Physics foundation series 2026

本站添加: 2026-06-03

[00:00:01]Okay, everything is looking jaundist, right?

[00:00:08]How did it all become yellow?

[00:00:13]Is it yellow for you also? So, it is not jies for me.

[00:00:19]Yeah. Okay. So what I did yesterday was the template for studying the rotation groups and in fact we can use the template to study rotation group in any dimension.

[00:00:41]Okay. So let me define this is all water now. It should dry I think a little bit.

[00:00:53]Okay.

[00:00:57]So let me define a vector R.

[00:01:04]I won't specify any dimension or anything. I'll do it in general in the beginning and then specialize to whatever we want.

[00:01:12]So it has components x1, x2, xn.

[00:01:25]Okay. So it's a vector in some n dimensional space. So I define rotation on this vector that it is rotated and I denote the rotated vector R prime as a rotation on R.

[00:01:55]So what are the properties of this rotation?

[00:01:59]So before that I'll generalize the concept of the invariance of the scalar product between any two vectors. I'll demand that I'll construct a set of transformations which leave the scalar product of any two vectors invariant. So let me consider a vector A which is a function of all these variables and it is rotated and CA and similarly vector B.

[00:02:46]So I can say that the components of a prime are given by say some n byn matrix.

[00:03:02]Okay. And this is and similarly BJ prime is Yeah.

[00:03:40]Okay.

[00:03:45]Where this R now is an n byn matrix and it transforms a to a prime and b to b prime.

[00:03:54]So a and b are some vectors in this n dimensional space.

[00:04:02]Now I say that this trans transformation should be such that it preserves the scalar product.

[00:04:11]That means if I have a dob as the scalar product I will not write the vector notation because I'm using the matrices. So you should any vector is just a column uh matrix with n entries n real entries.

[00:04:34]So this is a real there r is a real n byn matrix.

[00:04:49]It could be any general transformation but I want those transformations which preserve the scalar product.

[00:04:59]That means after the transformation the scalar product remains the same.

[00:05:05]If you if you say B is the same as A then this is this means that it preserves the length of the vector.

[00:05:13]Okay, that's the kind of transformation I am interest.

[00:05:19]So if that is the case then what is this a prime dov prime?

[00:05:27]This we have done yesterday. I just repeating it. So this is sum j is = 1 to n a j b okay writing it in terms of the components.

[00:05:49]So if you think of r² r dot r it will be x1 2 + x2 + x3 2 up to xn square this is simply the length of the vector square of the length of the vector.

[00:06:05]Okay. So now this I know I have defined the transformation here.

[00:06:12]This is J and sum over K sum over L.

[00:06:40]that should be equal to a dob that is what I'm demanding.

[00:06:50]So that is that's possible if k is equal to l. So therefore the property of these matrices should be such that summation / j or j k or j l should be delta k.

[00:07:19]This is still not a matrix multiplication type but I will do that because there is a summation over J. I will write it as KJ and in order to do that I have to transpose this matrix.

[00:07:34]So therefore this is nothing but a element of the transpose of R multiplying R J.

[00:07:56]Since this J is summed over, this is a proper matrix multiplication and that should be equal to delta K.

[00:08:10]Okay.

[00:08:12]Now remember this J now can go all the way up to the dimension of the spaced.

[00:08:19]So the invariance of this scalar product or the inner product in n dimension inner product is a very general you can define it in any vector space but for ordinary vectors we simply call it as the scalar product. The invariance of the scalar product implies that the transformation should have the property ranspose r.

[00:08:52]Remember this is valid for each element k. So if k is not equal to l it is zero.

[00:09:00]If k is equal to l it is one. So it's an identity matrix.

[00:09:05]So R transpose R is identic.

[00:09:23]Okay.

[00:09:27]Now I also assume that this is a continuous transformation.

[00:09:36]And therefore the set of R are real N byN matrices with the property R transpose R is equal to I. Therefore identity matrix therefore R transpose is equal to R inverse and because the determinant of R transpose R is equal to 1.

[00:10:25]This implies that determinant of r is equal to plus or minus one.

[00:10:35]The same result that we got for the rotations in the plane. So all this is reproduced here.

[00:10:44]And so identity matrix means that this is simply there's no change in the vector and the inverse exists that is also a transformation and of course the operation is matrix multiplication. It's an n byn real matrix. And this transformation also is associative in the sense that if you take r1 r2 r3 is same as r1 r2 r3. Why is this so?

[00:11:31]Because matrix multiplication is associative. So once you specify that these are n byn matrices it's given. So the set of all matrices this real n byn matrix matrices with this property which leaves this scalar product invariant they form a group.

