 I don't know what he means by this. Is that? I don't know what he means by this. Oh, for the sake of this. There you go. There you go. Well, it tells you where I strapped such a sentence. It was deeper than the outside. But I was just trying to suck that. It worked. It worked. If you get by reviewing just some things that we worked out for the hydrogen or hydrogen like atoms in the last lecture, here's the Hamiltonian. For simplicity, I've replaced the reduced mass by the electron mass because they're almost the same. The energy levels are minus 1 over 2 in square times the characteristic energy I'm calling k, which is expressed in terms of physical constants to see over here. And the degeneracy of these levels is in square. If I count the levels on a vertical energy scale with energy zero at the top, then the ground state is at minus 1 half. The next one is minus 1 half times 1 over 2 square. That's 3 quarters of the way up to the e to the 0 level, which is the ionization level. The next one, being equal to 3, is minus 1 half over 3 square and so on, as a result of the infinite number of levels that accumulate by this. Now, last time in the last lecture, I could also resolve these levels by a regular, random common number, but I'm not doing that here in this diagram. I'm just giving energy levels. Now, the degeneracy of these levels is n squared, so for the ground state, the degeneracy is 1, for the n equals 2, it's a degeneracy of 4, n equals 3, it's 9, and it goes on up like that, or the order of the degeneracies. A valid question to ask at this point is how well does this, the predictions of this theory, agree with experiment? And the answer is, it agrees pretty well that it could not perfectly, and I'd like to indicate to you what the differences are. The first thing to say is that experiment doesn't directly measure energy levels, so at least spectroscopic measurements don't. What they measure is the energy of photons that are emitted in radiative transitions, and these energies of the photons are proportional while they're equal to the energy differences between levels in the atom. So there's some disentangling that's involved in taking experimental data to spectrum of the atom and deducing from it what the energy levels of the atom are, and I'm going to gloss over that and imagine that that work has been done and for simplicity, just speak of the experimental energy levels. In any case, if you do that, what you find is that is that they aren't exactly the same as this picture that's given there. In particular, the n equals 2 level is found to be split. It's actually two very closely line levels. The n equals 3 is three levels that are closely lined, and it's all enough like this. The ground state is just on a single level. This is splitting, and these levels is called the fine structure. There's a certain energy scale to it. Let's call it that. Don't need stuff that fast for fine structure. And the basic question is what is the energy scale of the fine structure compared to, let's say, the energy scale. Let's just call it E of, I don't know what E of, don't be well, let's just say, of the model we're using. And the answer is that the ratio of these two energies is about one part in 10 to the fourth. So I have 10 to the right is four. At least that's what it is for hydrogen. It actually scales as E squared if you're going to talk about hydrogen like atoms. So it becomes more important to see if it's larger. But in any case, for hydrogen, it's quite a small spinning relative to its feet length. Nevertheless, spectroscopic data is available at quite the, the spectroscopic measurements can be made with quite, quite precision. And this splitting, this fine structure splitting was known experimentally good many years before modern quantum mechanics matured. And so to understand it was one of the major problems that was faced with the development of quantum mechanics. What it means for our particular study so far is that this model, this Hamiltonian that we wrote down here, does not incorporate all the physics. There's some extra physics involved. This is in fact only a model, meaning that it has just part of the physics in it. And to give it a name, I'll call it, I'll call this model, I'll call it the electrostatic model, or electrostatic spinless model, because the interaction between the electron and the nucleus is described by the electrostatic interaction between the electrostatic field produced by the nucleus, and also we're obviously ignoring spin in this. We know electrons do have spin. All right. So some physics is missing, and in particular it's obvious that the spin is missing because we know about spin. And it turns out that spin and other similar effects explain the fine structure. We'll go into this in more detail later on, but the main thrust of my lecture today is going to revolve around how do we incorporate spin as a new degree of freedom into the structure of the system? How does it come into wave functions? And in particular how does angular momentum interact with spin and the extra degrees of freedom that are being introduced? All right. One basic question here is what justifies an electric spin? Part of the answer is that we're neglecting degrees of freedom that we know about. You may recall earlier in the semester that I explained that when we're looking for complete sets of communion observables, you may stop before you get to the end of the list. As I said, you may know that there are other things around that you could measure, but you just choose not to measure them. In that case, this is equivalent physically to the neglect of certain degrees of freedom. And it may be justified if those degrees of freedom do not interact strongly with the ones that you're interested in. So in