 So, get back to the slide here. So, if you see the slide, the SO 4 generators, you can try to rewrite a linear combination with the rescale Runge-Lenz vector with L such that you see the I is exactly like an SU 2 algebra. There is another SU 2 algebra with the k and these two do not talk to each other. What do I mean by do not talk to each other? Any of the components of I with any of the components of k, the commutator is 0. What such a thing we call it in discrete groups? We call it as like a direct product group. If you have G 1 and G 2 elements of G 1 and G 2 commute order really does not matter, you call it as a direct product group. Similarly, the SO 4 algebra involves two subalgebras which are SU 2 and SU 2. The corresponding group SO 4 will be a direct product of two SU 2 groups. I think one of the assignment problems had to do with this. If you have done the assignments, then you will appreciate what I am saying. Now, tell me if I have given this information that Hamiltonian commutes with I, Hamiltonian commutes with k where I has I x I y I z, k has k x k y k z. Now, what do I write for a state? State should be a simultaneous eigenstate of Hamiltonian I z and k z. Of course, you can also say because SU 2 you know that it should also be a eigenstate of I squared and k squared right. Or you can write the states of SU 2 as highest weight comma the weight vectors right this is what we do ok. So, hence we can write the states which are simultaneous eigenstates of Hamiltonian because it commutes with I hat and k hat the Hamiltonian. You can write it as a simultaneous eigenstate of Hamiltonian I z and k z. Is everything clear to you? So, that is what I have done here. I have written the corresponding highest weight I z, highest weight k z and it should be a simultaneous eigenstate of Hamiltonian also. So, I put the energy on the states. What is the meaning of this? I am sure somebody can help me out. Hamiltonian on E I I z k k z will give you E k z and if you take I z operator on E k z then it will be I z h cross. I am putting h cross because I am using them even in the definition of the Runge-Lenz vector k z. Similarly for k z. So, I satisfy this algebra right I wrote you also already I x I y is I h cross I z k x k y is I h cross k z and you also have I x with k y or any of the components. In fact, I will call it as I and k j this is going to be 0. That is why this is s u to sub algebra s u to sub algebra then the total algebra of s. So, 4 how did we write i and k? We wrote i and k as think up to some constant it is l plus m prime. m prime is the rescaled Runge-Lenz vector up to constants and k s m prime ok. So, these are the SO 4 generators and you took a linear combination of the SO 4 generators. So, that you get the familiar s u to algebra which I know how to work out and then I play around only with the familiar s u to algebras. So, if you want to determine I there are ways of defining I is such that any of those I plus I minus which you generate with I x plus I y if I z is I then it will be 0 I plus will be 0. Similarly, if it is k z is k I plus on that state will be 0 that is one way of determining what is k and what is I in the l algebra language, but you could also define a Casimir. What is the Casimir? I define that also a Casimir operator is one which commutes with all the generators quadratic Casimir. So, you can define a Casimir which is I dot I what is that I x squared I y squared I z squared. You can define one Casimir like this, you can define another Casimir which is k dot k and if you take these Casimir's and operate it on the same states what will you get? Someone what will you get? I into I plus 1 h cross squared it is an eigenstate. Casimir means it commutes with it is a linear combination of quadratic bilinears in the generators it commutes with all the generators. In fact, this commutes with each of the components of I x I y I z it also commutes with the Hamiltonian because of that it also commutes with k x k y k z because this commutator is 0. So, that is why it is also you can write an eigenvalue equation involving this. If you did it with C 2 what will happen? C 2 will give you k into k plus 1 h cross squared, clear. So, these informations we are going to exploit to do further evaluation and determining what is this energy. So, coming back to the slide I am going to take an interesting combination of I dot I which was one of the quadratic Casimir, k dot k is one of the quadratic Casimir. I am going to take a linear combination with the plus sign I am going to call it as C 1 that is also going to be an eigenstate of this operator. And another one C 2 which is difference of the