 So, now I am doing SU 2. So, I take this combination this anyway we did it is anyway associative it does not really matter. Let us take the bound state of 3 spin half particles. So, in dimensions it is 2, this is 2 dimensional, this is 2 dimensional and then another 2 dimensional. So, this part we have already done there the first bracket. The first bracket turns out to be this and then you take it with a box because I am going to look at a composite made of 3 fundamental objects with hindsight that it should not give me anything other than spin 3 by 2 or spin half that is the experimental require ok. So, now you can redo this. So, you can combine with this box a single box along with this. Then this box can go below not violating any symmetry. So, let me write that that is from the first box multiplying this one. Similarly, I should do it with this the box can go below and it can also be can just attach itself without violating the symmetry. You do not violate any of the symmetries by putting this below and do it. Because you are doing SU 2 spin half particles the group is SU 2 the fundamental objects. If you put an up spin here you have to put a down spin then you cannot repeat anything because this is totally anti symmetric. So, that is why I said always a vertical diagram with more than two boxes is not allowed in SU 2. So, this is not allowed in SU 2 such a state does not exist in SU 2. So, what have we got? I have got this, this and this. This vertical 2 box is like the singlet or a trivial representation. It does not do anything the ladder operation any group operation will remain keep the state itself. So, this can be removed. This piece is like attaching multiplying all your results with identity or multiplying all your characters with A 1 representation. E cross A 1 is E, E cross A 2 is E. So, it is like an A 1 representation or a trivial singlet representation and it is not going to harm your you can just forget about it. It is like identity you can remove it. So, this piece can be removed and I can redraw these diagrams as if it is this is for this and similarly here. So, I have shown you that a composite bound state involving three spin half particles finally, for the SU 2 can be redrawn as this is true in general for SU 2 it will be. So, that you can say it to be for SU 3 if you want. Single boxes repeating where is it repeating this one down yeah good. So, it is a good point. This single box directly in has an information that there are three spin half particles which are composed to give you a spin half right. You can do it in two ways you can make. So, it is three objects right. So, this one I would say that it is actually going to be spin of particle 1 and then you can have spin of particle 2 and spin of particle 3 to have some symmetric combination. Here you can have. So, you can call it spin 1 ok. So, there is some mixing where the two spins you take a symmetric combination and multiply with the third spin. Here you do anti-symmetric combination. The square bracket means anti-symmetric combination. So, there are two possibilities exist when you are combining three spins you can have some different symmetry like this one is totally symmetric in all the three spins and this one has some mixed symmetry. So, let us let me do it here. What will happen here? There will be a q 1, q 2, q 3 which is symmetric because the diagram forces you it should be a symmetric. If you interchange the object which I put in this box and this box you do not get anything new clear. So, similarly here you could view something like it is the symmetric object here and then you add a piece here or you have an anti-symmetric object here and add a piece here that is called mixed symmetry. I am sure you have done mixed tensors. Symmetric tensors, anti-symmetric tensors you can have mixed tensors where some indices will be symmetric, some indices will be anti-symmetric and those are the mixed tensors. Similar thing can happen here, but as a diagram this is the meaning that there will be a subset which will be you know which will have this S 1 has no symmetry, S 2 and S 3 you can make it symmetric or S 2 and S 3 you can make it anti-symmetric and write the states. Yeah, both will be, now if you see here, if you see here when I compose this with this the these two boxes are symmetric. When I compose this with this you see that these two boxes are symmetric, you can call S 1 and S 2 symmetric. I just call this as 2 and 3 and that one as 1, but it is not the because of associative property, but technically you can call if you want maybe this is a better notation as far as the diagram is concerned you have an S 1, S 3, S 2 symmetric and then you add an S 3 that is what you did here. S 1 and S 2 are symmetric and then you added an S 3. If you see here it is S 1 and S 2 were anti-symmetric and then to that you added this which is S 3. So, this is another way of seeing it. So, these two have different tensorial properties. So, the next question you can ask is this proton belong to this or will belong to that or a linear combination. All these questions starts coming up you know. One of the ways in which they fix all these things is by somehow getting the magnetic moment result experimentally seen out of which linear combination gives me the result that is the way they fix it. We do not know why that linear combination. Nature seems to project one linear combination of these two pieces such that your magnetic moment in your experiment can be matched for proton and neutral. So, I do not have any handle which one I cannot say that this is proton. I cannot say this is proton even though they are spin half particles. Linear combination of it what linear combination what coefficients I should choose are all dictated by experiments, but it does not give me anything other than spin 3 by 2 and spin half. So, I play around with a bound state of three quarks, three fundamental objects. To understand spin I should take them to be fundamental objects to belong to SU 2 representation. To understand the quark content I need to see it as a SU 3 representation because I am just working with quark content which is only the three objects which is UDS which is sometimes called as a flavors. So, is this clear? Now what are these dimensions can someone tell me for SU 3? What are those dimensions? So, this one is dimension is 4 plus 2 plus 2 this is for SU 2. So, what about here SU 3 by your hook numerator? This is anyway I can see it to be a trivial representation. So, that is 1, 10, 8, 8 and 1. Is it adding up to 27? 