 Let's get started. At the end of the other last time, it just began the beginning of a new topic, which concerns the bigger record-care of the iridescent offensive operators. It's kind of a long story. This is the most mathematical part of the whole course. So please bear with me. In any case, the context is, if you have a unitary representation of rotation acting like this case, which I call E, by means of a little accessible orientation, we can find angular momentum operators from then we construct J squared and J z, and from then we construct the simultaneous eigenpaces, which we call standard angular momentum basis, and that's the basic setup for what happens with rotations act on ket spaces. Now, let's let psi be some ket, and let's let rotation operator act on it. We can call this thing older than other rotations in state, but let's call the new state psi prime, or call it the rotated or new state called psi prime. This is how the rotation operators act on ket factors. Now, if A is an operator, let's say A is standard for some operator, we like to find a rotated operator just like we have rotated states. Let's call a rotated operator A prime, and the question is what should be the definition of it. It's going to be something that attends to the old operator. What's the meaning of a rotated operator? Well, the definition of this is that we're going to require for the definition that the expectation value of the rotated operator with respect to the rotated state should be equal to the expectation value of the unrotated operator with respect to the unrotated state. So the requirement is that it would be sandwiched psi prime around A prime. We're not going to get the same thing as psi sandwiched around A. This is just a definition. In fact, it's a definition of A prime. It's a requirement we're going to impose, but it's a reasonable thing to do. Now, if we do this on the left-hand side, since psi is U of R times psi, the left-hand side becomes psi sandwiched around U dagger H prime U. And that is equal to the right side, which is psi sandwiched around A. And since psi is arbitrary, this is supposed to hold for all psi, we obtain that U dagger H prime U is equal to A. And if we then solve this for A prime, we can fill in the question mark over here, which is that the rotated operator should be defined this way. It should be U of R dagger times A times U of R. And this will be our definition of rotated operators that we'll use. Now, it's customary in classical tensor and vector and tensor analysis to classify physical quantities by how they transform their coordinate transformations, such as the rotation as an example. For example, we talk about scalars, vectors, tensors, and things of that sort. And also, quantum mechanics is a similar classification scheme for operators. It's a classification as to how do they transform under rotations. And so I'd now like to tell you about that. In the first place, let's take the simplest example, which is that a scalar operator. A scalar in classical tensor analysis is a form that's invariant under rotations. And we'll make the same definition here. We'll call an operator K a scalar. If it's the same operator we get after we rotate it like this, we rotate the operator. The terms are back to the same operator. This should hold for all rotations R. And I forgot to mention that these are proper rotations only. We'll deal with improper rotations a little while down the line. So if the operators are invariant under all rotations and we call it a scalar, it's a logical definition for a scalar operator. The most important scalar operators from a physical standpoint are Hamiltonians for isolated systems. They're invariant under rotations because the energy does depend on the orientation of the system. And this has certain implications for the spectrum of the operator and the degeneracies and so on, which we'll talk about for a little bit probably getting into the next lecture. But in any case, this is the definition. Now the next thing we'd like to define is a vector operator. Now a vector operator in quantum mechanics really means, we say a vector operator, but what it really means is a vector of operators, a collection of three operators that have certain transformation properties under rotations. It's not sufficient just to have three operators that've got to transform properly. And for example, in hydrogen anatomy, you've got three operators, H all squared and LZ, these are certain nice operators, and I can put them in a column like that, but that's not a vector operator because it doesn't transform as a vector. So in general, let's call a vector operator over what it is, V here, like this, and you can think of it as having components, V, Y and DZ. And the question now that we need to address is what should be the proper definition of a vector operator? Well, once again, let's refer to an unrotated state psi and a rotated state psi prime. If we take the expectation value of these three operators with respect to any state, of course what we get is a collection of three numbers. You can think of it as a classical ordinary vector, a classic C number vector. Well, our requirement for the definition of a vector operator shall be that the expectation value, which is a vector of ordinary numbers, should transform as a vector of numbers does in ordinary tensor analysis. So the requirement is this. It