two quadratic Casimir. I will leave it you to check that C 2 will turn out to be 0. If you substitute explicitly the values of I in terms of l plus m prime and k as l minus m prime, C 2 will turn out to be a 0 operator. What does that mean? If C 2 is a 0 operator, null operator, C 1 eigenvalue is this, C 2 eigenvalue is this. I want to define a C 2 which is C 1 minus C 2. If this is equal to 0, the physics of the system shows that this is 0 ok. The hydrogen atom spectrum or hydrogen atom operators Rangel and Srescale explicitly if you substitute the C 1 and C 2, you find that the difference is 0. This is a constraint from the hydrogen atom system. Just looking at SU 2 cross SU 2, you would not have this constraint. If you fall back on the Rangel Lenz vector, you see that this is an additional information I get. This information forces, this is the information pertinent to the hydrogen atom problem which forces that my states have to be E, I, I z, k equal to I, k z. This is a subspace. The system gets into a subspace where this eigenvalue should be same as this eigenvalue which forces that the highest weight has to be I. These two are two SU 2 algebras, but the eigenvalue of the second SU 2 algebra highest weight should be same as the highest weight of the first SU 2 algebra because this constraint comes from my system under investigation. Hydrogen atom where the Rangel Lenz vector scale satisfies this condition. So, looking at this, you can say that this I has to be 0 ok. So, that is a subspace. Is this clear? So, once I put this condition, then I can write the C 1. C 1 is what will this be? It will add up now. C 1 will give you I into I plus 1 h cross square. C 2 will also give you k into k plus 1 h cross square, but k is I. So, you will get this to be twice I into I plus 1 h cross square on because C 1 will give you I into I plus 1 h cross square. C 2 will give you k into k plus 1 h cross square, but k into k plus 1 h cross square because of this condition which came out from this input will force you and so you have twice I into I plus 1 h cross square ok. So, this C 1 getting back to the slide, the C 1 which I have written C subscript 1 that turns out to be write out in terms of we wrote it as I dot I plus k dot k, but rewrite it in terms of angular momentum, orbital angular momentum and Rangel Lenz vector reskid. Please do that exercise. It turns out that it is going to be L squared plus M prime squared and you can write the quantum mechanical form of M squared to see the expression in terms of the Hamiltonian or energy ideal. This step I am not doing it here, but please check it out. So, it is going to be the C 1 Eigen value which I wrote has to capture the energy of the state. So, this expression tells you that there is a relation which captures the energy of the state ok. So, let me get to that state on the screen. So, C 1 on I I z k, k is equal to I is 2 into I this I have already explained. Here what do you have to do? You have to equate h cross squared 2 I into I plus 1 has to be equated to the right hand side of this operator which operates on the state. So, the corresponding Hamiltonian will give me energy. So, the Eigen value which you find here should be related to this expression. You all agree? Is that clear? C 1 operator the operator form in terms of the Hamiltonian is this. So, you have another Eigen value for it and that Eigen value must be same as this expression. From here you can determine for a state with I highest weight here, you can determine the energy Eigen value. Please simplify this. Please simplify and see what you get. It will turn out to be your expected answer provided you call this 2 I plus 1 as principal quantum number n. So, this simplify please simplify this and see that you will end up getting this. I did not do anything other than exploiting Rangel N's vector is a conserved quantity and tried to recast. So, that it looks like an SO 4 algebra and SO 4 can be seen as 2 SU 2. And then I had this additional information using the Rangel N's vector and orbital angular momentum it satisfies this condition and that forced me that I have to work with I and k are not independent k has to be equal to I. Once I have that condition I end up getting this Eigen value, but this Eigen value should be same as minus h cross square minus mu kappa square by 2 e 4 e. I am going to put from here you can determine what is e and I am putting a subscript to remember that the state on which it was operating had this I and this turns out to be proportional to minus 1 over 2 I plus 1. So, far so good. So, we are going to use the fact that C subscript 1 is also in terms of the orbital angular