 So, just as a warm up what we did discussed in the last week's lectures, the confinement was on to the character table and I said that we will use this Mullican notation, Mullican symbols to write down the irreps by now it should be clear what I mean by irrep. So, that is a short form I am going to use irreducible representation which cannot be further broken down into any block diagonal form by any similarity transformation. So, those are the irreps and you denote for one dimensional irrep either by a letter A or a letter B. Letter A is used if the character of the n fold axis rotation that particular element C n element if it is plus 1, you use the letter A if the character is minus 1 you use the letter B. For two dimensional irreps we use the letter E 3 and three dimensional representation we call it as F 40. For all your point groups you do not go beyond 3 usually it is all the groups which we are studying will be up to three dimensions ok. So, then we also have some more more finer details we introduce a subscript G for Gerard or U for Angerard depending upon sigma h character is plus 1 or minus 1 respectively ok. So, just to give you an example we have done this many times C 2 V the character table we worked it out right E C 2 sigma in the X Z plane and the other mirror plane which is also vertical plane both are vertical plane because it has the Z coordinate in it the plane contains the Z axis ok. So, these A 1 and A 2 are corresponding to say a two different one dimensional irrep the first one is the trivial unit representation right we call it unit representation and the C 2 element the C 2 is the principal axis here that element is plus 1 here that is why we calling it as A these two and the C 2 element for the B's are minus 1. So, the last two characters we call it as minus 1 and we have already seen that if you had four conjugacy class you will have four irreps and by using those postulates as a consequence of great orthogonality theorem there will be four one dimensional irrep for the Abelian group of order of four. A question for you if instead of C 2 V if I replace it by C 4 ok, if I replace it by C 4 then can this character table be used or can we have another character table ok. Think about it and Thursday we can discuss. So, is C 4 and C 2 V character table are they same or there is something which you need to work out ok. There is a slight difference right C 4 has C 4 times C 4 squared is C 4 cube there is some subtlety is there. So, you need to worry about that subtlety and see whether the character table you can also look up the literature and see how the C 4 table is written ok. C 2 H is again principal access is two fold and then you have a mirror which is in the xy plane ok. So, if you take that then you see that the E C 2 sigma H and I if you look at the sigma H eigenvalue the sigma H eigenvalue is plus here. So, that is why the subscript G is written if it is minus then you write it as an gerard and this A and B nomenclature is with respect to the principal access. If these two are plus then it is A and the subscript depends on the sigma H value. So, it is G for this and U for this and similarly this one is B both of them are B because the C 2 is minus 1 are you all with me are you all able to see. And if you look at the sigma H eigenvalue you call that as G or call that as U. So, this is the nomenclature and I will come back to this basis we have already seen that the z axis rotation about z axis does not alter any of these elements. So, the unit representation the basis is z axis ok. So, when it comes here there will be some subtleties because when you do z axis what happens sigma H will take it is in the xy plane. So, it takes z to minus z. So, your trivial representation or unit representation cannot have z axis as the basis. So, it should be a vector in a xy plane right cross product of a vector in a xy plane having components x and y that is what we call it as a axial vector. You all know what is an axial vector a general vector when you do an inversion ok. If you took a vector when you do an inversion what happens it becomes minus the same vector right. So, this is what will happen for polar vectors. Axial vectors axial vectors under inversion will remain as axial vector does not change some sometimes called a pseudo vector or axial vector both are means the same. You know what is the example? Simple example is your angular momentum r cross p cross product of two polar vectors will behave like a axial vector right. So, this is this is an axial vector. What about p cross L? p is the momentum L is the angular momentum p cross L behaves like a polar vector ok. So, this one is behaving like a this is a usual conventional cross product which you know in your three dimensions. So, in this case when you see the unit representation or the trivial representation in the slide ok. So, this slide shows that sigma h Eigen values plus 1 right all the Eigen values have to be plus 1 which basis will have all the Eigen values of plus 1. If you want to look at it you will be able to show that it should be an axial vector only the z component of the axial vector roughly you can see that L is z if you had taken x will change and under c 2 x p p x right let us write it L is z is x p y minus y p x right. If you do a sigma h operation on this it should be applicable not only to a position vector should be applicable to any vector components. So, sigma h on this is in the x y plane right. So, I have to do sigma h on this one. So, what will happen? x and y components do not change. So, it gives you a plus 1 times similarly c 2 on L is z will be c 2 on x p y minus y p x c 2 on x will change sign and c 2 on p y will also change sign. The y