 So, now let me get to the problem of quadrupole momentance ok. So, the Dutron has been one used a Wigner-Ickardt theorem to find the ratios of the expectation values of electric quadrupole moment operator. Specifically I have taken the component it is a rank 2 tensor I have taken its q value rank 2 tensor with q equal to 0 for the 3 orientations of Dutron ok. So, that is the question. So, for this CG coefficient is not really applicable here because this one was a linear combination of one was L equal to 1 and the other one was L equal to half. So, let us not use this ok, but what will this be is the question using Wigner-Ickardt theorem. So, how do you go about this? So, the Dutron is given to you as j initial equal to 1 and m and you have an operator which is the quadrupole moment tensor operator and you are asked to find the expectation value. Expectation value means it is the same state ok. So, I can forget about this i j equal to 1 m I want to find what is j equal to 1 m. So, to find this using Wigner-Ickardt theorem using Wigner-Ickardt theorem you can work out this expectation value to be some constant as I said which depends on the total j 2 and 1 which we call it as a reduced matrix element. So, j equal to 1 q operator which depends on 2 j equal to 1 which is some constant I do not know what it is, but I have a CG coefficient. The CG coefficient is going to combine this angular momentum with this angular momentum. So, it is 1 comma m semicolon 2 comma 0 right and the final state should also be 1 comma m. So, this is the expectation value. This coefficient is always same irrespective of what value of m is there. Things which are going to change is only the CG coefficients for different amps. So, if I ask you the ratio what is the ratio for m equal to 0 plus 1 and minus 1. So, that ratio will be what? So, you can write it out expectation value between 1 plus 1 q 2 0 1 plus 1. What will this be? This is going to be the ratio of your CG coefficients. This is 1 1 for these two orientation. Similarly, you can find it for the other one and you can write it as a all the possible ratios. This CG coefficients just like the table I gave you. I explained to two of the tables for you. This table will be given to you and you have to only figure out what is that coefficient from the table and substitute it. So, the coefficients actually turns out to be. So, let me just give you the answer. It turns out to be this ratio turns out to be 1. So, essentially it is minus root 3 by 10 and this is also minus root 3 by 10. But if you want to find 0, if you put in 0 then you get that coefficient CG coefficient is this CG coefficient is minus 2 by root 5. So, the table will be given and you can look at the table and find out what the coefficients are. So, if you want to find the ratios of all the three orientation, it is like 1 is to 1 is to this fact. This is clear? So, experimental is if they give you 1, if they give you this result to you, then you do not even need the experimentalist to determine the coefficients for the other orientation. Wigner-Rickardt theorem will fix it. Is that clear? If he gives q 2 0 expectation value for a neutron with m equal to plus 1, then the rest of the two orientations you do not need the experimental data. In fact, you can make sure that you can validate the experimental result by showing as Wigner-Rickardt theorem has actually given that ratio correctly. So, that is the power of the SU 2 group symmetry of most of the spherical systems. Any spherical systems if you are looking at, they are given by a potential which is like a 1 over r potential, right. All these systems have rotational symmetry. If it has rotational symmetry, the generators of rotational symmetry in the position space would be just orbital angular momentum, but if you include internal spin space, then it is a total angular momentum. Once you put in total angular momentum, it is called the SU 2 set, because j can be half odd integers or integers. Is that clear? So, most of the system with rotational symmetry Wigner-Rickardt theorem is really powerful to give you the matrix elements or selection rules of transition from an initial state to a final state due to some operator like electric dipole moment or quadrupole dipole moment also. So, now let us see this interesting thing which I have already given to you in the next problem on the slide. Use the Clebsch-Gordan coefficient. So, this notation is equivalent to this notation which I have already explained, but let me put that notation also for you. Sometimes people write this also as a inner product. People do write this this way, both are equivalent. The CG coefficient which I have written, sometimes they write it as an inner product of two states. So, that is the coefficient which I have given in your problem. I use the Clebsch-Gordan coefficient which is given to you and you are asked to verify dipole moment, quadrupole moment and so on. So, you have to verify that the static 2 k pole moment of a charge distribution has 0 expectation value in any state with angular momentum j less than k by 2. How many of you have already verified this? So, what is to be done here? You have to find the expectation value. So, let us do the same quadrupole moment tensor which we did here. So, if you take the quadrupole moment, it is 2 to the power of k equal to what? k is 1. So, k is 1. If I put that in, so if you take j to be less than 1, you have to get it to be 0. Clear? You have to get it to be 0. So, let us check it. If you put this to be half, you use this coefficient. The coefficient which is given to you has to be used. If j is half, then you can see that that coefficient itself vanishes. Clear? The CG coefficient vanishes. So, to do an expectation value of q to 0 between instead of deuteron if you take spin half particle, that expectation value will be 0. So, this is what you can check for one or two cases from using this coefficient and make the statement that, a powerful statement that to find the 2 k pole moment expectation value, it is going to be 0 if the particle has j less than k by 2. Is this clear? Now, I also want you to appreciate not just seeing it as a problem. It is just that you know without even doing computation, you can tell the experimentalist these things you do not even need to measure. You will never see it in the lab. It will be 0. That kind of a thing you can give as an input to the experimentalist. That is why the Wigner-Rickard theorem is really powerful. It is not just a passing theorem just stated. It is just working out the CG coefficient and looking at the ratios. So, this is the statement saying that 2 k pole moment expectation value for j less than k by 2 will vanish. So, you do not even look for all the j's which is not going to show any signature in the lab. You will say if you try to do that expectation value or any