[00:11:55]So it is the orthogonal group.

[00:12:10]Oh yes.

[00:12:14]Okay.

[00:12:16]Now if in particular you choose determinant of R is equal to + 1.

[00:12:23]So let determinant of R = to + one.

[00:12:36]Then it's called as a special orthogonal group.

[00:12:59]S.

[00:13:02]So how do you define this? that it's a set of elements R which are N byN real matrices real elements real matrices such that R transpose R is= to identity and determinant of R x = + 1.

[00:13:39]Okay. Now if you put n = 2, you will get so 2 which is what we did yesterday.

[00:13:47]Okay. And so is a special uh has a special property that it is a billion.

[00:13:54]So which means on a plane you can rotate and uh you can change the order of rotation you will still get the same uh overall rotation that is not you can just try it with anything in three dimensions for example it is not going to work. So therefore except in two dimensions this s is not an group. It's in general there are no commuting elements but they could have subgroups which are a billion. Okay. We'll see that pretty soon. Okay. So this is a very general definition of the group s.

[00:14:42]Okay.

[00:14:48]But we are not interested in this group in general.

[00:14:56]We are more interested in the rotations in three dimensions that is of relevance to us in quantum mechanics. Okay.

[00:15:13]So rotations in three dimensions that is the group SO3.

[00:15:37]Yesterday we saw in the case of group SO2 there was the parameter of the group is theta which is the angle which varies continuously and corresponding to that theta there was one generator of the rotations.

[00:15:52]Okay, we constructed the generator also.

[00:15:55]So in three dimensions an arbitrary rotation needs how many parameters?

[00:16:06]Can I three correct what about four dimensions suppose I want to do it's a get an example I want to do rotations in four dimensions how many parameters are needed how do you calculate the number of parameters sorry I did I hear a Number six. Why?

[00:16:51]>> Huh?

[00:16:54]>> Yeah. That is just using the property of the matrix. You can say there are six independent parameters. Okay. But physically, what does it mean?

[00:17:04]Huh?

[00:17:07]No. Four dimensions.

[00:17:10]There are four coordinates, right? There are four coordinates. So, it's uh in in terms of the vector itself, there are four degrees, four independent coordinates. Any vector is specified by four coordinates.

[00:17:34]Right? So this is where the number of suppose you make an arbitrary rotation.

[00:17:45]Okay. In general how many rotation independent rotations you have to do in order to take the vector from one orientation to another orientation.

[00:17:58]Okay. So you said it is three in three dimensions.

[00:18:03]That oiler theorem tells you that the theorem of oiler tells you that any arbitrary rotation in three dimensions.

[00:18:36]involves three distinct rotations.

[00:18:52]That means the all the the three have axes which are independent.

[00:18:58]Okay. So three distinct rotation.

[00:19:02]Of course you may rotate it in a plane.

[00:19:05]It will involve only one angle and the other two might be zero. But what is the maximum that is required for an arbitrary rotation? You have to specify three angles involves three distinct rotations along three independent axis.

[00:19:37]Okay. So that's how three angles come.

[00:19:40]So how many independent rotations you can do in four dimensions?

[00:19:46]H six. That's what you said. So that comes from the number of planes.

[00:19:53]So a plane will involve two coordinates.

[00:19:57]And in four dimensions you have overall four coordinates and 4 C2. Okay, that is 4 into 3 2. So that is six. So if it is some n dimensions what it is? What is it? Yeah, number of planes. Huh? Uh nc2 right. So it is n into n - 1 by 2 number of parameters. You can check that orthogonal matrix in any n dimension would involve so many parameters. That means if you impose this condition R transpose R is equal to A I. So take a matrix with all the elements which then impose this condition and that condition the reduces the number of it's a real matrix. So in four dimension it has some 16 entries and this condition will reduce it reduce the number of independent parameters. That is what he said from the orthogonal matrix point of view that it all six. So just that condition will give you this. So theorem tells you that you can of course these three independent axis can be chosen in many different ways.

[00:21:08]Okay. So yesterday I mentioned that you can write an arbitrary rotated vector a prime as a times something which is a function of rotation along an axis which is given by n a * n plus b c * n cross.

[00:21:39]Yeah. Okay. So there is this angle of rotation and the orientation of the axis. The orientation of the axis requires two angles. So there are three angles that you need for any arbitrary rotation.

[00:21:57]So I am not going into any particular uh representation except the one in which I choose the independent rotations along the xyz axis.

[00:22:12]Okay, the choice is up to me. So I can write for example if I'm rotating about rotation about Z axis of course I'll call XY Z as X1 X2 X3 in order to uh implement the summation signs but we understand the Z axis is same as X3 axis.

