 Now, I am getting on to one of the climax of group theory today the hydrogen atom or Kepler problem both are 1 over r potentials right. The planetary motion which you are all familiar and also the hydrogen atom problem only thing is the kappa coefficient in the Hamiltonian will be different. It will be GMM if you are looking at the center as the sun if it is circular orbit otherwise you have to put it at one of the what is that point called? Point O is called focus of an elliptic orbit no and what is P and A called Perilian and Apilion of an elliptical orbit. So, you know all these things semi major, semi minor axis this is the way the planetary motion is right. So, for such a situation whether it is planetary motion or a charge particle in a coulombic potential Hamiltonian I can compactly write it as this is like a reduced mass of a system mu and kappa will be different depending on whether it is coulombic potential or it is a Kepler problem ok. What you all know already from classical physics is that angular momentum is conserved for the system right. Angular momentum is conserved because it is a spherically symmetric potential V of r. So, for hydrogen atom you can show that the Hamiltonian for the hydrogen atom it commutes with all the components of the orbital angular momentum. So, equivalently we say that the angular momentum is conserved in classical mechanics by showing that d L by d t which is proportional to the Hamiltonian with L which is 0 implies that L is a constant of motion is that right. You would all not commutator, but a Poisson bracket in classical mechanics, but if you go to quantum mechanics what is the difference you have d by d t of expectation value of L. So, you have done it in quantum mechanics. What is this going to be proportional to commutator of L right expectation and this is going to be 0 that is why you say that the rate of change of angular momentum is 0 means this is a conserved quantity. So, L is a expectation value of L is going to be constant. So, associated with every symmetry if you have a symmetry in the system the generators of that symmetry in the context of continuous groups and group elements automatically is taken care of right. Once you put the generator, exponentiating the group generators are going to commute with the Hamiltonian there is nothing new, but in the context of discrete groups you put commutator with every group element because there there is no concept of you can have generators, but then they are not like exponentiating the generators to decide all the elements which is infinite. There are only finite elements a finite number of generators and each one has some order right. If you had a cyclic group and if I give the order of the group is 11 you know that a to the 11 is identity that kind of a constraint is not here. Here you can see the difference between what you have learnt in the first half of the semester and the second half of the semester that here we concentrate on the generators and Lie algebras even though we concentrate the group elements should also continue these properties because you do exponentiation of these generators ok. So, that is why I am going to concentrate only on the generators which are orbital angular momentum which is familiar to you. Other thing is the hydrogen atom hydrogen atom Hamiltonian which you are writing is does not have any time dependence right it is independent of time. So, we always require that this one is going to be 0 always there is no time dependence in the Hamiltonian. So, there is a this being 0 because this should be proportional to Hamiltonian right. What is the generator of the time translation? That is also Hamilton. So, it is trivial statement. So, you have you have the fact that the Hamiltonian is also a conserved quantity or the energy is conserved in these systems ok. The total energy of the system for example, the hydrogen atom problem and this being conserved is due to. So, energy is conserved due to what symmetry? Time translations this is conserved due to time translation symmetry just like L is conserved due to rotational. So, these are the conserved quantities which you have in your problem and you can associate some kind of a meaning a geometric meaning in time coordinate there is a translation some geometrical operation. Similarly, you have a rotation in your physical x, y, z space which also has a meaning ok. So, this is why the hydrogen atom has these two symmetries time translation symmetry as well as the rotational symmetry. On top of it, there is some kind of a new conserved quantity you see in these Kepler problem or the hydrogen atom problem and that is what I am showing it on the screen here. So, what you see here is that there is a Range lens vector lens vector is what we call is another conserved quantity. So, it is an another conserved quantity it does not have this kind of a geometrical way of looking at it. The way we look at here as time translation