 We're working on finding the standard angular momentum basis for wave functions in three-dimensional space, representing a particle without spin, a single particle moving in three-dimensional space. The standard angular momentum basis is otherwise the union basis of the operators L squared and Lz. And so we took those operators and applied them to a wave function called psi LL. The strategy here is to first find the stretch state in which the magnetic bond number is equal to L. It's maximum value. So the Lz equation just says that Lzx on psi LL brings out the quantum number L. There's also an L squared equation. However, we replace that with a different one, which involves the raising operator L plus. The logic is that the stretch state is annihilated by the raising operator. And it actually gives us an equation to work with that's actually easier than the L squared equation, although it's actually equivalent. In any case, we solve these two equations for psi LL, the stretch wave function. And we found that it was an arbitrary function of the radius multiplied by sine L theta times e to the i of 5. In other words, the theta in 5 dependence is determined but not the radial dependence of the wave function. This makes sense because the angular momentum operators are remarkably not involved in the radial derivatives. In fact, they don't involve radial coordinate at all. They're purely angular. And so they determine the angular dependence but not the radial dependence. Now, what this means is that the operators L squared and Lz, taken by themselves, do not form a complete set of commuting observables for a wave function in three dimensions, that's because the g of r here, since this is arbitrary, means that the simultaneous eigenstates are degenerate. We can choose any radial wave function we want. Another way to say that is that if you want to have a unique specification basis space between an additional index or else an additional observable would make a complete set. One way of doing that is just to take this radial wave function g of r and replace it by, just call it u n of r, or let's say u n of r is a, let's just say it's a basis, a normal basis of radial wave functions. If we do this, then we now have the wave function side, which has three indices on it, try n and ll. And this gives us a state we can find in rack volumes. This way is n and ll like this with three quantum numbers. If we hadn't got that, we could apply the lowering operator l minus, which will lower the magnetic quantum number. l minus, let's say, the power l minus m applied to side nll will give us side nlm. If you, two times the constant, there's a factor in this and saw it in here. But then the constant lowers it down to an arbitrary magnetic quantum number, side nlm. And this can direct the ket line from my right to this way, it's nlm. And this way, we would have a unique specification of basis state. This is an example of a standard angle in that basis. The general notation that we developed earlier was gamma jm, or gamma just stood for any extra quantum numbers needed in the present cases, the radial quantum number, and the j is the same as l in this context. All right. Now, there's a slightly easier approach to this whole subject, which is just to ignore the radial variable altogether. And instead of talking about functions on three-dimensional space, let's talk about functions just as a function of the angle of theta and phi like this. So I have two functions, f and g of theta and phi. These can be regarded as functions on a sphere. The radius of the sphere doesn't matter. We can just take it to be a radius of 1, talk about the unit sphere. Only the angular dependence matters. We can find a scalar product of such functions, f and g, as the integral over a solid n, d omega times f of theta and phi complex conjugate times g of theta and phi. So there's a Hilbert space of wave functions on the unit sphere like this. The d omega, of course, is the angular part of the normalization integral that you use for the wave functions in three-dimensional space. It's also a radial part, but this is the angular part. And if we do this, then we can now ask for what is the standard angular momentum basis for functions on the unit sphere? That's to say, the simultaneous eigenstates of L squared and Lz. Well, the equations are going to be the same as we have here, except it's going to be, let's call it fLL. So the stretch state is a function of theta and phi. We'll first of all be annihilated by L plus. And secondly, if we apply Lz to it, we're going to get a factor of L coming out, fLL and theta and phi. And it's really practically the same problem as we just dealt with, except there's no radial coordinate. So we know what the solution is. It is that fLL and theta and phi, and I'll leave this space for a normalization constant here, is sine of theta times e to the i l phi, same calculation as it has on Monday, like that. Now, if we normalize this, integrating the overall solid angles to get the norm, it's an integral to do. And if you do, you find the answer is this. It's 1 over 2 to the l times n factorial. And then there's the square root of 2 l plus 1 factorial divided by 4 pi, just do the integral when we get that. So this is the normalized stretch state. And then finally, if I introduce a conventional phase factor, which is minus 1 to the l, which I can't explain, it's just a convention, then instead of calling it f, it better be called it y, because this is actually the stretched ylm, the stretch state for the ylms. In other words, the simultaneous eigenstates of l squared and lc in the unit sphere on the ylms for the Sperma-Mormonis. Now, once we have those, so we can also think of these in ket language this way, just right at ll. Notice on the unit sphere there's no degeneracy. There's a normalization in phase, but to specify l squared and lc, their quantum number is going to be a weak weight function. All right. So by the way, another thing to mention