 Well, we were doing at the end of the hour last time. We're talking about magnetic resonance experiments. These involve typically a uniform magnetic field, which are called B0, pointing in some direction to B0 at. This is constant and tidal in space. It's assumed to be in a certain sense of a large field. Then we have a time-dependent field, which is called B1, which is placed in an orthogonal direction. And it has a periodic time with a periodic frequency, which is a mega one, is the frequency of the small field. The small field B1 is regarded as a perturbation. That's how it's applied in practice. We're denoted by the frequency of procession in the background field. We're calling that the omega 0, which is gamma times B0. Gamma is the quantity minus g times charge provided by 2mc. It has a minus sign in it. For simplicity, we'll assume here the charge is negative, or else the g factor is negative, so that gamma is positive. We'll have to go through this again, this business again, and rethink it in the case of the case in which gamma has the opposite sign. In any case, notice the difference in the notation because omega 0 is the frequency of procession in the background field B0, whereas omega 1 is the frequency of the time-dependence of B1 and is therefore independent of the magnitude of B1. Now, the interesting case, well, actually, we're going to solve the system in all generality for a certain time-dependence of B1. But the interesting case from an experimental standpoint is one in which the magnitude of B1 is much smaller than B0, so that it plays a role in small perturbation. And moreover, one in which the frequency of the perturbation is comparable to the procession frequency in the background field. This is a resonance condition, and what it means is that over time, even though the perturbation is small, its effects can build up to have a larger effect. And in particular, as we'll see, it will use a spin plus of the spins, which is possessing the background field. So this is the basic setup. And in the Schrodinger equation, which I wrote down here in the hour last time, it's just simply this right here, where the total magnetic field B0 equals B1. Notice that Hamiltonian now has an explicit time-dependence because B1 is time-dependent. Chemical problems with explicit time-dependence are usually hard to solve. You need to solve a time-dependent Schrodinger equation, which is what this says. But this turns out that if a particular time-dependence will consider, this is the case where the exact solution can be given. Now in particular, the time-dependence for B1 that we're going to consider is one in which the B1 vector is rotating on a counterclockwise direction in a plane, in a plane perpendicular to B0. So maybe just to rewrite it a little bit, if I call this B1 and 0 at the initial time, and then here's a B1 in a later time, a time t like this, then the angle here is omega 1 times t. So just picture that it's turning around and around in a perpendicular plane. Let's take this as an example of a time-dependence of frequency of omega 1. All right. If we do this, then this vector B1, which appears here, can be written in terms of the rotation matrix about the axis of B0 by an angle, which is omega 1t. So let's write this this way as r of B0 at omega 1t applied to the initial conditions for the B1 vector, which I'll call just B1, 0. That's a constant vector that you see here. And in fact, this rotation matrix appears often enough to allow me to make an abbreviation for it. r1 and t is defined to be the rotation matrix of the axis B0 with the frequency of omega 1t. OK, so that's just a way of rewriting this time-dependence. Now, the way we solve the Schrodinger equation is that in fact, we go over to a rotating plane. The rotating frame is going to be one that's rotating exactly the same way that B1 is rotating in a counterclockwise direction around the axis B0. Because the idea is that if we do this, then the time-dependence of B1 is canceled out. This is the whole point, is to get rid of the time-dependence of the Hamiltonian. Because in the rotating frame, B1 is in a constant vector. So here's what we'll do. As we'll take our state vector psi of t, and we'll write it as a unitary rotation operator with the same axis B0 and half and the same angle, omega 1 and t, as our rotation, multiply it over to a new ket vector, what we'll call 5t. So at any fact, 5t becomes the ket vector for the system in the rotating frame. And then we just take this transformation and substitute it into the Schrodinger equation. This u that appears here is the same axis and angle as this r1 that I defined. So the same writing, we just call this also u1 of t as the u of the same axis and angle of angle 1 of t. These are related to having the same axis and angle of the classical component of rotations here. So now we just need to do the algebra of substituting this transformation into the original Schrodinger equation. And that's just a bit of differentiating it. So i h bar d dt acting on psi is, first of all, i h bar d dt acting on u. This is the same as u1. But this u1 here, we write it out in exponential form, is e to the minus i of omega 1t, and e0 hat dotted into the spin as these are spin rotations. This is just the exponential form of the rotation operator. And so differentiating with respect, this is over h bar. And so differentiating with respect of t, we can just bring down the x called at minus t. And so the differentiating is u here. Oh, and by the way, we can bring down the x while we either to the right or to the left of the operator at all manner. Because if you need to do the exponential. So allow me to write this this way, as this is the same thing as