[00:22:50]So rotation about Z axis means that the Z coordinate will not change. So therefore the change the rotation occurs in the XY plane. But we have already know the planer rotation. Yesterday we have done that. So therefore we can write this as cos I'll call this as cos theta 3 minus sin theta 3 because rotation around the z axis changes only x and y. So and that rotation below already is sin theta 3 cossine theta 3 and no change along the Z axis.

[00:23:40]Similarly, I'll call this as R3 where this three denotes the axis, the third axis.

[00:23:53]Then similarly I can write cos theta_2 0 sin theta_2 0 1 0 - sin theta_2 This is denoted as R2.

[00:24:35]So this is rotation in the XZ plane or if you want to use the right hand sum rule it is the Z X plane. That's why the sign comes here minus sign. So this is rotation in the XY plane. This is the rotation in the Z X plane. Then no this is rotation. Yeah, rotation in the Z X plane. And then the rotation in the YZ plane.

[00:25:04]And that relieves X-axis invariant.

[00:25:10]And I'll denote the rotation as cos theta_1 minus sin theta_1.

[00:25:32]R3 of theta_3 R2 of theta_2 theta_1 okay but remember each rotation involves three angles so therefore when we say I'm rotating theta 3 you can think of this as theta as a vector where theta 1 and theta_2 are zero and theta 3 non zero similarly Here theta 1 is 0, theta 3 is 0, theta_2 is non zero. And this is theta_1 and theta 2 and theta 3 are zero. Each one of these planer rotations.

[00:26:15]Okay, defines because this is restricted to the plane xy.

[00:26:24]So this defines an SO2 subgroup of SO3 because this by itself if you look at this leave the Z axis invariant consider all the rotations with Z ais invariant as we saw yesterday it's just a planer rotation and it forms a group so in fact it is an SO2 each one of these and these are continuous parameters So therefore each one of these defines an SO2.

[00:27:01]Note it's an abilion subgroup.

[00:27:04]SO2 is a billion. So it defines an SO2 subgroup of SO3.

[00:27:21]So in fact since there are three independent axis you can define three planes. So it has three a billion subgroups. All of well they are all SO2.

[00:27:34]Is this uh clear?

[00:27:37]Okay. So then what I am going through the same steps as we did yesterday. Next I want to consider the infinite decimal generators.

[00:27:48]Okay. But one thing that you notice is that even though the subgroups are a in general the elements of group SO3 are not commuting. So therefore it's not an ailian group that you must that you should keep in mind. So any general rotation any general rotation r of theta I can break it up into rotations about these three independent axis.

[00:28:35]If you were to do the same thing in oiler resolution, you would say that all there you are looking at the rotation of the axis. You are going from a frame s to s prime by rotating the axis themselves. So that's uh what's called as the passive rotation there. First you rotate it along the Z axis and then you perform a rotation along the X axis and then you perform a rotation along the new Z axis.

[00:29:09]Since you have already rotated here when you rotate it along the x-axis zed goes to z prime and with respect to that zed prime you make another rotation and that is the oiler's decomposition which you study in rigid body I can choose what I want here it is very simple I choose it this way so that I can write any general rotation as a product of three separate rotations s along axis which are independent.

[00:29:44]Okay.

[00:29:46]Is that uh clear? Okay. So, so I have now defined the rotation group in three dimensions fully rotation transformations and how to get them by using some decomposition oiler or otherwise you can do that. So next thing is I want to look at because this is a continuous group every group every operation every transformation can be generated continuously from an identity that's a property of the groups that we are considering and in quantum mechanics those are the types of groups that we are interested.

[00:30:40]So following the same steps as yesterday, my R3 for an inf infinite decimal transformation I can write this as R3 3 of delta theta 3.

[00:31:01]This is equal to when I expand it up to order delta 3.

[00:31:10]This is 1 minus delta 3 theta 3 0 one. Essentially whatever I did yesterday to that you add one row and column with a unit element and along the diagonal.

[00:31:36]Okay.

[00:31:38]So now this is equal to I plus this is a unit matrix 111 I take it out and then write it as delta theta 3 * L3.

[00:31:57]So this matrix L3 is then given by 0 -1 1 0.

[00:32:16]Okay.

[00:32:19]This is the infinite decimal decomposition of the rotation along the third axis.

[00:32:25]You can do that for R2 what will it be I will write again same thing it's a unit matrix plus delta theta_2 * L2 so what is my matrix L2 These are all traceless matrices.

[00:33:10]-1 0 0 1 trace plus antiymmetric matrices. Well, antiymmetric it's enough. Okay, is this clear? I'm just repeating the same procedure.

