 Last time I explained very briefly how in that classical letter, electromagnetic theory, a system of charged particles in some state of motion have a magnetic moment vector and an angular momentum vector, which in general have no simple relationship among themselves. However, in some simple circumstances, such as a single charged particle moving in a circular orbit, those two vectors are proportional. And the proportionality factor is q over 2 and c, where q is the charge of the particle. I also explained that if you go over the quantum mechanics, then this relationship, which gives the proportionality between the magnetic moment and the angular momentum, actually carries over in the quantum mechanics without modification, as long as you're talking about orbital angular momentum, which is what all usually means, orbital angular momentum. This is something we'll see in more detail later on when we consider the Zeeman effect. We'll see this come out explicitly. However, in the case of particles with spin, the relationship between magnetic moment and the angular momentum, which is traditionally called the S, is the spin or angular momentum vector for the intrinsic magnetic moment of a particle. One still has a proportionality between the two, but it's necessary in addition to this factor q over 2 and c, it's necessary to assert a bunch factor or a so-called g factor, which is a dimensionless number that compensates for the, that makes proportionality come out right. So different particles have different g factors, which I'll say more about in just a moment. Now, before I do that, however, I should say something about spin and the intrinsic angular momentum particles. It's generally true that elementary particles have an intrinsic angular momentum. That is to say, an angular momentum in addition to the so-called orbital angular momentum that they have because of their motion through space. The fact that there isn't such an angular momentum there is revealed in many different ways. But one of them is that in order to get conservation of angular momentum for systems that are efficiently invariant, it's necessary to include the spin angular momentum as well as the orbital angular momentum. Otherwise, you don't have conservation of angular momentum. In addition, the one sees it through the magnetic effects with the magnetic moments. Taking the electron for an example as a very crude model, if you want to think about spin, you can think of the electron not as a point critical, but rather as a small sphere, charged sphere, which is spinning on its axis. This would clearly give rise to an angular momentum. And because of the charge, you would also involve a current and the current would give rise to magnetic moments. So this little crude model of that allow you to see why there's both magnetic moments and angular momentum. However, this model is extremely crude. I want to warn you, and it's not really suitable for anything quantitative, just as a way of giving you a picture of thinking about what spin means. Now, that's how it is for elementary particles. As I explained at last hour, if you have a composite particle, such as a proton, which is made up of quarks and made up of quarks, then what we call the spin of such a particle, the proton, is really the sum of the orbital length as well as spin angular momentum of the constituent particles. Same thing can be said for other composite particles, such as nuclei. For example, a deuteron is a bound state of a proton and a neutron, each of which has a spin of 1 half. And in addition, there's some orbital angular momentum. It's a two-body problem, a proton and a neutron orbit around each other. And in addition, there's some orbital angular momentum. And the sum total of all of this gives an angular momentum of 1, which is considered the spin of deuteron. You could just say, with atoms, if you have an atom, for example, a hydrogen atom in the ground state has a spin of 0. That's because the proton and electron spins are aligned opposite one another in the ground state, giving the total spin of 0. And the electron orbital angular momentum is 0 as well. So the ground state of a hydrogen is really a big problem, is a spin 0 particle. All right. Now, when we're talking quantum mechanics, of course, the magnetic moment and the angular momentum have to be reinterpreted as operators. They have to act on some nilburn space, because they're observable, both of them. And the question is, what is the nilburn space? Well, in the case of a solar atom, we discussed this earlier in connection with the Stern-Kerlach apparatus. We might analyze the experimental results. We decided that the nilburn space was two-dimensional by called nilburn space b. This is a way of saying that it consists of two-component complex vectors, or in mathematical language, this is a space c2, like this. And it turns out that for particles, whether they're elementary or not, it turns out that the nilburn space corresponding to the spin degrees of freedom, that is to say, the nilburn space upon which the spin operator acts. In general, this is a space consisting of what we call spiners. These are complex vectors with 2s plus 1 components. We can write this this way as c to the 2s plus 1, where s is a lowercase s, not to be confused with capital S, the spin operator. The lowercase s is what you call the spin of the particle. It's an angular momentum quantum number. And therefore, it's either an integer or a half integer. And it characterizes the particle. So the nilburn space upon which the spin operator