particular, in the case of a so-called spinless electron, which is what we seem to have here, the basic physical fact is that spins interact with magnetic fields. And if you have a problem that doesn't have magnetic fields or only weak ones, then it may be appropriate just to ignore the spin and treat the electron as if it were a spinless particle. This is a very common practice because electrostatic fields with no magnetic fields are very common situations for electrons moving in condensed matter physics and tommy physics all over the place. And that's just what we did here. We looked at the electrostatic interactions and ignored any magnetic fields. All right. Now, at this point, I'd like to shift your attention to this table, which I prepared here. This table includes a bunch of stuff we've done already in the course that summarizes it in a particular way. On the first slide here, I have a Hamiltonian for a charged particle in a given external electric and magnetic field, which we've described by the potentials a and 5. We've seen this before. It was borrowed from classical mechanics. In particular, it does not include the spin. And as I just explained, it may be appropriate to ignore the spin even if we know the particle has spin and using them for other likeness. Now, if we do this, then a complete set of community observables is made up by just the position of the particle. That's to say the x, y, and z components are three commuting operators. And if we do this, then the position eigencats form a basis for the Hilbert space of the system. Position eigencats physically represent the state of the system after a measurement has been made finding the position of the particle in some small region around a 0.3-dimensional space. And these position eigencats are parameterized by the position which is applied in a three-dimensional space. It's a continuous index. Now, the orbital Hilbert space, which is a Hilbert space for this problem, is, in a sense, the span of these basis factors. For making statements like this, I'm glossing over an awful lot of mathematics, but the basic idea is that this is the basis. This constitutes the basis for the orbital Hilbert space. Now, if phi is a state vector that lies inside this orbital Hilbert space, then its wave function is nothing but the scalar product of the position eigencats of the state vector. In other words, the wave function is the expansion coefficients of the state vector with respect to the basis states, which in this case are the position eigencats. So that's what happens if we ignore the spin degrees of freedom. In another class of problems, however, we keep the spin degrees of freedom and ignore the spatial ones. We did this for the spins in the magnetic field, but a Hamiltonian is taken to be the minus magnetic moment dotted in the magnetic field. We allow the magnetic field to depend on time, but not to depend on space, because we're ignoring spatial degrees of freedom. This would be appropriate if the magnetic field is uniform and does not depend on the space or give its uniform over the spatial extent of the wave function of the particle, but it doesn't matter if the next dependence of v doesn't matter. In that case, they may be reasonable just to concentrate on the spin degrees of freedom as we did, and then you get this Hamiltonian. Now what is the complete set of commuting observables in this case? It can be taken as s squared and s sub z. In fact, this is just what we did in our analysis of the Schrodinger-Lach apparatus in which we constructed, actually constructed the Hilbert space out of the physical measurements. It's really s sub z in that case, and we concentrate on it. The operator s squared is actually a superfluous here. The s squared operator is capital s squared. The reason it's a superfluous is because it's a constant. It has the value of a lowercase s times s plus 1 times h bar squared. Here, the capital s squared is the operator, and the lowercase s is the number, which we call the spin particle, which is just a characteristic of the particle. So this operator is really multiple of the identity, and that's why you don't really need it. In any case, the basis states in this case are the eigenstates of s squared and s z, which we denote by s and m. s is just fixed. m, however, runs from minus s to plus s. The spin-Hilbert space itself is the span of these basis vectors. If we have the state chi, which belongs to the spin-Hilbert space, we hope we may call the wave function. The spin-wave function can be seen as, call it chi sub m, which is the product of these basis states with the state vector. In other parts of this course or other lectures, instead of writing chi sub m for the spin-wave function, I may write it as chi sub m, like this. Moreover, we oftentimes think of chi sub m as being arranged in the column vector. So with chi sub s at the top, chi sub s minus 1 next, like this, this is a column vector with 2s plus 1 components, and they're all complex numbers. And as I'm sure you know, such a column vector is called a spinner. In this case, it's a spinner of complex numbers. However, for the discussion for the thing, I'm not going to use this chi sub m notation because I want to emphasize the similarity between the chi of m and the chi of r in terms of a wave function. In both cases, the wave function is just the expansion coefficients of the state of the system with respect to the basis states. All right. Now, on the third row, in some cases, we may need to include the interaction between the spatial and the spin degrees of freedom. This was the case of the Schrodinger-Bella apparatus because it's precisely this interaction between the spatial and the spin degrees of freedom, which causes the medium of atoms to split into