or rotation, we really do not have a geometrical way of looking at what is responsible, what symmetry it is responsible for giving this conserved quantity, but Range lens vector is conserved. dm by dd in classical mechanics is 0 you all know that. If you do not know please go back and look at Goldstein to check whatever I am saying ok. The Range lens vector is a constant of motion and classical physics it is a conserved quantity even in the hydrogen atom problem, but we do not know what is the geometrical interpretation. So, in that sense the symmetry is sometimes said that you do not only have geometrical symmetry, you can have something more other than geometrical symmetry. So, we say hydrogen atom has a dynamical symmetry. In fact, this vector will be the vector from in the directed along O to P, P is the Perylian and what happens is that this vector as it undergoes the in the planetary motion the revolution that is also moved. So, it is curious to see that that vector is a conserved quantity or a constant of motion. So, this is all is the information we have for the hydrogen atom or planetary motion. We can associate generators for these L vector is having 3 components of generators and Range lens vector is conserved. And we say that it is dynamical symmetry. Now we are going to use that there are 3 generators here, there are 3 generators here. We want to construct the algebra of these generators which is the symmetry for the hydrogen atom problem. This is a rotational symmetry, but this does not have any symmetry, but you can still say the algebra of L with L, M with M all the components, algebra mixing L and M the that will give me the algebra of generators which corresponds to conserved quantities for the hydrogen atom problem. Do you all agree? Is that right? Approach is very straight forward. So, once I have this, the next thing is I am going to exploit this symmetry algebra Lie algebra generated using L and N and I want to see whether I could get the energy spectrum. So, I want to find the energy spectrum for the hydrogen atom problem if you know that the Hamiltonian on an energy state will give you E n where E n is proportional to. So, at the exact value also you all know N squared and you went through the derivation using Schrodinger equation in spherical polar coordinates separating things and finding them to be special functions and you worked out what are the energy eigenvalues given by those differential equations satisfied by those special functions. What are the special functions? Lug of polynomial right. So, you have these and the spherical harmonics Y L M they are the solutions and then you found those energy eigenvalues. It was not like you cannot just say this is the answer, but you had to do some work. Now, what we want to do is that we want to achieve this using the algebra satisfied by L and N. This is all we are going to do. There is something more here the energy levels are not they are degenerate right. Each energy level has an N has an L which is running from 0 up to N minus 1 correct or for each L you have M which is going from minus L to plus L right. What is the degeneracy? If you include the spin it is 2 N squared if since I am doing non-relativistic quantum mechanics I cannot detect the spin I will only multiply by a 2 factor. Each level has a degeneracy and that degeneracy in the absence of spin is N square. All these information you derive by solving your spherical harmonics splitting it up and show that you have an N squared degeneracy for every level right. S shell is non-degenerate, p shell is degenerate and so on you know that right. So, now what I want to do is I want to see whether I can get this degeneracy and get this energy without going through so, to show the power of my Lie algebra. Only thing I am going to see is first given a system I find the conserved quantities and then I look at the algebra of those conserved quantities. This is a motivation clear ok. So, let me get to the slide. So, just for completeness the Rangelens vector involves p cross L is it an axial vector or a polar vector L is an axial vector polar vector because the second term is a polar vector. So, you cannot add if it was actually ok. So, this has some properties. Classically what we see is because it is p cross L you can show L dot m is 0. Similarly, you can show that L dot arc hat is also 0 right because L will be like r cross p is that right. So, the two terms of the Rangelens vector dot product is going to be 0 and the square of this Rangelens vector interestingly can be rewritten just try and work this out. You will see that it is proportional to the Hamiltonian and this term ok. So, this I will leave it you to check I will put the slide today, but please check it. This is classical, but if I go to quantum mechanical I need to make sure that the operators are Hermitian. Whenever I write a dot product x dot p I have to make sure that I add x dot p plus p dot x divided