acts in the language I ended this last time is what you call an irreducible subspace under rotations. Reminded to an irreducible subspace, is a space which is spanned by the set of vectors jm, where the m-values go from minus j to plus j. They're connected by raising and lowering operators. So it's a space that has dimension 2j plus 1. Now, I mentioned that in the subgeneral system, it's necessary to introduce, in general, an extra index gamma in case there's a degeneracy. The j and n-values don't uniquely specify a state. For spin systems, the extra index gamma is not necessary because there's only one irreducible subspace. In the case of the electron, we just have two component spinors. We don't have pairs of two component spinors with different other quantum numbers lying around. And the same thing is true for spins of other values of the spin quantum numbers as well. So for the view around, for example, three is becoming C3, so three component spinors, because the spin is equal to one. All right. And so a basis for the spin in Hilbert space is basically a standard angular momentum basis, except we don't need the index gamma, so I'll cross it out. And also, we use a change in notation. We indicate the angle of momentum by a symbol s because it's spin, and otherwise the magnetic number of the quantum number is still down. Even if it has a subscript of 100 to 1, we indicate that it's the magnetic quantum number plus quantum spin angle of momentum. All right, so these are basis vectors in this open space. And therefore, the matrices, for example, the represented spin operators are the matrices in the standard angular momentum basis, which I went over in lecture last time, I believe. For example, for a Duberon, which has a spin of one, these would be three by three matrices acting on the space that's spanned by these vectors. They're still half part of the two by two matrices. Those are proportional to all the matrices. All right, so that explains, at least that's a basic fact about the Hilbert space for spin systems. Now, let me switch to a new board and I'll say something about the key factors. I'll just do this by running through some examples. Let's take, first of all, the case of the electron, which is probably the place to begin, most important particle. The electron, it has a charge q equals minus e. So the rule is that the magnetic volume is the g factor times minus e over two mc, multiplied times the spin operator like this. This is the electron mass here, and it's the electron g factor, so I'll put e subscripts on that. The spin operator is, of course, has dimensions of angular momentum, which is the same as h bar. So it makes something which is dimensionless, I'll divide that by h bar and multiply the pre-factor also by h bar. And if we do this, then what we get is a factor here of e h bar of two mc like this. This has, if you see, must have units of magnetic moment, since it's not going with the g and the s over h bar, part of dimensionless. And this is, in fact, called the, this is a standard unit of magnetic moment, e h bar of two mc, and it's called the form of magneton, and the v subscript here is verse four. It's a standard unit of angular momentum when talking about electron systems. So we can write this then as minus g e times the form of magneton times the spin s over h bar. Like this. Now, about the g-factor for the electron, the g-factor is equal to the best-following value that's been originally explained two times the number, which is 1.0011 dot, dot, dot. It's very closely equal to two. It's two of the small correction. In fact, the g-factor equal to two is something that emerges automatically from the Dirac equation, which is the relativistic version of the Schrodinger equation for an electron, which we'll study in the next semester. And the fact that it automatically gives you the g-factor of two is considered one of the greatest successes of the Dirac equation. That's where it's very interesting how it comes out. Of course, you can also measure it, and when you do measure it, you'll find that there's small corrections to this that's not exactly equal to two. These small corrections can be derived by theory. This is actually the beginning of a perturbation expansion, and it looks like this. It's 1 plus alpha over 2, 5 plus higher order terms, where alpha here is the fine structure constant. It's e squared over h bar c, which is about one over 137, approximately this. And this is the small parameter, which is used in quantum electrodynamics for carrying out perturbation expansions. The extra corrections which appear here are not a part of the standard Dirac equation. They involve physics, which is not contained in the Dirac equation. And what that physics is, is the interaction of the electron with the quantized electromagnetic field, and also with the quantized positive electron field. When all those things are taken into account, then you can actually calculate these corrections. These corrections are also called radiated corrections because they involve the emission and absorption of virtual photons, and also the creation of virtual electron pairs. All right, so to summarize this, from the electron, there's good theory for predicting or calculating the g-factors, and it works very well. The factor has been a competition between theory and experiment to push this series, or this number, measure the experimental of this series theoretically onto higher and higher accuracy. I don't know what the score is right now, but I think it's approaching, if not more than 10 