two beams which are labeled by their spin states. One goes up and one goes down depending on the spin state, the mu sub z, which is proportional to s sub z. And so an appropriate Hamiltonian for a problem like this is the one I've written here, because we have both the interaction of the charge with the external electromagnetic field, but also that of the spin of the magnetic field. Now, actually, in the case of the silver atoms, the charge is q of zero because of the neutral atoms. So the qa and q5 terms don't appear. We just get a simple kinetic energy p squared over 2M minus mu dot b for the interaction of the spin. But more generally, for a charged particle, this would be the Hamiltonian. All right. Now, what is the complete set of commuting observables in this case? Well, it's the position of the particle of s sub z. It's a combination of the two previous ones. The basis kets are the simultaneous eigenstep in states of these two observables. There's a position. It's rm, the position of the magnet and the s sub z eigenvalue. This basis kets physically represent the state of the system after a measurement has placed the position of the particle in a small region around a point r in which the z component of the angular momentum has been measured and had a value of m. Such as, if you take one of the two beams coming out of the steric arrow, how can you make a measurement of the position? That would be the state that you'd get. Again, the r parameter runs up to all three-dimensional space and the m parameter goes to minus s to plus s. The Hilbert space in this case, which I'll call the total Hilbert space, is the span of these basis states. And if we have a state psi, which is the state vector representing the state of the system, which lies in this total Hilbert space, then its wave function is defined as the scalar product of that basic state of, excuse me, of that state vector psi with the basis states rm corresponding to the complete set of commuting observed lists. And so I want you to understand that this is the, this is the last column that shows you how wave functions are defined in these three different cases. All right, so the point of this is to understand the wave functions for spatial span and combined spatial plus span and degrees of freedom. Now, now, I have the big feeling that I'm evading something that I mentioned, but I'll go a lot anyway because I can't think what it was. So let's talk about wave functions for particles with spin as I've just indicated these are psi of r comma m like this. Now it may happen that a wave function like this can be factored into a product of a, let's call it an orbital wave function phi r times the spin wave function phi m. It's possible it could have this form and I want to emphasize that this is a special case and that a general wave function for particles with spin cannot be factorized in this manner. Yes, a question. So, for example, in number three, line number three in the table, we don't have x square because are we assuming it to be a real, I mean, an elementary particle with a single s? Yes, because it's a constant operator, it's really proportional to the identity. So, if it were a composite particle, then we could also say we need to include a square then. That's a somewhat more complicated question for a composite particle like, well, the hydrogen atom dissolves a composite particle. You really need to take into account the degrees of freedom of both the proton and the electron. So then it becomes actually a product of two spaces of the type three. See, because each of them has an orbital spin. Does that help? We'll come back to that point later on because as I said, the spin of a, what we call the spin of a composite particle is really the total angular momentum of that composite particle. That's an actually important point. All right. Okay, so the return to the point here is that this is a special case of a wave function in the composite system in the total global space. And a general wave function is not in this form. However, a general wave function, although it cannot be represented, is a single product of wave functions from these two different spaces. It can in general be represented as a linear combination of such products. It's easy to see this. Let's suppose we establish a basis of spatial wave functions. Let me call it psi a of r, where a is an index that labels the basis of one, two, and so on like this. So let's say this is a basis in our orbital orbit space like this. Likewise, let me define a basis of the term of spin wave functions where a is equal to one, two, and so on like this. And so this is a basis in which to spin. It is possible to take an arbitrary wave function of both the space and spin, and to write it as a linear combination of products of these basis wave functions. In other words, the sum of a and b of some coefficients, let me call them cab, times the twice of a, or r times the twice of b, or m, like this. Now, how do I know this? Well, really all you're doing is just using these two bases to expand the two different dependencies of the wave function on a position of r and m. In particular, the expansion coefficient c, a, c, and m are given by integral over space d cubed r and the sum of magnetic quantum numbers m of the basis pi a of r complex conjugate, the basis chi b of m complex conjugate times psi of r common m. This is just using, I'm assuming these bases here are orthonormal and this is just using the orthonimality of these bases to carry out the expansion. So the point of this is that although a general state in the combined space is not to be represented as a product, it can always be represented as linear combinations of such products. All right. Now, now, I'm not sure if I've done this table yet, so