 I have just given you a comparison between SU 2 and SU 3. So, SU 2 had 3 generators one of them is diagonal, SU 3 will have 8 generators the lowest non-trivial dimensions of those generators has to be 3 cross 3. Inside that the SU 2 algebra should sit in. So, you still have one diagonal generator which is coming from the SU 2, because this part is sitting inside that and you also have one more which you can construct. So, there are two diagonal generators for SU 3 I did find out lambda 8 I just said I want to construct with integers put on it real numbers and in such a way that the traces 0 and Hermitian what does the multiple basis can do? You should not increase just arbitrarily high numbers I want the numbers to also be low you are saying I will put 150 yeah not that way you have to put in with lower number of integers to start with and see whether you can construct with lower integers I do not think you can do any other thing can you try it out overall negative sign is not a new one. Relative numbers if you can put something else, but they have to be integers because they are some things which is capturing for me some kind of quantum numbers in some fundamental units and I want them to be integer. If you put in this constraint and you do not want to the way you are saying as I will put 500, 500 is minus something right, but then you can still see that there will be just a 500 scaling factor out of it. Scaling factor is unimportant what you have to get is a non-trivial. So, you have only two diagonal generators in that sense and the number of diagonal generators for the Lie Algebra is what we call it as a, this is what we call it as a rank of the Lie Algebras ok, rank of the group is also another way of saying. So, if I say a rank of the group is 2 then you should know that the Lie Algebras should have two diagonal generators. What does that also mean? If I write a state in the case of SU 3 just like I wrote it for SU 2, I should be able to write the eigenvalue corresponding to lambda 3, let me call it as M 3, eigenvalue corresponding to lambda 8 ok. And you will have some analog of the Casimir operator, let me not try to say what it is, but definitely the magnetic quantum number will have two values depending on lambda 3 acts it will give you M 3 right, lambda 3 will give you M 3 times M 3 M 8, lambda 8 will give you M 8 times M 3 M 3, do you agree? This is the slight variant it starts happening once you go to the states in the case of SU 3. So, many times this M 3 and M 8 together they write this as a is denoted as a M vector. So, it is M 3 and M 8 are the two components of the M vector ok. So, if you operate lambda 3 on the M vector it pulls out the first component, if lambda 8 acts on the M vector it is a two dimensional vector you get the second component, this is a short hand notation of vector and this is what we call it as a weight vector is called as a weight vector. What happens in SU 2? SU 2 it is just a one component vector because it is only picking up one of the diagonal generator it has only one diagonal generator. So, that is one component vector, if you had two diagonal generator the weight vector will be two components, if there are three diagonal generators you will have weight vectors to be three components. So, this is the formal notation the first non-trivial group is SU 3 where you will start seeing that you can represent the analog of your magnetic quantum number by your magnetic vector where the first component will be like your magnetic quantum number because that is the lambda 3 eigenvalue. The second component will be like the new diagonal generator for SU 3 ok. Diagonal generator because the diagonal generators commutative bracket will be 0 right because they are diagonal you can show that lambda 3 lambda 8 is 0. So, which means you can write a state to be a simultaneous eigenstates of both lambda 3 and lambda 8 and I am compactly trying to write the lambda 3 lambda 8 eigenvalues as a two component vector ok. So, this is just a two component vector weight vector belonging to SU 2 ok belongs to SU 2 the two components. So, let me come to the things which I am saying on the slide here. So, number of diagonal generators is called rank SU 2 has rank 1 Z component of J is diagonal and hence we can write the simultaneous eigenstates of Casimir operator and J z in the case of SU 2. There is an analog of Casimir operators, but let me not get into it for SU 3 again you can construct bilinears here also. You can take lambda 1 square lambda 2 square up to lambda 8 square and you can show it to be commuting with lambda 3 and lambda 8, but I am just going to confine myself to the magnetic one ok. So, recall your SU 2 algebra where for comfort you define raising and lowering operators which is Hermitian conjugates of each other. So, if a J 1 is Hermitian, J 2 is Hermitian, but J 1 plus i J 2 is not Hermitian, but J 1 plus i J 2 the Hermitian conjugate of it is J 1 minus i J 2 these things you know and we could rewrite the same algebra which I wrote in the beginning for J x J y J z right. You can rewrite it in terms of J plus J minus and you know this is a closed algebra again all this equations together. There is some slight probably a different in notation from quantum mechanics there is a 1 over root 2 here some books follow without a root 2, but then you will get a twice J 3 and so on. So, that is a matter of normalization ok. This is the normalization I am following and J 3 anyway I have said that J 3 will give you I am suppressing the h cross I am putting h cross to be 1. So, this gives you some m which is some value which could be half odd integers in the case of SU 2 right. So, spin j spin half it is half if it is spin 1 it is 1. So, it could be integers or half integers and then the ladder operation will take you from that is why the definition of a ladder operation you can prove all these things using this algebra. I am sure you would have done it as a quantum mechanics course here I am just trying to give you the final result which you have learnt. Basically, if you take this magnetic quantum number m the plus raising operator will take it to m plus 1 minus will take the magnetic quantum number m to m minus 