significant digits. So there's a very good agreement to the carrying experiment. The reason for doing this is to push QED, quantum electrodynamics, as far as possible, to see hopefully one would find some breakdown at some point and learn about new physics. As far as I know, it hasn't actually happened yet. All right, in the case, this is the basic story about the g-factors of electrons. Notice that there's a minus sign here. It comes from the fact that the electron is a negative particle, and it means the electron magnetic moment and it's in plenty of opposite directions. One final remark is that s divided by h bar, in the case of an electron, is the same thing as one half of the polymagnes in sigma. So this can also be written as gE over two times the four magneton times the polymagnes. This is another way of writing an electron magnetic moment. I better put a minus sign on it, too. And since gE is approximately equal to two, if we make the approximation that g is equal to two, then this becomes very nearly as the four magneton is called the five times sigma. It's another simple result for the magnetic moment of the electron. Now, the next particle I'll talk about is positron. The positron is the anti-particle of the electrons in e plus. And so it has a charge q is equal to plus e. And so the relation is that the magnetic moment of the positron is equal to g times e h bar over two mc times s divided by h bar like this. And this is the electron mass because the electron and positron are the same mass. They also have the same g factor. They also have anti-particles as one another. The only difference is the sign. So in the case of the positron, all the electron formula is applied, except for the plus sign instead of the minus sign. In that case, the negative element of the magnetic moment point in the same direction. All right, so that's the story of the positron. Let's move on to the proton, the next most interesting particle. Proton is the same one half particle. And the magnetic moment following the general formula is going to be equal to g factor in this case for the proton. Since the charge q is equal to plus e, we get e h bar over two mc times the spin divided by h bar, to make the spin part dimensionless, except now this is the proton mass which appears there and not the electron mass. This quantity e h bar over two mc is called, is known as mu sub n and it's called a nuclear magneton because it's a convenient, I can't spell things while I'm talking, it's a nuclear magneton and it is a convenient unit for discussing magnetic moments of protons, neutrons and nuclei which are made up out of protons and neutrons. Because of the proton mass which appears here in comparison to electron mass for the bore magneton or this right here, the nuclear magneton is about 2,000 times smaller than the bore magneton. One of the consequences of this is that if you have an atom which has unpaired electrons, so that there are some magnetic moment of the electrons which is the remaining, then the electron magnetic moment is going to dominate the total magnetic moment of the atom, the nuclear magnetic moment would be a very small correction to it at a level of about 1,000 or less. If you want to really see the magnetic moments in a clean way, it's best to use an atom in which the electron spins or pair because then there is no, and the orbital angle of them is two zero because in that case, the nuclear magnetic moment is the only thing left. All right, anyway, that's the nuclear magnetic moment. Now, about the g-factor for the proton it has a value of about 5.588, some number like this and the accuracy of that number is due to the fact that people measured it experimentally. There is no theory of present time which is able to predict this number out to in high accuracy as there was for the case of the electron. And the reason for this is that it involves the strong interactions and it's very difficult to do calculations in the strong interactions. There are simple models of quarks and any quarks put together that go back to the earliest days of the quark theory back in the 1960s which are able to predict the magnetic moments of the proton and neutron and the other varions to within about 5% accuracy. So it's sort of not bad for our first step. However, to improve it beyond that is very difficult and involves nowadays people are doing enormous computer calculations with lattice QCD and so on. One of the things they want to get is a more accurate prediction of the magnetic moments as tests of QED. But the basic fact is that these numbers are fundamentally experimental. All right. Now, by the way, since this magnetic moment does not have the Dirac factor of two, the two that comes out in the role of the distance Dirac equation, people call these magnetic moments, sometimes they call them anomalous meaning not predicted by Dirac. Of course, one could say that even the electrons anomalous to include the regulated corrections is not exactly the value of two either. But anyway, it's very close to that. The case of the neutron is also interesting. Neutron is a neutral particle. So Q is equal to zero. And nevertheless, it still has a magnetic moment even when the charges cancel out from the quarks and their motion. The magnetic moments don't and there's a net magnetic moment left over and the magnetic moment of the neutron is written this way. It's mu is equal to the g factor of the neutron. By convention, we use the nuclear magneton, that is to say, with a proton mass in here. So it becomes dh bar for twice the distance of