I'll try not to erase it, but I want to say some more things. So, I'll do it in the space down here. So, when we have a situation like this where a wave function in the combined space consists of linear combinations of products of the constituent spaces, we write this in a certain way. We say that the combined space, which in this case is what I'm calling e total here in this table here, we say that e total is a tensor product of the other two spaces, the constituent spaces e will only be spin in this case, which is these two here. And this x symbol here is a symbol for a tensor product. Now, this is an example of a tensor product with Hilbert spaces, and I won't define the tensor product in a rigorous or mathematical way because if I did, I think you'd find it rather unknowable that's the kind of thing mathematicians love and everyone else hates. But the basic idea is that the wave functions inside the tensor product space are all possible linear combinations that can be made out of products and wave functions from the two constituent spaces, just like you see up here. Okay. Likewise, let me go to the top of the board there where I have a particular wave function in the internal space, which is a product of wave functions that can be used in constituent spaces. And allow me to write this now in a cat language. It would look like this. It would say that side is equal to the state five tensor product with the state chi looking like this. This is a different use of the tensor product symbol because it applies to cats and not to spaces. What it means simply is just multiplication of wave functions in fact I'll say usually in physics literature people leave out a tensor product symbol and they just write this as five times chi as if it's a product. It is a kind of a product but in fact it corresponds to the work of two more linear wave functions so it's easy to remember. Also, as I mentioned, this is really only a special case that you get a product like this with the sum of products. This would look like this. You could say the size of the sum on the indices A and B of expansion coefficients C and A and B and then we have phi sub A tensor product chi sub B like this and as I say people would usually leave out the tensor product sign that they're writing a physics paper but this is a cat language for the same statement that's made up of both of the terms of wave functions. This is a definition in effect of the tensor product notation for both spaces and also for state factors. Right. Now let's see what it connects. Yes. So this is the mathematical notation for tensor products and how you define spaces. The idea now is to return to the specific problem of hydrogen and to address the issue of including the spin or hydrogen like atoms including the spin could be as free as. So we know that this is, we're talking electrons we know that electrons have a spin and we know that the spin interacts with a magnetic field by a term which is minus mu dot B. There's an interaction in Hamiltonian the however if we oh and I'll tell you what I'll make this even more general instead of making this hydrogen let me erase the hydrogen spectrum because I don't want to talk exactly about hydrogen anymore and allow me to take this potential energy and replace it by let's call it the charge of the electron times the phi of r like this because this includes now not only the hydrogen like atoms but it allows us also to talk in the same graph in case of the alkali atoms I'll remind you that the alkali atoms have a core a rotation of the giant core of electrons giving us some electron density as a function of radius and then you've got a symbol valence electron out here that you're interested in that has some position vector r and may have some velocity B like this and the Hamiltonian is describing the dynamics of this symbol out of valence electron in the combined field of the nucleus which has got a charge plus z e of the center and as well as that of the inner core electrons providing this approximate rotation of the invariant potential so in any case we have a Hamiltonian that looks like this where it flies electrostatic potential of the combined nucleus as well as the core electrons the point is that it's a function of only the radius so it's a central force problem that has a core problem let's talk about this problem and let's include the spin degrees 3 now as I just started to say this then of course interacts with magnetic fields however the field that we see here if we work in a frame which is attached to the nucleus I'll assume the nucleus is infinitely heavy a good approximation so that the inertial frame is attached to the nucleus then in that case in this inertial frame or nucleus frame the electric field is just given by minus the grain of electrostatic potential and then magnetic field is zero that's what we'll call maybe we'll call this the lab frame it really means the same thing as the nucleus frame so it looks like there is no mu not B term for the electron it isn't quite right however because the electron is moving and thus the electron we thought of as being that rested in a frame which is moving with respect to the laboratory frame and if you're on a moving frame then you don't see the electric magnetic fields of the lab frame instead you see fields that are Lorentz transformed by the motion of the electron and so this leads us to consider Lorentz transformations getting us to feel E prime or D prime when the prime refers to the moving frame and the moving frame is one moving with the velocity of the electron so I guess you've been doing this in physics 209 you can work out the formulas to look them up but the formulas for this are Lorentz transformations don't actually care about E prime D prime is what we're