here is that the l, the eigenmetric quantum number, only takes on integer values. There are no half integer values that appear. That's because in solving the lz equation here, the lz is a dv phi somewhere here. I have summarized the operators, the lz is minus i dv phi. We're solving the lz equation, which gives us the e to the l, a to the l phi dependence. If you wanted the answer to be single value, l has to be an integer. Although I mentioned in our last lecture that we could see that the only integer values would appear anyway, way back at the level of the rotation operators, because those were single value representations of the classical rotation, it's not double value as you would have for a half integer angular momentum. So the Sperma-Mormonis, the Sperma-Mormonis do not have a half integer angular momentum. This is the Sperma-Mormonis. Now, if you want the other spherical harmonics, the y lm's of theta and phi, these will be obtained just by applying lowering operators. This will be sub-constant, and will apply the l minus to the l minus m power applied to the y l l of theta and phi. And this will give us all the rest of the Sperma-Mormonis going from m equals plus l down to m equals minus l, which is 2l plus 1. And the l minus operator is summarized here. It's a differential operator. Like this, we need to raise some power and let it act on this expression here for the spec state. Now, if you do that, and you go through the algebra, what you'll find is that this operator raises this power by playing games so that you can convert it into an expression involving the Rodriguez formula for the associated Legendre functions. And when the smoke clears, what you end up with is a few basic facts about the y lm's, which I'll summarize here. As a principle, y lm of theta and phi turns out to be a constant times the associated Legendre function indicated p l down to cosine theta times e to the i m phi. Now, the details of this are presented in the notes, including all the constants that come out. So I won't go through that. But this is one of the main results. The associated Legendre functions are normally thought of as only being defined for positive m, whereas the m here can take on negative values. And so one of my writing is to put an absolute value around that m, and this formula then also applies to the negative values of m as well. It's worthwhile remembering that y lm's goes either the i m phi and the dependence on the azimuthal angle is particularly simple. Another thing that might be worthwhile remembering is that the associated Legendre function goes like the sine to the absolute value of m power theta times a polynomial in cosine theta. That's why a polynomial functions, because they aren't really polynomials in cosine theta. The sine goes like, of course, 1 squared with 1 minus cosine squared theta. In any case, what this shows is that these are equal to 0 at the north pole, where theta is equal to 0, that's in the left pole, unless m equals 0. The special case of m equals 0 is that, given a separate name, what it's given is indicated this way, is p l without any m index on it. These are the ordinary Legendre polynomials. In this case, they genuinely are polynomials. And either the i m phi becomes just 1, so it disappears. So for the special case of magnetic quantum number 0, the y l m is proportional to the Legendre polynomials. All right, I think those are the basic facts about the y l m. So most of the facts will be important for us. The rest of them, you can look it up to the tables. Actually, the truth is, generally, in quantum mechanics, it's easier to work with than raising lowering operators, than it is to work with the wave functions. This is certainly true for the harmonic oscillator, but it's also true for y l m, so people can do this. All right, now, the y l m's are, obviously, given a privileged role to the z direction. That's partly because the spherical coordinate system does one of the angles. It has a mutual angle. There's about the z axis, and there's no analogous angles around the axis of the y axis. For this reason, in physics problems, we oftentimes make the directions of beams and magnetic fields and things like that along the z direction. However, there are lots of problems where things are not pointed along the z direction, or maybe you have two beams, and they can't both be in the z direction. For that reason, it's often times in practical problems, it's often times necessary to rotate y l m's in effect to refer them to a different axis than the z axis. And so for that reason, as well as others, I want to tell you now how I'm rotating y l m's. So before I do that, however, let me introduce you to some slightly new notation. Let's draw the axes, x, y, and z like this, and let me put in an octane of the sphere. And let's draw a unit vector, which I'll call r-hash, just to indicate some point on the sphere here, unit sphere. That point, of course, can be specified by its, let me draw it better. That point, of course, can be specified by its spherical angles. So here's an angle theta, and then here's the angle, again, like this, here's the angle phi. And if you'll allow me, I will write r-hat, it's just an alternative notation for the spherical angles theta and phi, just to indicate a point on the unit sphere. And so for example, in some function theta and phi, I'll also write this as f of r-hat in an equivalent notation. That's supposed to be an r-hat. We can also write this in a direct notation, if you like, like this, just as we do in ordinary wave functions, where the ket-f represents the function f of theta and phi, and the ket-r-hat, which has been turned into a draw there in the matrix element, represents a delta function, which is concentrated on the sphere at a particular theta-phile location, so that when we do the integral of the scalar product, it just picks off the balance of a given point. Anyway, it's a direct notation for functions of the unit sphere. Now, let's take a