momentum and the scaled Rangel N's vector is related to the Hamiltonian and the corresponding Eigen value will be the energy e as I said on the board. So, that is going to give us I am just summarizing this on the slide. So, the C 1 Eigen value will be it will have Eigen value I into I plus 1 for I squared and k squared is again I equal to k. So, you have this and substituting and equating both of these Eigen values which is what I have done here. You end up getting an expression for the energy Eigen value which is what I explain now. So, clearly you see that I can be half odd integers in general because they are SU 2 quantum numbers 2 I plus 1 will always be integers. So, replace 2 I plus 1 to be N which is nothing, it is similar to your principal angular momentum, principal quantum number which you see in your hydrogen atom. So, essentially the energy Eigen value replacing 2 I plus 1 as N you get energy Eigen value to be your familiar energy level stationary states of your hydrogen atom which is proportional to 1 over N squared. Remember this N is 2 I plus 1 and N be in the hydrogen atom context we call it principal quantum number and it is integer and I could be half integers, but 2 I plus 1 will always be an integer. The next thing which we have to which I already promised you that I will do is how to find the N squared fold degenerate. How can you show that each level is N squared degenerate and for that you need to see the physical orbital angular momentum. The orbital angular momentum as I was explaining here on the board if I hat and K hat is L plus M prime and L minus M prime the physical orbital angular momentum is actually a linear combination of I and K. So, this is going to be like an angular momentum addition of 2 angular momentum so you will end up having the Eigen values. So, these L is it will be on the Eigen states are L M with Eigen values M h cross where M will take values from minus L to plus L and what are the all values allowed for L is given by the angular momentum addition. So, it will be from I minus K to I plus K so you will have because I is equal to K is the constraint which comes from this condition. The L values take value from 2 I 2 I minus 1 up to 0. So, you do see that if you replace 2 I plus 1 as N then L goes from 0 1 to N minus 1 which is what you physically see the azimuthal quantum number given a principal quantum number N you will have azimuthal quantum number taking 0 1 up to N minus 1 which comes naturally from the angular momentum addition which allows the range of L to be 2 I 2 I minus 1 up to 0 ok. And from there you can determine the degeneracy to be N square. I am explaining this again on the slide that the orbital angular momentum L is given by addition of 2 SU 2 generators I and K. The Eigen states of L is it we denote it by L and L is it and the range of L will be from I plus K to I minus K with the condition which we have imposed that I and K are not independent, but I is equal to K that forces the L which is the orbital angular momentum quantum number to run from can be 2 I or 2 I minus 1 up to 0 replacing it in terms of N which is 2 I plus 1 we get L to be 0 1 up till N minus 1 directly giving us the degeneracy to be similar to L going from 0 to N minus 1 each of those magnetic quantum numbers will be 12 plus 1 dimension. So, if you sum it up you will end up getting N square which is the degeneracy of an energy level which is whose energy is 1 over N square proportional to 1 over N. So, I hope I have given you a feel of how group theory is actually useful to get the energy spectrum in a neat way exploiting the energy spectrum. The conserved quantities as generators and trying to write a algebra and find out the constraints and determine the energy Eigen values this is what has happened in the hydrogen atom spectrum, but the tool which I have taught can be used for any abstract system or any complex system where you may find a set of conserved quantities you can try to find the algebra for those generators and see whether they resemble some known Lie Algebras and you can exploit these power and get certain quantities in an elegant fashion ok. I hope you enjoyed the course and there are certain things like wave function in the hydrogen atom cannot be explicitly determined, but definitely the energy Eigen values you do not need to go through those complex equations to solve, but use the symmetries including the dynamical symmetry due to Rangel's vector where we have written it as SO 4 in getting the energies of the stationary states to be proportional to minus 1 over N square in a neat fashion including the degeneracies which is N square ok. Thank you.