u1 of t applied to phi of t. So I'm taking the derivative of u1, and we write it this way, as u1 times, it's going to be omega 1 v0 hat dotted into spin s times phi. That's just the first term. The i h bar here cancels the minus i over h bar there. t disappears in differentiation, and omega 1 v0 hat dotted into s is what's left over. And then in addition, we need to differentiate phi. So we've got the operator u1, and then this is multiplied by h bar d dt applied to phi itself, like this. So this is differentiating the left-hand side. Now, we'll plug this into the Schrodinger equation for psi over the right-hand side. And therefore, that's equal to our factor gamma. Gamma, by the way, is the constant that converts magnetic field strengths into frequencies. As you see here, the proportion of one another again is just a proportionality factor. In a gamma, it's in multiplied times v0 plus v1, which I'm going to now write as r1 times v1, 0. So r1 of t, it really depends on time, times v1, 0. Dot it into the spin s, multiply times psi. But psi is the same thing as u1 times phi. So u1, phi, like this, OK? Now, what we're going to be interested in solving for is the time dependence of the u cat phi. So we'll do this, obviously, and we'll do this to multiply both sides of this equation from the left-hand side by u1 inverse, which is, of course, the same thing as u1 dagger. So let's multiply u1 dagger to this expression above and this expression below, which are equal to each other. And if we do, on the above, we get omega 1 times v0 hat dot it in the spin s, multiplying phi. Then we get plus i h bar v phi dt with a second term up here. Now, multiply u1 dagger on the line below. Let's note this again as a constant, so you can just pull the u1 right through it. The vectors v0 and v1 0 are vectors, but they're just vectors of numbers. They don't depend on otherwise. They're multiplicative operators, but otherwise, they're not operators. So you can just pull the u1 right through them, too. And likewise, the r1 of t is just a matrix of numbers. So the effect is, u1 dagger can be pulled through all this stuff and just bring it next to the s, which really is a genuine non-trivial operator on the spinner, on the spinner phi. So the result is that this becomes equal to the right-hand side, this gamma copying these vectors into v0 plus r1 times v1 0. And then this is dotted into, because we've got u1 dagger s u1, u1 dagger s u1, like this, which is then multiplied by 1 to phi. This is what we get. And so now you've got a conjugation of the spin operator by a rotation operator. Well, this conjugation is exactly what the adjuvant formula is for. I'll remind you that in general terms, we've worked this out, this is the adjuvant formula. This is conjugating a, this is in general notation here, where j is any angular momentum for a system, and u is a corresponding rotation operator. We conjugate this equivalent to rotating the angular momentum by the corresponding classical rotation. With j replaced by spin s, that's just what we have here at z1. So the result of this is that this conjugated spin operator here is the same thing as r1, as a matrix, just multiplying a spin, or what's better, like that. It's equal to that. All right. Now, the right-hand side, we've got a dot product of two vectors, a magnetic field vector and a rotated spin vector. Here we can use a useful identity, which I guess I'll put back here, because it's kind of a little regression. A useful identity is this. If you have two vectors x and y, and there's a dot product except for y, let's say, is rotated by some rotation, you can transfer the rotation over to the other side of the dot product if you're replaced by its inverse. So this is r1 multiplied by y, not been to x. Now, the reason for this identity is that the dot product is invariant in the rotations. So if you take the two sides of the dot product and multiply both of them by the same rotation, it won't change the dot product. So allow me to multiply both sides of this dot product by r inverse, which includes the r inverse. Oops, it shouldn't be this long. This is x and y. The x and y won't change. Say it again, multiply both sides of this dot product by r inverse. You get the r inverse applied to the x, and it cancels the r and the y, so you get the identity. All right, in any case, let's just go through here, because we have this dot product. There's a rotation applied to s. I'm going to move this rotation over to the other side of the dot product, then I can replace it by minus 1 r1 inverse. So this is gamma times r1 inverse multiplying the vector, which is b0 plus r1 times b10. That creates one new vector, which I'll put in square brackets, dotted into the spin mass, then multiply by 1 to 5. Now, r1 is a rotation about the z-axis, the b0 axis. Excuse me. This is the word that's usually in the z direction. And that's just what b0 is here. So the r1, r1 inverse, doesn't do anything to b0, but it brings it inside the parentheses. On the other hand, for b10, the r1 inverse cancels the r1. But that's good, because that's where all the time becomes lies, than the original Hamiltonian. And so this thing becomes the same thing as gamma times b0 plus simply b10, dotted into spin mass, over flying 5, and to repeat the left-hand side, which you see up here. Now, when you bring the gamma inside, gamma times b0 is, what I say here, times its magnitude is the frequency of omega 0. So this can be written this way, bringing the gamma inside is omega 0 times b0 hat, plus gamma times b10, dotted into spin mass, over flying 5, like that. Now, I'm going to bring this omega 1 curve, which is multiplied by b0, dotted into s. Unless you've got omega 0 times b0, dotted