 So, now, I am going to get to the selection rule where the operator F can be any of those physical observables operators of the physical observables F could be an electric dipole moment or F could be a magnetic dipole moment or F could be a quadrupole moment tensor. So, these are some examples of the F which I have written formally can be any of these observables not just 3, but any of you can have more observables. Essentially what you have to remember is that the observables which I am going to look at should have some leg in one of the irreps. If it does not have we cannot talk say anything about it. If it has a leg then what we want to do is we want to do suppose I want to do a transition from one ground state level to another level ok. Just recall your harmonic oscillator 1 D harmonic oscillator and also the two dimensional particle in a box just for the sake you can recall those two examples. So, this is the energy line you have a ground state and it belongs to. So, let me call it as psi ground and it belongs to which irrep a 1 irrep right and then you can have a first excited state. Let me call it as 1 here to remember it is the first excited state and that will belong to it is an odd function this is an even function right. So, this will belong to b 1 ok it belongs to the irrep b 1. So, what is the group symmetry here? Group symmetry is C 2 1 D harmonic oscillator group symmetry is C 2 and I can start and you know every level is non degenerate because it is an abelian group each element is a class by itself and you can show that if it is a 1 D basis that is no way you can get degeneracy. So, this is going to keep going and so on. Question can be asked suppose I am in the ground state and I put in an interaction given by some electric dipole moment operate kind can the particle which is in the ground state can it go to let us say this one question mark can it go to let us say this one ok or both possible something is possible something is not possible is the questions we are asking you understand the question is there a transition allowed going from a initial state to a final state due to interaction by some operator associated with it to also you have an irrep ok. This observable also has a is also associated with the irreps ok is that clear? So, this is the question can we find a transition happen ok. For example, if I take the initial state to be ground state and if I put the operator as x operator let us look at only the x component of your electric dipole moment is non-zero and if I want to look at psi final dx you all know how to do this right all of you know how to do this which one will be non-zero which one will be 0 is the question. If this was an even function and if this one is also an even function will this be 0 or no 0 right. So, you know that. So, if this belongs to A 1 if this also belongs to A 1 x operator belongs to B 1 clear. So, operator belongs to B 1 you can show that product of A 1 with B 1 will be what A 1 is unit A 1 with B 1 is again B 1 and B 1 with A 1 is also A 1 B 1 sorry. So, ultimately this object can be treated as if there is some I do not care about the numbers it is something which depends on B 1 you all agree this is what I get out in group theory using tensor products states also belong to irreducible representations operators also belong to irreducible representation. If you take the x component of the electric dipole moment operator it belongs to the B 1 irrep of C 2 and if I blindly treat it like a tensor products of irreps I end up getting something which will give me some object which belongs to B 1 could be a tertiary basis. And by odd even argument you have already argued this is odd sorry this is even this is odd this is even that integral has to be 0 is what you have argued. Now, we want to see how to argue it is 0 from group theory. So, that is what is the theme now and there the great orthogonality theorem again comes into our rescue to say that it is going to be 0. If you go to other examples like particle in a two dimensional box the group symmetry is C 4 V ok. Again you can plot as an energy ground state will be the lowest energy which should be a belonging to A 1 ok. Then you can start looking at psi 1 comma 2 psi 2 comma 1 because you can start giving degeneracy because C 4 V allows for two dimensional irreps you can have a elements of V similarly other irreps ok. I am not really writing what the irreps are, but each thing will repeat this has happening for the ground state does not mean A 1 should not happen some other like here you know the second excited state psi 2 belongs to A 1 ok. So, the repetition is allowed. So, this belongs to this the energy scale decides for you whether it is a lower energy or higher energy. So, there could be repetition another one with A 1 belonging to A 1, but a different wave function and then another two fold degenerate the two fold degenerate you should draw it close. So, that you know that both have the same energy that is the meaning of drawing those lines very close. So, this will belong to E some other wave function and so on. There will be phi 1 and phi 2 because it is two fold degenerate. Now, our claim is that we again want to see that if there is a triggering by an external operator F corresponds to some measurable observable whether the particle