rotation, let's make it a proper rotation. And let's take a ylm, and let's say evaluated at a point r-hat in the sphere, or rather, let's do it this way. Let's take a ylm and let's rotate it. And then we'll evaluate the rotated ylm at a point r-hat in the sphere. Now, what I'm gonna do is to evaluate this rotate ylm in two different ways, which gives rise to an interesting result. The first way is to use the definition of what rotation operators do on wave functions in three dimensions. And now you rotate the argument by the inverse classical rotation. So this is ylm of the capital R inverse applied to r-hat. All right, that's one way of expressing this. Another way of expressing this is to say that this is the wave function evaluated at position r-hat in the sphere of the rotated state. So if I think of ylm of r-hat in the graph language, this way, it's r-hat and lm, where the ket lm just stands for the same as the ylm. Then on the left-hand side here, we have the wave function of the rotated state. That in the graph language is r-hat on the left, the rotation operator in the middle and then the ylm on the right-hand side. It means the same thing. Now, allow me to insert a resolution of the identity between the r-hat and the unitary rotation operator. So this becomes then the sum on prime r-hat lm prime, lm prime, u of r and lm. One pause because you can clearly see the other product which is a part of the resolution of the identity. But notice that I'm only summing on m prime. To get a resolution of the identity, you need to sum over a complete set. On the sphere, the complete set is all the ylm's including varying l as well as m. There's no extra quantum numbers needed because as we showed just a moment ago, there's elsewhere than lz to be your unique state if you're only in a sphere. But the question here is why didn't I sum also over l in order to get a complete set of basis states in this resolution of the identity? Well, the answer is I could have. I could have put this in l prime and then just like this in which case we really do have an insertion of the resolution of the identity. However, the matrix elements of the rotation operator are diagonally angular momentum. So this matrix element here goes like l to ll prime. And if I take that back into account, I can do the l prime sum that just replaces l prime and my l and takes me back to where I was a minute ago. So that's part of the explanation. But let me just remark about this a little more. The fact that this matrix element of the rotation operator is diagonally ll prime is sometimes expressed as a common language in quantum mechanics. It's expressed by saying the rotation operator does not mix as an expression, does not mix as a different l. That's just another way of saying it. It's diagonally l. There's another way of saying it, however. There's just to say that the states which appear on the right-hand side here, these are the standard angular momentum basis or the yl ones. If you fix the value of l, then allow them to be variable. That's to say you consider the span of states lm like this for a fixed element or m ranges from minus l over plus l. This is a space that we've been calling an irreducible subspace, more exactly or more precisely an irreducible invariant subspace. To say that it's invariant means that the rotation operator acts on a vector of the space. It produces another vector in the same space. In particular, if it acts on an lm, it produces a linear combination of these same vectors with that same value of l but not other values of l. And this is just another way of saying that the matrix elements are diagonal and l and l prime. Anyway, I just want to point out all those different ways of interpreting this. So when all is said and done, I can remove the l prime from the sum and take the primes off the l here. And then what we've got is this, but it's really just a resolution of the identity being inserted in here. Now, having done that, the first term, we can do some things with this. The first term is the ylm prime, evaluated in position r hat in the sphere. And the second term is the D matrix. This is the definition of the D matrix. It's the matrix elements of rotation operators between angular and basis states. So this is the D matrix with index l, m prime, m, and it's parameterized by the classical rotation r like this. So the result is just to repeat this, we have one version of a rotation operator here, one result, one rotating the ylm in one way, and here's rotating it in another way. Let me clean it up and write out the formula here. So ylm at the inverse rotated position is equal to a sum of m prime of ylm prime at the original position, both five times the D matrix, dlm prime of the rotation matrix r like this. So this is the basic formula for rotating ylm's. As you see, the rotated ylm is expressed as a linear combination of the unrotated ones. Now this formula gives us the value of the ylm at a rotated point on the sphere in terms of the ylm's at some other point. If there's other point, if we take that to be a reference point, then you see if you know the value of one point of the sphere, you can use this formula to get the ylm's anywhere on the sphere in terms of the D matrices. A convenient reference point is the north pole. If I make this, allow me just to use my eraser to replace r hat by the north pole here. Like this, the z hat like this. If we do this, then we're now expressing a ylm at an arbitrary point in terms of the value of the north pole. Well what is the value of the north pole? As I explained a moment ago, it may still be up there. So the north pole, all the ylm's are equal to zero unless m is equal to zero. And so this is proportional to delta m prime zero. And it turns out when you go through the normalization constants in solos, the proportionality factor is the square root of two l plus one over four pi. And so this sum can be done, and this becomes the m prime goes to zero. So what we get is the square root