interested in it turns out to be minus V over C across the electric field that's seen in the lab frame like that and so the electron in its own risk does see an electric field which is V prime and not V and so we have a guess that we should have a perturbing Hamiltonian called H1 which should be able to minus V it is this correct the answer is it's not quite correct because you need to multiply by a factor of a half like this, minus a half times this the half comes about because of Thomas's session and the Thomas precession is a big tangent that I can't go off into and explain in detail because I'm going to take two or three lectures to do it but very roughly it involves the fact that this frame of the electron not only has a velocity but it also has an acceleration it's an accelerated frame and what you find in special relativity is that the accelerated frames give rise to a rotation it's as if the rest frame of the electron is rotating as well as moving now we know that in a rotating frame there are corioles forces and we've also seen in the case of magnetic resonance experiments that in these rotating frames the effect of the corioles forces is to enhance or reduce depending on the direction the effects of the magnetic field and that this is what happened in going to our rotating frames we've reduced the effect of the background magnetic field in the magnetic resonance experiments what you'll find is is that the rotation through the Thomas precession in fact cancels out one half of the effective magnetic field that's produced by the Lorentz transformation I'm really glossing over an awful lot of details but that's the net result of this and that's where the factor of one half comes from there was a period in the history of atomic physics when experiment did not agree with theory because people didn't know about this one half in fact they knew by looking at the data that it was a factor of two but then Thomas wrote his paper on rotations generated by Lorentz transformations and accelerated frames and everything was cleared up so that was the for our purposes today let's just say that this is the Hamiltonian that we're going to be interested in giving the interaction of the electron with the fields, electromagnetic fields of the nucleus, electrostatic in the lab frame but the magnetic field is also the moving frame alright now we'll make space for the erase the Thomas precession part of this these two vectors we know things about the magnetic moment vector mu is the g vector of the electron times the v over 2 mc times the spin the minus sign because the charge of the electron is the area the magnetic field v prime I just explained what that is by Lorentz transformations that's minus v over c across into the electric field the electric field is seen in the lab frame the electric field v is equal to minus the gradient of phi but we're assuming phi is the central force potential so you can write this this way minus r vector over r times the v phi v r is differentiating the spectrum of radius upon which the potential depends so if you combine all this together you see I've got 1, 2, 3, 4 minus signs which all together give me a plus sign and if we put this together this becomes a charge v over 2 mc squared then here's the times the g factor g of the electron here's the 2 there's the v over 2 mc there's another vector of c there there's c squared then we multiply this times 1 over r v phi v r I'm taking care of all the scalars first then look at the vectors we have the spin s and then that's dotted in v prime v prime goes like v cross e goes like to the direction of the radius r so this turns into v cross r is the final stuff it's left for vectors and if you'll allow me to multiply this by the mass and divide by the mass then we're going to get it squared we can do that and oh I made a mistake there's actually two factors to here and here so this is 4 however let's approximate the g factor in the electron by 2 which is a very good approximation in which case that 2 goes out that 4 goes out and becomes a 2 but as far as the mv cross r notice that this is the same thing as the momentum cross in r and the momentum cross in r is the same thing as minus r cross in momentum you have to be careful on using identities of ordinary vector calculus in a situation like this because these are not vectors of numbers these are vectors of operators and in particular p in the position momentum operators don't communicate with each other so actually in general it's not true for operators that a cross v is equal to the minus v cross a in this case it works out all right because if you look at it in detail and see what the commentators are you'll find out they've managed in the case of the resultants this is minus the orbital angle and anyway with a small clear that's just what you end up with you should find that h1 is equal to minus v over twice m squared c squared times 1 over r times v phi dr times l dot into s like this so there's a constant factor there's something that depends purely on the radius and then there's a dot product of l dot s and as I'm sure you know this is called the spin orbit spin orbit term in the Hamiltonian so to improve our model our electrostatic spinless model to include some extra physics we include the the spin orbit term h1 there we call this copy minus v over 2 m squared c squared 1 over r and phi dr times l dot s and so this is an improved model for hydrogen for alpha y atoms including spin orbit term the main purpose of my lecture today is not to go into the spin orbit to find structure effects we'll do that later a few lectures on the line rather I want to exercise what's involved in the incorporated spin degrees of freedom in the wave functions in the mechanics we then from say line number 3 in