is that if you take psi prime, sandwiched around the expectation value of the vector operator in the rotated state, this should be the rotated version of the expectation value of any vector operator in the unrotated state. In other words, expectation value is transformed as ordinary vectors in three-dimensional space. See, this is a vector of numbers here. Multiply that vector, this is a column three vector multiplied by the matrix. This is another column three vector. That shouldn't be the expectation value in the rotated state. Now again, this is just a definition. This is a definition of a vector operator. Let's notice that since psi prime is u times psi, the left-hand side is the same thing as psi sandwiched around u-gag or pu like this. And since these two things, two sides have to be equal in size arbitrary, this implies that u prime times the vector operator times u must be equal to the rotation matrix multiplied onto the original vector operator. Or to put this into a slightly more convenient form, let's just swap R in our inverse and at the same time swap u and u-dagger. And if we do, then this becomes a lighter over here. This becomes u times the vector operator times u-dagger is equal to our inverse multiplied onto the vector operator. And this is a definition of the vector operator. Just as this is the definition of the scalar operator. Now in case this definition of the vector operator is not clear what it means rotationally, let me write it out into components. I'll just say more here. This is u. The u-dagger conjugates one of the components of the vector operator. This is the rotated version of that component. It turns into our inverse is the same as our transpose. It turns into the sum on J of v sub j times rj i. rj i is the components of our transpose. So this is an equivalent statement of the first one to the shorthand of the second statement there. So these are definitions of vector, scalar and vector operators. Similarly, we can go on and define a tensor operator. A tensor operator is what's called a Tij. It can be thought of as a matrix or a tensor of operators, nine of them, i and j 1 over x, y and z. And it has the property that if we rotate a component of the tensor operator by conjugating u, that's the rotated operator we've written over here. And we get a linear combination that looks like in which each index transforms in the same way as the single index does of the vector operator, basically by the matrix r So what that means is this must be equal to the sum of k and l of Tkl times rk i times rlj and that's the definition of a tensor operator. A tensor operator rank 2. You can go on to define a higher order tensor operators. The most important tensor operator for this course is going to be the quadruple moment operator, electric quadruple moment, so we'll have something to say about that before too long. But in any case, this is a definition. Now, these definitions can be cast into somewhat different forms. Let me first of all take the scalar which is the simplest case here for me to raise this in a little room. It's pretty obvious just from the definition that if I multiply by u of r on the other side, this is equivalent to saying k times u of r is equal to u of r times k. And so, in fact, it works both ways. And so this is another definition of a scalar operator. A scalar operator is one that commutes with all rotation operators, as you see, or if the rotated operator is the same as the original operator, it means the same thing. Here's another equivalent version of a scalar operator. If this equation holds for all rotations which it's supposed to here, it must hold in particular so let's look and see what happens in the case of infinitesimal rotations. I'll do this over here. The u of r for infinitesimal rotations can be written this way. One minus i times theta times n hat dotted into j. And I'm, by the way, I'll set h bar equals one today just to say writing. So that's the u of r factor here. And then I'll copy the k and then for the u of r dagger I just take the dagger of this so this becomes one plus i theta times n hat dotted into j. This is a rotation in axis angle four where we're assuming the angle is small so it's an infinitesimal rotation operator. And if this is a scalar it has to equal k. Now if we expand up the left hand side, multiply it out to first of all the k at lowest order of theta and the next order of term is i theta times the commutator of n hat dotted into the j with the operator k which is a quadratic term which I don't ignore because it's only lowest order but it's important here. So this has to equal k and the k's obviously cancel but it's obvious that the commutator must vanish. So it says that the commutator of k, our scalar operator with n hat dot j is zero. This thing has to equal zero. But n has a unit vector. It's an arbitrary unit vector. If we choose it to be x hat, y hat and z hat, the result is that k commutes with all three components of j. And so we get the condition then that if k is a scalar operator it commutes with all three components of the angular momentum. This is an alternative way of writing the definition of the scalar operator. Actually all I've shown so far is that if k satisfies this equation, it commutes with rotation operators then it commutes with angular momentum. But the converse is true too because if k commutes with angular momentum it