into s for this first term. So I'm going to bring this over, and it'll subtract the frequencies. And the result is going to be a simplified Schrodinger equation for the invariable phi. And it looks like this. It's i h bar d dt divided by the phi. It's equal to, I'll use the square brackets here. And then it's omega 0 minus omega 1, and y could be 0 hat, plus gamma times b10, dotted into the spin mass, whole phi over 5. Now, this is great progress, because we've eliminated the time dependence of the Hamiltonian. In effect, by going to this rotating frame, which you see here is the initial conditions for the perpendicular vector. But it's no time dependence anymore. And in fact, this is in the same form of the simpler problem that we saw in the last lecture, in which you just got a uniform magnetic field. In that case, you know that this spins, precesses, around the direction of the magnetic field. It's a rotation operator. His axis is the magnetic field, and this angle is gamma times the magnetic field strength times time, so this frequency is gamma times the field strength. All right, so this is what we've got here. So we can write this, and let's write this thing in the square brackets in various ways. This can be written as, let's call it, gamma times an effective magnetic field. And we can write that as gamma times the magnitude of the effective magnetic field times the unit vector of the effective magnetic field. And if we do this, then the solution becomes immediate. And this is at phi as a function of time, is the rotation operator. Oh, let me do one more thing here. Let's say gamma times b effective. Let's give this a name. This is a frequency. We'll call it a capital gather here. This frequency is given a name, by the way. It's called the Rabi-Flocking frequency. Rabi-Flocking frequency. I suppose I could call it omega effective. This really means the same thing. And so the solution then is that phi as a function of time is equal to a rotation operator with an axis which is b-hat effective, an angle which is capital of omega t, the Rabi frequency times time, multiplying by phi is 0. Now, that's just from the uniform field solution. Now, and then I'll define this with the transformation of the top of the board here, I don't know if your video can get up that high, connecting this i and the phi. This is the transformation of this thing. It's supposed to go to the rotating frame. But before I do that, allow me to interpret this effective magnetic field to the effective field, which is represented in various ways here. Let's draw a picture here. Here's the original magnetic field. If we think of it as being strong, b0 times b0 hat. And here is the perpendicular field, b1, 0, let's take the initial time like this, which is supposed to be much smaller. Now, this thing that occurs in the square brackets, first of all, omega 0 times b0 hat, that's the first term. This is just another way of writing gamma times magnitude b0 times b0 hat, which is just the same thing as gamma times the magnitude of b0. And so this first term here, well, if I want to plot magnetic fields, let me factor out a factor in gamma and just take it out. And so what we're doing is we're subtracting from b0, we're subtracting another field, which is omega 1 times b0 divided by gamma. So let's draw this attractive field like this. This is a field which is b0, which is omega 1 over gamma times b0 hat, that's the effective first term there. And so the vector sum of these two vectors is the effective magnetic field, which is then cut in line like this. This is the effective magnetic vector. And we'll stretch it out like this. There's a unit vector v effective in that direction. OK. Now, here's the physics of what's happening. It has to do with coriolis forces. You know, if you go into a rotating frame, there appear usually called fictitious forces and these are coriolis forces. And they go like proportional to the velocity crossed into the angular velocity, where the velocity is the velocity of the particle seen in the rotating frame. And the angular velocity is the angular velocity of rotation. These are coriolis forces. These have the same functional form as a force on a v cross b force on a charged particle in the magnetic field. Here, we're thinking of a uniform angular velocity, the uniform magnetic field. So the result is that if you go into a rotating frame in which the axis of rotation is the same as that of the magnetic field, you can get velocity dependent forces that either add or subtract one another. The effect of the uniform property state could be the effect of the magnetic field by means of coriolis forces. And this is what's happening here. It's that this term, which is minus 1 times b0 hat over gamma, this is the effective magnetic field, which is due to the coriolis forces with a minus sign. And it's canceling out the magnetic field of the original background field. And the way I drew it, it actually canceled out most of it. It's a large, largely cancel all of it. If you put it in this form here, where you see that it's omega 0 minus omega 1 times b0, you see that this near cancellation occurs when the frequency of the perturbation becomes comparable to the frequency, recession frequency in the background field. In other words, it's a resonance condition. If you go to the exact resonance in which you're, these are usually microwaves, by the way, this frequency of omega 1, the microwave frequency of omega 1 becomes exactly equal to the recession frequency, then you've got an exact cancellation. And this vector just drops to 0 and is left to the left of the b1 0 in the rotating frame. More generally, there remains a component of long hair. So I think I mentioned