Lecture 20: Central potentials and angular momentum.
L20.1 Translation operator. Central potentials (19:13)
L20.2 Angular momentum operators and their algebra (14:28)
L20.3 Commuting observables for angular momentum (17:17)
L20.4 Simultaneous eigenstates and quantization of angular momentum (24:35)
L20.1 Translation operator. Central potentials (19:13)
MITOCW | watch?v=sPsDI0dICtc
PROFESSOR: The fact is that angular momentum is an observable, and as such it deserves attention. There
is an active way of thinking of observables, and we have not developed it that much in this
course. But for example, with a momentum operator you’ve learned that the momentum
operator can give you the differential operator. It’s a derivative, and derivatives tell you how to
move, how a function varies.
So with the momentum operator, for example, you have the momentum operator p hat, which
is h bar over i d dx. And you could ask the question of, OK, so the momentum operator moves
or takes a derivative, does the momentum operator move a function? Does it generate a
translation? And the answer is, yes. That’s another way of thinking of the momentum operator
as a generator of translations.
But how does it do it? This is a Hermitian operator, and it takes a derivative. It doesn’t
translate the function. But there is a universal trick that if you exponentiate i times a Hermitian
operator, you get a new kind of operator that actually, in this case, moves things. So we could
think of exponentiating e to the i p hat, and for purposes of units I have to put a constant with
units of length, and an h bar here. And now you have the exponential of an operator.
That’s good. That’s a very interesting operator, and we can ask what does it do when you act
on a wave function? It’s an operator. And look, simplify by putting what p is going to do. P is h
over i d dx. So this is like a d dx exponentiated acting on psi of x. And as an exponential, it can
be expanded in a Taylor series with this funny object there, but it would be the sum from n
equals 0 to infinity 1 over n factorial a d dx. I will write it this as normal derivatives, because we
just have a function of x, a d dx to the n psi of x.
And you see that, of course, this is psi of x plus a d psi dx plus 1 over 2 a squared, d second
psi dx squared. But this is nothing else but the Taylor series for this. And there it is, the
miracle. The e to the i momentum generated translation. It really moves the wave function. So
that in a sense is a deeper way of characterizing the momentum operator as a generator of
translations.
With the angular momentum operators, we will have that they generate rotations. So I need a
little bit more mathematics here, because I have to deal with three dimensions, a vector, and
produce an exponential that rotates the vector, so that it gives you the wave function at a
rotated point. But this will be the same story. Angular momentum will generate rotations the
same way as momentum generates translations.
And there is yet another story that when you will appreciate the abstract properties of angular
momentum that some of them will appear today, you will realize that in addition of angular
momentum that represent rotations of objects doing things, there is another way of having
angular momentum. And that’s spin angular momentum. That mysterious property of particles
that have-- even though they have 0 size, they behave as if they were little balls rotating and
spinning.
That spin angular momentum has no ordinary wave functions associated to it, and it’s
fractional sometimes. And the study of angular momentum inspired by orbital angular
momentum associated with normal rotations, will lead us to understand where spin angular
momentum comes about. So it’s a gigantic interesting subject, and we’re beginning with it
today.
So it’s really quantum mechanics in three dimensions, central potentials, and angular
momentum. And let’s begin by mentioning that if we are in three dimensions-- and many
things with it so far in this course, we always took the time to write them in three dimensions.
So we wrote this, for example, as a generalization of the derivative form of the momentum
operator. Meaning there is a Px, which is h bar over i d dx, Py h bar over i d dy, and Pz equal
h bar over i d dz.
And we had commutators within Px and x, Py and y, and Pz and z. There were always the
same commutators of the form x Px equal i h bar. Similar things here. With this we wrote the
three dimensional Schrodinger equation, which was minus h squared over 2m, and instead of
p squared three dimensional, he would have a derivative if you were doing in one dimension.
For three dimensions you have the Laplacian. And this time you have a wave function that
depends on the vector x plus v of r-- v of x. Should I write r? Let me write r vector. V of r psi of
r equal e psi of r.
This is our time independent Schrodinger equation. This corresponds to the energy
eigenstate, but in three dimensions. So this is the equation we wish to understand, and our
ability to understand that equation in a simple and nice way rests on a simplification. That is
not always true, but it’s true under so many circumstances that it’s worth studying by itself. And
it’s the case when you have a central potential, and by that we mean that the potential is not
quite the vector function of r, but is just a function of the magnitude of r.