by 2 because when you do Hermitian it has to be Hermitian conjugate it has to be self conjugate right. When you have a cross product you have to do it with a relative sign negatives. This is the Hermitian operator quantum mechanical operator for the Rangelens vector ok. Again you can show that its order matters in quantum mechanics. So, L dot m equal to m dot L equal to 0 and m square when you try to write in the process of doing it you get an additional h cross square here ok. This also I will leave it you to check. This is the modification between classical and when you go to quantum because I want to solve the hydrogen atom spectrum I am going into quantum quantum mechanical operators. So, the conserved quantities will commute with Hamiltonian. The conserved quantities are angular momentum, orbital angular momentum and the Rangelens vector components. Li algebra involving them because L dot m is 0 its trivial for you to check that m i with L i is 0 ok, but you can show that m i with L j there is a non-trivial transformation. The non-trivial transformation involves the other components of the Rangelens vector. This is not enough one more you have to see is amongst the Rangelens vector components what is the algebra. So, please use the quantum mechanical form of the Rangelens operator to check that the algebra of the conserved quantities which are generators of the algebra they are closed ok. Is there a sub algebra L i L j I am not shown L i L j is a sub algebra on top of it you will have these algebras with the Rangelens vector. Make a scaling by an appropriate constant such that this algebra will resemble like an SO 4 algebra ok. So, we did this. So, basically this constant which I have here it is not exactly a constant, but it is dependent on the Hamiltonian. You apply on energy Eigen states on both sides it will pick up an energy of a particular state ok. So, I want to make this scaling such that I can make my algebra look like a SO 4 algebra. What is that scaling? This also I think you did, but for completeness let me say. So, the last equation on the state with energy E will resemble. So, you operate it on a state with energy E and then you can scale your M prime the scale factor will be mu by 2 E to the power of half ok. So, you can see that here because you have this factor it should become the inverse so that this cancels and I want this factor to get cancelled. We do this many times whenever you have x p commutator is i h cross is familiar to you when you have an a b operator with something if you wanted to resemble like x p you rescale ok. So, that is what we are doing here ok. Now, it should ring the bell after we did this that the hydrogen atom M i prime if you identify it as L i 4 and if you identify the L hat i with the remaining spatial components i j k are all 1 2 and 3. So, there are all the 6 generators I have compactly written using the L i 4 and L l j k. What does this tell us? I told you that SO 4 the 4 refers to this 4 spatial dimension and abstract 4 dimensional space where we do rotations which gives you generators which are going to belong to SO 4 algebra. Now, I kind of put in a fictitious 4th coordinate to account for the Runge-Lenz vector. So, Runge-Lenz vector mimics as if the hydrogen atom problem can be treated like a 4 dimensional space it is not exactly 4 dimensional space hydrogen atom is only in a 3 dimensional space ok. The potential is 1 over r where r is x square plus y square plus z square, but you are introducing this fictitious coordinate just to mimic this algebra of SO 4 as I have shown in the slide. So, that is why it is called fictitious coordinate omega. So, that the rotation in a 4 dimensional space any arbitrary rotation is exponentiating the generators. I use L i j here i n j will run from 1 to 4 there are 6 generators anti symmetric clear. So, now, we want to exploit the symmetry where the 4th component gives me the Runge-Lenz vector component up to scaling. I want to exploit this SO 4 symmetry to see whether we can obtain the energy eigenvalues of this type and degeneracy of this. I want to achieve that. Now, I put in the setting I have showed you how the algebra of Runge-Lenz vector and angular momentum with Runge-Lenz vector with an appropriate scaling is exactly like an SO 4 algebra that I have already convinced. Now, I am going to use that SO 4 algebra and see whether I can extract information about the energy spectrum. The reason why you can extract is because the rescaled Runge-Lenz vector has a 1 over 2 e. The e is showing up there that is why you are going to use that fact. I can find the energy from information on m prime and L vector L prime the L angular momentum orbital angle because there is a 1 over 2 e in the scaling fact. So, just remember that it is not that it has got in that scaling to get you to the SO 4 algebra where the scaling has an information about the energy eigenvalue of a state. Is that right? It is clear?