[00:33:33]Okay, if there's any doubt, please stop me.

[00:33:38]And R3 is similarly I L1 where this matrix L1 is again another anti-ymmetric matrix which is 0 0 0 1 - one sorry 0 1 0 So any general rotation or which is a product of all these things or which is a combination of all these three infinite decimal rotations.

[00:34:37]I can write this as product of these three I of course at this stage I can put plus dot dot dot for higher order terms.

[00:35:11]Okay. And therefore this you can write as the identity matrix plus summation over I delta theta I.

[00:35:28]You can simply write it as I plus a matrix A where this matrix A is simply 0 - delta theta 3 delta theta 2 minus delta theta_1 this is delta theta 3 0 delta theta 2 - which is simply this.

[00:36:14]Now at this stage as I said I need to put dot dot dot because there can be higher order terms. Now I take the limit. Now this delta theta I mean I going to zero but in such a way that I want a finite transformation.

[00:36:35]This is R.

[00:36:38]I will define theta I to be some n * delta theta i and take the limit n going to zero. So therefore limit so r theta i is equal to limit n going to infinity of i plus summation / i and this is 1 by n 1 by n theta I in fact whatever I'm doing in this part I could have done it for S so4 or SO5 just by changing the summations it would work.

[00:37:43]Now this to the power n that means I take this infinite decimal transformation and in the limit n going to infinity after defining theta i as n * delta i. This is nothing but the exponential.

[00:37:59]This is the same step as what we did yesterday. But we didn't have we had only one generator and one parameter.

[00:38:08]And here we have three parameters and three generators of rotation.

[00:38:15]So therefore I can now write this arbitrary rotation is exponential of summation of i l.

[00:38:40]Okay.

[00:38:43]So these LI are called as the generators of rotation and they are given by this whether it is infinite decimal rotation or otherwise they are given by these matrices.

[00:39:01]Okay.

[00:39:06]So therefore I have the following property of the group element general rotation where there are three independent rotations theta 3 and these are generated by the matrices li with the parameters theta 1 and theta 2, theta 3.

[00:39:55]Okay, one thing is you must be very careful here that this is not equal to these are all matrices. They are not numbers. Right? parameters of course take real values but these L's are matrices. Therefore you must be careful that this is not equal to right but then I will have a problem. The problem is the following thing that suppose I take r of someta it's a vector theta there are three independent rotations let's say r of some other rotation let's say five the closure property of the group says that this should be equal to R of S right some angle.

[00:41:21]Of course that angle will not be theta + 5. This is three three dimension. Okay.

[00:41:26]But it but it it should have this form of the group element.

[00:41:34]It should have a closure property. So if r of theta is given by this r of s should also be given by the same form but I know it is not I cannot just combine these two like this.

[00:41:52]Okay. If I do, if I can do that then this S will be simply theta + 5 vector that means theta 1 + 51 plus theta_2 + 52 theta 3 + 53 for s 1 s2 2 s3 okay so there is we need one more property in order to satisfy it so there I mean why is it why is this not true in general for any matrix that is because there is what is called as a Baker Hosdorf Baker Campbell. Familiar with this house dog identity.

[00:42:46]This identity combines the exponentials of matrices. I won't won't go into the proof of it. I'll just state it. It's a very important identity.

[00:42:58]Suppose you have an exponential of a matrix A times an exponential of a matrix B.

[00:43:10]This is equal to A + B.

[00:43:16]That is the kind of thing that we want here for this to be also to be rotation but it has additional terms.

[00:43:24]It has 1 by two commutator of AB a and b are some arbitrary matrices square matrices of the same dimension you're multiplying them. So then this identity tells you that the exponential of these matrices product of them is given by this plus I think 1x 12 I'm not sure let me check a comma a comma b plus you have infinite terms like Okay. So therefore there is no way that I can take this as this form this also in the same form replacing theta by five and I can get this unless I know what these commutators are and unless I know and how do you get see remember this is the exponent is linear in LA here also it will be linear in LI and we want this to be linear in LA.

[00:44:43]How do you get that? You will get it only if these commutators of these matrices L are also linear in L.

[00:44:56]Okay. If they are not then the whole uh presumption assumption that we have made about the rotation group goes away. So in order to do that we should look at the commutator of these matrices LI right. So let us look at that. So let us see the what is the commutator of let's say L1 L2 this is L1 L2 minus L2 L1 that's the definition of any commutator between two matrices can you tell me what this is just do it takes just 2 minutes, maybe 1 minute.

[00:45:51]You have the matrices L1, L2, L3 here.

[00:46:00]Tell me what you get.

[00:46:30]They're all mostly zeros and ones are minus.