That’s a little bit funny way of writing it, because I’m using the same letter v, but I hope there’s
no confusion. I mean that the potential just depends on the value of r. So what this means
physically is that over concentric spheres, the potential is constant. All over the surface of
spheres of constant radius, the potential is constant, because it only depends on the radius.
And this potential is there for a spherically symmetric. You can rotate the world, and the
potential still looks the same, because rotations don’t change the magnitude of vectors. If you
have a vector of some length, you rotate it, it’s the same length, and therefore you remain on
this sphere. So the central potential are spherically symmetric. By that we mean they’re
invariant under rotations.
So this is the reason why angular momentum will play an important role, because precisely the
angular momentum operators, in the fashion we discussed a minute ago, generate rotation.
So they will have a nice interplay, to be developed in the following lectures, with the
Hamiltonian. So at this moment we have a central potential, and let’s assume that’s the case.
And we need to understand a little more of this differential equation. So let me write the
formula for the Laplacian of a function.
It has a radial contribution. You know it’s second order derivatives. And it has a radial part, and
an angular part. The units are 1 over length squared. So you need, if you have an angular
part, all over here is going to be angular, you still need the 1 over on r squared here for the
units to work out.
So here it is, it’s slightly complicated. d d theta sine theta d d theta of-- well, put the psi, plus 1
over sine squared theta, d second d phi squared all acting on psi. It’s a complicated operator,
and here is some radial derivatives, and here there are some angular derivatives. So you see,
today’s lecture will have many steps, and you have to keep track of where we’re going. And
what we’re going to do is, build up a structure that allows us pretty much to forget about all this
thing. That’s our goal.
And angular momentum will play a role in doing this. So there are in fact two things I want to
justify, two facts to be justified. So I will erase this. The first fact is the relation between this
differential operator and angular momentum. So two facts to justify. The first is that minus h
squared 1 over sine theta d d theta sine theta d d theta plus 1 over sine squared theta d
second d phi squared.
This whole thing can be viewed as the differential operator version of angular momentum.
Remember, d dx was a differential operator version of momentum. So maybe this has to do
with angular momentum, and indeed this whole thing, remember, units of angular momentum
is h bar. Angular momentum is length times momentum. And from the certainty principle, you
know that x times p has units of h bar. So angular momentum has units of h bar. So there’s h
bar squared here. So this must be angular momentum squared.
In fact, if you think about that, angular momentum is x times p. So x times a derivative. So it’s
a first order differential operator, but this is a second order one. So this could not be just
angular momentum. Anyway, angular momentum is a vector. So this will turn out to be, and
we will want to justify L squared. The quantum version of the angular momentum operator
squared.
And the other thing I want to justify if I write-- call this equation one. So this is fact one, and
fact two, is that equation one is relevant, when-- let me wait a second to complete this. This
equation is an equation for a particle moving in a potential, a spherically symmetric potential. It
turns out that is relevant under more general circumstances. If you have two particles whose
potential energy-- if you have two particles you have a potential energy between them, maybe
it’s a electromagnetic-- if the potential energy just depends on the distance that separates
them, this two body problem can be reduced to a one body problem of this form.
This is a fairly non-trivial fact, and an absolutely interesting one. Because if you want to really
solve the hydrogen atom, you have an electron and a proton. Now it turns out that the proton
is almost 2,000 times heavier than the electron. And therefore, you could almost think that the
proton creates a potential in which the electron moves.
But similar analysis is valid for neuron orbiting a nucleus. And in that case, the neuron is still
lighter than the proton, but not that much lighter. Or maybe for a quark and an anti-quark
orbiting each other. Or an electron and a positron orbiting each other, and this would be valid
and useful. So we need to somehow explain that as well.
If you really want to understand what’s going on, is that equation one is relevant when we
have a two body problem with a potential function v of x1 x2. The potential energy given that
configuration, x1 and x2 of the first and second particle, is a function of the separation only.
The absolute value or the length of the vector, it’s 1 minus x2.
This far we’ll get through today. This will be next lecture still.
L20.2 Angular momentum operators and their algebra (14:28)
MITOCW | watch?v=0T83-47Vi-M
PROFESSOR: So I want to go a little further to try to put resonances in a more intriguing footing. That you
can play with and if you-- at some point interested.
So let’s think of discovering [INAUDIBLE] that we have. We had A s-- remember the scattered
wave was A s e to the ikx [INAUDIBLE] that divided 2. And what was A s? Well, A s squared–
the sine square delta. So if you remember this was sine delta e to the i delta. So let’s stick to
that and try to write it in a funny way. Certainly, A s is becoming large near resonance, so let’s
think when A s becomes large. Well, in another way let’s be a little creative about things, It’s
good sometimes not to be logical.
