Welcome back everyone. 
I'm Charles Clark. 
This is exploring quantum physics. 
And we're going to, finish up for this 
week with a discussion of angular 
momentum. 
Next week, you'll have a set of lectures 
by Ian Applebaum that introduce spin. 
And then we'll return to bring those 
parts together in the following week. 
So let's begin. 
We're going to say hello again to an old 
friend the hydrogen atom. 
where as we saw in the classical 
treatment, there was this remarkable 
unusual conserved quantity. 
The run that described in terms of the 
Runge-Lenz vector. 
and that as you will recall helps us 
completely define the orbit of the 
classical particle. 
It makes possible a classification of all 
possible, orbits of the electron in the 
hydrogen atom, in, in terms of the 
single, in terms of the angular momentum 
and the Runge-Lenz vector. 
Now as defined in classical mechanics 
this has some issues because, in 
classical mechanics P cross L is equal to 
L cross P and as we saw the, sorry, minus 
L cross B. 
There's no, there's no issue about 
commutation so in quantum mechanics we 
have to, we have to use a form that is 
symmetric, it makes sense. 
And so Wolfgang Pauli derived, this 
formula for the quantum mechanical 
counterpart of the Runge-Lenz vector, 
it's as you can see, it's symmetric. 
So P cross L minus L cross P, that's 
something which, it looks like P cross L 
in classical mechanics. 
So actually, why don't you just take a 
quick look at a comparison, and give me 
your opinion about the result in this in 
the video quiz. 
Well I hope it was clear to you that the, 
the form that Pauli derived is the same 
the same as this, this classical 
conventional classical form in the limit 
of Planck's constant going to zero, 
except that it involves a division. 
So basically M, M goes to A divide by Mu. 
So it's, it's just multiplied by one over 
mass, so it's changed only by a 
constance, that means it's still a 
constant emotion. 
Now we can actually find out a lot by 
working out the commutators of L and M, 
where M is this this operator this 
produced by Pauli that is the quantum 
mechanical analog of the run and lens 
vector. 
So, some of these you've already done, so 
you have already proved, I think, if you 
followed my advice this important 
identity for the angular momentum 
operator. 
Now, it doesn't take much of a stretch to 
see that this commutator must be the same 
as the commutator of any other truly 
vector quantity in the hydrogen atom 
problem. 
In other words, the commutator of L with 
and vector, gives, gives the 
relationship, that's described here. 
What is novel however, is the commutator 
of the Runge-Lenz vector itself. 
And the interesting thing about this is 
that it comes back, Mu, Mv is, Ih bar, a 
constant, which is minus twice the energy 
divided by the reduced mass. 
Epsilon U nu, mu V, UVWLW. 
So, now in some sense you see that the L 
and the M together are a closed set, with 
respect to the commutation relations. 
So this means they actually define 
[UNKNOWN] algebra which is, turns out to 
be a rotation group in a four dimensional 
space. 
And we're not going to go into detail on 
this, but just to say that by an 
appropriate choice of of scaling factors, 
basically compressing, compressing this 
factor into the definition of M. 
We can find, we can find all the bound 
state energies of the hydrogen atom 
without solving the Schrodinger equation 
via, completely algebraic approach. 
So were first going to look at the 
spectrum of the angular minimum operator 
itself. 
That is something that's of generic 
importance [INAUDIBLE] describes the 
motion of the particle of any central 
potential, not just that of the hydrogen 
atom. 
And it's widely used in scattering 
theory, and the calculation of bound 
states. 
And the expansion of wave functions and 
eigenfunctions of the angular momentum 
operator for a sort of general solution 
of vanisotropic problems as well. 
So, let's start. 
We're going to choose eigenfunctions of, 
both the L3, the projections of the 
anglar minimum onto the axis E3, or the 
Lz as it's often called, and of L 
squared, the total, the square of the 
angular minimum operator. 
