a molecule with T D symmetry which molecule has a T D symmetry tetrahedral methane has this which we saw in the video. So, there you could also do one more stuff that you can directly. So, the way I wrote here so, I said n c atom with 1 plus 2 cos theta is the character for a reducible representation for a specific rotation which you are doing ok. What you could do is that you can also remove the translation of a molecule that will be one the character for that character for translation is 1 plus 2 cos theta. Again you can write character for rotation can take R x R y R z basis how it transforms under rotation it will transform exactly like your position coordinates rotations are proper transformations right. If you take R x R y R z the operation which you are going to do will give you R x prime R y prime R z prime right. What will that be any change or same still the same 1 plus 2 cos theta. So, you could try and remove reflections do not add to your translation or rotations ok. So, instead of the earlier one where we wrote the gamma reducible wrote this as 9 minus 1 1 was 3 and the other one was 1 right or the other way round was 3 right. You could write the vibration alone reducible representation for vibration alone. So, there you can try and remove these two. So, if you try to do that chi vibration for rotation by theta will be n c minus translation minus rotation. So, it will be n c minus 2 that will be the expression for the vibrational it will only look at the vibrational degrees of freedom ok. So, what will that be? So, for theta equal to 0 all the 3 atoms 3 minus 2 is 1 1 into 3 that will give you directly 3 and you know vibrational degrees of freedom are 3 identity matrix has to be a 3 cross 3 identity matrix it has to be 3. What about this? That had one atom right this one atom when you do a theta rotation on the H 2 O molecule oxygen, hydrogen, hydrogen when you do a C 2 rotation about z axis these two get displaced only one atom is undisplaced. So, n c has to be 1 minus 2 and then 1 plus 2 into cos 180 degrees because I am doing a C 2 operation right. So, what will that be? Minus 1. So, this is this is plus 1 final these things will not change and now you can use the character table to see how the gamma vibration and claiming that it will be twice a 1 plus v 1 ok, but I want you to check it out ok please check it. So, this is the only non trivial operation in the water molecule, but if you go to the methane molecule you have other operations you can do a rotor reflection you can do a rotation by 90 degrees and then do a reflection right. If you have seen those slides and played on this animation you will see that you do have a rotor reflection. So, this character table when you start doing I also want you to check how the under rotor reflection the characters change how the characters change you I want you to fix it just like this expression this is for rotations for rotor reflection I am going to call it as s theta I want you to find out if n s atoms are undisplaced under rotor reflection what will be this factor I will leave it you to check it out and once you know that you can again play around also rotor reflection you will have translations and rotations to be removed out of it. Rotor reflection has a rotation component in it unlike your reflection alone ok. So, you have to remove those degrees of freedom and if you write down the vibrational degrees of freedom the character for that how many will be there in methane molecule 5 atoms 5 into 3 is 15 15 minus 6 is 9 9 vibrational degrees. So, the identity element has to be 9 which is true here you can see that the n c which is this n c minus 2 will be 5 minus 2 which is 3 3 into 3 is 9 for theta equal to 0 ok. So, that is why the identity element is 9 rotation by 120 degrees cos theta will be half minus half. So, this gives you a minus 1 and a plus 1 this will become 0. Is that clear you all with me? So, that is why the character for when you consider the C 3 operation turns out to be 0. C 2 can check it out how the C 2 axis is fixed on the T D plane ok and see which atoms are getting displaced which atoms are not getting displaced ok and then you can figure it out that this is 1. For the rotor reflection I have not really worked it out for you, but I want you to first do that and then you can figure that out it is minus 1. After you have written this writing this down is very straightforward you know how to find the number of times in a reducible representation each irrepacance for that you need the character table and using the character table you can fix it, but what do we observe from this final answer. There are the vibrational degrees of freedom will have one non-degenerate frequency, two degenerate frequency, two types of three four degeneracy is that clear? This is the interpretation. You have totally nine vibrational modes, but the number of frequencies are only four distinct frequencies in this case. In the water molecule all the frequencies were distinct, but here I am not finding the frequencies, but I can at least tell an experimentalist if you are going to make these atom interact with the electromagnetic field of certain frequency you start seeing spectra and you will be able to see that spectra from there they can find what is the frequency and you can tell them that you should be able to get two fold degenerate, you should be able to get three fold degenerate. I am not doing anything other than using only the tools of great orthogonality theory and of course, the symmetry of the molecule. So, this is what you saw in that video today maybe you can go on replay it again and see in a methane molecule they say that you have three different vibrations with which has the same frequency they can superpose and you get a superpose mode of all those frequency. Is that clear? Interesting you know I just want you to appreciate the fact that this is really use this information to write the net vibrational representation in TD which I have not worked it out here, but please try it out.