that table or actually line number 1 and then to the Hamiltonian the lower board here so we are using the same simple key but in the lower part we need kinetic momentum at the upper in the table entry we are talking about kinetic momentum well that was because up here we had an external magnetic field which is there and here we don't have an external magnetic magnetic field came from Lorenz transforming the the electric field of the of the Coulomb electric field of the nucleus and if it's not quiet also this is hand waiting this is just all we're doing here is making guesses about what the proper Hamiltonian should be to incorporate these effects the proper correct way to do this is to use relativistic quantum mechanics and we need to find all these things to come out very nicely we'll do that in the second semester in the meantime it doesn't hurt to practice with some things that are gorgeous in the hand waiting way but this does describe the physics of what's going on as to where this term comes from all right if there were a background on the electric field then you'd need to include that key dot to your field of C dot the time saver you'd need that but the next one will have an electric field by the way in deriving this I assume that the electric field in the left frame has a magnetic field with zero why don't I ask if this is really true is it really true that the magnetic field is zero than the left frame let's look at the physics of it so here's your little list of the charge plus the e and here's your electron if you're talking about the top of the line here's the valence electron which is what we're really interested in I'm recreating a picture I'll just erase like this well are there any magnetic fields magnetic fields are produced by moving charges so a professor is already moving charges here the answer is well yes they are, there's all these electrons that are swimming around with this core but it turns out they don't produce a magnetic field from their orbital notion because they're in closed shells as it turns out that cancels it all out and they don't produce the electrons of course also have a magnetic dipole from their spins, core electrons but those magnetic fields from the spins cancel out too because the spins are all here in closed shells this is what makes obliatus particularly simple so you don't get any magnetic fields from that here's something else where you might get a magnetic field from is the nucleus is not really an inertial frame because it has a finite mass so as the electron is ordered the nucleus is moving around and it creates a moving charge and so that produces a magnetic field well we're ignoring that because it turns out to be very small the nucleus may also have a magnetic moment itself because it has spin, in fact it usually does so there's a magnetic field produced by that that's a real magnetic field in the lab frame that I've been ignoring here that's actually important it gives rise to hyperfine interactions that we'll talk about later on but for simplicity I'm neglecting that as well so we're ending up with a in fact with a b equal to 0 in the lab frame it helps it helps cover some of the bases of the physical situation here alright alright so the exercise now is to examine this Hamiltonian to repeat to a certain extent what we've done earlier when we didn't have the spin orbit term but now the extra term but let me remind you that when we have just the first term the Hamiltonian might not call this H0 which was just the electrostatic model let me remind you that one of the first things we did we noticed that this H0 commutes with what I call orbital rotations which can be branched by a classical rotation the argument we gave for that was is that the H0 is made up of dot products of vectors d squared is p dot p the momentum is squared the radius r is the square root of r dot r those are those are position vectors it's logical that an orbital rotation position will never get rotated like vectors and so dot products are invariant that's not going through any rigorous derivation of this commutation relation in fact we'll study how things how operators transform the rotations more carefully in a couple of lectures but it's certainly plausible that since this is made out of dot products and dot products are invariant in the rotations that this Hamiltonian should communicate with all rotations in any case if it does communicate with all rotations let me remind you that these are orbital rotations my wife and I have this angle form they look like this as e to the minus i over h bar times theta times n hat dotted into l the orbital angle momentum by making this an infinitesimal rotation operator this implies that our Hamiltonian can be used for all three components of the orbital angle momentum of course you can check this directly by working out commentators but it follows some theory of rotations as well works the other way if h can be used in all three components of l in any function of those three components of l in particular the rotation operators this goes both ways now next given that h can be used for all three components of l a question is can we form a set of commuting operators well the three components of l don't communicate with each other but elsewhere in lz do they communicate with each other and so making that with h will begin as a set of three commuting operators by looking for the eigenfunctions so we've got the usual eigenfunctions r and l and r times y all in terms of theta and phi or the solution of the Hamiltonian h0 that's our central force that's a brief review of all the things in the force motion now what happens when we include this term again let's talk about orbital rotations does this fit an orbit term communicate with orbital rotations