somewhere that the experimentally interesting case is where omega 1 is close to omega 0. That means that most of this field has been canceled. That's the way I drew it here. All right. The perpendicular field remains, but it's now time-dependent to concern the rotating frame. You're rotating with the direction of that field. So this is the physics of this transformation. And it also gives you the meaning of this effect of that field. So here's what it is. And this is this frequency capital omega. Now, the frequency capital omega can be expressed in terms of these other frequencies here. And it's just doing the trigonometry. And what you get is for this capital omega, which is our Robbie floppy frequency. So omega 0 minus omega 1 quantity squared plus gamma squared times b1 0 squared. And again, if your near resonance, the first term is small, this gamma b1 0 would be the procession frequency if the b1 0 is the only field you had. It's much lower frequency than the background frequency. But the result is a near resonance omega is comparable to say it's much lower frequency than the background frequency. OK. So in any case, this gives our solution in the rotating frame. Now, to get the solution back in the original frame, all we have to do is to apply this transformation at the top of the board. And this then gives us that psi of t is equal to our u1, which is u0 hat times omega 1 t times this Robbie procession, which is along the axis of the effective and then capital omega t applied to the initial conditions, which is psi of 0. You see from the top of the board, the hat t equals 0, psi and phi are the same thing because of u operator's identity. So I just replace this phi of 0 here by psi of 0 over here. And this then is the solution. This is an explicit solution there for this problem at this particular time that I'm going to field. Straightforward to take the solution of Schrodinger equation and convert it into a solution for the expectation value of spin. And if you do this, what you get is this is exactly like what we did for the uniform field. And then you need to use the adjuvant formula to drive it. But the time dependence of the expectation value of spin is given by the initial spin, time t equals 0, multiplied by rotation matrices. These are now classical rotations with the same axes and angles that appear in the quantum solution. And so to draw a picture here, I take my original background field v0 like this. And let's say we've got a, let's see how it's going to do this. Let's say we've got a v effective that sticks out like this. So this is our v effective here in this direction. And let's say you've got an initial spin, which doesn't have to be pointing in any particular direction. So here's the ss times 0 like this. Then the effect of this motion is to, first of all, rotate the initial spin about the axis to be effective at the floppy frequency like this. So you're going around at this frequency capital T. However, at the same time, the axis of the spin sweeps out of color about this direction to be effective. At the same time, we then apply on top of that this rotation about the direction v0 at the frequency of omega 1. And what that does is take this whole column and makes it rotate properly like this around the axis v0 of the frequency of omega 1. Under the interesting case, which I mentioned up here, where v1 is much smaller than v0, and omega 1 is approximately the same as omega 0, it means you have a high-frequency precession around the background field, but at the same time, you've got a much slower frequency, most along this cone up and down like this. And so this means that if you take the direction of the spin on the sphere, more than south pole here, you get a band like this. There's an upper and a lower limit, which is determined by the size of the smaller cone. And the spin is basically precessing rapidly on the z-direction, but only going small, slower oscillations up and down. It does this. It goes up, and then it goes down like this. It goes back up again, and oscillates back and forth. And if you set this up right, where you exactly cancel it, where you exactly have resonance, then the effect is in the x, y plane. And what happens is that this band in the center and the equator, so you're going up and down between plus and minus the same z-goats, in particular, the spin started out at average value of spin, started out being exactly in the z-direction. Then what's going to happen is it will spiral all the way down to the south pole, and then spiral back again, up and down at the arachnid-floppy frequency. So that's the solution. Well, this is the basic mathematics of magnetic resonance type experiments. And let me just say some things about the many, many applications that these are quick to. First of all, in nuclear physics, this can be used to measure magnetic moments of nuclear particles as one of the first applications used, because it's much higher precision that you can get with the old fashioned String-Gerlach type apparatus. The other thing you can do is, because you see the point is, is that experimentally, you control the frequency of naval one by tuning your RF oscillator until you get a resonance, and frequencies can be measured rather high precision. This is much better than String-Gerlach, where you have to measure the positions of those blocks on the screen or the spot set. Anyway, so that's one application. Another application is that the energy levels of the state's SMS of a nucleus in the magnetic field and the background magnetic field, they, in general, don't depend entirely on the magnetic moment. That's all we've been taking into account in this discussion. But the nucleus is a charge, is a charge of