[00:46:40]Shouldn't take this long.

[00:47:34]Yes.

[00:47:38]>> Huh?

[00:47:44]>> Ah, I made a mistake. This is plus and this is minus, right? Because of the form of L2.

[00:47:56]Sorry, I made a mistake. But that's a very important uh observation because it is what is called as the right hand thumb rule that we have to follow. You have XY, YZ and Z X. Okay. So in the XZ plane it is not XZ but Z X. So that means this this portion will be uh where is this here? This is Z X.

[00:48:27]That's how it should come. That's the right hand thumb rule you have to follow. Sorry.

[00:48:32]My it's my mistake now. Can you tell me what it is? L3. Correct. It should be L3.

[00:48:45]Okay.

[00:48:48]Now can you guess what are the others?

[00:48:51]Suppose I take L2 L3.

[00:49:00]What is this commutator? Can you guess?

[00:49:04]L1. Right? You go cyclic order. So this is 231. So this will be L1. And the other one L31 is this. So you can guess that there is an algebra structure here where it is something like LI LJ is epsilon I J K LK okay so now every commutator yields an L which is linear in L Right?

[00:49:48]So that satisfies the criterion that we wanted for this. This is a + b and this is also linear because a commutator b that means some lj will give you lk and here this will give you a linear in l taken with a again whatever that is it is either uh zero or it is linear in l. Okay. So therefore you will get the structure that you need even in the product.

[00:50:26]Of course we can explicitly check this but it's important to know that it's not so obvious. See it was obvious in the case of SO2 because there was only one parameter and therefore we could easily do that. But here it is not so and then and there there was only one matrix. So it was the same matrix and the matrix commutes it with itself always. So in S42 this problem didn't arise but here it is important. So therefore this closure is satisfied because of the existence of this algebra.

[00:51:09]Okay. So we call these LK case as generators of the group of the rotation group in three dimensions. Proper orthogonal rotations are special orthogonal rotations in two dimensions.

[00:51:27]Okay.

[00:51:52]So I can write here this is equal to exponential of summation of i l i s I where r of theta r of phi are given by this form. So then what will be the relations between uh these angles and other things is a given by this uh you know very difficult relation. We will not go into that. Okay.

[00:52:28]The structure of this group and the algebra of the gener this is called as the algebra of the generators and these are the generators with this algebra we generate the group elements on.

[00:52:46]So groups with this structure are called as le groups.

[00:53:03]So this R of theta it forms a the rotation group SO3 is a le group Lee it's a mathematician susly so it's the named after the mathematician And associated with this flea group is the algebra of the generators LI LJ.

[00:53:42]So this is called as this is in general true of any s any s will have n into n minus one parameters and so many generators. So all rotation groups in n dimensions are le groups and they have an associated leal channel.

[00:54:17]If uh you go and do particle physics then you will come across more such groups like for example uh unitary groups s they are also le groups with associated leology. Okay. and one such group I'll be discussing tomorrow. Okay.

[00:54:38]So now next I want to consider the representation of these rotations.

[00:54:48]But before that is this this algebra is familiar to you right? You would have seen it somewhere.

[00:54:59]This is the angular momentum algebra. Of course, I don't yet have the interpretation of angular momentum, but it looks like the angular momentum algebra. But suppose I replace LI by - I H bar L I.

[00:55:18]Okay. So these LI are the angular momentum operators. I haven't specified what they are, but I'll just write it so that an arbitrary rotation can be written as e ^ I think it's plus i by h bar theta dot l.

[00:55:49]Okay, these l's obey the same angular momentum algebra.

[00:55:58]Uh I have to be careful with sign. I should put minus here minus here. So that this algebra of these LI angular momentum algebra is this is simply the angular momentum algebra that we are familiar in quantum mechanics.

[00:56:35]Okay, it physically makes sense.

[00:56:39]When do you get rotations? When you have angular momentum. So angular momentum generates rotations. So what are the generators of angular momentum? These are all are the generators of angular momentum. And what do they generate?

[00:56:55]They generate rotations.

[00:56:57]That's the physical interpretation. And that is what this equation completely encodes.

[00:57:05]Okay, but we can get the physical interpretation straight away by following the next step that we took yesterday.

[00:57:14]The subsequent one is how do the functions change under this group. So action of rotation on functions.

[00:57:44]So let's consider a function s call it x1 x2 x3 okay and you you now look at rotation in a plane for example just to make it simple okay so let let me look at the rotation along uh with z axis and then its effect on the function five.