Pc as a factor here and this is thin over h bar. I didn't mention this, but the proton has a spin of 1 half just like the proton does. So s over h bar in both cases is the same thing as 1 has sigma of both proton and the neutron. And the neutron g factor gn is equal to with a minus sign about 3.8 something, I think the exact number. And to say that it's negative means that the magnetic moment in the spin are in opposite directions from the neutron. And that's basically the story of the magnetic moment for the neutron, even though. We use the charge here of the unit of the quantum unit of charge as pi to the capital of the neutron is equal. That's just a set of conventions. So if you talk about the next case, if I talk about the deuteron, the deuteron is a bound state of the proton and neutron and it has a spin equals to 1. So in this case, the magnetic moment u is equal to the g factor of the deuteron times the charge, which is dh bar over, and again we use the warm, excuse me, we use the nuclear magneton here, dh bar over 2n Pc for the proton mass in here. This is the standard unit of magnetic moment that's all needed by the people used. And then what's left over is S over h bar, like this. However, now notice that since the spin is equal to 1, the spin operators are represented by 3 by 3 matrices, not 2 by 2 anymore. So for the deuteron, you have to take that into account. The g factor for the deuteron is about 0.8 something, 0.85, I think, the exact number. I don't understand the justification for using starting at the neutron to 0, and we started out with the formula for the magnetic moment depending on the cue of the article that you're talking about, and we don't have any justification for the deuteron. Why didn't I carry that all the way through? Well, if I did, I'd end up with a zero magnetic moment for the neutron, it doesn't work. So, that would seem to me like, oh, well, this formula must be wrong. Well, the formula in the first place was a French factor based on a formula bar of classical mechanics, which isn't even true all the time in classical mechanics. So we're feeling our way as we go. That's one thing to say. Another thing to say is that the nuclear magneton is used here, which involves a bunch on this. It's just used as a matter of convenience. It's a convenient unit for magnetic moment in the case of deuteron particles. So it's just a unit that's logical to use so that you can easily compare magnetic moments with different deuteron particles. And the fact is that the neutron has a magnetic moment in spite of not having a charge, because as I said, the charges cancel if the magnetic moments don't. It's a composite particle, and there's dynamics inside. That's a little bit of a watch about it. Anyway, there's a whole lot of conventions in how this is set up, but that's not how it works. Yes? In deuteron, why is the mass doubled? Well, again, it's the same thing. I could double the mass if I wanted to follow the original formula, but the reason that it's not done is because the nuclear magneton, as I said, is just chosen as a convenient unit of magnetic moment for all nuclear particles. That way, you can compare different particles without having different mass factors. Bohr magneton applies to an electron, but if you talk about magnetic moments of atoms, you're going to find that it's not necessarily the Bohr magneton. The atom has mass, of course, which is much more than electron. So again, it's a similar story. What it uses is Bohr magneton. We're talking about atomic magnetic moments due to electrons. But the g-factor is not necessarily equal to 2, because there's an orbital angular momentum that's coupling the electrons. It can be more complicated in the story. We'll get into that when I talk about the Zehran effect. We'll see that there's others. You probably have seen already the landout. When you get to be my age, you make your grade stuff so you remember your names. It's the affected g-factor, as I said, in the case of the atoms, the weak magneton fields. We'll come back to that when we look at the Zehran effect. All right, so this is what's going to be? Which? Landeton. Yeah, landeton. It's starting with itself, yes. Landet is the g-factor. Yeah, we'll come back to that. All right, so this is the basic facts about the geomagnetic moments. Now, what about the Hamiltonian, describing the evolution of magnetic moments in space? I mentioned the classical level of the force on the magnetic guide pole. It's given by the gradient of u dotted and the v. u is the characteristic of the particle, and v is the function of general space and time. And so if we write this as minus the gradient of w to make it look like a potential, then w, the potential energy, is minus mu dot v. And this is the potential energy, which is usually ascribed to a magnetic guide pole in classical magnetic theory, the energy of the guide pole in the external field. There's things about this formula that always boggled me when I was a student, and maybe by the time I'll tell you what the resolution of those things are, but in any case, this is what's usually done. And so it's a guess that we can take this over as a potential energy into a Hamiltonian, either classically or mechanically. If we do this classically, you'll get the correct force on the particles. So that certainly works classically. And remember when we did this in the case of the Stern-Gerlach apparatus to understand how the beam splits in two. In any case, this means that for a charged particle, the Hamiltonian, including the spin, ought to