So let’s write this as sine delta-- I’ll do it here-- sine delta over e to the minus i of delta . And
that’s sine delta over close delta minus i sine delta. That’s all good. A s-- let me divide by sine
delta both sides-- both numerator and denominator. So-- no divide it by cosine delta, so I’ll
have tan delta over 1 minus i tan delta. I divide it by cosine.
You want A s large? You really want it large, choose tan delta-- equals to minus i. Sounds
crazy, but it’s not really crazy. The reason it sounds crazy and it’s somewhat strange and not
very logical is tan delta is a phase and the tangent of any phase is never an imaginary
number.
So then I would have think of delta itself as a complex number. And what would that mean. So
things are weird. But it’s certainly the fact that A s will become infinite-- not just large-- but
infinite. A s will become infinite. And you say, wow, this doesn’t make any sense.
But maybe it makes sense in the following way. This is the line of real phase shifts.
[INAUDIBLE] are real. And here is the world of complex phase shifts. These are the real phase
shifts and there are the complex phase shifts. Maybe if the phase shift becomes infinite-- off
the real axis-- it’s just large on the real axis. So actually, if you wanted it to be very large you
would have to get off the real axis. If this sounds vague, it is still vague. But in a minute we’ll
make it precise.
So I suggest that we take this idea seriously-- that maybe this means something. And we can
try to argue that by looking back at what resonances do. So what I will do is look with
[INAUDIBLE] a resonance here-- tangent delta. So let’s look at what A s does. We have it
there. A s is tan delta-- well, tan delta-- we had it in the middle of blackboard is beta over alpha
minus k, 1 minus i beta over alpha minus k, again.
So that’s how A s behaves in general. That’s fine, there’s no – at this moment there’s nothing
crazy about this. Because this is something you all agreed, nobody complained about this
formula. So A s is given by that formula-- that’s also legal math, so far. So we’ll have this. And
then let’s simplify it a little bit which is beta over alpha minus k minus i beta. So this still beta
over alpha minus i beta minus k.
So we usually would plot A s as a function of k. That’s what we’re trying to do, it’s a function of
k. And now here is the formula for A s as the function of k. And here is k. But let’s be daring
now and not say this is k, this is the complex k-plane. And yes, you work with real k, but that’s
because that has a direct physical interpretation. But maybe the complex plane has a more
subtle physical interpretation and that’s what they claim is happening here.
This quantity becomes infinite near the resonance. Here was the resonance, what you call the
resonance. But this becomes really infinite not at alpha-- for when k is equal to alpha, but
when k is equal to alpha minus i beta. Beta was supposed to be small for a resonance. So
here is minus i beta and here is this very unusual point. Where the scattering amplitude blows
up. It has what is in complex variables-- if you’ve taken 1806 it’s called a pole.
In a complex variable when you have a denominator that vanishes linearly we call it a pole.
Things blow up. So this carrying amplitude has a pole off the real axis. And interpretation is
correct. At this point, this function becomes infinite. And what is happening on the real line that
A s is becoming large is just the remnant of that infinity over here that is affecting the value of
this point. So in the complex plane you understand the function a little better. You see why it’s
becoming big and you can see also with a little [INAUDIBLE] why the phases shifting very fast
because you have this point. And that’s called the resonance. And this is the mathematically
precise way of searching for resonances.
If you want to search for resonances what you should do is you have your formula for delta as
a function of k. I mean, it’s a complicated formula, but now try to solve the equation tan delta
of this is equal to minus i because that’s what guarantees that you have a pole that indeed it
blows up at some value. That’s where A s blows up which we see directly here-- it’s this value.
Alpha minus i beta, so alpha minus i beta is a pole of A s.
And therefore, you must be happening when tangent of delta is equal to minus i. So you have
a very complicated formula maybe for tangent of delta. But set it equal to minus i and asked
mathematically to solve it. And a number will come-- k a equal 2.73 minus 0.003. And you will
know-- oh, that’s a resonance, it’s off the axis. And the real part is the value of alpha. And
since this is beta the closer to the axis – if you find more-- the more resonant it is. And by the
time it’s far from the axis, some people call it the resonance-- some people say, no that not the
resonance. It’s a matter of taste. But there are important things which are these poles. So I will
not give you exercises on that, but you may want to try it if you want to have some
entertainment with these things.