of two l plus one over four pi times the D matrix Vl m prime is zero, zero m to be evaluated at notation matrix r like this. Now allow me to, let's think of this r inverse z hat here as some position on the sphere which we wish to evaluate the ylm's. It actually might be easier to understand this if I replace r by its inverse. So let me copy the formula down again on the left hand side here, r inverse z hat. And now if it'll allow me again to use my eraser, I'll replace r by r inverse. R is a dummy variable here, so I can replace it by its inverse. So the inverse goes away there and it pops up over here, like that. Now let's think of r z hat as a point on the sphere. In particular, let's let r in the Euler angle form have Euler angles pi, theta, and zero. These are otherwise what we call the Euler angles I'll elevate again. So this is a rotation about the z axis by angle pi multiply times the rotation about the y axis by angle theta. The reason I do that is because if we applied this rotation in the raw picture you can see here's the x, y, z if I go a better picture. If we start with a vector of z axis, the unit vector, when we first rotate by y, about the y axis by angle theta, you see it swings it out by angle theta in the xc plane, then we follow it by rotation of the z axis by angle pi and it swings it out like this. And what it does is it produces a unit vector pointing in the direction theta and pi. In other words, we have this, is that this particular rotation pi, theta, zero applied to the z axis is equal to a vector r hat which has coordinates theta and pi in the sphere. And so if we interpret the rotation in the violin formula, as this particular one in the Euler angle form, then on the left-hand side, we've got a position theta and pi. In the right-hand side, we've got a D matrix which can be expressed in Euler angle form. Actually, allow me to take this r inverse and use the unitarity of the D matrices. The inverse of the unitary matrix is its Hermitian conjugate. So I can take this minus one out or replace it by a dagger, that's to say it's the D matrix, it's the D matrix dagger of this component. But the dagger is the same thing as the transverse complex conjugate. So it's the same thing as taking a unit of zero and replacing it by zero m and putting a star here like this. And now finally, if I plug in this particular Euler angle rotation into this formula, then here's what we get. L m's connected to the rotations. This is what we get is this. As we get Y L m of theta and pi is the square root of two L plus one over four pi times the D matrix, D L m zero of five theta and zero, those are the Euler angles, complex conjugate. So this reveals a relationship. It reveals that the Y L m's can be expressed in terms of the D matrices. There are special cases of D matrices in which one index is zero and one of the Euler angles is zero. In fact, if you really want to develop the theory of the Y L m's, it's best to do it in terms of the D matrices. It's the most elegant way of doing it. Now, this is one result about Y L m's that I want to mention. There's another one I want to mention which is the addition theorem for the spherical harmonics. This is discussed in Jackson's book for without the use of rotation operators. Let me show you what the reason for it is. Is that in dealing with problems in atomic physics or in the units matter physics, one oftentimes has to deal with this expression where r and r are primary two vectors. The idea is that these are two vectors indicating the positions of two charged particles, and so the coolant potential between them is proportional to this difference. This is also, of course, the Green's function for the Poisson equation. And it's oftentimes necessary to expand this in terms of the angular dependence of the two vectors. Let's let r vector in the spherical coordinates be r theta and phi. Let's let r prime vector in the spherical coordinates be r prime theta prime phi prime. The expansion works like this involves solving the Laplace equation and what we do is we define, let's call it r less than is the minimum of r and r prime. And r greater than is the maximum of r and r prime. And if you do this, you can show that this coolant, this coolant denominator here can be written as a sum from the L equal to zero to infinity of angular momentum quantum numbers times the expression r less than the L divided by r greater than the L plus one multiplied by the Legendre polynomial of an angular law called gamma, cosine gamma, where gamma is the angle between the vectors and r prime. So imagine this, you've got a coordinate system, x, y, and z. You've got two vectors here with one might be short and another one long, r and r prime. They represent the positions of two charged particles and you've got a distance between them, which is in the denominator here. And then the gamma then is the angle between these two vectors like this. All right. Now, it's oftentimes necessary to express this to expand the angular dependence in theta and phi as well as theta prime and phi prime in terms of y l n's. Just by looking at the picture, you might imagine you could use rotation operators to rotate one of the other of these two vectors onto the z-axis. And then the angle of gamma would become the theta coordinate of the other vector and then somehow use rotation operators to bring it back to its original shape. And that's, in fact, true. And if you do this, what you find is the following. As you find that this is equal to pi divided by 2l plus 1 times the sum of magnetic quantum numbers of a y l n of one of the positions on the sphere theta and phi times the y l n complex conjugate of the other one theta prime phi prime like this. Oh, excuse me. Here's what I mean to say. This is the same thing as the Legendre polynomial of the dot product of the unit vectors. This is the expansion I want. It's the expansion of the Legendre polynomial part of this earlier sum. So the r dot r hat dotted and r hat prime, this