be the usual Hamiltonian we've discussed so far, which is a p minus q of s to the a, thought in the square plus q phi. This is in an external electromagnetic charged particle in an external electromagnetic field divided, described by potentials a and phi. But if we want to include the spin as well, we need to put it with a minus sign mu dot v. v, of course, is the general function of position and time, as are the potentials of a and phi. In the view itself, it's proportional to the spin operator. So the ultimate Hamiltonian depends on the fundamental observables of a position hidden in the phi's and the x's and the a's and the phi's and also the b's. That's the x-perimple, the momenta, as well as the spin, which is hidden inside here. Now, if the particle is neutral, then it simplifies quite a lot. If you just get p squared over 2m minus some mu dot b, so this is, for example, the Hamiltonian for a neutron moving in a magnetic field. Now, for the rest of the lecture today, we're going to simplify this even further and ignore the spatial degrees of freedom. This would be appropriate if the particle in question is, since they've got a wave function which is localized, and the magnetic field is slowly varying, so the magnetic field is nearly constant over the spatial extent of the particle. So there's an x and t here. And if that's true, then we can, in effect, ignore the spatial extent. And we just have a magnetic field which depends on time. Of course, it doesn't have to depend on time, but it may. And also, likewise, we can ignore the effects of the kinetic energy because there's no x dependence that becomes effectively a free particle. So let's decouple from the spin. So just to concentrate on the spin degrees of freedom, for the rest of the day, we'll take the Hamiltonian to be a magnetic, all the q dotted over the magnetic field, which will allow the general to depend on time. Now, plugging in the expression for the magnetic moment in terms of the spin, this becomes minus the g factor times the charge of the two-dimensional city times the spin operator dotted into the magnetic field. And I'm going to take this entire quantity here and define it to be gamma, just to give it a name, a brief name. I'll summarize it. Your gamma is equal to minus g times q over 2xc like this. And so the Hamiltonian then becomes h is equal to gamma times the magnetic field b, which in general is a function of time, dot, n is spin. This is the appropriate Hamiltonian for discussing spin evolutions in cases in which the particle is spatially degrees of freedom of particle are not important. All right. Now, so the moment space operator on the right-hand side of the spin b is just a set of effective ordinary numbers, so is gamma. So h is an operator of the same kind. And in particular, the Hamiltonian now acts on this moment space for the spin, which is this space c2s plus 1, but it's to say the wavelength, which are 2s plus 1 component complex numbers. All right. Now, this is an example of a time-dependent Hamiltonian. And I'll remind you in general that the unitary time evolution operator depends on two times. I mean, Hamiltonian depends on time, the initial and the final time. Let me call this unitary time evolution operator t, t comma t0. I previously used a symbol mu here because these operators are unitary, but I'll call it t0 because I want to use u for rotation operators. So the Schrodinger equation is this. It's i h bar times the particle with respect to t, t to t comma t0 is equal to the Hamiltonian back to where it's all indicated as a function of time, back to where t comma t0 is like this. And this becomes the same thing as gamma times b dotted into s, t of t comma t0. That's all right. In general, it's not easy to solve. This is the time-dependent Schrodinger equation with a time-dependent Hamiltonian. And in general, it's not easy to solve these equations. In fact, you obviously can't solve it unless you know explicitly how b depends on time. But there are some general statements that can be made about that, regardless of how the magnetic field depends on time. In particular, let's consider this as a Schrodinger equation with a time-evolution operator. Let's consider solving this over a small time step, which is, in effect, a Taylor series expansion to make a small time step. So here's the idea. We'll take t of t plus delta t comma t0, where the delta t is small. And this is going to be equal to t comma t0 plus delta t times the derivative. The derivative is divided by i h bar. It's given by the Schrodinger equation here. So I need to multiply by the right hand side by i h bar. So this is going to be minus i minus i over h bar times n times b, which is a function of time in general, dotted into the spin s, all times the t of t0. So this is a functional initiative of an old-point tense delta t. So in the first order of delta t, that's the small time step. Now, t comma t0 is a common factor here. We can take out to the right, and we can write this this way. It's identity minus i over h bar gamma delta t. And by the way, for the magnetic field, let me break this up into the magnitude, which I'll call b of t, times its correction, I'll call b hat of t. So the b hat of t would be a vector. And we get b of t here times b hat of t dotted into the spin. And this whole thing will multiply as t of t comma t0. And this identity, I should just write as 1, because it's the unit operator, well, unit operator. If you have a time dependence on b vector, and you're going to write it in terms of a scalar and a vector, doesn't