Transcription
Okay, so, uh, we have been working on the angular momentum eigen states. And we have these nice relations which say that since L-squared and L-z commute, I have these eigen states which have eigenvalues h-bar squared L(L+1) for L-squared and also for L-z, the same eigen state gives me h-bar M L. Okay, so that's how, uh, these things look.
Now, obviously, these things also have a representation in coordinate space. Okay, remember that, for example, when we looked at the general wave state, okay, in Hilbert space, we called it psi, but when we looked at it, okay, how it looks in coordinate space, it was given by psi of x, okay, a function. So what do these things look like in the coordinate space? So these L-squared and L-z, remember, these are operators which involve derivatives with respect to angles. We worked one of them out explicitly, okay? So remember that, for example, we had this L-z which was equal to h-bar over i del d phi. The L-squared operator itself is more complicated. In fact, it's just the angular part of the, uh, Laplacian, okay, in the Hamiltonian or the kinetic energy, okay? But so if I now want to find the eigen vectors of these things, L-z you can see it's going to be simple, but for L-squared we are going to get somewhat more complicated functions, okay?
So what these things correspond to, okay? So the L M eigen states correspond in coordinate space to a set of functions. I just briefly mentioned that when we were doing separations of variables, but these things which are called the spherical harmonics, okay? So these are a set of very useful functions if you want to do expansions in terms of angles and such things. So they have these obviously orthogonality relations, which means that if I just find the inner product of this Y L M with another Y L prime M prime, okay, that corresponds to, okay, in fact, in this case, it becomes equal to, because this is a number, okay, so this is equal to Y L M star, because it's on the left-hand side, right? Theta and phi, and Y L prime, I should be consistent, I put the M on the top, L prime M prime, okay, Theta phi, integrated over all angles, which means I have a sin theta d theta, theta goes from 0 to pi, and d phi, which is integrated from 0 to 2 pi, so it covers the whole sphere, okay? This due to orthogonality gives me L L prime delta M M prime, okay? So that's a property of these functions, okay? So, uh, that's not obviously surprising. These also form parts of the eigen states of the, uh, Hermitian operator.
So what are these things? I'm just going to write them once, and we'll not really use them that much, uh, until we come to the hydrogen atom. Uh, then we might use them a little bit more, but, uh, at this point, let me, for completeness, because we are discussing the angular momentum, at least write the angular momentum eigen functions, uh, just once on the board. Okay, so these Y L M's of theta and phi have some normalization so that we get these nice orthogonality. And there are these functions P L M of cos theta, and there's an e to the i M phi part, okay? So the e to the i phi part, you can sort of guess, uh, because we know that the L-z operator is h-bar over i del d phi, so you can easily see that L-z operating on Y L M is going to give me, okay, h-bar M times Y L M again, okay? So it's just saying that L-z operating on Y L M gives me h-bar M times Y L M, okay? So you see the phi part is easy, but the theta part involves a little bit more, uh, algebra to find out, uh, these things. P L M's form some sort of function, okay? So in place of cos theta, if I write x, this is going to be something. Let me just check my notes so that I don't make mistakes, okay? So this is going to be equal to, uh, 1 minus x squared to the power absolute value of m over 2. You see, since I am going to put x equal to cos theta in my equation, 1 minus x, 1 minus cos squared, this thing is going to be sin squared theta, uh, so the square root means it's going to be sin theta. So this whole thing is going to be, uh, sin theta to the absolute power of m, okay? But that's not all, obviously. Uh, then I have, uh, d dx to the, uh, again, absolute of m. You see, m can be plus or minus one, right? Or plus, it can be plus or minus positive or negative. It ranges between minus L and plus L, okay? So, and then I have these P L of x, okay? So this P L M of x is called the associated, what how did we read this name? Legendre, okay, Legendre functions. Whereas this P L of x is called, uh, the, let me write it over here, Legendre polynomials, okay? And so that is given by, you can find it in various methods, but the, uh, one way of finding it is through these formulas known as the Rodrigues formula, d dx to the power L, x squared minus 1 to the power L, okay? So this is the Rodriguez formula. Okay, so this is the solution. Obviously, to get the solution, you have to solve the differential equation, which I'm not going to do here. It doesn't really, uh, contribute much to the physics, but, uh, it's all in your textbook. I recommend, well, make, let me assign you, assign it to you as a reading assignment. Just read through the derivation. It's a, uh, differential equation that you'll, uh, need to solve, and this is the solutions, uh, that you get, okay?