3 into 3 into 3 is 27, 10, 18, 26, 27. So, the young diagram actually helps you without doing a projector how the breaking happens. You can do the projector by writing the club squad in matrices even before you do that you can say that you can find this breaking. So, this was a crucial part which Gelman did. The reason why he did let me get to the slide. So, as I saying one set of baryons of spin 3 by 2 another has spin half why such a pattern then the irreducible decompositions of the tensor products of SU 3 will explain this ok. So, what he saw in the observation in the experimental lab was that later on you can put it in the H 1, H 2 plane. What is this called? This is called the weight diagram. Weight diagram for an irrep and you have used fundamental objects as UDS. So, it is an irrep of SU 3 ok. So, what happened was that the charges along these lines are you know this one has charge plus 2, this one has charge plus 1 and so on ok. So, there is some nature seems to have some kind of a pattern. So, you wanted to account this as a weight diagram of SU 3 irrep. Interestingly or experiment did not have this particle this omega particle at that time when Gelman was doing his SU 3 group to fit in this data all of these particle had spin 3 by 2, he wanted to fit this data by using SU 3 group. When he was trying to do it he got 10 that is this 10 on the board, but he did not the experimentalist did not have this omega particle and he said that you have not detected it, you better go and check for it. It is beautiful actually. Gelman recently passed away, did you did you see his profile? You should see his profile very nice. Very simple way of looking at SU 3 and he accounted that there should be 10 particles which belongs to an irreducible representation of SU 3. Whereas, only 9 particles or 9 states with spin 3 by 2 were seen in the lab. Interestingly after 3 years they detected it, they detected that particle. So, that is why he got his Nobel Prize. So, it is very beautiful to see that you know completely it may look very mathematical whatever I was saying is mathematical, but ultimately nature seems to have some kind of a symmetry and he was able to show it to us. Is that clear? So, this diagram is sometimes called decimate diagram because there are 10 possible states which is the highest weight state here, highest weight will be this corner state and then you can use all the possible ladder operation of SU 3 right alpha 1, alpha 2 and alpha 3 and you can generate all the 10 states by the SU 3 generators. So, this is a irreducible representation of SU 3 whose young diagram is this and whose dimension can be computed to be 10 dimension. Those 10 are the 10 states of the decimate of spin 3 by 2 baryons. Why baryons? Baryons are the ones which are composites of 3 quarks. These are all baryons proton is a baryon, neutron is a baryon. So, all of them are baryons and you can explain from group theory as an irrep and you can also account for the 9 already present 9 states before when Gelman proposed this and one more was seen after 3 years and validating that SU 3 quark model is the right approach to understand the set of particles which has spin 3 by 2, set of baryons which has cement. Is this clear? Similarly, the spin half baryons also had a this is very different from the earlier one. This is another irrep and this irrep had 8 particles and that is your 8 dimension. The spin half baryons which includes protons, neutrons belong to a different irrep from this. This irrep was totally symmetric. This irrep is partially symmetric. So, you can write this as if it is q 1, q 2 is symmetric and q 3 is has no symmetry. Similarly, here you can say q 1, q 2 is anti symmetric, anti symmetric I am denoting it by a square box. You all know by now what is anti symmetric symmetric right. If you interchange the two content, if it is anti symmetric the wave function will pick up a negative sign. Symmetric means if you interchange it will remain the same. So, q 1, q 2 and then q 3 here all of them are anti symmetric clear. That is why it is a trivial representation of SU 3 and these are non-trivial representation. As I said that whether the proton will belong to this or this, whether the proton in the SU 2 language will belong to this or this is dictated by some linear combinations. Because whenever we have to write a wave function, you have to write your wave function. Suppose I want to write a wave function for a proton ok. This will depend on the position coordinates, it will depend on the spin, it will depend on the quark content ok. So, sometimes that is called as the flavor ok. So, you will have a piece which is psi which is a function of r, then you will have you have to multiply it with chi which is a function of spin, you will have another phi which is a function of flavor. You have to multiply them. Why multiply? Are independent spaces spin space, position space and the flavor space, they are independent space. Only thing is this is totally a fermionic object. The proton is a fermion, it is a fermion. So, what you have to remember is that a fermion which is a bound state of 3 objects should be totally anti-symmetric. If you swap the spin content, flavor content. So, some things could be symmetric, some things have to be anti-symmetric, so that the product remains anti-symmetric. See these are things which you will all do not to violate policy exclusion principle. So, what I am trying to say is that you could take the spin wave function to be symmetric, flavor wave function to be anti-symmetric plus a linear combination where spin wave function is anti-symmetric, flavor wave function is symmetric, all these playing around you can do. Finally, find a linear combination of this total wave function for which you will find the magnetic moment which is unique. So, there is a lot of mixing going to happen between this and this and so on. So, I am not going to get into it, but that way you can uniquely pick a combination ok. So, as of theory group theory, it tells you that all the elementary particles which they call it as elementary is not exactly elementary, variance or composites and to account for the composites with spin tree by 2 and half, you can allow 3 fundamental objects and once you allow 3 fundamental objects, you can say that in the flavor space it allows for 3 flavor states which is Su 3 and then you start accounting for this 8-fold path or the octet diagram ok. So, the 8-fold path of the octet diagram you can account and the previous diagram which I showed is a decimate diagram and you can account them by treating it as a Su 3 group circuit ok. So, I hope I have given you some flavor of how your group theory language actually speaks what you see in your physics which were proton, I just wanted to call that as a for a proton ok. So, I can say this for the proton, for proton the wave function ok. So, this p has nothing to do with momentum, maybe you confused it with moment. So, p is just the proton ok. So, to summarize the two sets of baryons which I showed you, one as a decimate diagram minus the omega particle initially can be viewed as tensor product of 3 quarks as follows ok. So, this we have done it now, spin 3 by 2 baryons or 10 of them according to quark model proposed by Gelman and you can check this dimension which we have already checked. Similarly, mesons are bound states of quarks and anti quarks, mesons are baryonic or sorry fermionic or bosonic, it is bosonic, it is a bound state of a particle with an anti particle ok. Now, I will leave it you to see what should be happening and what is happening in the experimental scenario for mesons.