I want to say one more thing about this. And it’s the reason why this viewpoint is interesting, as
well. We already found that if we want to think of resonances more precisely. We can think of
them as just an equation. You solve for the equation, so that it gives you the resonance. And
this is the equation you must solve and you must admit complex k. But now you can say, look
actually you have e is equal to h squared, k squared, over 2m. And we have real k’s-- this is
the physical scattering solutions, complex k’s, also resonances.
How about imaginary k’s? If k is equal to i kappa-- kappa belonging to the real numbers-- then
the energy becomes minus h squared, kappa squared, over 2m and its less than zero and it
could represent bound states. So you’ll be then discovering solutions of real k representing
your waves. Now mathematically, you are led to resonances understood as poles in the
scattering amplitutde we did here. We see that k’s in the imaginary axis would represent
bound states. So the complex k-plane is very rich. It has room for your scattering solutions, it
has room for your resonance, it even has room for your bound states. They’re all there. That’s
why it’s a valuable extension. I have now proven for you that bound states correspond to
poles. It’s a simple calculation, and that I would assign it to you with a little bit of guidance. And
you will see that also for the case of bound states, you get a pole in the scattering amplitude,
and that will complete the interpretation of that.
Now people go a little further, actually, and they invent poles in this part and they’re called
anti-bound states. And you’ll say, what’s that? If you have a bound state you match a solution
to a pure decaying exponential for the [INAUDIBLE] region. In an anti-bound bound state you
match your solution to a pure increasing exponential. A pure one. Does that have an
interpretation? It actually does have interpretation. Some nuclear states are associated with
anti-bound states.
So the mathematical description-- the rich complex plane is ready for you if you just do
scattering amplitude k, resonances-- complex k. Normal bound states, imaginary k-- positive.
Anti-bounds is negative k. It’s a nice start.
L20.3 Commuting observables for angular momentum (17:17)
MITOCW | watch?v=Mh8vUEStCQ8
BARTON
ZWIEBACH:
We want to understand now our observables. So we said these are observables, so can we
observe them? Can we have a state in which we say, what is the value of Lx, the value of Ly,
and the value of Lz. Well, a little caution is necessary because we have states and we have
position and momentum operator and they didn’t commute and we ended up that we could not
tell simultaneously the position and the momentum of a state. So for this angular momentum
operators, they don’t commute, so a similar situation may be happening.
So I want to explain, for example, or ask, can we have simultaneous eigenstates of Lx, Ly, and
Lz? And the answer is no. And let’s see why that happens.
So let’s assume we can have simultaneous eigenstates and let’s assume, for example, that Lx
on that eigenstate phi nought is some number lambda x phi nought, and Ly and phi nought is
equal to lambda y phi nought. Well, the difficulty with this is essentially-- well, we could even
say that Lz on phi nought is equal to lambda z phi nought. So what is the complication? The
complication are those commutators. If you do Lx, Ly and phi nought, you’re supposed to get i
h-bar Lz and phi nought. And therefore, you’re supposed to get i h-bar lambda z times phi
nought, because it’s supposed to be an eigenstate.
But how about the left hand side? The left hand side is LxLy and phi nought minus LyLx and
phi nought. When Ly acts, it produces a lambda y, but then phi nought, and then when Lx
acts, it produces a lambda x, so this produces lambda x lambda y phi nought minus lambda y
lambda x phi nought, which is the same thing, so the left hand side is 0. 0 is equal to lambda z
phi nought, so you get a-- lambda z must be 0. If you have a non-trivial state, lambda z should
be 0.
By the other commutators-- this can be attained or applied to phi nought-- would be 0 again,
because each term produces a number and the order doesn’t matter. But then it would show
that lambda x is 0, and this will show that lambda y is 0. So at the end of the day, if these three
things hold, then all of them are 0. Lambda x equals lambda y equals lambda z equals 0. So
you can have something that is killed by all of the operators, but you cannot have a non-trivial
state with non-trivial eigenvalues of these things. So we cannot have-- we cannot tell what is
Lx on this state and Ly on this state simultaneously. Any of those two is too much.
So if we can’t tell that, what can we tell? So what is the most we can tell about this state Is our
question now. We can tell maybe what is its value of Lx, but then Ly and Lz are undetermined.
Or we can tell what is Lz and then Lx and Ly are undetermined, incalculable, impossible in
principle to calculate them. So let’s see what we can do, and the answer comes from a rather
surprising thing, the fact that if you think about what could commute with Lx, Ly, Lz, it should
be a rotationally invariant thing, because Lx, Ly, and Lz do rotations. So the only thing that
could possibly commute with this thing is something that is rotationally invariant.