will undergo a transition to this or a transition to this and so on ok. So, this is the theme of selection rules I cannot get you a number I can always say whether it is 0 or non-zero. So, I can say whether this transition is allowed or forbidden this transition is allowed or forbidden clearly you know which one is allowed which one is not allowed now in this example right. If you go from let us say ground state to the first excited state it is allowed for the exact operator to go from ground state to the second excited state is forbidden from odd odd even you can argue. Now, we are going to argue from group theory and see that this is exactly what is going to happen. So, selection rule will not give me a number for the non-zero answer it will only tell you whether the transition triggered by an observable operator F is going to happen or not happen. Not happen means it is 0 happening means I cannot give you the exact value for it, but I can say this will be C is that clear I am going to do that right now I am not done that, but what I have said here is that just the odd odd even you can argue it is 0 when you have to do an integration, but now I am going to do the group theory. Before you do the group theory I am warming you up what to expect from the algebra that is why I am trying to give you this examples. So, that you are all alert to see how the great orthogonality theorem gives you the answer I want you to appreciate it ok. No, no, no. So, this turns out to be this I am going to argue that this is 0. No, no, one particular example I am taking you can do magnetic dipole transitions if you do a perturbation by an external electric field you can have the speed or e kind of term in the Hamiltonian you can start doing these kinds of transitions you understand what I am saying. So, these are the matrix elements which you will compute and explicitly you can do the computation but from group theory and group symmetry you can at least without even doing the computation you can say it is 0 or non-zero. That is the power of loop. You do not need to compute the matrix element between an initial state and a final state due to this operate. You can do a elaborate quantum mechanics like the way this one you can write the explicit wave function do the integral and show it to be 0. This is simple thing you know or even and all you have learnt but in a complex situation you may not have these information without even doing can you say it is 0 it cannot be 0 it can be 0 is itself a powerful tool and that is what I want you to appreciate from group theory yeah. Suppose this is a dx integration which is going from minus infinity to plus infinity if you do any change of that x to minus x this answer will become minus of itself that is all I am trying to say. But this is odd even argument but I am also trying to tell you how to see from group theory here I have still not proved to you that this is 0 which I will prove now yeah. Any clear what we are going to now look for? So, essentially I am saying in any complex system with some group symmetry there could be three dimensionally rep two dimensionally rep the closed lines means they are degenerate. There will be so many wave functions I am just calling it as psi 1 psi 2 psi 3 ok and you can have for f a 1 you know all these possibilities. So, this I am calling it as a different function phi 1 and so on. So, I am just trying to say a hypothetical situation where you have the energy levels and you want to look at the matrix or this left hand side due to the initial state phi 2 the initial state phi j in an irrep nu due to the operator f going to a final state psi j in the irrep mu and this matrix element is it 0 or non 0 ok. If it is 0 you say that going from the state phi j in the irrep nu to this one triggered by this operator or this interaction is not possible that is the elaborate meaning of this expression ok. So, for operator f whether this is non 0 or 0 gives allowed or forbidden transitions ok. So, now I am coming to the proof. So, suppose I want to work out what this is I want to work this out clear. So, to work it out that integral regions and all I do not need to mention let us keep it as formal integration it could be in d x it could be d x d y. If it has other internal quantum numbers you may have to sum up over those quantum numbers also, but as of now let us keep the integral as formally an integral let us not worry about it. It belongs to an irrep which is given by alpha in this particular example it belongs to b 1 ok, but I have also taught you that whenever it has a group symmetry like c 3 v or any group symmetry c 2 you are allowed to do a group operation on a state belong to an irrep. If it is one dimensional irrep it will be constant times itself. If it is two dimensional irrep it will become a linear combination of the basis which in that irrep which is the degenerate lines which I have drawn right. If it is let us take this two dimensional irrep if I do a if I take phi 1 from that if I do a group operation on phi 1 it can in principle give you a linear combination of phi 1 and phi 2 you all agree. So, it will always be a linear combinations of them that linear combination coefficient depends on what is your group operation. If you do a identity operation it will be phi 1 if you do let us