commutes with any function of angular momentum and that's just what the rotation operators are. So it works both ways. So these two things are always drawn by a line here. These two things are actually equivalent. There are equivalent definitions of a scalar operator. So in particular if you want to find out if you're held, tell yourself if you're holding the angular momentum that's one of the ways of doing it. In order to apply the answer to zero it means that it commutes with all rotations. Now let me do a similar calculation for vector operators which are here. Let's take a look at what happens with a vector operator and if the rotation is infinitesimal. So it's a very similar story. I'll plug in, I'm looking at this equation right here for the testable form. One line is sine theta and n hat dotted into J. This is multiplying on the v. And then we have one plus sine theta and n hat dotted into J for the u dagger on the other side. And this should be equal to r inverse v. Well I'll remind you of infinitesimal rotation. An axis angle form has to form an identity plus a theta times n hat dotted into these spring J matrices which to remind you that now here is a summary here they are 1, 2, 3. That's the basis of any symmetric matrices in three dimensional space. So that's what this is. And if I take the inverse of this then all I have to do is change that pluses from minus. And so this has got to be equal to an identity minus theta times n hat dotted into J, multiplying our vector operator v like this. Now again I'll span both sides out and multiply both sides out. The first term here is v. The next term is minus sine theta times the commutator of n hat dotted into J for the vector operator v. On the right-hand side you get v of lowest order and then you get minus theta times n hat dot J dot n multiplying over v. But n hat dot J multiplying over v is the same thing as n hat cross v. It's a cross product. So this becomes minus theta times n hat cross into v. This is using the properties of the script J matrices. But again the leading term cancels. And you can see the theta cancels as well. And I can bring the minus i over the other side and I multiply both sides by i. And so we get n hat dotted into J, commutator with vector v is equal to minus i times n hat cross v. If v is a vector operator then we get minus i by definition. Now allow me to transform this into somewhat more useful form. Let's take the J component of both sides like this. So on the left-hand side it becomes n hat dotted into J and then V is the J for the J component. And on the right-hand side it's minus i. And then it's let's call it J i k times n i times n i v J like this. So n hat dot J can also be written as n i times J i. Again I'm using summation convention here. This is a nice way of writing it because I've got the arbitrary vector n i on both sides. And since it's arbitrary it can be cancelled away. And what's left then is a commutator between J i and v J. Let's take this x1 J i k and write it as minus x1 i J k by swapping the first two indices. If we do this to summarize the results it says that the J i with v J, the commutator is equal to i times x1 i J k times v k. And this is a condition in which vector operators must satisfy. So in other words what I've shown is that this original definition of a vector operator implies these commutation relations with angular momentum just like this definition of the scalar operator implies those commutation relations with angular momentum. We might say that there's two different versions of the definition. There's one involved with finite rotations that you see here and these commutation relations are effectively what happens when the definition is infinitesimal. Likewise for a vector operator it turns out it's for finite rotations and for infinitesimal rotations it turns into commutation relations. By the way this condition also implies that these are actually equivalent definitions. Likewise it's possible to express the definition of the tensor operator which is way up at the top of the board in terms of commutation relations with angular momentum although I won't bother you. Now more about vector operators let's let me first mention that at this point there's a couple of simple theorems you can prove what you've got this definition of a vector operator. You can show for example that if you take two vector operators A and B and your partner dot product it results in a scalar operator. This makes sense because it's just what you have in the classical tensor analysis or if you take the cross product you can show that that's a two vector operator you can show that that's also a vector operator. It follows the usual rules on the vector analysis. What about some examples of vector operators? To make a specific context let's suppose we've got a particle moving in three dimensions so the wave function is psi and the radius r and let's suppose there's no spin like this then in that case we know the angular momentum l is equal to r cross p. Now the position vector r is really a vector of operators with components x, y and z and you'd think that it really ought to be a vector operator by the official definition. Well one way to check is to compute its commutation relations with the angular momentum. So another way to compute its commutators the j here is a general notation for angular momentum which in this context is l