Let me attempt to question. Yes, sure. Um, so probabilistically, the probability distribution then doesn't depend on the sign of M, right? Sign? Yes, it doesn't depend on the sign of M. Okay, in fact, uh, the, you see, this is now a complex function. The, the polynomials, these Legendre functions are all real, but there's this complex part, but it's really quite simple, okay? So you have either the cosine or the sine phi dependence coming in. Uh, so in terms of magnitude squared, that will not contribute for a, for an eigen function, okay? But if you have linear combinations of these things, for example, if you have Y L plus M and A Y L minus M, when you add them up, you are going to get a cosine M phi, obviously, that will give you some angular dependence, okay? But if you have the solutions, you can see, uh, are of two types. For M equal to zero, all these derivatives drop out, and you are left with this polynomial, the Legendre polynomial, okay? So that's the M equal to zero case, in which case this e to the i M phi also becomes zero, so this factor becomes one, and you have a completely as, smoothly symmetric distribution, okay? So it doesn't depend on, uh, the angle phi, okay? Uh, let me just, uh, try to attempt to, uh, show these things in. Let's see. I am hopeful that this will turn on. At least it's making a sound. So they're okay. Okay, try to reach this. Okay, so let's see what we have. Uh, okay, so these are, maybe we can, well, maybe they are visible. Okay, so we have a bunch of functions here. This is Y 0 0, okay? So this is L equal to zero, M equal to zero, zero angular momentum. You see, it's just a constant. And the, because of the normalization requirement, you see that the square of this is 1 over 4 pi, uh, so when you integrate over all angles, you get one, okay? So again, you see that, for example, Y 1 0, which one is that? That's L equal to 1, M equal to zero. There's no phi dependence. But for M equal to 1 versions, okay, you get these, okay, e to the i phi plus or minus i phi. Okay, by the way, this is from Wikipedia, so we should also mention them in the, uh, license file, okay? For the recording. Uh, okay, so similarly, Y 2 0 is something like this. So you have these sine and cosine dependences. So you see, uh, for example, for Y 1 0, you have something which is positive for theta between 0 and pi over 2, and negative for pi over 2 to pi. Okay, so these things have different symmetries depending on, uh, okay, what level of, uh, L that you have, okay? Let me just try to show these things to you pictorially, okay? So let's see. Okay, so this is, these are the pictures of these things, the real part, okay? So we take the real part of these functions, and this is the Y 0 0 function, okay? So it's constant over the surface of a sphere. So what we do is we take some, uh, spherical geometry and we look at the sign of the Y L M. So for Y 0 0, you get a constant function. That's what we discussed last time. This is just equal to 1 over, uh, 2 times square root of pi. Now, this is Y 1 0. You see, again, it has axial symmetry. This is Y 1 minus 1 and Y 1 plus 1. So I have these things. The, this bluish part is positive, the yellowish part is negative. So I have positive on the Northern Hemisphere here, negative in the lower hemisphere. This is the structure for Y 2 0 and so on. And you get some axial, okay, dependence, dependence on angle phi, because this is the z-axis, right? Vertical. Excuse me, you get some dependence on phi, uh, for Y equal to Y not equal to zero, okay? So these things are useful for expanding any angular dependence, okay? Any angular dependence in terms of angles theta and phi.