just one of those units underpin this? Well, the magnetic field could have a constant direction of changing its magnitude if it were to change it here. Just crank up the current in the coils. Then the vector wouldn't change what the magnitude would. On the other hand, you could have it changing its direction as well as magnitude. So in general, we both function the same. The interesting thing about this operator in the square brackets is that it's an infinitesimal rotation operator. That's to say, we can write it this way as I can write it in the form of u of an operator about an axis b hat by an angle, which I'll call omega t, omega delta t, excuse me. Or omega is something that's dimensions of frequency while applying t times t to the other t's at all. And the reason I know that is that if we take delta t to be small, so the angle is small, we can write this rotation operator in small angle form. And that's the same thing as 1 by this higher h bar times omega delta t times the axis b hat not into the angular momentum, which in this case is spin. And so by comparing this small angle expression for u with the quantity of square brackets up here, what we see is that omega is equal to gamma times t to the magnetic field. So there's a frequency which appears, which is gamma times t, which is spring to the magnetic field. And the result is that the result is that the order to go from the time evolution operator 1 time t to the next successive time step, we multiply by a rotation operator, which uses the instantaneous axis and the instantaneous frequency omega for the magnetic field. I couldn't fully rewrite this to emphasize that omega, since d depends on t in general, omega does also count up to theta t. Omega here in general is a function of time. And so is the axis. In fact, you can multiply the two together. Let's call omega t times b hat of t. That's defined as to the omega vector of t, the time to get the angular momentum back first. You find time to get the angular momentum back first in classical rigid body theory, for example. What it means is the evolution of the system can be thought of as a composition of a large number of small angle rotations, where at some time you've got an axis and a small angle of omega dot t. And you apply that small angle rotation, then at a later time, the axis and angle change a little bit and frequency change a little bit. You apply that rotation, the next relative to another one, you apply that one. And so by multiplying operators like this, these small angle, unitary rotation operators, you can build up t at some final time in terms of what t was at the initial time. And because the product of rotation operators is always a rotation operator, the net effect of this is that the solution of the hydropenetraining equation for a still in the magnetic field is always a rotation operator. It doesn't mean that it's going to be easy to figure out what the axis and angle is, but the final rotation is a composition of things that are changing in time, but at least we know it's a rotation operator. And it does, in fact, have some axis and some angle, also has smaller angles if you want to compute that. All right. If you remember rigid body theory in classical mechanics, there's a specific point where you can solve the Euler equations and you get the angular velocity as a function of time. Once you've got that, then you can want to find the orientation of the rigid body. You have to do an integration, which is exactly like this. It's the composition of large numbers of small angles in which the angular velocity vector is changing both in direction and in magnitude as a function of time. This is exactly the same picture. In fact, this is simpler because here we don't have to solve the Euler equations in order to find how the angular velocity vector changes as a function of time because it comes directly from the magnetic field, which we're assuming is given. All right. So within the proportionality factor, which is gamma, the magnetic field itself gives you the angular velocity, you see, as a function of time. All right. So that's a basic fact that the solution of all of these problems involving the evolution of spins and magnetic field is always a rotation operator about this one and some other cases where we can actually solve things. Simplest case where you can really solve it is one in which the magnetic field is constant. Let's call it constant in time, space in time. In this case, the Schrodinger equation for the time evolution of the property. Now, since it's constant in time, it means Hamiltonian is independent of time, instead of writing a two-time evolution operator. This term is in a single time, which is the elapsed time. So in this case, the Schrodinger equation for t is this ih bar of the time derivative of t of t is equal to gamma times v zero times v cap divided by this n s times t of t like this. Now, this equation is now easy to solve because this prefab here does depend on time, what you get is that t of t is the exponential of minus ih bar times gamma t, times t times gamma v naught times v cap divided into s, this is assuming that t of t is identity of t equal zero. And if we write omega naught is equal to gamma v naught, that's the same inversion factor between magnetic fields and frequencies that I was using in the previous board. Then this can be written this way as a unitary rotation operator with an axis of v naught, that's what this v is, the axis of v naught, and an angle which is omega naught t like this. It's a very simple result. It