the current distribution, and in general, it has higher multiple moments, such as an electric quadruple and it turns out to be important. That actually affects the energy levels of these states in a given magnetic field. And so when you make measurements of these higher order moments, the measurement of the non-zero electric quadruple moment of the Deuterons is the first great success of this type of experiment and gave a good deal of insight into the structure of the Deuteron. I may say more about that later in the course. So this is a use as a nuclear. This technique is also used in chemistry and defense matters. The reason is this, is that this background field that you apply here from some external coils is not necessarily exactly the same magnetic field as seen by the nucleus. The reason is that the nucleus is in an environment of electron clouds and some atoms somewhere, or a molecule. And that electron cloud partially shields and modifies the external magnetic field. The effect isn't very much. I think the numbers are something like 1 part and 10 to the sixth. But nevertheless, the result is that, for example, if you're looking at a hydrogen atom or a proton or something, the exact magnetic field at the location of the proton depends on the chemical nature of the molecule that the hydrogen atom finds itself in. And so this gives a good deal of information that can be compared to theory and so on for understanding electronic structure in chemistry and condensed matter physics. That's another application. This is something that's called a chemical shift or some chemical shift that makes the name of the use for this frequency shift because of this effect. Related to this is, well, there's also, I guess you know, there's magnetic resonance imaging, so-called, this is medical, medical imaging. There they measure, amongst other things, the density of protons in the biological sample, which usually is a person. And they are using, actually, the background magnetic field has a spatial gradient. So it's stronger if you wind up at your toes than if it's at your head or vice versa. And what this does is it means the resonance frequency, the omega-zero here, is a function of position. So by changing the microwave frequency, they're looking at it and get different slices through the body. There's quite a bit more to the MRI imaging of this, as a matter of fact. In fact, the chemical shifts play a role as well and have to do with how they can tell differences between different types of tissue that's become the same chemical shift I just mentioned. And then, finally, another application is an atomic box. There are various designs for atomic clocks, but the most accurate ones use a cesium atoms. And they use a transition, a spin transition in cesium. So the first thing you do is polarize the cesium. This will be done with the Stern-Gerlach apparatus. Then you run it through a magnetic resonance type apparatus, with the attempt of making a single spin flip from one side up to down. And then you run it through another Stern-Gerlach apparatus, which if you really flip them all down, you just get one down beam and no up beam on the output. Well, if the frequency of your RF frequency starts to drift a little bit, then you're going to start to get some up beam as well, some up spin. And so what you do is you have to feed that mechanism that looks at which spin is coming out through the second Stern-Gerlach apparatus. And you feed it back in to lock the frequency so that you stay on a single out-of-beam. And what that does is it gives you this RF generator here. It's now a lock step with the frequency of spin transitions, and it gives a very accurate plot in that way. You put counters on it to count the number of oscillations and divide it up to get seconds and so on. This is the atomic plots, I might have mentioned, are the basis of the GPS system. They put these things in geosynchronous orbit and they send out timing signals down to the Earth, which if you apply it like between several of them, you can find out where you are on the surface of the Earth. They're so accurate that it's necessary to correct for both special and general relativity, I think, motion of the satellite in order to get the desired accuracy of position on the ground. OK. And the lack of time, I won't tell you about spin echo. That's quite important. It's a technique developed by Irwin Hahn who's now a retired professor at the Department of Defense. All right. So that's all for magnetic resonance. And I'd now like to go on the loose topic, which is that field of angular momentum. This is in general a continuum with a general topic of rotations in angular momentum theory and pausing for applications. So I want to turn now to the topic of orbital angular momentum. So for this purpose, let's consider a single particle without spin moving in three-dimensional space. So the wave function is, of course, psi of x, y, and z. And allow me to write this as psi of r vector, where r is just a position vector where the wave function allows. Now, let me remind you that way back when we were doing translation operators in three dimensions of translation operators, pram, prism, y, and displacement a, that we found that this acts on a position eigencat and moves it over to a new place, which is just a displaced position. This makes physical sense, because if you cat r, a position cat, represents the result after one has made a measurement and found the particle to lie in some small region around this position vector r. Then if you want to apply a translation operator, and this is in an active sense, it's possible the new state should be localized around the translated position. That way, this equation indicates. This