[00:58:14]So that means x3 is not changing. Only x1 and x2 are changing. So therefore x s of x1 + delta x x2 + delta x 2 and x3. Let me consider this change corresponding to a plan of rotation with Z axis as an invariant axis. So this we have done yesterday and then we get this to be S of X1 X2 X3 and you expand tailor expand in two variables. The third one is remains constant remains unchanged. So then you will get this is d / d x1 delta x1 + v by dx2 times s and we noticed this yesterday delta x1 and delta x2 delta x1 is minus y * delta theta_1 delta theta and this is y * x * delta theta. There's only one angle which is changing here which is theta 3. So with that you can put this equal to sign.

[01:00:22]Okay.

[01:00:24]So just as we did yesterday, this is nothing but the angular momentum.

[01:00:30]Okay. This is sorry I mean x1 x2 x1.

[01:00:43]Okay.

[01:00:44]This is the angular momentum in the third direction.

[01:00:49]Right.

[01:00:57]So therefore we can write this Yes.

[01:01:22]L3 where this L3 angular momentum is a generator of rotations with Z axis.

[01:01:36]Right? So this L3 is minus.

[01:01:55]Okay.

[01:01:58]This is the generator of rotations around the set axis. So similarly you can do this with the other two variables.

[01:02:09]Okay. So you can define L2 which is minus I H bar Z delta by delta sorry X3 delta by delta X1 minus X1 D / D X and similarly L1 Similarly, L1 which is which is uh X2 - x3 you keep the order cyclic you know that's like the right hand thumb rule that you have to do is this okay right so you can write all the generators this thing now the interesting thing is that these are really the angular momentum components.

[01:03:37]Okay, I will leave it as an exercise for you that you can write you can take any LI LJ.

[01:03:49]Yeah.

[01:03:51]Okay.

[01:03:52]Commutator is I H bar epsilon I J K M N M N MK.

[01:04:08]Here we know directly that these are angular momentum operators because if you take R cross P in quantum mechanics and write the operator form for P you precisely get these L1 L2 L3 and they obey this algebra.

[01:04:28]Okay, it takes time for me too much time for me to do this algebra. So therefore please take this as homework. We check all you have to do is that take this commutator acting on some function s and then show that it reduces to this acting on some function s. Okay, this is normal thing that you do in quantum mechanics to verify operator algebbras.

[01:04:59]Okay, you can do that. So please check it yourself and it's the same algebra that you have here except that I have changed the notation here from these lis that means this anti-ymmetric matrix to by introducing an I what I have done here is that these are now herian matrices L dagger is equal to L because I have introduced I roo<unk>US1 these are real and therefore So they're antiymmetric matrices. These all are hermesian matrices. So these are all are the capital allies are the hermesian generators of the rotation group.

[01:05:45]Okay.

[01:05:49]Is that clear?

[01:05:52]So normally you do all these things in quantum mechanics course but there you will be doing directly the angular momentum algebra. Whereas here we are starting from the rotation group developing the algebra of the generators which is actually a le algebra and then identifying that algebra of the generators with the angular momentum operators.

[01:06:17]Okay. And it physically makes sense that angular momentum should be the generator of any rotation. If the angular momentum is zero then there is no rotation.

[01:06:29]It is very similar to for example translation.

[01:06:32]Generator of translation is momentum. If momentum is zero, there's no translation. Okay. So one must associate these mathematical identities with physical quantities.

[01:06:45]Okay. So now I'll change I have another 15 minutes. Let me see what I can do.

[01:07:03]Yeah, I will introduce I mean there I didn't assume any function I just said any general sigh. I want to introduce what are called as spherical harmonics.

[01:07:19]These are functions that we normally use in quantum mechanics and they play a special role in quantum mechanics. So I want to introduce these functions and then look at their properties and the rotations and I'll also tell you that the spherical harmonics are the tensors which transform under the rotation group. So they belong to the rotation group. So spherical harmonics.

[01:08:12]Where do you come across these functions? spherical harmonics hydrogen atom also 3D harmonic oscillator right they appear in all these problems so why do they appear in all these problems that's a question we cannot try to answer so this spherical harmonics suppose you take the vector I'll now use the notation XY Z. Please don't get confused. It's a when I want to write a summation, I use it as X1, X2, X3. Otherwise, in three dimensions, it's XY Z.

[01:08:57]So the vector has components XY Z. This is R.

[01:09:05]Okay. I prefer to write this as x + i y z and x - i y and change to spherical polar coordinates. So in spherical polar coordinates what is x h?

[01:09:37]If uh the orientation is theta 5 x is r sin theta cos 5. Correct? So this should be internalized because one keeps doing this in quantum mechanics.

[01:10:01]Y is good.

[01:10:13]Okay.

[01:10:15]So now this express I y is r sin theta e ^ i5 and this is r cos theta and this is r sin theta e ^ - i.

