Simultaneous eigenstates. So let's begin with that. We decided that we could.
pick one l and l squared, and they would commute. And we could try to find functions that are eigenstates of both. So if we have functions that are eigenstates of those, we'll try to expand in terms of those functions.
And all this operator will become a number acting on those functions. And that's why the Laplacian simplifies. And that's why we'll be able to reduce the Schrodinger equation to a radial equation.
This is the goal. Schrodinger equation has r, theta, and phi. But theta and phi will deal with all the angular dependence. We'll find functions for which that operator gives a number acting on them. And therefore, the whole differential equation will simplify.
So simultaneous eigenstates, and given the simplicity of Lz, everybody chooses Lz. So we should find simultaneous eigenstates of these two things. And let's call them psi, l, m of theta and phi, where l and m are numbers that at this moment are totally arbitrary, but are related to the eigenvalues of these equations.
So we wish that Lz acting on psi lm is going to be a number times psi lm. That is to be an eigenstate. The number must have the right units, must be an h bar.
And then we'll use m, that we don't say what m is yet, m, where m belongs to the real numbers because the eigenvalues of a Hermitian operator are always real. So this could be what we would demand from Lz. From L squared on psi LM, I can demand that this be equal because of units and h squared. And then a number, lambda psi lm. Now this lambda, do I know anything about this lambda?
Well, I could argue that this lambda has to be positive. And the reason is that this begins as some sort of Positive operator is l squared. Now that intuition may not be completely precise, but if you followed it a little more with an inner product, suppose we would have an inner product, and we could put psi lm here, and l squared psi lm. From this equation, this would be equal to h squared.
Lambda psi lm psi lm, an inner product if you have it there. And then if your wave functions are suitably normalized, this would be a 1. But this thing is lx lx plus ly ly plus lz lz. And Lx Lx, you could bring one Lx here and you would have Lx psi Lm Lx psi Lm plus the same thing for y and for z. And each of these things.
It's positive, because when you have the same wave function on the left and on the right, you integrate the norm squared, and it's positive. So this is positive. This is positive. So the sum must be positive, and lambda must be positive.
So lambda must be positive. This is our expectation, and it's a reasonable expectation. And that's why, in fact, anticipating a little the answer. People write this as l times l plus 1 psi lm, and where l is a real number at this moment.
And you say, well, that's a little strange. Why do you put it as l times l plus 1? What's the reason? The reason comes when we look at the differential equation. But the The reason you don't get in trouble by doing this is that as you span all the real numbers, the function l times l plus 1 is like this, l times l plus 1. And therefore, whatever lambda you have that is positive, there is some l.
For which l times l plus 1 is a positive number. So there's nothing wrong. I'm trying to argue there's nothing wrong with writing that the eigenvalue is of the form l times l plus 1. Because we know the eigenvalue is positive.
And therefore, whatever lambda you give me that is positive, I can always find, in fact, two values of l for which L times L plus 1 is equal to lambda. We can choose the positive one, and that's what we will do. So these are the equations we want to deal with.
Are there questions on the setting up of these equations? This is the conceptual part. Now begins a little bit of play with the differential equations, and we'll have to do a little bit of work. But.
This is what the physical intuition, the commutators, everything led us to believe, that we should be able to solve this much. We should be able to find functions that do all this. All right, let's do the first one. So the first equation. The first equation is, let me call it equation 1 and 2. The first equation is h bar over i dd phi.
That's Lz psi LM equal h bar M psi LM. So canceling the h bars, you'll get dd phi of psi lm is equal to Im psi lm. So psi lm is equal to e to the Im phi times some function of theta. Arbitrary function of theta at this moment.
So this is my solution. This is a psi lm of theta and phi. We've determined the phi dependence, and it's not that complicated. So at this moment, you say, well, I'm going to use this for wave functions.
I want them to behave normally. So if somebody gives me a value of phi, I can tell them what the wave function is. And since phi increases by 2 pi, and it's periodic with 2 pi, I may demand that psi lm of theta.
And phi plus 2 pi be the same as psi lm theta and phi. You could say, well, what if you could put the minus sign there? Well, you could try. The attempt would fail eventually. There's nothing obviously wrong with trying to put a sign there.
But it doesn't work. It would lead to very rather inconsistent things soon enough. So this condition here requires that this function be periodic.
And therefore, when phi changes by 2 pi, it should be a multiple of 2 pi. So m belong to the integers. So we found the first quantization.
The eigenvalues of Lz are quantized. They have to be integers. That was easy enough.
Let's look at the second equation. That takes a bit more work. So what is the second equation?
Well, it involves a slightly complicated differential operator. And let's see what it does. So l squared, well, we had it there.
So it's minus h squared 1 over sine theta dd theta sine theta dd theta. plus 1 over sine squared theta d second d phi squared psi lm equal h squared l times l plus 1 psi lm. One thing we can do here is let the d d phi squared act on this because we know what d d phi does.
d d phi Brings an im factor, because you know already the phi dependence of psi lm. So things we can do. So we'll do the d second d phi squared gives you im squared, which is minus m squared, multiplying the same function.