and the v here let's replace that by r already some components is x of i or x of j every one. So you compute the commutator of l i of x j and if you just work on the commutator what you find is that it's i epsilon i j k times x k it's a simple commutator to do. So you see it does satisfy these commutational relations and that shows that the position operator actually is a vector operator according to the official definition. Likewise it's pretty easy to compute the commutators of the components of angular momentum to a linear momentum but if you do that you'll find you get i epsilon i j k dk showing that the linear momentum is also a vector operator just as you would expect. Given those two facts you can find some other facts by the way the dot product of two vector operators is a scaler that means that the dot product of the position vector times itself which is r squared is a scaler and it's also very logical but it means that the potential energy in the central force problem is a scaler operator. Likewise the dot product of the momentum operator with itself is a scaler operator and that means the kinetic energy in the central force problem is a scaler and therefore the central force Hamiltonian is a scaler and these are pretty obvious conclusions but this is in detail how you show it in all greater Now again I'm using these these theorems that I quoted here that for example the cross problem of two vector operators is another vector operator once you've shown that the position of r and the momentum p are vector operators it follows that angular momentum is also and so it must satisfy the same commutation relation with j replaced by l now v replaced by l so that's these commutation relations l i l j is equal to i epsilon i j k l k and you recognize this as being just the standard commutation relations for angular momentum in fact this holds for any angular momentum on any system not just orbital and so you see by this definition you see that the angular momentum operator is always a vector operator according to this definition or equally that would appear this would not appear in fact if we apply if we substitute v equals j for this equation right here the definition of vector operator what we obtain is the adjoining equation for how angular momentum transforms under conjugation by rotations so the adjoining formula is actually a statement that the angular momentum operator is a vector operator that's another way of doing that now I'm going to change the subject slightly for the next topic the spherical basis I want to tell you now the spherical basis the spherical basis is a basis of unit factors in ordinary three-dimensional physical space but it's not the same as the original Cartesian basis what I'm going to do first is just pull the definition out of the air with no motivation and show you what the properties are then I'll show you what it's good for and finally I'll give you some deeper insight as to where it comes from and what it gains so first of all let's talk about the Cartesian basis which is our usual basis let's call these unit vectors c1, c2 and c3 those c stands for Cartesian and this is just the same thing as x hat, y hat and z hat like that the spherical basis is a different sort of basis factors I'll use the e for them there's one of them is e1 which has an overall minus sign and it's x hat plus i y hat over the square root of 2 look at this, e hat 0 which is z hat and then there's an e hat minus 1 which is x hat minus i y hat over the square root of 2 with an overall plus sign like this so this is just the definition we can denote these collectively by e sub q if it relates on the value 0 and plus and minus 1 you see that the index q looks sort of like a magnetic common number for something that's got an angle of 1 goes from minus 1 to plus 1 okay like I said this is just a definition pulled out of the air however let me show you some of the properties of the spherical basis vectors one of them is that they're orthonormal well before I get into that it's obvious that spherical basis vectors are complex vectors at least of what minus 1 are so this is a complex set of basis vectors for dealing with ordinary three dimensional space which is usually a kind of real vector space this means that it has a vector that has real components with respect to the Cartesian basis that it's going to have complex components with respect to the spherical basis seems strange but it's useful alright so here's some of the properties of this basis and it's all orthonormal in the following sense that if we take eq and dotted in the eq prime with a complex conjugate we've got a chronically delta qq prime that's just the orthonomality of complex vectors you have to put a star and one of them here's another property let's suppose v is a vector and let's define v sub q to be the component of v with respect to the spherical basis vector so it's eq dotted with v without the vector v can be represented as a linear combination of the vectors eq complex conjugated with coefficients which are vq you can verify that directly just by plugging this in but that's the same thing then as the sum of q of e hat complex conjugated multiplied times e hat sub q dotted in the original vector v since the original vector v here is arbitrary you can kind of cancel it from both sides and you end up with actually a resolution of the identity that looks like