equation implies that the translation operator acts on the wave functions in the following way. The translated wave function evaluated in position r is equal to the original wave function evaluated at the inverse translated position. This is something we went through earlier with translation operators. Well, now what I'd like to do is to extend this to the case of rotations. So let's let r be a proper rotation that belongs to SO3, and the one to figure out what is the rotation operator that corresponds to a u of r. The logical thing to do on the pattern of the translation operators is to say that u of r acts on a position eigencat and a position vector lower case r and just moves it over to the rotated position. It's the obvious thing to say from an active standpoint of rotations. And if you do this, you'll find the corresponding wave function, you act on the wave function like this. You evaluate it in a position r, but this is the original wave function evaluated at the inverse rotated position. This is just exactly in a pattern of what happened with translations. Nevertheless, this minus 1 that occurs here can be a confusing part to remember. So let me show you a picture to make it clear that we really need to have this here. Let's draw it in a plane. So let's say I've got the xy plane. And let's say there's some wave function here, which is like a wave packet like this. And let's call this the only wave function, only wave function psi. Let me choose a point on this. It doesn't have to be in the center. Maybe it's on the tail of the wave function like this. And I'll call this point r0 like that. Now, I want to apply a rotation in an active sense. That means this wave packet is going to get moved over to a new place. Let's draw it over here. Here's the new wave packet. Let's call the new wave function psi prime. Let me take an equivalent point on this new wave packet on the same tail and call this new vector over here. Well, the new vector should be the rotated version of the old vector. That's just what we mean by applying a rotation. So this psi prime is the same. It's going to be this u of r acting on psi 0. Excuse me. Psi prime, the new rotated wave function is our u of r rotation operator acting on the old wave function, which we call the psi like this. Well, from this picture, this is what r does to rotate the old and the new. Well, from this picture, the value of the new wave function at this rotated point is the same as the value of the old wave function at the unrotated point. In other words, psi prime evaluated at r times r0 is equal to the original psi evaluated at position r0, the original position like this. However, position r0, which occurs here, is just a w variable. So let's do this. Let's let r0 be able to rotate an operator r times another position r, excuse me, make this inverse like this. And if we do this, then we get psi prime at position r is equal to the original psi at the inverse rotated position. And this is just the rule, repeating the rule I had over here, but just showing you the false depiction that's all. So this is the geometrical meaning of this effectively a definition of rotation operators now. So by doing this, we have the most logical definition of rotation operators on spinless wave functions in three dimensions. So we know what the rotation operators are. One thing to notice right away is that the representation of a single value that is to say, for a given classical rotation, there is only one rotation operator, not two as we found in the case of spin 1 1⁄2 particles. And what that means right away is that the angular momentum values which are contained in this rotation action can only be integers. As we saw earlier, the half integer angular momentum are always double value representations of the classical rotations. By the way, we changed notation when we were talking about orbital angular momentum. The general notation of j and quantum of the j, which we applied for any kind of angular momentum we're not going to replace by L and L. This is a standard convention of the physics literature anyway, for the orbital angular momentum that's part of it. But in any case, we can see now in quantum number L they only take no integer values, 0, 1, 2, 3, and so on. What about angular momentum? We define in this course angular momentum is the generator of rotations. So if we call the angular momentum L, it means that our rotation operator U of r, which we'll just define, must be possible to write. Let's suppose I take it in an axis angle form. Let's take a U of n hat on the theta. This must be equal to e to the minus i over h bar times theta times n hat dotted with L. That's effectively the definition of the angular momentum of the operator. And if the angle is small, which becomes 1 minus i over h bar theta times n hat dotted with L, just expanding out to the first term of the Taylor series. And then it's small. All right. Now let's use, it's in the box right here. Let's use this definition of U up there. It expresses U in terms of the classical rotation. Let's make the angle small and plug it into that expression of the bottom. In order to do that, I'll need something for r inverse. So if we take r at the same angle in an axis, this is the same thing as e to the theta n hat dotted into script j. I'll remind you that's for study of classical rotations. If I make an inverse on it, it's the same as changing the sine of theta, so I get a minus sign there. And expanding that out to the first order, we have the identity minus theta times n hat dotted into these script j matrices, which are this vector of 3 by 3 matrices. So both of these equations apply, which is much less than 1. Now, let's plug it into our formula