2 ok. So, there are operators which helps you to go from one state with magnetic quantum number m to m plus 1 or m minus 2 and then this coefficients also can be determined purely from this algebra and they turn out to be related to J minus m J plus m ok. So, this is something which you all know not when you derive this. My only requirement is that I want you to understand the fact that you had a diagonal generator and then the remaining generators you made it into a complex conjugates of each other ok which are off diagonal, but you try to make it into a complex conjugates. So, this is the theme which is very important in root k ok. So, let me try and say for the SU 3 algebra whatever I said for SU 2 there are 8 generators the explicit form you do not need to memorize you can write it down on a sheet you know to see that lambda 1, lambda 2, lambda 3 is diagonal and similarly lambda 8 is diagonal. These coefficients are you know put on purpose to take care of some of the quantum numbers which are seen experimented ok just like I wrote for the poly matrices half h cross just to take care that the experimental value of the J z quantum number is half h cross or minus half h cross. Similarly these coefficients come because of some experimental detail, but as of now even if you keep this to be you know some kind of a constant these are the 8 matrices where lambda 8 and lambda 3 are diagonal matrices. So, the rank of the SU 3 algebra is 2. So, I have confined to SU 2 and SU 3, but whatever I am saying for SU 2 and SU 3 goes systematically to SU 4, SU 5 and so on. I hope you appreciate that fact. So, the SU 4 you will have how many generators somebody 15 generators ok you can check it out 1, 5 generators 15 of them and then how many diagonal will be 3 of them. So, you will always have for any SU n n square minus 1 generator and you will have n minus 1 diagonal generator. So, the rank of the SU n will be n minus 1 ok. So, just like I wrote the up spin and the down spin ok. So, I am going to do this here for you. The lower dimension which I am going to write is called sometimes as a defining representations of fundamental representation. So, let me put both the cases here SU 2, SU 3 you can have states which is up spin which you denote it as 1, 0 and down spin which you denote it as 0, 1 and this is also equal to m equal to half I am suppressing h cross this is equal to m equal to minus half this is SU 3. SU 3 you will have m 1 vector which is which I am going to denote it as what is m 1 vector the two components of the m 1 it has two components. First component will be the eigenvalue of lambda 3, second component will be the eigenvalue of lambda 8 ok. Here this m was an eigenvalue of j z alone or sigma z. Now, it will be that it is a simultaneous eigenvalue of lambda 3 and lambda 8 clear. So, m 1, m 2 this is what I am calling it as mu 1 in the slide I am calling it as mu 1. Let me just follow the nodes which I have mu 1, mu 2 and mu 3 ok. Now, try to find operating your lambda 3 by 2 on mu 1 do that matrix what will that give you are you all with me mu 1 denotes 1, 0, 0. So, the mu 1 on the mu 1 if you operate this I am going to use lambda 3 by 2. So, you will get a half just a convention this is because of SU 2 sits in that and I am looking at the fundamental representation which should also have the same half eigenvalue that is the reason. In fact, all the generators I will scale it like that this is just a overall scale should not really matter. What is this going to be lambda 8 I wrote somewhere, but then there was also this convention of putting a normalization here let us follow the normalization. So, tell me what happens here that will give you. So, what is the mu 1 vector explicitly is that ok. So, this is nothing, but half it is half root 3 is what I get right, but this is just a matter of normalization we fix the normalization and we will get going. What I am trying to give you the fact is that formally I am writing it as a weight vector they express it two components of the weight vectors are the eigenvalues of lambda 3 and lambda 8 up to some normalization. If you use that then you define your mu 1 vector to be this ok. So, this is all I am trying to get any questions on this it is fine. So, one way of seeing is that if you square this you do get 1 plus 1 plus 4 which is 6 right. So, I have to put a root 6 and then I also have another half coming up. So, there is some reason which is put in we will come to it or another way of saying is I want to get my proton charge to be plus 1. You understand what I am saying if I want the proton or an electron, electron of course is fundamental it is not going to be a composite which is going to be by tensor product. If I want my proton charge to be convention with my experimental evidence, I need to fiddle around with this normalization constant and Gell-Mann has put in this ok. So, I am not doing anything ok. This is still ok. So, the three basis states of the lowest non-trivial representation of SU 3 algebra which is called also as a defining representations ok. All these things which I am writing are sometimes called in the literature as fundamental or defining. So, fundamental or defining that is the lowest non-trivial representation for SU 2 it is 2 cross 2 SU 3 it is 3 cross ok. So, let me just try to say what I am trying to put in here. This lambda is analog of your quadratic Casimir will come to it at some point. Mu 1 is bold phase here which refers to two component vector in this context and explicitly the corresponding basis state in the three dimensional vector space which I take is just the usual 1 1 0 0 0 1 0 and 0 0 1. Corresponding to this you try and find out what are the Eigen values of lambda 3 by 2 and lambda 8 by 2 and you can show that the mu 1 vector is nothing, but half thumb or root 3 by 6 which I have explained it now, but you can show mu 2 will be minus half a minus half is coming because of this and root 3 by 6 and the last one mu 3 has 0 as the first component and the second component is 0. So, this is minus 2 by 3 by half. So, it is minus 1 by root 3. So, that is what we get as minus root 3 by 3. So, these are the analog of your magnetic quantum number in the