this identity is the sum of q z hat q star times z hat q this is the resolution of the identity in the ordinary three-dimensional space this jumps the position of two vectors next to each other like this is not a dot product it's rather an outer product but sometimes called a tensor product you should do a quantum mechanics when you write q over product q like this it's the same idea if we allow both sides to act on some vector here we have a dot product e hat eq dotted with v and the right inside the identity multiplies with v itself so it just reproduces this equation here alright so that's the properties of the spherical basis now what is it good for example where the spherical basis comes in it turns out to be very handy let's consider radiative transitions such as a hydrogen like atom or an alkali I'll remind you that these are central force problems if we ignore the spin of the electron the wave function has the form of psi nlm of a position r and this has the form of a radial wave function psi nl of radius r times theta times pi now let's suppose and the energies I'll remind you have the form e nl like this and some of the force problems energies depend on these two quantum numbers hydrogen, pure hydrogen is an exception because it doesn't depend on l but as I pointed out that's characteristic of the very special of the Coulomb potential that if you make any central force perturbations small even to the Coulomb problem you start to get a lot of dependence so in any case let's talk about two levels let's say an upper level which is calling n prime l prime and a lower level which is labeled by n and l like this and radiative transitions going from one to the other I'm calling these two levels levels but actually they're not just single levels there's multiple levels in there because there's also a magnetic quantum number n prime above which means there's really two l prime plus one levels indicated here by this line and they count below there's also a magnetic quantum number n so there's two l plus one lower levels down below the energy is not dependent on your primary so these lines that I've drawn here represent what they call manifolds of levels or collections of levels that are generated what you observe spectroscopically if you're looking at a spectral line is basically the frequency of the magnetic photons and that depends on the energy difference between the two levels but that energy difference doesn't depend on the magnetic quantum number so actually what you're really getting is a superposition of lines for all possible initial magnetic quantum numbers and all possible final magnetic quantum numbers they all produce photons of the same frequency and if you're interested in the intensity of the spectral line you need to add all those various sub-lines together because they'll all be the same frequency now it's shown in the theory of emission of radiation that the intensity of a given spectral line taking you from an initial state to a final state is proportional to a square of a matrix element in which you've got the initial and final states on the two sides and the position operator in the middle in the present case the initial state is going to be n prime l prime m prime and the final state is nlm like this actually what you're really doing the intensity is the square this is a vector of course because it's an expectation value of vector operator but you need to square it to get the intensity in the case of these are the matrix elements that need to be computed now by the way what's really going on is that the position operator is multiplied by minus e which is the charge of the electron this is the dipole operator for the electron this is the matrix element which is involved in dipole transitions alright now because of the common frequency for all the different transitions we need to compute this matrix element for all possible values of m prime all possible values of m as well as three components of the position vector so let's say for example this were a 3d state and this were a 2p state this is an l equals 2 and this is an l equals 1 the number of matrix elements that you would get would be 3 for the position vector r times 2l prime plus 1 for the possible n prime values times 2l plus 1 for the possible final m values and in particular for 3d to 2p this would be 3 times 5 times 3 which would be 4 to 5 you'd have to calculate 45 matrix elements to find the transition from 3d to 2p it looks like a lot of work most of the matrix elements vanish as it turns out in fact the vanishing of matrix elements is not really by what's called selection rules what we're driving for in this lecture and probably on Wednesday also is a more systematic theory of selection rules so this is a simple example of what we're getting at but there's still a lot of them that don't vanish I think there's 9 of them that don't vanish and so even if you calculate the ones that don't vanish it's still a lot of work alright now so how can we simplify this well in order to simplify this let's begin by making some observations about the yl elements so this is a little bit of a tangent now we'll put this on hold and then go on a tangent here are the 3 yl elements to the case of l equals 1 essentially we had a homework problem and I guess you've seen these already if we multiply these 3 yl elements by the radius r what we'll get is this r times