component will change sign x component is that right c 2 180 degree rotation about z axis will change both x and y components. So, it is minus into minus which is plus. So, that is why it is going to be plus 1 times x p y minus y p x. So, if you are given a axial vector the z component of the axial vector is the basis for the unit representation of c 2 h. But whereas, if you look at c 2 v the z component itself of a polar vector itself is a basis vector ok. So, I leave it you to check the other basis vectors in the same argument on the slide if you see c 2 v I have said that the axial z component is the basis. What do I mean by this? If I do sigma on that r z ok. So, I need to show that if you had r z. So, this is for c 2 v this I did it for c 2 h if I take c 2 c 2 of course, you have argued here what did we get it is plus 1 right. So, that is plus 1 sigma v y z component someone do it on this and sigma v x z component and what else we left I did it anyway we know. So, if you see here the character table the r z basis is the right basis because it will be an eigenvalue equation of every group element acting on r z because one dimension it is also trivially an eigenvalue equation. So, every group element the character in the a 2 representation for every group element on r z turns out to be. So, I should say here the group element the group element on this is the character on a 2 for the g element on r z is that ok. This is true for all one dimensional irreps ok. So, I did this for a 2 you can try and do this for b 1 and b 2 check it out the x y we have already discussed right when we wrote the rotation matrices we discussed this, but for r y and r x also you can check b 1 there can be 2 possible basis it can be the polar x component or the axial y component should be one dimensional basis. Similarly, if you look at b 2 you can show that it is polar y component or r x. This equation is like the group this is what I did know the group element if you apply on r z that is the group element then it takes up the corresponding character times the spaces. So, this is like an Eigen value equation the number which you get should match with your character table number. If it is not happening then you have to find out which irrep it is happening this should be for all g's for all g's element of g. In fact, this g is actually c 2 h c 2 v sorry c 2 v. This is c 2 v check that this is satisfied ok. Then you know that r z is the basis vector for the irreducible representation ok. So, let me write that also clearly for the 1 d irrep p 2. Any questions on this? Same thing I want you to do for the other rows I have done explicitly now for the first row of c 2 h and I have done it for the second row of c 2 v ok. Please try it out and make sure that you understand the conventional basis in the position vectors or axial vectors. Position vectors are like polar vectors and axial vectors are going to have this pseudo operation under inversion that is right correct ok that is the notation I am going to follow ok. So, I have added one more column in the c 2 h which will come back to it ok. This is called binary basis. X is a basis which you can call it as a primary fundamental basis. You can take powers of those basis and you can start looking at what are the binary basis will come to it and sometimes it is not always possible to write the basis. So, if you do not find a basis you leave it blank ok in the convention position vectors ok. So, since I have put this in let me also give you a small example ok. So, c 2 character table ok c 2 is isomorphic to also the permutation of two objects right. It is also isomorphic to just only mirror and so on ok. So, c 2 what is the number of elements there is only one element e under c 2. I can even put this to be some g such that g squared is e that is the condition order to group we are all isomorphic. You can treat it like a permutation group or a c 2 group or other mirror groups also e and c v e and sigma v or e and sigma h. This in the permutation group what are the diagrams identity element is 2 1 cycles. How will I draw this this way and g is 1 2 cycle the square of that element is identity right you all with me. So, in that notation how many irreps are there how many irreps are there 2 irreps first irrep will be unit element second irrep will be 1 minus 1 this cannot change this has to change for orthogonal. According to our thing we will write a unit representation in the permutation group as if it is a symmetriser and this will be anti symmetriser. If you are looking at it as a c 2 group what are the index we will use this will be a c 2 element then this will be a and this will be b. There are no b 1 b 1 because there are not more than one. There is one irrep with mullican notation a one reberep with mullican notation b. So, what are the basis vectors you can ask you can treat as if you are looking at a one-dimensional problem. Let us take our simple harmonic oscillator 1 d harmonic oscillator what is the symmetry of this v of x equal to v of minus x. I can define a g operation on x takes it to minus x identity operation does not do anything g operation changes to minus x. So, the x is a good coordinate. In fact, y can also be used z can also be used by the same kind of, but I am confining myself to a 1 d problem. So, this is x no that is not x someone this is not x g has to g on x has to be minus x which one. So, this is the character table which will help me to look at the harmonic oscillator problem. So, tell me I have a Hamiltonian for the harmonic oscillator what is the Hamiltonian it is the total energy which I can write it as v of x and when you write the Schrodinger equation you write high psi of x equal to e psi of x, but