say a c 2 operation then something else can happen and so on. C 3 v if you are doing you can have a linear combinations of mixing between phi 1 and phi 2 is that clear? So, that is what I am doing now. Apply a group operation and sum over all the group elements ok. So, group operation psi i alpha for a particular group element there is a matrix for each of the irreps. If it is a e dimensional irrep it will be a 2 cross 2 matrix. If it is one dimensional irrep it will just be the character depending on which irrep you are looking at it will be in general a matrix and you get a linear combinations within the same alpha irrep. It does not move from alpha to some other irrep. If it belongs to the e dimensional irrep the rotation operation given by the matrices 2 cross 2 matrices will only mix these two. It would not give you anything from here is that clear? And then I am summing up over all the elements also. Now, tell me if I just look at this thing this one you can use this great orthogonality theorem taking the other irrep to be your unit representation then you will get delta alpha a 1. Why? Because the unit representation matrix elements are all 1 right or you all find with that you remember. So, you can take this gamma alpha of G then gamma a 1 of G it is same as that summation over G. So, it is exactly same as that this is anyway 1 by 1 matrix you can show that this is going to be delta alpha a 1. So, which means alpha if it is not belonging to a 1 it is always 0 is that clear? So, that is what is the last line that this bracketed 1 piece this piece by great orthogonality theorem will be non-zero only for a 1 it will be 0 otherwise clear? Here we got b 1 which is not a 1 by doing this tensor product I said it belongs to b 1 go to the top step here psi i alpha when alpha is not equal to a 1 is 0 is that clear? So, what have I proved? I wanted to find what is psi i alpha and I proved using great orthogonality theorem it is 0 if alpha is not equal to a 1. So, that is why this combination when I do I get a b 1 and invoking that line I will say that this will be 0 because b 1 is different from a 1. So, b 1 is 0 because b 1 is different from a 1 right. Let us do for the sake of doing this instead of doing this we did this for ground state to this excited state that is 0. What about this one to this one? Can you check what happens there? Suppose I take first excited state x operator psi ground this one belongs to b 1 this one also belongs to b 1 this one belongs to a 1 a 1 times b 1 is b 1 b 1 times b 1 what is it? You just have to multiply the characters right the characters multiplied will give you that is a 1. So, this multiplication of b 1 times b 1 is a 1 a 1 times a 1 is a 1. So, essentially it will give you a piece which is a 1 and by this argument it will be non-zero. So, what have I tried to prove for you that there is a transition allowed to go from ground state to first excited state triggered by the x component of the dipole moment operator, but there is no transition possible to go from ground state to second excited state for example, triggered by the dipole moment operator explicitly you can do, but now I have invoked this argument to say that anything which is non-zero is happening when after you have done all the tensor product that alpha turns out to be a 1 it will be non-zero the alpha turns out to be not a 1 you can blindly set it to 0 you do not need to do any computation. No, no nothing to a 1 is a trivial representation whose character is always 1 you can always insert it any time. Yeah, if you take b 1 then that expression which I have written is same as this expression only when I put a 1 because these are just 1. I am just inserting see this expression which I have it on the screen did not have that gamma a 1 and I am saying that this expression is same as multiplying with gamma a 1 because gamma a 1 does not do anything to it. It is like inserting something just for the for convenience, but by inserting that I can do this argument of great orthogonality theorem to argue it is 0. Is that clear? So, also one of your problems which I gave it in your exam. Now, second which expression this is together say I am just saying apply the group operation and after that you do a summation over all the elements and average it over it that is all. So, I am not doing anything more than that any regular representation can be written as a direct sum of the yes if it is regular representation I am not doing any regular or anything I am just looking at a state which belongs to one irrep. I am just doing a group operation if it belongs to a degenerate irrep then there is a group operation will actually mix them and I am just trying to see what happens under this operation that is all. Yeah, any other question? See both sides you can put that summation it is just that 1 over g will make it 1 because it is not really acting on that ok. Is this clear? So, now I am just trying to add it to the system I am just summarizing whatever I said now with these two examples. Yeah, in this particular harmonic oscillator x belongs to the basis here will be z and this will be x because I am looking at x going to minus x the x 1 will change sign I am doing a 1 D harmonic oscillator. So, I am just I know that this belongs to the primary basis belongs like this ok.