So, uh, one standard example that I always give is, suppose you want to describe the temperature distribution on the surface of Earth. Uh, so how do I expand that in terms of these functions? Well, I can say there is some average temperature of the surface of the Earth. That's the Y 0 0 component. Then I say, well, it is, if it is, say, uh, summer in the Northern Hemisphere and winter in the Southern Hemisphere, it's going to be hotter, okay, in the north and warmer in the south. But then you can say also that it's warmer near the equator and colder near the poles. So you have all these contributions that come in that give you more and more detail about the angular distribution of that. But if you now want to put in some, also, aximuthal dependence, so you say that, well, during the day, the sun-facing side is hot and the night side is cold, then you have to put in one of these M equal to M not equal to zero terms. Okay, so these are basically the shapes of these spherical harmonics, and you can expand any function because you have this orthogonality. You can expand any function of theta and phi, uh, in terms of these Y L M's. They form a nice complete set which you can do that with. So, uh, they also turn out to be these solutions to the Laplace equation. Excuse me, and will give us the angular part of the wave function. The radial part we have not yet discussed, okay? The radial equation we have not, uh, gotten into, okay? So let me turn this off. Excuse me. All right. Okay, let me see if I want to say anything else about this. Okay, so, uh, again, if you have any function, F of theta and phi, a general function, you can always expand this in terms of these, uh, eigen functions, okay, Y L M of theta and phi, okay? And you have these expansion coefficients that you have to find. L will vary from zero to infinity, and then M, we know, has to go from minus L to plus L, okay? So they have to, these things obviously have to obey the same sets of rules that we derived, uh, for the, uh, for the states. How do I determine these C L M's? Well, orthogonality just tells me, okay, which is over here. Orthogonality tells me that if I just multiply both sides of the equation by some Y L prime M prime of theta and phi, the complex conjugate, F of theta and phi, and integrate over all angles, okay? So d theta is from zero to pi, and d phi is from zero to 2 pi, and I need a sin theta to get the proper angular, okay, differential element, then this is going to give me a delta, Kronecker delta, for L equal to L prime and M equal to M prime on the right-hand side. So I'm going to get just C L prime M prime, okay? So this just allows me to expand any function in terms of those. Okay, again, just the angular part, just the angular dependence. The radial dependence depends on the solution to the radial equation, which depends on obviously V of R, okay? Depending on what V is, what the potential is, I'm going to get different types of radial, [Music] solutions. All right. Now, when you go through these solutions, you have to go through a procedure in which, okay, you have to make sure that the functions are reasonable, physical, etcetera, which we have not done. But certain things you can see very easily. For example, remember that I have this L-z is equal to h-bar over i del d phi. So if I'm looking for eigen functions of L-z, what am I going to get? L-z operating on some function of phi is equal to some eigenvalue times F of phi, right? Yeah, so e to the i something. So this F of phi is going to be proportional to, because there has to be a normalization, but e to the i, uh, h-bar times this lambda. Well, not really h-bar, h. It should be i h-bar, I think, okay? i over h-bar, thank you. i or h-bar. Accept it. I can't write it. i or h-bar. Uh, times lambda times phi, okay? So I have this type of behavior, and then I say that physically, okay, that I want functions which come back to their own values. So remember what phi is. Phi is this angle here from the x-axis to. If this is the position vector, remember this is the angle theta. If I make a projection of that on the xy plane, okay, this is the angle phi. And I am now pointing at a certain point where I'm determining the wave function. And what I do is I say, well, for this thing to be physical, if I go around this once, I must come to the same position. So this object here, lambda over h-bar, must be equal to what? H. Okay, it, when I increase phi by 2 pi, I want to get to the same thing. So this lambda over h-bar must be equal to something like 0, 1, 2, 3, etcetera. So this lambda and the eigenvalue must be equal to M times h-bar, okay? If it is not equal to an integer, then if I go around the z-axis once, okay, I do not have the same value for the wave function, which looks strange, actually, at this point, but we'll be working with such systems. Similarly, for the case of theta, if L is not equal to 0, 1, 2, 3, 4, etcetera, you get unphysical solutions. So when you solve the differential equation, you are forced to use L equal to 0, 1, 2, 3, etcetera, and M is equal to minus L, okay, minus L plus 1, dot dot dot, up to plus L, okay? So exactly what we found using the operator methods. So it is nice that we, okay, convincing that we see these things going through like this, uh, the same algebra, and the same results for the, uh, eigenvalues, okay? So L becomes that, and the, uh, if you just look at L-squared times Y L M, you indeed get h-bar squared L(L+1) times Y L M, okay? So exactly the same thing as L-squared Y L M is equal to h-bar squared L(L+1) Y L M. So it's not really surprising because we have solved the problem by some other method, by