The thing that could work out is some thing that is invariant and there are rotations. Now we
said, for example, the magnitude of the vector R is invariant under rotation. You rotate the
vector, the demanded is invariant. So we can try the operator L squared, which is
proportionate to the magnitude squared, so we define it to be LxLx plus LyLy plus LzLz. And
we tried, we tried to see if maybe Lx commutes with L squared.
Remember, we had a role for L squared in this differential operator that had the Laplacian, the
angular part of the Laplacian was our role for L squared, so L squared is starting to come
back. So let’s see here-- this is Lx-- now, I’ll write the whole thing-- LxLx plus LyLy plus LzLz.
Now, Lx and Lx commute, so I don’t have to bother with this thing, that’s 0. But the other ones
don’t commute. So let’s do the distributive law. So this would be an Lx, Ly Ly plus Ly Lx, Ly–
this is from the first-- plus Lx, Lz Lz plus Lz Lx, Lz. You know, if you don’t put these operators
in the right order, you don’t get the right answer. So I think I did. Yes. It’s correct.
Now you use the commutators and hope for the best. So Lx, Ly is i h-bar LzLy. Lx, Ly is plus i
h-bar LyLz. So far, no signs of canceling, these two things are very different from each other.
They don’t even appear with a minus sign, so this is not a commutator, but anyway, what is
this? Lx with Lz. Well, you should always think cyclically. Lz with Lx is i h-bar, so this would be
minus i h-bar LyLz, and this is again Lz with Lx would have been i h-bar Ly, so this is minus i
h-bar LzLy, and it better cancel-- yes. This term cancels with the first and this term cancels
with this and you get 0.
That’s an incredible relief, because now you have a second operator that is measurable
simultaneously. You can get eigenstates that are eigenstates of one of the L’s-- for example,
Lx and L squared, because they commute, and you won’t have the problems you have there.
In fact, it’s a general theorem of linear algebra that-- we’ll see a little bit of that in this course
and you’ll see it more completely in 805-- that if you have two Hermitian operators that
commute, you can find a simultaneous eigenstates of both operators. I mean, eigenstates that
are eigenstates of 1, and eigenstates of the second. Simultaneous eigenstates are possible.
So we can find simultaneous eigenstates of these operators, and in fact, you could find
simultaneous eigenstates of Lx and L squared, but given the simplicity of all this, it also means
that Ly commutes with L squared, and that Lz also commutes with L squared. So you have a
choice-- you can choose Lx, Ly, or Lz and L squared and try to form simultaneous eigenstates
from all these operators. Two of them. Let’s study those operators as differential operators a
little bit.
So x, y, and z are your spherical coordinates and they are r sin theta cos phi, r sin theta sin
phi, and r cos theta. We’re trying to calculate the differential operators associated with angular
momentum using spherical coordinates. So r is x squared plus y squared plus z squared.
Theta is cosine minus 1 of z over r and 5 is tan minus 1 y over z. And there’s something very
nice about one angular momentum operator in spherical coordinates, there is only one
angular momentum that is very simple-- its rotations about z. Rotations about z don’t change
the angle theta of spherical coordinates, just change the angle phi. r doesn’t change. The
other rotation, the rotation about x messes up phi and theta and all the others are
complicated, so maybe we can have some luck and understand what is d/d-phi, the d/d-phi
operator.
Well, the d/d-phi operator is d/dy dy/d-phi plus d/dx dx/d-phi-- the rules of chain rule for partial
derivatives-- plus d/dz dz/d-phi. But z doesn’t depend on phi. On the other hand, dy/d-phi is
what? dy/d-phi, this becomes a cos phi-- it’s x. X d/dy. And dx/d-phi is minus y. And you say,
wow, x d/dy is like x py minus y px, that’s a z component of angular momentum!
So indeed, Lz, which is h-bar over i x d/dy minus y d/dx, h-bar over i is because of the p’s-- x
py minus y px. And this thing is d/d-phi, so Lz, we discover, is just h-bar over i d/d-phi. A very
nice equation that tells you that the angular momentum in the z direction is associated with its
operator.
So I have left us exercises to calculate the other operators that are more messy, and to
calculate Lx Ly as well in terms of d/d-theta’s and d/d-phi’s. And as you remember, angular
momentum has units of h-bar and angles have no units, so the units are good and we should
find that. So that calculation is left as an exercise, but now you probably could believe that L
squared, which is LxLx plus LyLy plus LzLz is really minus h squared 1 over sin theta d/dtheta. No, it’s-- not 1 over sin theta-- uh, yep. 1 sin theta d/d-theta-- sin theta d/d-theta plus 1
over sin squared theta d second d phi squared.