is the same thing as calling cosine gamma as the angle between the two radial vectors. And here it is fully expanded in terms of spherical harmonics. I won't go through the derivation of this. This is in the notes, and you can read it there. But it's an important result to use it later in dealing with perturbation theory and atomic physics. You may also want to compare the derivation of my notes, which uses rotation operators to the one in Jackson's book, which is hindered by the fact that he can't use rotation operators in that book. It's a different course. OK, so those are the basic facts about the y l m's. Now what I'd like to do is to turn to central force motion, which occupies several lectures. To begin with, let's suppose we've got some object that creates a central force field on rods that they've got there. And let's suppose there's a simplification that's mass goes to infinity, so it's a very massive object. And this allows us to attach a frame to an x, y, and z, which is an inertial frame. And let's suppose there's another particle out here of mass m, which is finite, and it's moving in the force field. And let's suppose the force field is described by a potential, which is a function only of the radius of the distance between the two particles, like this. Under these circumstances, the motion of this particle's lowercase n becomes effectively one body problem. It's just moving in a fixed force field created by the other physical object of origin. It does rely on the approximation, or the assumption that the mass is infinite, of the object that's creating a force. In many circumstances, this is a pretty good approximation. For example, in atoms, the nucleus is several thousand times heavier than all the electrons combined. In the solar system, the sun is like 1,000 times as massive as Jupiter, which is the most massive planet. So there's a lot of cases where this holds true. In any case, we then end up with effectively a one body problem. And Hamiltonian for this is the sum of the kinetic and the potential energies. The kinetic energy is p squared over 2m, and the potential is v of r. The kinetic energy, by the way, is the square of the momentum of three vectors. So if I want to emphasize that, I'll write it this way. It's the square of the vector. Now, about this Hamiltonian, the first thing to say is that it commutes with rotations. If I take a classical rotation r and then a corresponding rotation operator, exactly as we've been talking about this week, you'll find that it commutes with a Hamiltonian. It's easy to see why this is so. It's because this Hamiltonian is invariant under rotations. The kinetic energy involves the dot product momentum with itself, and dot products are invariant under rotations. And the potential energy is a function of the distance only, which is obviously invariant under rotations about the origin. And this means that h commutes with all rotation operators. Now, in particular, it will commute with infinitesimal rotation operators that have the form of 1 minus i over h bar, an angle theta times n hat dot into l. This is sort of the case with an angle of small. And if we plug that in there, it follows from this that h commutes with all three components of the negative momentum. Now, you can directly use the definition of orbital angle of minimus r cross p and use this Hamiltonian to work out the commutators. You may have done a calculation like this already in your undergraduate course, and you'll find they're equal to 0. But what I want to point out here is the geometrical meaning of the vanishing of those commutators is that the Hamiltonian is rotationally invariant. And if you're in a hurry, you want to see if something commutes with the components of angular momentum. The way to do this is to look at it and see if it's rotationally invariant. This is made up of dot products. The answer will always be that this will be one of these vanishing commutators. Notice that if h commutes with all three components of angular momentum, then it commutes with any function of those three components of angular momentum, which include rotation operators. So the implication works the other way as well. All right. So since h commutes with all three components of L, we start to think about forming a complete set of commuting observables. However, the components of L don't commute with each other. So one strategy is to do this, is to consider the second consistent h and L squared and Lc, because L squared and Lc are convenient pair of operators constructed at an angular momentum, which do commute with each other. And in fact, we already know what their eigenstates look like. The eigenstates look like this. It's called psi Lm of r theta and phi. We know that it's equal to an arbitrary radial wave function, which I'll now call capital R of r, multiplied times the y Lm of theta and phi. So as far as finding the eigenstates of L squared and Lc, we've done that part already. I'm sorry, don't confuse capital R for the radial wave function with capital R for the rotation matrix. I think in context it will be clear which one is the tenet. In any case, the idea here is that when we, in addition, require the wave function to be an eigenfunction of the Hamiltonian, that will determine the radial wave function and will have, overall, a unique wave function to within normalization of phase. In other words, h L squared and Lz will form a complete set of commuting observables. So to see how we determine the radial wave function, all we need to do is to take this form for the overall wave function and plug it into the Schrodinger equation, which is going to be h psi equals e psi. Let's consider the time independent Schrodinger equation. And there's two terms in the Hamiltonian. There's the kinetic potential energies. The kinetic energy p squared will turn into a Laplacian operator, of course, when you write it down as a differential operator. And the Laplacian operator can be converted into a spherical