y1 1 that's equal to this normalization minus the square root of 3 over 8 pi and then r sine theta times e to the i5 is the same thing as x plus iy likewise if you multiply r times y10 you get r cosine theta which is z so this is the square root of 3 over 4 pi times c and if you take r times y1 minus 1 we get now a plus sign square root of 3 over 8 pi times x minus iy like this on the other hand the spherical basis here here's the spherical basis up there and here's the definition of the components of the spherical basis with some vector let's say the plus 1 component of the position vector it's going to be 1 over the square root of 2 with a minus sign of x plus iy so those components of the spherical of the position vector with respect to the spherical basis are appearing here this can be summarized this way by saying radius r times y1 q theta over i is equal to this normalization which is the square root of 3 over 4 pi times e to the i sub q dotted into the position vector r this is what we might call r sub q the spherical basis component of the position vector r alright so I have to I've got it up here so here's what we'd like to do let me take this back down instead of working on the Cartesian basis x, y, and z of the position vector let's replace this by the spherical basis what we call r q which is the same thing as e hat q dotted of the position vector r in other words let's look at the let's look at what I'll do here let's look at this matrix element n prime, l prime, m prime on the left r q in the middle and n l in the middle of the right it's the same 45 matrix elements to do but now we can use this relation here that connects r q with the yl x in fact r q becomes equal to this normalization upside down square root of 4 pi over 3 times the radius r times y1 q of theta in phi like that alright and so if I if I look at this matrix element why did I notice the integral we bring it over here there's first of all a radial part of the integral 0 to infinity of r squared v r and there's a radial part of the function on the left that's r in prime, l prime of the radius r complex conjugated there's the radial part of the eigenfunction on the right which is r in the l of r then there's the radial part of the operator in the middle which is just a factor of r that's the radial part of the integral now let's do the angular part now we'll first I'll pick up this constant factor here square root of 4 pi over 3 which is the integral of the solid angles so we get a yl prime in prime from the wave function on the left which has to be complex conjugated we get a y1 q from the spherical components of the position operator in the middle and then we get a ylm for the the angular part of the wave function on the right and the result is is that this integral breaks into a radial part times an angular part and all the multiplicity that's occurring here of integrals that we need to do come upon the m prime and q that are effectively three magnetic quantum numbers and those three magnetic quantum numbers appear only in the angular part not in the radial part so our 45 matrix elements are really 45 angular matrix elements times one single radial matrix element that only depends on the else as far as the angular integral is concerned in the last lecture I explained you the three ylm formula which is used for doing integrals just like this and the formula itself was reasonably complicated with constant factors but the basic fact is it becomes proportional to a clutch particle efficient and it's this it's all prime m prime scalar product with l1 mq like this the stuff that doesn't depend on the magnetic quantum numbers you can look up in your notes and see what the other factors are but the point is the dependence on the magnetic quantum numbers now appears here at this clutch-gordan coefficient well one thing to say right away is that there's a selection rule for the clutch-gordan coefficient on the magnetic quantum numbers this is simply zero unless m prime is equal to m plus q in fact this explains why the magnetic solvents that are zero because they don't satisfy this condition this condition might not have been so obvious if we'd worked with Cartesian components of the position vector up above but by going with spherical components we get this selection rule so right away a huge number of magnetic solvents is zero more than that is that we really need to evaluate all of them I think there's nine that are left in this example that I talked about that are non-zero if you really need to evaluate all of them what you need to do is just do the one single radial integral there's nothing to do except just to do it here the integral whatever means you can but once you've done that by using the three-by-all-in formula the rest of it becomes proportional to clutch-gordan coefficients that you can look up or you can calculate them but in any case this greatly simplifies the problem of combining those magnetic solvents requiring a radiative transition this is the second step of the spherical basis first I'll show you what the definition was pull it out of the air and the second one will give you an idea of what it's good for now let me turn to the third step which is to show you the deepest significance of the spherical basis I began this lecture reminding you of how what happens when we have rotations acting on a catch space you've got rotation operators