we also know that the Hamiltonian this Hamiltonian does not change under the group symmetry. This group symmetry let me call it as the order to symmetry which are calling it as C 2, but you can call it also as an inversion right x going to. So, what is the meaning of it? So, typically when I see the Hamiltonian I have to list out what are the group symmetries. So, what are the group symmetries here? The non-trivial element of that group is only g which takes x 2 minus x does not change does not change Hamiltonian. You all agree because of this symmetric potential if you had some asymmetric potential that condition will not be true clear. So, if I do h g on psi of x what is g on psi of x you are going to do psi of minus x. So, in this Eigen value equation if I replace x 2 minus x. So, this is actually dependent on x right. If suppose this is this coordinate is a dummy coordinate you could write h of minus x psi of minus x equal to psi of minus x, but h of minus x is same as h of x right h of x is same as h of minus x you all agree. So, which means this I can replace it again as Hamiltonian on psi of minus x is E on psi of all you all with me. So, then what happens h on psi of minus x is again E of psi of minus x. So, what have I shown? What have I shown? This is an Eigen value equation I also see that there is a group symmetry the group is the group with this character table with the non-trivial element g that group symmetry that non-trivial element of that group symmetry does not alter my Hamiltonian. So, that is why I am calling that to be the symmetry of this Hamiltonian. Once I put this in what am I getting? I am showing that psi of x and g psi of x have same energy you all with me both have same energy. What is this definition? What is this called? It will be degenerate if psi of x cannot be written as constant multiple of g of psi of x otherwise it will be non-digital. And g of psi of x I am going to write it as psi of minus x allow. So, psi of x and psi of minus x psi of x and psi of minus x share same energy. What is the next step? You know from harmonic oscillator solving have you seen degenerate energy Eigen values in the 1D harmonic oscillator? You know why? Because of the same group symmetry which you mechanically use right. Equivalently look at this character table. The character table tells you that your wave functions which you write must be either an irreducible representation of A or B or this representation symmetric or anti-symmetric. This is the statement ok. So, let me try and say this a little more clearly. So, psi of x there exists a projector a symmetrizer associated with one of the C reps. What is the symmetriser mean? I told you already symmetriser mean that all the group elements should not change the configuration. What will happen is that it has to become the projector will project my wave function any arbitrary wave function into a symmetric combination. By this what I mean is if I do a group operation on that projected state g on psi of x right projected state is psi of x plus psi of minus x. So, g on psi of x plus psi of minus x will turn out to be same. You will remain in the same space irrespective of the group operation. So, there should be a projector a projector this is ad hocly given now, but I will systematize it at some point right now you take it and then the group operation does not take you from this space to this space. Similarly, if you do p on psi of x what will this be psi of x minus psi of minus x and whenever you do a group operation on psi of x minus psi of minus x it will be minus of psi of look at this relation the g. G gives you a plus 1 eigenvalue here and g gives you a minus 1 eigenvalue here and let us go and look at this character thing. The g gives me a plus 1 eigenvalue means this is a good candidate because g operating on this is plus 1 times this. If you look at minus 1 if you want to get it is psi of x minus psi of minus x. Your harmonic oscillator wave functions should either belong to this irrep or this irrep it cannot be dangling because the group symmetry of the harmonic oscillator is respected by this order to group. It has to be necessarily what is this wave function property it is even function it is odd function. This is what you see in your harmonic oscillator. Harmonic oscillator without doing a calculation just purely from group symmetry I can say that the wave functions have to be either symmetric wave function or anti symmetric wave function it cannot be both it cannot be a function without any symmetry is purely from this group symmetry one. How we get these projectors I will explain, but as of now once you get into this thing you can see clearly that g if you write a function and this has to be psi of minus x it has to be proportional to it I should say. So, it should be sums I should put here it is not exactly equal to I should say some c eigenvalue c. This breaks it into either the wave function should be symmetric or anti symmetric and it will be only a non degenerate wave function. Once I put this in you can show that with psi of x equal to some c psi of minus x where c is plus or minus 1. This condition that it is proportional these two are not independent if it is not equal to it then you would have said it is degenerate wave function. In this case since it is proportional to this, proportionality constant c you can fix because the group is of order 2. So, you if you take square of it c square have to be plus 1 the only option is c has to be plus or minus. So, the this implies non degenerate odd or even functions. So, this is what you get without solving the problem and you can determine the actual wave function with the symmetry properties required by your irreps.