coordinates. And this is a standard thing to do. Laplacian involves actually quite a lot of algebra, if you do it in a brute force way. But I've summarized the results for us in the form that we needed here. Here's the momentum operator squared. It's minus h bar squared times the Laplacian. And that, in turn, involves minus h bar squared times the radial part of the Laplacian, which is here. And it turns out the angular part of the Laplacian is l squared over r squared, where l squared is the square of the angular momentum. Here we just worked out. There's obviously a fair amount of algebra in deriving these results, but there's standard results that you can find in the Laplacian typical book on E and M. In any case, the angular part is proportional to l squared. Now, l squared and r squared here are both operators acting on wave functions. And I've written in this funny way that l squared divided by r squared, operators and general don't commute. But remember, l squared only involves angular derivatives. So it commutes with 1 over r squared. And it doesn't matter which of these two factors I put first. So this is actually a meaningful expression, even though they're closer to operas. All right. So the idea then is to use this, divided by 2m, and convert it into kinetic energy. The enemy has been covered up now. And the kinetic energy part of the Hamiltonian have then included this term, the l squared over r squared. And we will then get the explicit expression of h at l and psi. So let's do this. So first of all, we have minus h of r squared over 2. And there's the radial part of Laplacian, which is 1 over r squared d dr times r squared times d dr. And this acts on psi, which I'll write as capital R of r times ylm of theta in phi. And then there's the second term of Laplacian, which is l squared over r squared. I have to divide that by 1 over 2m. So I get plus 1 over 2m r squared times l squared acting on r of r ylm of theta in phi. Then I have the potential energy plus v of r, and I'm going to space it over my capital R of r times ylm of theta in phi. And then the right-hand side, we have the total energy times r of r times ylm of theta in phi, just plugging this eigenfunction of l squared and lz into the eigenvalue equation for the Hamiltonian. Now, the only place an angular derivative is a curve here is in the l squared operator. And in fact, l squared, as I keep saying, involves only angular derivatives. So in particular, it just shines right through this r of r, and it acts only in the ylm. And when it does, it brings out the eigenvalue, which is l times l plus 1 h bar squared. And having done that, the operator disappears and is now replaced by this constant. And so now, all four terms, 1, 2, 3, 4, involve the same ylm factor, which we can cancel out. And we end up with an equation for the radial wave function alone, where it becomes this. It's minus h bar squared over 2m, 1 over r squared d dr times r squared times d capital R dr, plus l times l plus 1 h bar squared over 2m r squared times capital R of r, plus d of r times capital R of r is equal to the energy times capital R of r. In fact, it's customary in this business to define what I call u of r is equal to the potential d of r plus l times l plus 1 h bar squared over 2m r squared. This is just a definition of u of r, u of r. But what we say is, is that d is the true potential. The l times l plus 1 h bar squared over 2m r squared is a centrifugal potential, and u of r is said to be the effective potential. Notice that, physically, the centrifugal potential is actually part of the kinetic energy. In fact, the truth is that it's the angular part of the kinetic energy, whereas the first term here, involving the radial derivatives, is the radial part of the kinetic energy. The classical picture of this is that if you cut, here's the radius vector r from the origin, and you've got a particle which is moving at some velocity like this, let's say here's the velocity vector like this. You break the velocity into a radial part and into an angular part. And the radial part of the velocity gives you a radial kinetic energy. And the angular part gives you an angular kinetic energy. And that's what these two terms are. Nevertheless, from a mathematical standpoint, the centrifugal potential is a function only in the radius that is for fixed value of l. And so it looks like a potential. And so people usually throw it in with the true potential to create the effective potential. It's a standard to the view. In any case, to repeat then, we can write the radial Schrodinger equation like this. 96 bar squared over 2, we have one over r squared d v r. r squared d of the radial weight function with respect to r plus u of r times the radial weight function is equal to the energy times the radial weight function. This radial weight function is normalized by using an integral 0 to infinity of r squared v r times r of r squared. This is a normalization integral. The reason we put an r squared in here is because this is really the radial part of the normalization integral of the weight function in three-dimensional space that this thing came from. As far as the angular integration, we took care of that already with the YLMs where we put in a d omega. Let's call this version 1 of the radial Schrodinger equation. There's another version which is obtained by making a definition called f of r is equal to the radius times capital r of r. So if you plug that in to the algebra, then what you get is for the function f is you get minus h bar squared over 2m e squared f d r squared plus u of r times f of r, easy energy times f of r. And let's call this version 2 of the radial Schrodinger equation. Version 2 is easy to remember because it was just like the Schrodinger equation for a particle moving one-dimension except you call the variable r here instead of x. And remember that r only goes from 0 to infinity. And the u of r is the effective potential which includes the centrifugal potential. But