 coming back to some connections with what we were looking at in discrete groups, let us take a simple free particle system, simple system which is a free particle. Free particle is just given by there is no potential energy just a kinetic energy and the momentum p could have any value ok. It is not confined to a box can have any value. What you can show here is that if you take the commutator of Hamiltonian for a free particle with momentum p x, p y, p z that commutator is 0. You all agree? In the discrete group what did I say? If the Hamiltonian is respecting some discrete groups symmetry Hamiltonian should commute with every element of that discrete group, is that right? Now Hamiltonian should commute with the generator of that symmetry. Suppose free particle, this free particle you can see if you do a translation x to x prime p squared over 2 m does not change. You all agree? If you take x to x prime del by del x is going to remain as del by del x prime ok. It is going to be same. Free particle Hamiltonian respects translational symmetry. If it respects translational symmetry Hamiltonian should commute with the generator of that symmetry and the generator of that symmetry is momentum, linear momentum. So, let us take rotational symmetry. Does this free particle system respect rotational symmetry? Yes or no? What will you do to check that whether it respects rotational symmetry? Find the commutator with respect to the generators. If it is 0, you know that should happen. But of course, you also know if you do a rotation, it is like p dot p. Under rotation p dot p is same as p prime dot p. So, it is rotational invariant. So, your p squared should commute with l x l y n. Is that right? If you do a rotation, it becomes p prime dot p prime divided by 2 m, but this is same as p dot p right. It is a dot product of two vectors. So, similarly under translation it is p dot p nothing happens, it is invariant actually ok. So, this implies this Hamiltonian respects translation symmetry and rotation symmetry. Now, you can check that the Hamiltonian with p. So, consequence of any symmetry of that system requires that the generators have to commute with the Hamilton. In the discrete symmetry, it had to commute with all the elements. If it commutes with generator, it is good enough because elements of the group are constructed using, exponentiating 1 minus the generator. So, all the group elements will also commute. Like what I am saying is that the Hamiltonian with exponential of i by h cross a mu p mu, this will also be 0 right, group elements will also ok. So, it is good enough to see the commutator with the generator to get the implication about what is the symmetry of the given system described by some Hamiltonian. Hydrogen atom, is it translation symmetry is there? Is there a translation symmetry? Think about it. Harmonic oscillator, does it have a translation symmetry? No, hydrogen atom you have to still think. There are two particle system, if you do a translation relative coordinate what happens ok. So, please think about it, but harmonic oscillator, 1 D harmonic oscillator, the potential energy if you shift x to x plus a, it is going to give you a new, it is not going to remain in it. Or equivalently, if you take the commutator of the Hamiltonian with p, will it be 0? Will it be this is the next way of indicating whether it is respecting rotational symmetry or translation. Is this clear? There is a parallel whatever you have learned in discrete group is not completely disjoint from what you are doing now. It is just that it is going to be connected by a continuous parameter and we are looking at the essentially the generators and once we know the generators we can exponentiate it and find any arbitrary group element ok. Because the parameter is continuous, you will have uncountably many number of elements that is why it is going to be a continuous group or we call it as a leak. So, this I have already explained to you under infinitesimal rotation the change in position vector I put a bold face to remember it is a vector with a rotation by an angle theta cross r and then you try to use how to write the relation between the wave function obeys this. Two similar steps which we did I have already done it for you and shown you what is what is happening to your generator for rotations and then you have to do n number of steps of infinitesimal transformation and you will get the exponential. So, this also I have explained that if you take infinitesimal rotation and then you can rewrite it as I have put the h cross to be 1, you can rewrite it as if it is by an arbitrary angle n hat magnitude is delta theta times l. This is something which I want you to check can you please check this that if you do two consecutive rotations about x axis and y axis infinitesimal rotation by the same magnitude angle ok. Do the other way round the difference between these two you know it is not 0, but it is going to be for the delta theta squared. So, the difference between these two rotation is going to be for the delta theta squared and it is as if it is a rotation about the k axis. If it is x axis and y axis this result is about the k axis. So, please check this and from here you can explicitly use the first line determine that L x L y is I is L z. So, these two line implies that it is an non-abelian group rotations if it was abelian what would have happened right hand side would have been 0. So, then fact that the right hand side is not 0 at order delta theta squared ok. If you are doing so many books they take infinitesimal and ignore delta theta squared. If you ignore delta theta squared and look at it as a very small angle transformation you can still treat it to be closest to abelian, but if you keep up to the order delta theta squared then the right hand side is not 0 and this deviation is what gives you automatically the algebra of orbital angular momentum. So, I have just summarized whatever I did on the board for completeness on the slide and please check this part verify this relation ok. So, now, coming into the jargon of what we call the rotation groups as you all know that this rotation matrix which takes you from r to r prime this rotation matrix is a 3 cross 3 matrix and they are orthogonal matrices. Why orthogonal matrices 10 dot r prime which can be written as r r transpose r r right am I right r transpose r has to be identity the set of matrices which satisfies that property r transpose r to be identity are orthogonal matrices. So, the corresponding group is what we call it as an orthogonal group and minimal non-trivial dimension which we have to look at is this rotation in 3 spatial dimensions is denoted by O of 3. So, this is set of orthogonal matrices in 3 dimensional space. The determinant can be plus or minus 1 in general need not be only rotations it