[01:10:42]Okay.

[01:10:46]And in this basis we call this this is apart from normalization factors this is simply our y11 theta 5.

[01:11:04]This is r y 1 0 theta y and this is r y1 - 1 theta y.

[01:11:22]This is the tensor notation.

[01:11:25]A vector is really a tensor of rank one.

[01:11:29]So these are y why why these y's are called as the spherical harmonics.

[01:11:34]This y1 m of theta 5 called as spherical harmonics of rank one.

[01:11:57]It's a tensor of rank one. And any vector in three dimension is a tensor of langu.

[01:12:04]You're familiar with this right? Spher spherical harmonics. So this is and why do we write it in this form?

[01:12:14]That is because given the representation of L on the function space given them uh as operators like this you can express these things in terms of L3 and L².

[01:12:33]So L has components like this and I can express all these things in the spherical polar coordinates.

[01:12:44]This you will do in quantum mechanics.

[01:12:47]So I will just write the operator forms.

[01:12:53]Okay, you can check it. Okay, these are all very straightforward relations that you can check. So you can write that L3 in spherical polar coordinates L3 is equal to - I H bar D by D5 note that because it's X and X there is no dimension so it depends only on the angle and it's just D by D5 if you write it in the spherical polar coordinates and then if you take L² operator which is just L L1 L1 L1 L2 into L2 L3 into L3 from the operator form and then the L² is - HR² 1 by sin theta.

[01:14:00]The partial derivative with respect to theta sin theta d by d theta that is the operator.

[01:14:30]L².

[01:14:32]All I have done is I have taken these forms and I have written them in the spherical coordinate form. Why do I do this?

[01:14:43]That is because I can now express this violence as states of L² and L3.

[01:14:56]So you can check it's now straightforward to check because this is d by d5 for example what is l3 operating on let's say y11 y1 is given here right which is sin theta e ^ i5 and then minus - i h bar. So this is simply h bar * y1 and l3 operating on y10 is there is no five dependence in this so it is zero and similarly l3 on y1 -1 is - h R 1 - okay so therefore if you put all these things together you can write this as L3 Y1 M is H bar M 1 M so these are all states It's of the operator L3 which is the third component. Remember this is H bar minus H bar and zero.

[01:16:38]Okay. This M takes value + one 0 and minus one.

[01:16:54]And if you look at L², I'll just give you the result.

[01:17:04]I think you would have many of you would have done it in your quantum mechanics course. L² operating on this Y1.

[01:17:12]What is it?

[01:17:15]L into L. There's no L. This one, right?

[01:17:23]L into L + 1 is two because it's rank one Y1 M. So therefore this is 2 H bar² sorry L² operating on Y1 M is equal to 2 H bar Y1.

[01:17:44]So these are simultaneous states of L3 and L² and this two is nothing but 1 into and it's the same for all N.

[01:17:56]So therefore the quantum number corres the value corresponding to L square denotes the representation and within the representation L3 operating gives you the various components of these spherical harmonics of rank one.

[01:18:17]Okay. Now you can construct higher I'll just mention this I think it will be done in quantum mechanics uh lectures if not I think we can have a tutorial also so in general you can construct higher rank tensors.

[01:18:42]Okay. Now for example, if you want to construct a second rank tensor, how would you do it?

[01:18:49]If I was dealing with XY Z, this is rank one, which is the components of the vector itself. I can construct a rank two tensor which is simply I can take x² y² z² xy y z and z x you can call as a cart cartisian rank two tensor they're constructed from the xyz component itself But this is not a convenient notation because under rotations this length of the vector is invariant that doesn't change.

[01:19:43]So there is one condition that x² + y square is equal to r² and that remains constant under rotations. So therefore out of these six there are only five independent uh second rank tensors you can construct. Okay. So how do you construct these second rank tensors?

[01:20:02]The algorithm is the following. You can start with x + i y whole square instead of this.

[01:20:12]Okay. So this x + i by square is simply the square of this which is equal to r² sin² theta e ^ 2 i and similarly you have x - i y² which is r² sin² square theta e ^ - 2 i there are only five okay so this when l3 operates on this this brings down two instead of one and here it brings down minus2 these are states of the operator L3 so therefore the m value goes from + 2 to minus2. So there will be other tensors in between 2 -1 0 sorry 2 + 1 -1 and -2.