it has the same mathematical form as the one-dimensional Schrodinger equation. And many of the results that we know about one-dimensional Schrodinger equations can be applied here as well. So for example, if the weight function vanishes at infinity, if you're talking about a bound state at will, it means that it's non-degenerate. So anyway, these are these two different versions. And different versions are useful in different circumstances. By this substitution, f of r equals r times capital r, you can see that the normalization integral in the second version of the Schrodinger equation is simply integral from 0 to infinity d r of the absolute value of f of r squared. Normalization also looks like what we did in one-dimensional quantum mechanics except that the limits of integration are only 0 to infinity. All right. Now, one thing to notice about the radial Schrodinger equation is that since the effective potential up here depends on the angular momentum quantum number, it means that the weight functions in the energies will also depend on the angular momentum quantum number. And if we want to make that explicit, let me put an L on both the energy and the radial weight This is going to be true, neither of these two versions. It's going to be true. It's always going to be true that the energies and weight functions depend on L. Now, in addition to that, if there are, well, one thing to say is that the energy in question can either be a continuous or discrete. You can either have bound states or you can have a continuous spectrum. It depends on the circumstances. But if you're talking about bound states, then, of course, will be discrete energies, the energy eigenvalues. And so you need an extra index to index those. So if you call a number which you need to introduce a another quantum number, let's call it N, it's kind of a radial quantum number. And in general, for the bound states, which have the discrete index, this energy, as well as the radial weight functions, really, we need to put a second index on there. We call it NL, run through here and right like this. So the E and L's, then, are the energy eigenvalues of the radial weight equation, which is indexed by L. It kind of works like this. If we take the case L equal to 0, and I draw the energy axis going like this, we've got L equal to 0 down here, and I draw the energy eigenvalues, they can be something like this. They're going to be, the points won't lie on top of each other because there is no degeneracy as the weight function dies off to infinity. But you might get some spectrum like this. And for L equals 1, you're going to get another spectrum. And the basic idea is that these different spectra for different values of L normally don't talk to each other. That's to say, one of these weight equations doesn't know about the others. And so it isn't very likely that any of these energy eigenvalues, but one of the L's will coincide with the energy eigenvalues for any other L. This is a normal situation in some randomly chosen central force potential. Notice, however, that the energies do not depend on the magnetic quantum number. You see the total weight function in three-dimensional space is side nLm of r theta n5, which would be the radial weight function, r nL of radius r times the yLm of theta n5. But the energies depend only on nL. The energies in the radial weight functions depend only on nL, but the full weight function in three-dimensional space also depends on the magnetic quantum number. And the result of this is that if you're taking to account the magnetic quantum number, you see that each of these enL's, which is each of these spots that I drew here, has a degeneracy, which is 2L plus 1, because of the possible magnetic quantum numbers. This is a generic degeneracy, which occurs in all central force problems in three dimensions. Physically, the reason why the energy does not depend on the magnetic quantum number is that that is the quantum number that indicates the orientation of the system. It's the projection of the angular momentum vector over the z-axis, and that's something that depends on the orientation. But the energy itself does not depend on the orientation because the system is rotationally invariant. And that's the physical reason for this degeneracy. Now, here's another basic fact about radial weight equations. This has to do with the behavior of the radial weight function near the origin. Many problems, you need to know how the weight function behaves near the origin. So we're interested in what happens as r goes to 0 of the radial weight function, rl, rl. I'll leave off the independence here because it won't matter. There's some angular momentum quantum number l. For example, in atoms, the size of the nucleus is something like 10 to the fifth times smaller than the size of the atom. Many problems require knowing the electron weight function in the neighborhood of the nucleus, perturbation theory, and various other things. And so in comparison to the scaling of the weight function, which is determined by the size of the atom, you're really looking at a very tiny part, the weight function, way down near r equals 0. That's a typical problem when one needs to know this. Let's suppose that the radial weight function at small r goes as a power wall. Let's say it goes as r to some power k, plus higher order terms, which is really lower here. And the problem is to find out what is the power k. So to answer that question, we just simply substitute this into the radial Schrodinger equation, which is up there at the top of the board. If r goes as ar to the k, then r prime, it's first derivative, and it goes as a times k times r to the k minus 1. If I then multiply by r squared on the following formula there for the radial part of the kinetic energy above, multiply r squared, then I get k minus 1 goes to k plus 1. If I take r squared r prime and take this derivative again, that goes as a times k times k plus 1 times r to the k. And then if I then