means pictorially that if I have a magnetic field v naught which is constant in magnitude and direction pointing like this, then the time of evolution of the system consists of the rotation about that axis by an angle which is omega naught t. So, for example, it means that the state vector at some time t is equal to this rotation, v naught comma omega naught t applied to the rotation time, but it's applied to the state vector at the initial time. Let's take a look at the expectation value of spin as a function of time. This is something that is seen experimentally, frequently seen experimentally, expectation value of spin because you have a large number of atoms and you're really averaging over all of them. And let's just call this for short, let's call this the expectation value of s as a function of time. And let's call this u thing there, let's just call this u naught like this. So this becomes u naught applied to psi dot. The lot of the u naught refers to the axis of magnetic, let me just call it u instead of u naught. This is the time of evolution of the wave. So, plugging in for psi t in terms of psi zero, this becomes psi zero sandwiched around u dagger spin s u psi zero, referring to this back to the initial time. Now what you've got is a rotation operator contributing to an angular momentum operator. And this is why I was telling you about this adorned formula once because that's just what they tell you, it's what happens when you do such a conjugation. This was written in general form, which is any angular momentum or unity, rotation operator generated by the angular momentum. When you do this conjugation, it's equivalent to just rotating angular momentum vector as a three vector by the classical rotation. Why exactly here? So that this product of three operators, so this matrix element turns into the corresponding classical rotation with the same axis we've been able to add and the same angle of angle we've been able to add applied to the spin on winner s. Now this classical rotation is a three by three matrix of numbers, they depend on time, but they're just numbers, they're not operators, they don't act on spin states or anything. Whereas acid sources that are operative, possessive operators that do act on spin states. So this rotation matrix, which is appearing inside this matrix, it can actually be taken out. The result is a vector right in this way, it's that same rotation matrix, d hat zero, negative zero, t, applied to the expectation value of spin in the initial state. Expectation value is of course just a vector of ordinary numbers, c numbers. So now you've got just a three by three matrix of all of my three by three matrix order numbers. Anyway, we can summarize all of this by saying that the expectation value of spin is a function of time, is the classical rotation r about the axis d hat with the frequency or angle saying zero, t, acting on multiplying onto the expectation value of spin at the initial time. What it shows us is that if your initial spin, expectation value of spin is some vector pointing off in some direction like this, when the effect of the solving Schrodinger equation is this thing goes around, sweeps out a cone, like this moving in a counter clockwise direction. The direction is counter clockwise if gamma is positive. Gamma will remind you it's minus g times q of two m c. So if we're talking about an electron so that q is also negative and that makes gamma positive. So for an electron, the expectation value of spin rotates in the counter clockwise direction about the direction of the magnetic field. If there were a positive particle, we would rotate the other way. Just changing the silent count. Okay, so let's see a simple case. The simplest possible case of the motion of spins and magnetic fields. Let's not turn to a more complicated case which is the one that appears in magnetic resonance experiments. The idea of magnetic resonance experiments is that you have a strong background magnetic field which I'll call v zero, which is its magnitude v zero times its unit vector v zero hat and this is the constant in the uniform. It's considered to be a strong background field, yes. I understand that when we have these expectations, you know, in that box equation on the top board, expectation and then we have a matrix multiplying the expectation and the expectation can be considered as a three-toppel, you know, across the numbers. So that equation, I sort of understand what you're saying, but what does the matrix R multiply and operate as actually, what does that really mean? Well, that what this means is that if I want to take a component of this, this is the same thing as the sum on j of R i j times S j. That's the height of component of the result. So it's a linear combination of the spin vectors. It's still a vector because there's an index i here. So this part doesn't bother you. It was a part up here I had, right? It's a part. So look, I have an equation that has vectors on both sides. Here's a spin inside and here's this protein. It's good. Let me write this out more explicitly. S i of t, S i, S i of t, equal to S i 0 sum on j, R i j, S j, S i 0. Does that help? So the R i j's are just numbers and they are not operators. They don't act on the state. So this sum on R i j can bring out. And then this formula I have here is just an abbreviated version of that. So it's just a vector of operators and then you want to bang by matrix. Yeah, but it's interesting you see when you conjugate with rotation operators, these are really operators. So these act on, these act