no the angular momentum vector is a spatial vector it's r cross p so if you rotate orbital rotations will cause that vector to rotate but the spin vector depends on the spin which is not affected by orbital rotations so you've got a dot product rotating one of the vectors and not the other therefore this dot product is not invariant therefore because of this term the Hamiltonian no longer communicates with orbital rotations and so it doesn't communicate with L squared and Lz anymore either at least there's no reason that it doesn't communicate with L vector anymore so what do we do well the first thing to notice is there's actually two kinds of rotations that appear in this Hamiltonian there's orbital and spin rotations spin rotations one is defined in terms of we've seen spin rotations but we're talking about purely spin systems if you just use a spin angle you're going to run like this now does this Hamiltonian communicate with spin rotations well the answer again is no because spin rotations will rotate spin vectors namely that one but they won't do anything to orbital vectors so now what you're doing is you're rotating the other half of this dot product and again it won't be invariant of those rotations well clearly if you want to make this dot product invariant of the rotations you need to do both orbital and spin rotations at the same time so that's both sides of this dot product are rotated by the same rotation and so this suggests that if you look at what we're called a total rotation operator which is probably right in an axis angle form there's nothing but a product of the orbital and a spin of the orbital with the same axis and angle and the spin of the same axis and angle because if we do this then the final turn is invariant as well so the Hamiltonian now connects these two rotations but not with either orbital or spin as we see there are rotation operators corresponding to different subclasses of degrees of freedom of the system in this case orbital spin in total and we can rotate one and the other but not one about doing the other necessarily alright now this product of these total rotations if I write these in exponential form the first one is even the minus i over h bar times data times n hat dotted into l and the second one is even the minus i over h bar times data times n hat dotted into this ns so this product rotation has the general form of e to the x times e to the y that these are operators no it's kind of the next product of exponentials we've seen before remember when we were talking about Glauber's theorem we worked out some cases where x and y don't commute well if x and y do commute this is actually then the combination of exponentials are based the same rules as ordinary numbers it just turns into e to the x plus y this is true again x and y commute but in the present case the x and y do commute because the essentials here are the operators l and s in particular l and s write it like this actually l and s are vectors what this really means is it means that every component of l in the s l y s j is equal to zero the reason that these two operators commute l and s is because they act on different spaces in fact there's even a basic question of what do we even mean by an orbital angular momentum acting on a wave function of a combined system what does that even mean it means this if I have a state psi then I decompose it from the products of states from the two constituent spaces like I explained a minute ago by a basic state like this and suppose I have an operator let me call it a that acts only on the spatial orbit space but we can extend this definition to act on the combined space by just saying that we get this same coefficient CAB in a act on the first term similarly if you have an operator B that acts only in the spin part you can write it this way it's something like AB with the same coefficient CAB by A and then the B acts on the spin part like this so for operators that are either purely orbital or purely spin their definition is easily extended by using linear superposition into operators that act on the combined space it's kind of obvious actually it's an obvious thing to do but it's also clear for these formulas that operators of these form purely orbital and purely spin operator communicate with each other because it won't matter what order you apply them in they act on different spaces and that's the reason for this commutator being 0 and therefore these exponentials can be combined and so this thing turns into this combined rotation operator turns into e to the minus i over h bar theta n hat this way it's n hat dotted into j where j is now defined as l plus s j is an operator which is the sum of the two angular momentum if we do this the total rotations can be written as e to the minus i over h bar theta times n hat dotted into j the total rotations are generated by the sums of the angular momentum to the constituent systems and in particular they have a total, a full Hamiltonian now it's going to communicate with the rotation operators of this form so let me try to make some space for that so here is what, here is the argument involving h0 if I make the argument that the full h is not the orbital spin anymore it's the total rotations and this is going to imply that h commutes with the total a over m of j and in particular that would mean that h will also be j squared in jz that doesn't mean that this is a complete set for getting observables but at least it's agreed they could commute and it raises the question of what are the eigenstates of j squared in jz what are the eigenstates of the whatever system you have if you have an angular momentum in axon the eigenstates of j squared in jz are what we're calling the standard angular momentum basis for that