transition elements you will find it to be 0. That is all we have tried. Is this giving some flavor of I am sure you have done Wigner-Rickard theorem in quantum mechanics code and Wigner-Rickard theorem does not stop there. It is going to be useful as I said that when I do a decay process or a scattering process to go from initial state A to a final state which is mediated by a tensor. Typically the scattering matrix is a tensor of rank 0. It is like a scalar operator. If you have a scalar operator those expectation values if you want to see you can still invoke the Wigner-Rickard theorem. So, the next problem decimate Baryon wave functions. So, what was the decimate? How did we find the decimate Baryon? So, Baryons are made of three fundamental quarks which come in three flavors up, down and strange right. And then we denoted the Baryons to belong to a tensor product of three fundamental representations. And if you try to compose it in two ways the first you compose these two and then this one you get up this is only three boxes. I am going to put this separately just to remember that the first two objects are symmetric, first two objects are anti symmetric and then you have one more which is the trivial difference. So, you have to write the wave function for this tensor product which gave you this one had what is the dimension of this? This we worked it out this was 10 and they represent the decimate Baryons, Baryons in the decimate. They all have spin to be 3 by 2 and then you have a singlet singlet or trivial representation of SU 3. See all these things which I am doing is SU 3 Baryons, ok. What is the meaning of SU 3 Baryons? This has to be three dimensional. This is three dimensional and we work out the dimensions of them and as I said this one is d equal to 8, this one also d equal to 8 and this one is d equal to 1 and you have to get 3 into 3 into 3 which is 27 is to be 27 is a reducible representation that is decomposed into reducible components 10 plus 8 plus 8 plus 1 which adds up to 27, clear. And this some combinations of this will represent your octet, octet Baryons in a 8 fold path, ok. So, this is what is the particle content which is seen. In fact, this corresponds to j equal to half. This includes the Baryons and the new sorry the protons and the neutrons. This one has the other Baryons. In fact, as I said one of the Baryons which was omega was detected after Gellman's quark mode. So, now what is the question? I have asked you to find the just like we found the JST, JM states which is in the binary basis. Now, you have to find the states for them which is tertiary basis, right. You compose fundamental objects free of them and you should get the tertiary basis wave functions for the same procedure of finding CG coefficients, but this CG coefficients will be the SU 3 CG coefficients. You should find out of J and M, you will have highest weight vector lambda and the corresponding weight vector mu. You have to play around with that. As a nice trick where we connect the symmetric group of degree 3. So, you remember the symmetric group of degree 3, what is the character tables? Sure now even in the middle of the sleep you should know this, right. E, then you have 1, 2, 1, 3 and 2, 3 cycle and then you have 1, 2, 3 and 1, 3, 2. These two belong to the same class, am I right? And then we had irreducible diagrams. So, when I want to write the projection operator for this one, I can write it as summation over characters for that representation every element g times the operator. The g is running between 1 to 6. This is a symmetric group of degree 3, permutation of 3 distinct objects. Order of the group is 3 factorial which is 6. What I am going to make a modification here is that this projector forces that these three boxes are for distinct objects. But now when I go to SU 3, I am going to say that these three objects could be repeating. So, what is the modification? If it was distinct, you will get only one dimensional representation. Because it is repeating, you get 10 dimension, right. What are the things possible here in this 10 dimensional state? You can put u, u, u, d, d, d, s, s, then you can put two of them u, d, u, d, d and so on. So, list it out. You will find that there are only 10 possibilities where you have one distinct object u, d, s, but the remaining 9 will have something repeat, ok. So, when I do this projection, I am still going to use the character table. Character table was 1, 1 and 1 for this representation and I can find out what is the state corresponding to that diagram, ok. So, let us write it out. Suppose I start with a state u, u, u and if I do a projector of this, what will I get? This g operator will interchange the first position and second position when g is 1, 2, first and third position and so on. All the three positions have identical particles. What will happen? It will give me back the same thing, but if you do it on, if you do the 2, 3, it will do what? It will interchange second particle with third particle, third particle with second particle. If you do 1, 2, nothing happens. 1, 2, 3 become 1 will become 2, 2 will become 3, 3 will become 2, 3 will become 1. So, write it out and see what you get. Can you write it out? So, the first element is identity element. It does nothing. So, you will have u, u, d, 1, 2 also does not do anything. It will be again u, u, d. 1, 3 will be d, u, d, 2, 3 will be 2, 3 on this, u, d, u and then 1, 2, 3 will be u, d, u and then 1, 3, 2 will be d and then you put in the normalization. So, what it tells me is that you have to symmetrize completely. If you interchange any two, you should get back the same state, totally symmetric. So, the 10 decimates states which I have here is a totally symmetric state where you could use this projection operator to get your results for the symmetrics. That also tells you how you do for the anti-symmetric case. What will happen? The 1, 2 will change it with a negative sign. So, essentially you will find if you add up everything, you will get it to be 0. So, this is one way of determining. This you could have done it for even SU 2. What I did by 1 over root 2, 1 over root 2 can also be done by using the symmetric group of degree 2 character table. But this is going to give me all the symmetrics. Symmetric wave functions which corresponds to the decimate baryons. I have not done everything, but you can do for all the 10 of them. And each one will be having a different quark content. Their states are all orthogonal. All the 10 states have different quark content. Is this clear? So, that was the question. Construct the decimate baryon wave function obtained from quarks transforming as fundamental representation of SU 3. So, the u u u is one state. Then you can show that it is a linear combination of u u b which will be the other state and so on. All the 10 states you can write it out.