case of spin half particle the magnetic quantum number there are two values minus half or plus half or my down state and up state. Now, you have three fundamental states ok. There are three fundamental states and this is what he attributed to saying that nature has three fundamental quarks. He called it as u quark just like this is called as up spin and down spin. He called it as u quark which is also up quark, down quark and strange quark each one is a is that clear? So, there are three fundamental states of SU 3 which is attributed to and the corresponding values are going to give you something which is physical will come to it, but there are two diagonal elements. So, you will have two eigenvalues which is put together which is called as a weight weight ok. So, I have drawn it also for as a diagram the three fundamental states this blob the circle is what we call it as a up quark state another one is called as a down quark. Basically it is a coordinate in the weight vector space ok. Weight vectors has two components that is why I am drawing a two dimensional plane. If it was SU 2 what is the situation? SU 2 does not have the second one. So, this one is not there and only half and minus half the second component will not be there. Is this clear? So, it is a weight vector concept clear that you as you increase the number of diagonal generators that weight weight weight vector diagram will become three dimensional four dimensional so on, but at least algebraically I can write down the plotting I could do this way definitely for SU 3 in a two dimensional plane. The axis are the H 1 eigenvalues and the H 2 eigenvalues and H 1 and H 2 I am calling lambda 3 by 2 as H 1 and lambda 8 by 2 I am calling it as H 2 ok this is the notation yeah. So, see the thing is most of your protons and neutrons were conventional then they started seeing new particles and they wanted to give some name to it and then they said oh it will involve some new exotic particles and they probably called it. So, that brings me to a summary of what I want to say for the SU 3 just like you had J plus and J minus, J plus and J minus the role is what if I do a J minus on this the M quantum number shifts by one unit plus or minus one unit depending on whether I do J plus or J minus you all know that. So, I am going to call that as some kind of a shift which is just a single number one and J plus and J minus I will call it as plus or minus on this is a single number in the case of SU. In a case of SU 3 what should J plus J minus do? It should take from here to here or here to here and you know that this is a weight vector I would like to write that as a plus or minus corresponding number which is a two component vector or in other words when this acts on mu 1. So, let us do that. So, if you have J plus minus on let me do a J minus J minus alpha on M will give you M minus alpha is that clear? Now, I am going to say that E minus alpha on mu 1 is going to give me a single number is promoted to a two component vector for SU 3 and then these operations there we just wrote J plus minus because it was just increasing or decreasing by one we did not even need to put up last one, but in general here it could be a non trivial vector which will take you from mu 1 to mu 1 minus alpha and this should be some definite one of these three states because J minus on this will give you this. Similarly, J minus on that should give you one of those states I will come to it, but this is the formal ladder operator notation where the weight vector decreases by some unit and what is this vector we need to figure it out two component. How many alphas are there is what? There are remaining how many generators are there? Out of 8 lambda 3 and lambda 8 are diagonal, the other generators are 6 of them are there. Now, I have to make just like I take J 1 and J 2 which was remaining I made J 1 plus or minus I J 2. Now, out of the 6 I have to take lambda 1 plus or minus I lambda 2, lambda 4 plus or minus I lambda 5, then you will have lambda 6 plus or minus I lambda 7. So, there will be three such ladder operators going only in this direction, clear? You can only go in this direction. If you are here with plus half the ladder operation J minus takes you to minus half, but now SU 3 you saw that there are three states. You can have one ladder operation this way which is the conventional SU 2 1 which takes you from half to minus half. So, you can see that this vector alpha has to be 1 comma 0, understand? So, this decreases by 1, but the other one remains the same. The root 3 by 6 there could be another ladder operation which is going this way which is your alpha let me call it as alpha 1 this is alpha 2 and another one which can go this way which is alpha 3. So, one of them here is going to be the corresponding E plus or minus alpha 1 is nothing, but lambda 1 plus or minus I lambda 2. This will turn out to be E plus or minus alpha 2 will turn out to be proportional to lambda 3 sorry lambda 3 is not there lambda 4 plus or minus I lambda 5 and one more which is the does not involve lambda 3 and lambda 8, but the remaining 6 generators will form Hermitian concept and it is very beautiful seeing it in that. Is this clear? We will come to it I will repeat it again, but I am not going to redo the lambda 2 and lambda 3 what I did here, but do not forget this there is one more week next Thursday only, but this is what we will continue from. How to do this and then generate for a generally algebra how people look at these things. So, these are also in some sense some kind of a vector and this vector generates for you to go from one weight vector to another weight vector. So, the one which generates for that is called as a root vector. So, this is what we call it as a root vector. I explain today weight vector. I am also saying that to go from one weight to another the ladder operation or this Hermitian conjugate operators which are raising and lowering operators can be constructed for SU 3 and there will be 3 of them and they are going to be given by these root vector. So, I have to give you what is the root vector, what is the weight vector and then you can play around in doing all the matrix representations and so on.