system this line of reasoning started in this Hamiltonian it leads to the question is what is the standard angular momentum basis for the combined system of orbital and spin in fact what are the eigenstates of j squared in jz okay so that's the question that I could turn to next but in a general context not just this orbital and spin one because we get a lot of of other cases of we don't want to be too specific on this point that's the more general so to be so much general about this let's suppose we have two Hilbert spaces let's call them b1 and b2 and let's suppose there are angular momentum operators j1 and j2 which we find on these on these two spaces but as a result of this there are standard angular momentum bases that you can call gamma 1, j1 and m1 on the first space and gamma 2, j2 and m2 on the second space I'll remind you that gamma is a general notation for any extra indices that are needed to resolve the genesis now let's also then define the total space e which is a tensor product of b1 cross e2 in this case we have an angular momentum j, total angular momentum which is defined as j1 plus j2 and this will satisfy the standard angular momentum commutation relations so we have an example of a space e with an angular momentum vector on it and we can define standard angular momentum commutation relations and therefore it is possible to find a standard angular momentum basis which I'll call gamma and jm on this space and the question is what is the relationship between the standard angular momentum basis on the product space in terms of the standard angular momentum bases on the two constituent spaces now to analyze this question what is the space in which these two spaces consist of just a single irreducible subspace under rotations an irreducible subspace I'll remind you is a space that's created when you apply raising and lowering operators to some stretch in particular you would fix the gamma 1m1 and just let gamma and j1 just let the m1 vary if you did that that gives you a single irreducible subspace in the case of art if you were to identify this where the orbital and spin example about a minute ago the spin example already is a single irreducible subspace there's no need for gamma and j is the same as the spin and s which is just constant however the orbital space in that case we look like this nLm that's the notation we use for that and to talk about a single irreducible subspace it means we can fix the n and the l and let the n be a variable in other words this is the looking at the set of wave functions that look like this and then we look at the yl and gamma and theta and phi in which l and m are fixed and we just let m vary from minus l to plus l so in an L fix so this is the space that's mentioned 2l plus 1 and of course the spin space that's mentioned 2s plus 1 so because these are irreducible in particular in varying subspaces it suffices in taking in any spaces like e1 and e2 and not into their irreducible in varying subspaces and then just look at them one at a time so for the orbital degrees of free let's just look at one space for nLs fixed and the spin is already a single space what this means is that we can get rid of the index gamma1 and gamma2 in this notation because we don't need that if we're talking about a single irreducible subspace moreover the indices j1 and j2 are fixed so let me erase my example here and just say that j1 and j2 are fixed so here this is a space where m1 is variable this means this is a space this means that the dimension if I make a table of dimensions here here I don't need a space here's the space here's an angular momentum here's a standard angular momentum basis and if I make the dimension the dimension of this one is 2j1 plus 1 and the dimension of that is 2j2 plus 1 so the product space is going to be the product of the dimensions it's going to be 2j1 plus 1 times 2j2 plus 1 how do I know that? I know that because the basis of this space is made up of the products of basis vectors in the constituent spaces alright, let's talk about the basis and it is j1 m1 to be explicit about it I'll put an offensive product sign j2 m2 is a product of the constituent spaces usually I'll leave out the offensive product sign and just write this as j1 m1 times j2 m2 because that's what it is for wave functions it's just multiply the wave functions and in fact I'll write this in yet another form as j1 j2 m1 m2 like this means the same thing I'll call this the uncoupled basis it's the uncoupled basis in our product space alright now we also like to find the standard angular momentum basis in this space that's what we're really aiming for and as far as we know at this point this could involve an extra index of gamma because the standard angular momentum basis is the eigenbasis of j squared and jz that's the total j squared and jz and for the work we know whether we need an extra index to resolve possible degeneracies actually as it turns out this problem this extra index of gamma is not needed here what that means is that if each anchor and min value of j occurs the ones that do occur for all at once so as it turns out we won't do the index of gamma here either but this gives us a second basis so let's call this base c's in the end it's the first basis the second basis is the basis of jm which is the eigenbasis of total j squared and jc and this is what we call the coupled basis and now if we want to work with the eigenstates of total j squared and jz which as we've seen we need to do if we want to talk about spin orbit terms we need to get a transformation from the uncoupled basis to the coupled basis this is just a change of basis in this product space that involves you probably know clutch burden coefficients we'll talk about that next time thanks for watching