these operator methods, which allowed us to make certain shortcuts, and we obtained, uh, all of those results without going through to these, uh, complicated functions. But now there is one thing when we are, we were discussing the operator method that I sort of, okay, skipped over, okay? So what was that? We obtained these eigen states. So let me remember here, we had these eigen states. We said, well, okay, we can go down by h-bar, go up by h-bar in angular momentum in L-z, okay? So the values of L-z were M times h-bar, M plus 1 times h-bar, etcetera. So we had all this spectrum, and we said we can go up to some L and down to some minus L, okay? And, okay, so we said this can be 2, 1, 0, minus 1, minus 2, for example, for L equal to 2. But there's another possibility, okay, which we sort of did not discuss, but now we'll discuss it now, okay? Because there's this possibility that I could start with 1/2 h-bar and have minus 1/2 h-bar. Okay, now this thing does not arise in the solution of the differential equation because, as I said, uh, you have these boundary conditions. If you go around the z-axis by 2 pi, okay, you want to find the same function, okay, same value of the function. With this, you cannot do that, right? Because this corresponds to L equal to 1/2, okay, and M equal to 1/2 or minus 1/2, which means you are going to get eigen functions which look like e to the i 1/2 phi, and if you increase phi by 2 pi, you get the negative of the function, okay, which doesn't look nice. Okay, but on the other hand, if this corresponds to some type of angular momentum which I do not have to express in coordinate space, okay, then they are completely valid. And it turns out that these things correspond to a different set of angular momenta. Okay, it is the angular momentum with, with respect to the central mass of the object. So what we have done up to this point, okay, so, okay, so everything to the left of this line is called orbital angular momentum. So we have been studying orbital angular momentum, which is given by L equals R cross P. Okay, so I have some particle coordinate at R, okay, and some momentum variable. So R cross P is the orbital angular momentum. Now, what we are going to discuss is called the spin angular momentum. Okay, so what is spin angular momentum? It turns out that particles come with, okay, particles have some, okay, we call intrinsic. So intrinsic means that it is built into the particle. So a particle comes with this intrinsic angular momentum, which we call the spin. Okay, so each particle itself alone, even if it's not moving any place, has some angular momentum. And those angular momenta are given again by the same types of, uh, angular momentum structure, except that now these fractional cases are also allowed. Okay, so we still allow things like 1 h-bar, 2 h-bar, 3 h-bar. So a particle can have a total angular momentum corresponding to one or two, as in the orbital angular momentum, but it can also have half an angular momentum. Okay, so a half an angular momentum will mean that this M can be plus a half or minus a half. In fact, the particles are separated into two parts. There are two types of particles. Okay, so these, this intrinsic angular momentum can have values like L equal to 0, L equal to 1, L equal to 2, so just the same type of things that the orbital angular momentum has. Okay, so these types of particles are called bosons, or they can have, okay, L equal to 1/2 or 3/2 or whatever. These things are now called fermions. Okay, so these are after two physicists. Bose is really after the name of Bose, who was a statistician who worked with Einstein. So these, uh, particles, then, are called Bose-Einstein particles, and these things are called Fermi particles. So they obey different types of statistical behavior. So these particles, bosons, obey Bose-Einstein statistics. Fermions, uh, satisfy Fermi statistics. Okay, so we'll see what those things are. So they, they, these two different types of spin result in very different types of behavior, uh, in particles. Okay, particles like, for example, photons are bosons, okay? And all of the particles that we have been studying, things of mass, things like the electron, the proton, the neutron, they are all, uh, fermions, okay? So they, uh, have different types of behavior compared to photons and etcetera. So now, what do we do? We have found out that there is this new type of angular momentum, and we have to, uh, work with that. Well, the analysis is exactly the same, because we start with the same commutation relations. Sure, angular momentum is speed. Take integer. But why spin angular momentum doesn't take integer? Just I mean, why like that? There's a difference like that? No. Okay, the spin can also have L equal to 0, 1, 2, 3. Okay, so this is also for spin. But, uh, some particles have fractional spins also. Why? Because, okay, that's the way nature is. Thankfully, because if we didn't have fermions, uh, you would not have any materials. Okay, the reason, for example, my hand is not going through this table is because the fermions in the table don't want the fermions in my hand to approach them that much. Okay, but if I was made up of photons, okay, so if I was a boson object, uh, then everything would nicely want to collapse into one another. Okay, so that's the way it goes. Uh, you see, it is possible that there's this angular momentum solution, uh, that comes out of these, this analysis, and then it's not really too surprising that some particles would take advantage of such a possibility. Okay, so this is possible. So there are such particles that have, uh, fractional spin. Okay, so