for rotated wave function. So the U of r at the psi is going to be 1 minus i over h r theta times n hat dotted into any momentum l. This adds on the psi to the rotated wave function, which is evaluated in the position r vector. And on the right-hand side, this is going to be psi evaluated in the inverse-located position, which is this matrix here, i minus theta times n hat dotted into script j, all applying the position to f r, like so. Let's expand the right-hand side first. This is the same thing as psi evaluated at position vector r minus theta n hat dotted into j multiplied r. But I'll remind you that n hat dotted into j acting on a vector is an alternative notation for the cross product. So this is the sine of this n hat crossed r. So this is psi evaluated at position r minus theta times n hat crossed r. Since theta is small, we'll expand this out. We get psi r minus theta times n hat crossed r dotted into the gradient of psi. Meanwhile, on the left-hand side, the identity operator 1 just does this psi r. And the correction terms are just going to be the numerical factors minus i over h r theta. When there's an operator, n hat dotted into l. And this adds on psi. And so this thing over here has to be equal to that thing over there, the leading term psi r, as you might expect. The next term has a common factor of theta, both cases. So I'll drop the theta, which is minus i's, I could cancel them too. I'll bring the i over h r over to the other side. And if we do, we obtain this. Is it n hat dotted into l, acting on psi, is equal to minus i h bar times n hat crossed r dotted into the gradient of psi. Well, allow me to rearrange this whole product so that it becomes n hat dotted into r crossed gradient of psi. So it's minus i over h bar, n hat dotted into r crossed gradient of psi. Now, the n hat is an arbitrary unit factor. So we can drop that to both sides. When we get l acting on the side, the ordinal angle of momentum is equal to minus i h bar times r cross gradient of psi. However, minus i h bar times the gradient of momentum. So this is the same thing as r cross p acting on the side. And the result of this is that d, that appears at the ordinal angle of momentum, l is equal to r cross p. The usual way, this is the other result, of course, the usual way of deriving this in your vectory course is to say that r cross p is the definition of angular momentum in classical mechanics. And we'll just borrow it in quantum mechanics and call it the angular momentum there. But you see here, this is being derived as a generator of rotations, which are defined in the most obvious way, or how rotations are going to be defined in wave functions. And it gives the same answer. There's something else that comes out too. We showed quite generally that if you had representation of rotations, that the generator satisfied the angular momentum commutation relations. And so that has to apply to this analysis as well. And so if you have confidence that everything you've done so far is correct, then you must have the standard angular momentum commutation relations for horrible angular momentum. Of course, you can check that by using the Heisenberg quantum commutation relations in a position of momentum. And we calculate this to check to see if you're doing things right. And it has to be right. Of course, it comes out from my point of view. These are the commutation relations. Is it really currently, right? Yeah, with Poisson bracket? I'm sorry, I made it Poisson bracket. I meant to make a commutator. But then you make too many Poisson brackets, like this. All right. Now, so the result of this is that we have now both rotational operators and their generators, which is the angular momentum for spinless particles moving in three dimensions. The next thing we'd like to do when pursuing angular momentum is to set up the standard angular momentum basis. I'll remind you that, and the general notation we used previously called this gather jm. This is a simultaneous eigenbasis of the operators J squared and Jc. And in case their simultaneous eigenstates are degenerate, we have to introduce an additional n index to resolve the degeneracies. Talk about this before. Well, in order to find the standard angular momentum basis in the present case, we would like to find the simultaneous eigenvectors or eigenfunctions in this case of the operators L squared and Lc. And they might be degenerate. We'll see whether they are or not. Let's call these things psi Lm of position vector r. So we want L squared acting on this is equal to L times L plus 1 psi Lm of r. And we want Lc acting on this psi Lm of r is equal to m times psi Lm of r. And I'm going to, I think for the rest of the lecture today, I'll set h bar equals to 1 because it saves some writing. But these are the equations we want to solve to find these simultaneous eigenfunctions of L squared and Lc. Now, the way we do this is we take these operators, the angular momentum operators L which is r cross p, there's only three operators. The three components is angular momentum. And write them out, because differential operators, which I've summarized for you here on this board, here they are. Here they are written out as r cross p with momentum operators written out as derivatives. Then what we do is we take these three differential operators and we transform them over to spherical coordinates. This involves the chain rule. You don't bother to copy this down because this is kind of complicated and it's in the notes. But if you do this, then you get differential operators and r theta and phi. So it's just a coordinate transformation from x, y, z over to r theta and phi as far as transforming these operators. And this is what you obtain