can be rotation times reflection inversion ok. So, the O 3 means it could be orthogonal matrices determinant could be plus or minus 1, but S O 3 S O 3 requires that the determinant has to be plus 1. So, this has proper and improper rotations possible S O 3 will allow only proper rotations and you can show that O 3 is nothing, but S O 3 times an inversion direct product clear. So, the CS group will have an identity element and minus 1 minus 1 minus 1 diagonal element. So, any element which you find for S O 3 with identity will be S O 3 with the inversion element will give you the elements which does not belong to S O 3, but belongs to O 3. This is the same notation which I use for the discrete. So, each element of S O 3 will be specified by axis and the rotation is what I was saying and special orthogonal groups rotation is by an angle psi about a direction which I am calling it as hat n given by theta and phi unit vector. Couple of observations you can see if you take a rotation about z axis let us say by an angle opposite clockwise or anticlockwise rotation, you can show that the axis can be inverted and the rotation angle can be made if it is clockwise it can be made anticlockwise that is a symmetry like this. So, what does this mean? If you do a rotation by pi, pi and minus pi are equivalent. So, what you see is that the rotation by pi about positive hat n axis or minus hat n axis gives you the same value, but for arbitrary psi it is psi goes to minus r ok. Why am I putting all these things? I was always saying that I need to look at the parameter space right. When I did the space translation I said the parameters a x, a y, a z are from minus infinity to plus infinity. Now, this rotation I am saying you choose an axis the axis can be on a sphere ok. Let me take a sphere and let me call the z axis as psi ok. So, I am going to use sorry the radial vector to be psi. So, any vector which I am going to take so, this part let this radius be pi ok. I am going to fill the whole sphere it is a solid sphere with the maximum radius being pi. Any particular point I take I will call that point as psi theta and phi. I have phi specifies for you the direction of your axis. The radial distance from the origin this is psi ok. The radial distance from the origin specifies the angle the magnitude of the angle by which you are doing a rotation. So, the parameter space analog to a x, a y, a z for the rotation can be compactly put in on a solid sphere with psi magnitude of psi should not exceed pi right. This angle always from minus pi to plus pi, but this magnitude will only can go up to pi right. So, this is the parameter space which is compact unlike your space translation where I said that it is r 3, a x, a y, a z can be anywhere from minus infinity to plus infinity is the parameter space for space translation. What is the parameter space for rotation? You can nicely draw it as a solid sphere with the radial coordinate ranging from 0 to pi and every point in this corresponds to a rotation element. Suppose I take this to be psi prime and theta 1 and phi 1, this will give you a new rotation. So, the parameter space is compact that is the first observation unlike here where the parameter space is non-compact. I want you to think about what happens for Lorentz transformation. It has phi x t, phi y t, phi z t and phi x z, x y z and what happens there you please think about it. So, group manifold the parameter space which defines for you the group SO 3 is a sphere solid sphere of radius pi. So, once you have this parameter space to be compact, you say the corresponding group is a compact group. Compact means it is all within this finite region whereas, here there is no bound. There is also one more small thing you can observe here. Yeah, it is not infinite it is only goes up to pi, psi only the psi is maximum value of psi is only pi. No, no it is a sphere, but the one which is applicable to rotation I explained it on the screen for you. In the screen I have explained that psi has to lie between 0 and pi. We will talk about a spherical coordinate system it has one radius and two angles. So, angles would lie between pi that is fine, but the radius also should be bounded radius could go from 0 to infinity. So, if you try to match every element of rotation to a point in this space, the point in this space the radial coordinate is taken by the amount of rotation magnitude of an angle rotation about an arbitrary axis. So, this radial coordinate is replaced by your magnitude of rotation angle and the theta phi gives you the direction about which you are doing the rotation. So, this psi is not exactly the radial coordinate its spherical polar coordinates. This is a way of mapping it onto a region to show that it is a compact bounded space. Every point here corresponds to associated with this point you will have a rotation defined ok. If you want to go from point A to point B if you want to do a rotation psi and then rotation another psi 1. So, you can keep doing that and then you can start you know going around and coming back to the same point also. This will give you at every point you will have rotation multiplied and if come back to the same point you expect that you have done nothing right. It is like an identity operation. Interestingly this point and this point diametrically opposite points are all identified. This is one of the properties that if you do a rotation by an angle pi, rotation by an angle pi about the opposite diametrically opposite axis are one in the same ok. If you do a series of rotations you could either do a series of rotations and come back to the same point which is like as if done nothing. You could also achieve that you go from here to the boundary, go to the diametrically opposite point and do a series of rotation and come back clear. So, there are actually non-trivial paths of achieving an operation either you can be inside the sphere this is the group manifold. Manifold or the parameter space with the requirement that diametrically opposite points are identified. So, you can have various paths. This path you can make it smaller and smaller and make it as if it is identity. This path you cannot break it into smaller and smaller into a point. There are two paths ok. So, these are the ones which tells you that the parameter space is a doubly connected ok. So, let me stop here Lorentz transformation also I explained it to you that you can write it in a compact notation as L 0 I for your k x t k y t and this is what I said and you can write your L x L y L z in terms of L x y L x z and L y z ok. So, this I will leave it you to do the algebra of these generators there will be one more assignment which will come to you and that will be the algebra for Lorentz transformation. So, we will come to the Lie algebra of formal definitions in the next class.