and planetary rotation operators and a catch space E like this from the planetary rotation operators we get angular momentum operators from those who construct J squared and Jz which can have simultaneous eigenstates and then we can construct the standard angular momentum basis like this this general program here doesn't this general program can actually be applied to any vector space upon which the rotations act it does not have to be a space of wave functions that's to say a catch space there's actually interesting possibilities another possibility instead of a catch space is this is to replace E by ordinary physical space ordinary three-dimensional space so in ordinary three-dimensional space we set up an inertial frame that's Y and Z like this and if we do this then the points of three-dimensional space can be identified by their coordinates and so they become real three-vectors R3 like this in mathematical rotation so we can say that E is replaced by R3 now do the rotations act in ordinary physical space well sure they do, they act by these classical rotation matrices so in other words U of R in this general notation above it's just replaced by R itself these rotation matrices are orthogonal and a orthogonal matrix is actually a special case of the unitary matrix so this actually is a unitary representation of rotations yes question when you said that U R acts on the physical space you just mean that by association is really the R that is acting on it it's not really the U of R is it well the general context is that you have a unitary representation of rotations acting on some vector space now when it was a catch space we call those U of R because those were ordinary operators like you might have expected but the space in question the vector space does not have to be a catch space so if it's going to be ordinary physical space then E gets replaced by R3 and now the question is do rotations act on R3 in a unitary manner and they do indeed by just by the usual rotations those are in fact unitary so in comparison with the notation we've been using we'll replace E by R3 we'll replace U of R like that capital R alright so if that's true then what we ought to be able to do is by looking at infinitesimal rotations we ought to be able to get a commision matrix excuse me a vector of commision matrices for operators that act on R3 that's satisfied by the angular and commutation relations and then from then we ought to be able to get a j square and a j z and from then we ought to be able to get a standard angular momentum basis I won't write it in the form of a catch but still it's going to be a basis which is an angular basis of j square and j z it's the same mathematical structure in different space so the first step is to ask what is the angular momentum of the commision matrices so to do that we say here's the definition of the angular momentum it's that U of R which in this case is the same thing as R is equal to 1 minus i theta get that out of the j this can be taken as the definition of j this is the form of the applies with theta small however on ordinary free dimensional space we know that an infinitesimal rotation can be written this way as identity plus data times that get that out of our script j matrices like this and one here really means the same thing as identity so these two are equivalent to one another if Roman block j is just equal to i times script j if you look at your script j matrices here there's a basis of any symmetric matrices if I multiply them by i then they become the basis of the mission matrices and in fact it's just this I'll write it up j s is equal to 0, 0, 0 0, 0, minus i 0, i, 0 you get j y is equal to 0, 0, 0, minus i 0 and you get j z is equal to 0, minus i, 0 i, 0, 0, 0, 0 0 like that and those are the three matrices now from them we can construct j squared and j z we got j z already we also construct j plus and minus let's look at j squared if you take these three matrices and square them it's an exercise for you here's what you find you can get 2's down to diagonal or you can get 0's something off diagonal in other words the operator will matrix j squared that is in fact just 2 times the identity now this means that every vector in ordinary three dimensional physical space is automatically an eigenvector of j squared an eigenvalue of 2 however we know the eigenvalues of j squared have to have to form j times j plus 1 where j is an integer or a half integer and if j times j plus 1 is equal to 2 the problem is that j is equal to 1 so ordinary three dimensional physical space is a j equals 1 space in terms of how it transforms in the rotations in fact j equals 1 space has to be at least three dimensional and if it goes 0 plus or minus 1 it has three values in fact this space is exactly three dimensional so in fact this physical space is actually a single very visible subspace under rotations with an angular momentum value of 1 now this is associated this is responsible for the fact why and we'll elaborate on this as we go ordinary vectors or vector operators are often times associated with an angular momentum quantum number of 1 just to give you an example of this the photon which is associated as a particle is associated with the vector potential the electromagnetic field is considered a vector operator excuse me a vector particle and another way of saying that is that it's a spin 1 particle