what happens is, we are going to, we have practically the same analysis that we did for the, uh, orbital angular momentum. So let me just put here spin angular momentum. I am not going to obviously go through it in as much detail as we did with the orbital angular momentum, because it's just the same thing. The only difference is that instead of the L, we now start using S. Okay, so S is the angular momentum operator. Okay, if I take any component of this, this is this operator corresponds to a measurement of the angular momentum in the x direction. Okay, so I have, if I just find the commutate all those things, I find that it's equal to i h-bar S y. Now, you see, in the case of orbital angular momentum, we started with the orbital angular momentum operators R cross P, and then we could write this. Okay, yes. Can you explain again why we have angular momentums in chunks of two, but not but 1/3 would not work? Right. If I had, uh, h-bar over 3 here, if I apply the lowering operator, I'm going to end up with minus h-bar over 2 h-bar over 3, which is lower than that. I have to start from some upper level, okay, so upper angular momentum level and go to minus that. Okay, so if the upper one is equal to h-bar over 3, it doesn't fit. So it has to be h-bar over 2. Okay, so this doesn't, doesn't work. Okay, so I have again the same set of operators. These are really derived from, okay, the properties of rotations. Remember that. Okay, maybe I'll open a parenthesis and discuss that, because that's also interesting. Let me just complete. This is, uh, z S x is equal to i h-bar S y. Okay, so we start with these, and obviously, it's not surprising that if we start with the same equations, the only thing we have called to done is replace L with S. Okay, so we are going to get the same results. Okay, but, uh, I'll just write the results here now. Okay, let me open a, uh, parenthesis here and show you something that's interesting. Okay, in fact, let's start with just translation. If I have an operator like this, if I have e to the, uh, i, uh, or h-bar, some constant distance x0 times Px, okay? So this is some operator. Suppose it acts on some function f(x). Now, if I just put in what the operator is, this is h-bar over i. So the h-bar over i is canceled. I have e to the x0 d dx, right? f(x). So this is exponential of a derivative operator. So it looks perhaps a little scary, but we know how to exponentiate operators. We just expand this, right? So this is 1, 1 plus x0 d dx, plus 1 over 2 factorial, x0 squared, d squared dx squared, plus dot dot dot, f(x). So does that look familiar? No, no. Go back. Real back. Yes, all the way to freshman. Okay, so this is really the expansion of f(x) near x. So this, what do I get? What do I get? No, you should recall your h(x+x0) right? Okay, x plus x0. So what did this do? It shifted the function by x0. I can try the same thing with, for example, an angular momentum operator. The only one that we know well at this point is the L-z operator. So if I take e to the i or h-bar, phi0, L-z. Now this will operate on some function of phi. So it's going to be e to the i or h-bar. Well, i or h-bar cancel because I have h-bar over i in L-z. So it's going to be del d phi, f of, okay, phi, right? Okay, so this is going to be phi plus phi0. Well, we have another operator, uh, which is the, uh, okay, so i or h-bar. Let me put here some t times the Hamiltonian H, which will act on S at, okay, t equal to zero. What will this give me? T. T. Well, okay, it will give me S at T, right? Remember we discussed these things. So this structure here says, if you have these two somehow related quantities, x displacement and momentum, okay, so F0 and L-z, time and energy, these things cause displacements in whatever they are operating on. Okay, so for example, the technical H for this, let me just, we say that the angular momentum is a generator of the translation, generator of, uh, translations in angle. Okay, so for this one, we would say the momentum operator is a generator of translations in position. The Hamiltonian is a generator of translation in time. Okay, so this is another starting point for quantum mechanics. So if you take a quantum mechanics course in the graduate level, the first things that you perhaps study are these things, and then you study how these rotations interact with one another. How does a rotation along the z-axis and the rotation along the x-axis, okay, interact to another? Even classically, okay, the rotation around different axes are not, okay, do not commute. Okay, so if you have, if you take the Earth, for example, okay, and you have this perhaps North Pole, and if you just rotate it around this axis, axis looking towards me, it will come perhaps to this point, and then rotate it around this axis, it will move to that point. But if you now rotate the object first around the z-axis and then rotate it with respect to that one, it will go to a different point, right? So the rotations do not commute with one another, and their commutation relations will give you this. Okay, so the reason I have went through this is because I have just written these things down in some ad hoc manner. Okay, we know why these things should be so for orbital angular momentum, but for spin angular momentum, obviously, uh, we do not have an R cross P type of term. So what you do is you either say, well, okay, let me just assume that these commutation relations are exactly the same as the orbital angular momentum case, or you can really, uh, do a more fundamental analysis and obtain this. Okay, so what's the summary? We have these relations. Okay, so perhaps we have a break now and.