on the part. S sort of has two kinds of characters. It has, first of all, an index i equals 1, 2, 3. It's a vector and we're in space. But it also has, it also has an operator. So if you regard it as put it in a basis of the matrix, that doesn't say it's a vector of matrices. That's how the polymatrices are viewed. And so if the matrix aspect of this is the x of the wave functions and then it's a vector index, which is happening in one of my days in the morning, does that help? So to go back to magnetic resonance experiments, the basic idea is that just at the end of it, is that you have a strongly magnetic field so that spin is recessed in this magnetic field with a corresponding frequency. We can call it a mega-node as defined as gamma times v-node. And the gamma-node here as a four is minus q times minus g times q over 2 and c. And we'll assume that this is positive just for the sake of, if you want to worry about, it would be positive for a negative particle and if you want to worry about the other side, it's best to go through the equations again and rethink the picture because it changes the directions in various factors. All right, so we've got this magnetic field and here's the unit vector vz of the half of the same number. Now, typically in this frequency of omega-zero, you can think of this being a high frequency in the gigahertz range, something like that. Maybe 100 gigahertz either depends on the, depends on the circumstances. The, and that's because this v-zero is strong, Tesla field, something like that. Now, then what you do is you impose another field which is a, which is perpendicular to the back-dog field. And I call it v1, so v1 is perpendicular to v-zero. That's part of the requirements. Also v1 is time-dependent and it has time-dependent in such a way that it has a frequency which I'll call omega-1. I'll show you explicitly the time-dependence I have in mind in just a moment. But in any case, it's got a, it's a time-dependence with a time-dependence periodic with a certain frequency of v1. Now, this notation is tricky and I think this part of the next is hard to follow which is that omega-zero is the precession frequency in the back-dog field that depends only in the magnitude of the strong background magnetic field. Omega-1 is the frequency of the time-dependence of v1 and it's independent of the magnitude of v1. In practice, what one does is for v1 is you apply some RF oscillating frequency to the system. At the same time, there's a strong uniform background static field. So this is a lower frequency, in fact, this is a, well, this is an RF frequency. It's quiet here with the frequency of v1. All right. So this is the v1, this is the basic set up here. Now, the other thing on the magnitude of v1 is normally taken to be much less than the magnitude of v0. So that the time-dependent field in the perpendicular direction is a small perturbation on the magnetic field, the background field. What one looks for, however, is one looks for resonance. Let's just say you're trying to make a resonance between the frequency of the perturbation which is omega-1 and the frequency of the natural motion of the background field which is omega-0. But if you're on resonance or near resonance, then the small effect builds up over time and you're gonna actually get a large effect. So this is why the whole thing is called nuclear magnetic resonance, or NMR for short, that's the general technique it's used. And it's electrons that call electron magnetic resonance, nuclear magnetic resonance for nuclear particles. All right. Now, the particular time-dependent magnetic field, v1, that I might consider today, is one in which v1 rotates in a circle and the plane perpendicular to v0. So if this is the initial value of v1, we call it v10 times zero, then at a later time, we have a v1 of time t and this angle here is omega-1, this angle here is omega-1t. Let's make it rotate in a color clockwise direction like this going around and around with this frequency of omega-1. We're going to be interested in the case in which omega-1 is at the same point or magnitude of omega-0, you're assured because this is the resonance condition, all right? Now, you can solve the equations for any values of the parameters, but these are the parameters that are most interesting for an experiment standpoint. So the Schrodinger equation now looks like this. It's i h bar d t of the state factor I'll call psi. And somebody tell me the time that I could develop just now, okay? Let me write out the Schrodinger equation. I'll stop that and we'll lost my watch. So it's pointing to the gamma times the total magnetic field which is v0 vector plus v1 vector, which is a function of time, got it into this spin, everything with psi and t. And this v1 vector, which is rotating in this kind of clockwise direction, we can write this as a classical rotation about the axis v0 hat and then angle, which is omega-1t, applied to the initial conditions of v1-0. So this classical rotation is putting this, like this, rotating, rotating from v1-0, swinging it around. So we capture the time dependence of the particular field by this classical rotation. And so this is the Schrodinger equation which next Monday I'll solve for you. Turns out it was the exact solution. Right, I was gonna ask you something. I need to do another makeup because I have lecture because I have to wait for two. I missed two. Is a Tuesday nine at seven to eight is still, is that still a suitable time for everyone? It's all right. Okay, I'll take a week and I'll come back some day.