so the fact that the photon has spin 1 is closely connected with the connection with the ordinary free vector potential in fact that's a vector operator alright in any case so this space has three dimensional space has j equals 1 and therefore there should be a standard angular momentum basis which is the sublinus eigenbasis of j squared and jz well j squared is no problem every vector in space is an eigenvector of j squared so you only need to look at jz and compute the eigenvector of jz with eigenvalue plus 1 let's say let's call this e hat sub 1 and if you do you find that it's really a phase vector of the eigenvector of the phase because if you normalize it from the phase of all this left you find that it's equal to x dash plus i y half of the square root of 2 in fact it is becomes you see the top vector of the spherical basis so you choose the phase of the minus 1 but then if you apply j minus to this you add lowering operators and what you get is the other three vectors so the spherical basis is a standard angular momentum basis in this under space and understanding this structure about how rotational operators act on vector spaces it's a good idea to have a liberal line in terms of what the interpretation of the space is it doesn't have to be a space of wave functions I've just illustrated in the ordinary physical space it could also be a space of operators which we'll talk about in a moment if there's time it could also be a space of classical electric or magnetic fields in other words classical fields and if you do this and you go through the decomposition in the standard angular basis what you find is the multiple expansion of electric and magnetic fields is usually presented in courses on ENM as being power series expansion of the distant field in terms of a localized charge and current distribution and it is indeed that but it actually also is a decomposition of the electric and or magnetic field into its different angular momentum columns it's another way of doing the same thing so for this reasons as I say it's helpful to have an open line to have the interpretation of that space easy alright I hope for I hope I have time to do this because there's one more thing I want to say about this spherical basis that needs to occur which has to do with transformational properties under rotations if we go back to the case of a ket space when we take a standard angular momentum basis vector J, M like this all cylinders don't count as needed let's take a look at what happens if we rotate one of these vectors basis vectors by rotation to it we actually went through this in the case of the YLN it was just a couple of lectures ago but I'll do it again we work this out by inserting a resolution of the identity right before the view so we write this as the sum on M of J, M prime power 5 and then just copy the rest of R, J, M and the result of this is this becomes the sum on M of the basis vector J, M prime and then what's left over is the D matrix P, J, M prime of the corresponding rotation so if you rotate one of these basis vectors it should be a linear combination of the basis vectors of the same J but other values of M with expansion coefficients which are the components of D matrix now by the exactly analogous line of reasoning we're just different notation if we take the rotation and apply it to one of the basis vectors of the spherical basis the answer can be written as a linear combination let's say Q plus 1, Q prime we had Q prime, the vectors of the spherical basis with expansion coefficients which are just the D matrices in this case it's D1, J equals 1 and it's Q prime Q R, the derivation is just the same just insert a resolution of the identity this however has an interesting consequence if I take this equation the local point by Q by D prime Q complex conjugated on the left-hand side in the dot product in other words we will pick out one term with the sum of the right-hand side and if we write this dot prime this way it's B had Q prime complex conjugated comma R, B, Q are using kind of a roundly parenthesis notation for a dot product but R acting a Q I think the dot product of star then on the right-hand side we pick out the one term and it becomes D1, Q prime Q of the rotation matrix R this is an interesting result because it shows you that for the case J equals 1 the P matrix is actually the same thing as the classical rotation but it's expressed in terms of the spherical basis instead of the Cartesian basis and besides it's more if I use Cartesian vectors over here then what you get is the beautiful components of the rotate class of the rotation matrix so these two matrices D1 and R are the usual our familiar rotation matrices are just related to one another by change of basis that's all that is going from Cartesian to spherical this is all leading up to the bigger red card term which is used to mean applications in quantum mechanics especially useful for analyzing matrix elements or finding out very quickly without doing any work when the matrix elements vanish those are the selection rules I gave a brief example already where this has been erased now this radiated transition in hydrogen like atoms we can also reduce to a greater reduced amount of labor that's involved in actually calculating those matrix elements if you have to end up getting numbers and so on Wednesday's lecture we'll cover the interaction