divide this by 1 over r squared, then I get a part of the k minus 2 here. Then I divide it by minus h bar squared over 2m to complete the room here for this, minus h bar squared over 2m. Put that in here. This then is the leading behavior of the first term in the radial Schrodinger equation when r is 0, just like that. Now moving on to the second term, we've got the u of r. The u of r has the centric potential and the true potential. And as far as the centric potential goes, that's l times l plus 1 h bar squared over 2m r squared times capital R of r. But if capital R of r goes like a times r to the k, this thing just becomes l times l plus 1 h bar squared over 2m, and then we have an a times r to the k minus 2, minus 2 constant dividing by the r squared here. What about the true potential d of r times the radial wave function of r of r? How does that go? Well, the answer is it depends on what the true potential is. In particular, it depends on how singular the true potential is at the origin. In practice, the most singular true potential we ever deal with, almost ever, is the Coulomb potential, which goes like 1 over r. And so if the radial wave function goes like r to the k, the worst that this term could ever go as is r to the k minus 1, and just say that's the worst case that we'll ever encounter. But you see, this doesn't diverge as rapidly as r to the k minus 2, or r to the k minus 1 is much smaller than r to the k minus 2. So the true potential is actually negligible compared to the centric potential. Likewise, if I multiply the energy times r of r, which is the right-hand side over there, this is going to go like r to the k. And so the dominant terms of small r come strictly from kinetic energy, the radial part, as well as the angular part. These two things have to balance each other. The true potential of the total energy don't matter. Now equating these two, setting the sum of these two equal to 0, you can see that the a and h bar squared are the two names that cancel, and so does the r to the k minus 2. And what we're left with is a k times k plus 1, which is equal to l times l plus 1, simple equation. Remember here, we're talking about a fixed value of l, because we're thinking of a particular value of l for the radial Schrodinger equation. And k here is an unknown that tells us how the wave function behaves near the origin. This equation is a quadratic equation in k, and it has two roots. One of this k equals l, and the other one is k equals minus l minus 1. So given that l can grow 0, 1, 2, 3, this means k is, in the first instance, is 0, 1, 2, 3, and so on. And in the second one, it's minus 1, minus 2, minus 3 is going down the line like this. These negative powers of k would imply a radial wave function, which diverges at the origin. And for minus 2, minus 3, et cetera, on down the line, those are non-physical, because it tells up an infinite amount of probability around the origin, and the wave function is not normalized. Also, the case minus 1 has to be excluded as well, because if the wave function psi goes as 1 over r, then the little plosium, which appears in the kinetic energy, is going to go at a delta function of r. And that, of course, is singular at the origin. And the only way you could have this is if it were balanced by potential energy, which was also a delta function. Since we're limiting ourselves in the worst case to pull up potentials is the most singular, that minus 1 has to be excluded also. Anyway, the result of this is that only k equals l survives. And the main conclusion, simple conclusion, is that the rate of wave function r over r near the origin goes as r to the l. It's an easy rule to remember. If I plot it, it looks like this. If I take r of r as a function of radius r, then for l equals 0, you start with a constant value. So you can have a function that you can do like this, but it has a generally non-zero value of the origin. This is l equals 0. If you take l equals 1, it's going to go linearly in r, and look like this, l equals 1. If you take l equals 2, it's going to get parabola to do this, l equals 2. For l equals 3, we get cubic polynomial that lays down like this. And the result is that the wave function lies down ever more and more flatly as the angular momentum of the number increases. If you think about a very tiny region around the origin here, like the size of the nucleus, you can see that it's only the l equals 0 wave function is going to have any appreciable value down there and all the rest of them will count. It's a question. So if you try to integrate to find the probability, and if I assume a 1 over r to the Pobsom power dependence of r of the radial function, wouldn't part of that get cancelled by the r squared in the differential volume that I would be using to integrate? Yes. In that case, the probability is normalizable, but the problem is a different problem, which is that now the kinetic energy term in the Schrodinger equation gives you a delta function. And that would have to be balanced by either the potential energy or the true energy term, which won't happen, because the radial wave function itself goes like 1 over r. And if we assume that the potential energy is no more singular than a coolant potential, there's no way to get a delta function in there. So that case is excluded for a different reason. But after all these different arguments, the result is just a simple one, because the k equals l is the only solution. Yes. Just because the energies are also observable, and so is the probability. That is why we need to make sure that the energy does not diverge. I mean, in other words, suppose if there was a quantity, it was diverging, but it was not observable. Well, if you have an energy unless you get an eigenfunction, the eigenfunction has to satisfy, as we normalize it, it would be a legitimate eigenfunction. So I don't know that you can even talk about the energy eigenvalue if you could normalize the wave function. I understand. OK, that's all.