You can cancel the h bar squared. Cancel h bar squared and multiply by minus sine squared theta. Just clean up things. So a few things.
So here is what we have. We have sine theta. dd theta. This is the minus sine squared that you're multiplying.
The h squared went away. Sine theta dplm d theta. Already I substituted that psi was e to the im phi times the p.
So I have that. And maybe I should put the parentheses here to make it all look. Nicer. Then I have in here two more terms.
I'll bring the right hand side to the left. It will end up with LL plus 1 sine squared theta minus M squared PLM equals 0. There we go. That's our differential equation. It's a major, somewhat complicated differential equation.
But it's a famous one because it comes from Laplacians. People had to study this equation to do anything with Laplacians and so many problems. So everything is known about this. And the first thing that is known is that Theta really appears as cosine theta everywhere.
And that makes sense. You see, theta and cosine theta is sort of the same thing, even though it doesn't look like it. You need angles that go from 0 to pi, and that's nice. But cosine theta in that interval goes from. 1 to minus 1. So it's a good parameter.
You know, people use 0 to 180 degrees of latitude, but you could use from 1 to minus 1, the cosine. That would be perfectly good. So theta or cosine theta is a different variable.
And this equation is simpler for cosine theta as a variable. So let me write that. Do that simplification.
So I have it here. If x is cosine theta, d dx is minus 1 over sine theta dd theta. Please check that. And you can also show that sine theta dd theta is equal to minus 1 minus x squared ddx.
The claim is that this differential equation just involves cosine theta. And this operator you see in the first term of the differential equation sine theta dd theta is this, where x is cosine theta. And then there's a sine squared theta, but sine squared theta is 1 minus cosine squared theta. So this differential equation becomes d dx.
Well, should I write the whole thing? No. I'll write the simplified version. It's not, it's only one slight. m of the x plus l times l plus 1 minus m squared over 1 minus x squared, PLM of x equals 0. The only thing that you may wonder is what happened to the 1 minus x squared.
That arises from this first term. Well, there's a 1 minus x squared here. And we divided by all of it.
So it disappeared from the first term, disappeared from here. But the m squared ended up divided by 1 minus x squared. So this is our equation. And so far, Our solutions are psi lms are going to be some coefficients, nlms, e to the im phi plm of cosine theta. Now I want to do a little more before finishing today's lecture.
So this equation is somewhat complicated. So the way physicists analyze it is by considering first the case when m is equal to 0. And when m is equal to 0, the differential equation, m equals 0 first. The differential equation becomes d dx.
1 minus x squared, the PL0, but PL0 people write as PL. The x plus l times l plus 1, PL equals 0. So this. We solve by a serious solution.
So we write PL of x equals some sort of a k sum over k a k xk. And we substitute in there. Now, if you substitute in it and pick the coefficient of x to the k.
You get a recursion relation, like we did for the case of the harmonic oscillator. And this is a simple recursion relation. It reads k plus 1. This is a two-line exercise. k plus 2, ak plus 2, plus l times l plus 1 minus k times k plus 1, ak. So actually, this recursive relation can be put as a ratio form.
The ratio form we're accustomed in which we divide ak plus 2 by ak. And that gives you ak plus 2 over ak. I'm sorry. All these coefficients must be equal to 0. And ak plus 2 over ak, therefore, is minus l times l plus 1 minus k times k plus 1 over k plus 1 times k plus 2. OK.
Good. We're almost done. So what has happened?
We had a general equation for phi. The first equation, 1, we solved. The second became an intricate differential equation. We still don't know how to solve it. m must be an integer so far.
l, we have no idea. Nevertheless, we now solve this for the case m equal to 0. And find this recursion relation. And the same story that happened for the harmonic oscillator happens here.
If this recursion doesn't terminate, you get singular functions that diverge at x equals 1 or minus 1. And therefore, this must terminate. Must terminate. And if it terminates, the only way to achieve termination of this series is if l is an integer equal to k.
So you can choose some case. You choose l equals to k, and then you get that p l of x is of the form of an x to the l coefficient, because L is equal to k, and ak is the last one that exists. And now, al plus 2, k plus 2 would be equal to 0. So you match this.
The last coefficient is the value of L. And the polynomial is an L-th polynomial, up to some number at the end. And you got a quantization.
L now can be. any positive integer or 0. So l can be 0, 1, 2, 3, 4. And it's the quantization of the magnitude of the angular momentum. This is a little surprising.
l squared is an operator that reflects the magnitude of the angular momentum. And suddenly, it is. is quantized, the eigenvalues of that operator that were l times l plus 1, that I had in some blackboard, must be quantized. So what you get here are the Legendre polynomials, the p l's of x that satisfy this differential equation are Legendre polynomials. And next time, when we return to this equation, we'll find that that m cannot exceed l. Otherwise, you can't solve this equation.
So we'll find the complete set of constraints on the eigenvalues of the op.