[01:21:26]Okay. So L3 operating on this and how do you construct the other tensors to you can uh you can construct them not with this L3 and L² because these are again vectors I mean again states of L3 and L² you can use the other components construct L+ and L minus which is L1 + IL L2 and L1 minus IL2 they are called as L+ and L minus operators what they do is they either step up the M value or step down the M value it's a long algebra to do that so I will not do it. So you can construct all the second rank tensors using this. So I will just say I'll generalize this that in in particular you can take any x ^ this which will be r ^ l sin ^ l theta e ^ i l This will always be proportional to y l of theta 5.

[01:22:58]This is what is called as the highest weight tensor.

[01:23:02]Okay. And in general all the other components I'm saying that not just square you can construct any L rank rank L tensor and then in general these are called as Y LN rank L and N is the number of components of that tensor. If it is L= 1 you have three components that's a vector. If L is equal to 2, you have five components. That's a rank two tensor. Okay. So basically you will have 2 J + 2 L + 1 components.

[01:23:39]Okay. So these are called as the YMs and you can generate them using this representation of this angular momentum operators in the position space. So Y LM with the property that L3 operator acting on Y LM is H bar M Y LN and L² operating on Y ln will give you H bar² L into L + one the thing I heard some time back okay so these are the tensors on the rotation group and why do we write it in this form?

[01:24:27]The interesting thing is that any representation of rotation on these functions spherical harmonics will not change the rank of the tensor.

[01:24:43]That means if you take the rotation of representation of rotation as some dl M M prime operating on some Y lm summation over M will give you Y L M prime Okay.

[01:25:13]And the representation of rotation will not change the rank of the tensor but it will only rearrange the M values.

[01:25:26]That means it will change it within this set that you can see in the cartisian tensors that XY Z any rotation will rewrite X prime in terms of XY Z Y prime in terms of XY Z prime in terms of XY Z so formally that statement is simply what we have in terms of these Y lms this is called as the D representations s of the rotational graph. Okay. So I I have fast forwarded too much here. Uh but most probably you'll be doing all these things in your quantum mechanics course.

[01:26:10]So I just wanted to summarize how to construct tensors on the groove and look at their properties. In the case of rotations, it is the spherical harmonics.

[01:26:23]Okay.

[01:26:26]I'll stop here. Uh uh most of these things are written in my notes which I think either you already have it or it will be given to you. So the details which I have sped you will find it written somewhere. So any questions?

[01:26:56]Tomorrow I'll show you an explicit example where all these things come together. Yeah.

[01:27:01]>> Sir, you talked about the right hand thumb rule.

[01:27:04]>> Uh could you please explain more about that? Oh, that is uh that we whenever you you have this XY Z you take it in the cyclic order that means uh this uh this is sometimes called right hand thumb rule because that when you use that in the case of currents uh in a magnetic field that is what happens but all it means that is that suppose you want to uh you have XY Z and then you want to rearrange them you arrange them using cyclic permutations.

[01:27:39]See given three symbols you can permute them in any many different ways. Okay, cyclic permutation will give you three different permutations.

[01:27:49]So XY Z if you permute it then it will become Y Z X and then it will become Z Y X. Okay, that's a psychic permutation.

[01:28:01]Huh? Z XY Z Y Z X and Z X Y. Okay. Not Z YX. That would be the odd one. Okay.

[01:28:12]Basically I mean cyclic permutation.

[01:28:16]Yeah.

[01:28:18]>> Yeah.

[01:28:23]>> So you were discussing about tenses.

[01:28:25]>> Yeah. So like uh in this context do we have any like metric kind of uh scale to change from co contravariant to covariant or something like that? Not here. Not here. Not that that will come.

[01:28:38]Uh here you can either use all covariant or contraari that contraar. See the because the metric here is uh dx² + d y square + dz square all have the same weight that is not so when you go to special relativity. So there you will have to you will have covariant and contraariant. Yeah you can use that notation but it will be simply a notation. Say for example I can write this as dl okay so I can use m prime uh as coariant and contraariant but it doesn't make any sense okay because the metric is unifor the same weight for all the coordinates >> so adding on to that that means we don't have any mixed tenses in rank two or something like that >> sorry >> we don't have any mixed tenses in rank two or something like that >> here no >> we don't have So the ind the index does not make any difference over here.

[01:29:36]>> Yeah, that's right. that co uh I'm not too confident in answering that there is that whether you can never construct such things but as far as I know it doesn't make any difference okay for example you can there are spherical tensors you can construct from rotation group they are also like DJ DL matrices these are sometimes called DJ matrices they in fact I didn't use the term What it means is that when it does not change the rank of a tensor at all under rotations, it provides a representation of a rotation and and that rotation is an irreducible rotation. Okay, it's called irreducible because it doesn't change the rank.

[01:30:25]Okay, so about your other question, I'm not I'll wait to answer that. Okay.

[01:30:35]Any other question?

[01:30:42]Okay.