for the three components of angular momentum. And then from that, you can construct the L plus and minus, which is the same thing as Lx plus and minus Ly, pi Ly. And likewise, you can construct L squared as well. I do bother to write down L squared in some of the notes if you want to see it. Before going on, let me make some remarks about this. Although all three of the rectangular coordinates, x, y, and z, appear on these operators, when you transform over to spherical coordinates, you find that only the angles, theta and phi appear. But there's no radial derivatives. There is no dvr that appears. So the three angular momentum vectors don't involve any differentiation of respect to r. And the reason for that is that the angular momentum is the generator of rotations. It's the correction term for when you have an infinitesimal rotation, the correction away from the identity. And if you're applying a rotation to a vector, you don't change its length. You just move it. It's an infinitesimal rotation. You just move it a small amount. If you apply one of these rotation operators to a function, you're going to be comparing the function at two points, which are at the same distance from the origin, because it's just a small rotation. And since they're at the same distance from the origin, that's why you don't see any hardware evidence. You're just sensing the value of the function over a sphere. That's the meaning and that's the reason for this. Another thing to point out is the Lz operator is particularly simple. X and Y are fairly complicated, but this one is quite simple. And the reason for that is clear also. It's because the spherical coordinate system favors the z-axis. It's not as simple as a new angle of the z-axis. Whereas the theta angle, there is no angle of the x and the y-axis in the usual coordinate system. So this one is pretty simple. All right. Now let's go back to the equations we want to solve. Turns out that what we really want to do, again, if you go back to what we did with the general theory, we spoke of finding the stretched eigenfunctions first and then applying lowering operators to get all the others. So allow me if you let me do this, use my fingers here and replace this term by an L here. So we're asking for the stretched eigenfunctions, psi LL, and then this gets replaced by L. So this would be the equations we need to solve for the stretched eigenfunctions. Actually, the L squared equation is harder to use and because it's a higher work, second-order derivatives. And it turns out that an equivalent thing to do is replace the state of using the L squared equation. This is to use the fact that if you're in the stretched state, then the raising operator annihilates it. In fact, the stretched state is the only state that's annihilated by the raising operator. So we can replace this by L plus acting on psi LL is equal to 0, and that's more convenient for the L squared equation. So let's solve these then. Let's start with the LZ equation. That's the easiest. So the LZ of the minus-reference of the minus i h bar, which I'm throwing away now, setting h bar equals to 1, minus i d d phi applied to psi LL of r theta in phi is equal to the right-hand side, which is L times psi LL of r theta in phi. So this equation is trivial to solve, but it is that psi LL of r theta in phi is equal to an arbitrary function of r and theta, because we're going to have only phi derivatives here. We call this f, so LL of r and theta, multiplied by 10 to the power of l phi. The LZ equation, in other words, gives us the phi dependence of the i d function. Now to that, then we need to apply L plus. Well, let's look at L plus here. L plus is minus i plus minus i phi plus minus i this, that potential theta, and so on. Let me write it out. What we want to apply this to is our function psi LL. I've memorized this now. So we're going to have to cover it up. So this equation becomes minus i, even the i of i, times it's i, i d theta, that's i d plus i d theta minus potential theta d d phi plus i d theta minus potential theta d d phi applied to our solution here, which is f LL of r comma theta times d to the i of phi, and the whole thing is equal to 0. That's the L plus equation. Now you can cancel the constants out of the run, because they go away. The d phi acts on the phi dependence here, and brings it down to i L. If you do this thing, you can cancel the i's, because it's coming back to the i. Then when you've done that, you can cancel to d i phi, because nothing depends on phi anymore. And what you get is an equation d theta minus potential theta times f LL of r comma theta is equal to 0. This is an elementary equation solved in the intelligent that f LL of theta is equal to the sine of theta to the null power. So the result that is for the stretched state, the sine LL of r theta phi is equal to sine L of theta times d to the i L of phi. And finally, it can be multiplied by any function of r in moments called g of r. And see the Lz equation determines the phi dependence. The L plus equation determines the theta dependence. And that's all there is. There are many more equations. The r dependence is not determined. The result is that this is an example in which we do need an extra index to resolve the genesis because also the Lz by themselves do not form a big set. Some other variable is needed. In fact, this g can be replaced here by a basis of radial wave functions called a un of r. And then we have that sine LL. We're just looking at the stretch tape so far. Again, it's like this index gamma in a general notation. OK. Well, that's the basic solution of this. We'll look around this next time. This leads to the theory of the y-axis and that's square root of r times.