So the claim that they had relating the angular momentum operator to the Laplacian is true.
But, you know, you now see the beginning of how you calculate these things, but it will be a
simple and nice exercise for you to do it.
L20.4 Simultaneous eigenstates and quantization of angular momentum (24:35)
MITOCW | watch?v=lWTUcojZ_gQ
PROFESSOR: Simultaneous eigenstates. So let’s begin with that. We decided that we could pick 1 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
dependents. 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 l z, everybody chooses l z. So we
should find simultaneous eigenstates of this 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 this equations.
So we wish that l z acting on psi l m is going to be a number times psi l m. That is to be an
eigenstate. The number must have the right units, must be an H bar. And then we’ll use m. 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 l z. From l squared on psi l m, I can demand that
this be equal because of units and h squared. And then a number, lambda psi l m. 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 can put psi l m here. And l squared, psi l m from this equation. This
would be equal to h squared lambda, psi l m, psi I m. 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 l x–
l x plus l y, l y plus l z, l z. And l x, l x-- you could bring one l x here, and you would have l x, psi
l m, l x psi l m. Plus the same thing for y and for z.
And each of these things is positive. Because when you have the same wavefunction on the
left and on the right, you integrate the norm squared. It’s positive. 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 l m.
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 is-- comes when we look at the differential equation. But 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’s 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 in 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 d d 5. That’s l z, psi l m, equal h bar m psi l
m. So canceling the h bars, you’ll get dd phi of psi l m is equal to i m, psi l m.
So psi l m is equal to e to the i m phi times some function of theta. Arbitrary function of theta
this moment. So this is my solution. This is up psi l m of theta and phi. With the term in the phi
dependants, and it’s not that complicated.
So at this moment, you say, well, I’m going to use this for wavefunctions. I want them to
behave normally. So if somebody gives me a value of phi, I can tell them what the
wavefunction is. And since phi increases by 2 pi and is periodic with 2 pi, I may demand that
psi l m of theta, and 5 plus 2 pi be the same as psi l m 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 the sine
there. But it doesn’t work. It would lead to 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 l z 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 is most 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 l m equal h squared l times l plus 1 psi l m.
One thing we can do here is let the dd phi squared act on this. Because we know what dd phi
does. Dd phi brings an i n factor, because you know already the phi dependents of psi l m. So
things we can do. So we’ll do the second d 5 squared gives minus-- gives you i m 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. To clean up things. So few things. So here is what we have. We have sine theta, dd
theta. This is the minus sine squared that you are multiplying. The h squared went away. Sine
theta, d p l m d theta.
Already I substituted that psi was into the i m 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 l l plus 1, sine squared theta, minus m
squared, p l m 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 [? Laplatians. ?] You know, people had
to study this equation to do anything with Laplatians, 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 [? close ?] and theta, in that interval goes from 1 to minus 1. So it’s a good parameter.
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 closer in theta, d d x is minus 1 over sine theta, d d theta. Please
check that. And you can also show that sine theta, d d theta is equal to minus 1 minus x
squared d d x.
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 is a sine squared theta, but sine squared theta is 1 minus cosine squared
theta.
So this differential equation becomes d d x-- 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 p l m 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 l m’s. Are going to be some
coefficients, m l m’s, into the i m phi p l m 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 d x 1 minus x squared d p l 0. But p l 0, people write as p l. The x plus l
times l plus 1, p l equals 0.
So this we solve by a serious solution. So we write p l of x equals some sort of a k-- sum over
k, a k, x k. And we substitute in there. Now if you substituted 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, a
k plus 2, plus l times l plus 1, minus k times k plus 1, a k. So actually, this recursive relation
can be put as a [? ratio ?] form. The [? ratio ?] form we’re accustomed, in which we divide a k
plus 2 by a k. And that gives you a k plus 2 over a k. I’m sorry, all this coefficient must be equal
to 0.
And a k plus 2 over a k, 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, one, we solved.
The second became an [? integrated ?] 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 recursive relation. And this 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 on 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 a k is the last one that exists. And now a l plus 2, k plus 2 would
be equal to 0. So you match this, the last efficient is the value of l. And the polynomial is an elf
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 quantized. The eigenvalues of that operator, where 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 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 operator.