So, we're just going to designate them 
for the moment in this form. 
Lm where L3 operating on Lm gives Mh bar 
times Lm. 
Now this, we haven't yet said anything 
about what the value of M is, except it's 
a pure number, and this comes from 
dimensional analysis. 
So Recall that L is equal to H bar over 
I, by definition H bar over I. 
R cross gram, this part, the operational 
part of the Hamiltonian contributes 
nothing in terms of its dimensionality so 
all the, all the units in the system come 
from Planck's constant. 
This is basically Bohr's idea, that the 
angular momentum is, is valued in units 
of Planck's constant, rather than, than 
that is, that's a fundamental property of 
quantum physics. 
So this just means that at the moment M 
is a value, a pure number that's going to 
be determined. 
Now how does that, how is that 
determination made? 
Well, it's handy to define two new 
operators, L plus, L1, which is L1 plus 
il two and L minus, L1 minus i L two, and 
it's straightforward to see that. 
When, you take the commutator of L3 with 
either of those, you get plus or minus H 
bar, times the, the thing that you 
started with. 
So, this is. 
You might say, this is suggestive that 
somehow, if you think about this, our 
previous geometric interpretation of the 
commutator with the angular momentum, it 
involves a rotation. 
So L plus is rotated a little bit in the 
positive direction about the the L, the 
L3 axis. 
And L minus in a negative direction. 
So that gives some sense of what we're 
going to find next, and that is, we're 
trying to see if, LM is an eigenvector of 
L3 is L plus of Lm. 
Also an eigenvector. 
And the answer is, it must be. 
And that we can see from the commutator 
because what we're going to do now is 
apply L3 to this new vector, L plus. 
Let's just use the plus sign here for a 
moment. 
L plus of LM. 
now we can do that. 
We can pass through the L3, if we add the 
commentator on the other side. 
So that's equal to L plus L3of Lm, plus H 
bar, l plus Lm, because this H bar is the 
commutator you see. 
So, then that means that when we apply L3 
to L plus Lm, we find that we get a new 
eigenvector. 
With the eigenvalue increased by H bar, 
so we'd write that as H bar times M plus 
one. 
And we get the same the same description 
for the L minus, except the the new 
eigenvector has a, has an eigenvalue of M 
minus one. 
So in some sense, if we start with, we 
start with we start with M then by, say 
Mh bar as an [UNKNOWN], then by 
sequential application of, of L plus we 
can increase, we create a new 
eigenvector. 
With an eigenvalue of, H bar greater each 
time. 
We just get a ladder of equally spaced 
rungs going up, and the same for a ladder 
of equally space going, rungs going down. 
And you can find that with fairly simple 
arguments the the the permitted range of 
M depends upon the value, the eigenvalue 
of L squared, and you get, so if the 
Eigenvalue of L squared of this form. 
L times L plus one H bar squared. 
And the the allowed values of this M are 
then, so these are integers, and the 
allowed values of M just range from minus 
L to plus L or 2l plus one, 2l plus one 
possible value. 
Now I'm not going to go into that 
argument in detail, but you might think 
about it and this is again, this is a 
similar type of approach that was used in 
the, in the harmonic oscillator to 
generate, using these raising and 
lowering operators to generate all 
possible eigenfunctions. 
So the harmonic oscillator of the ark. 
The argument is similar here. 
But there's a termination because the 
let's just say, because we've seen there 
is an energy associated with the value of 
an angular minimum, and you can't exceed 
the, that amount of energy. 
without increasing the toal angular 
momentum. 
So I've described to you how the, the the 
spectrum is generated now what about the 
eigenfunctions of the angular momentum 
operator. 
Now for those we need to find specific 
representations and 
So a well documented basis, called 
spherical harmonics, of which you can 
find details in the mathematical 
references. 
And the, the major thing to know, and I 
alluded to this previously, is that these 
are functions of angles only in the polar 
coordinate system. 
So in other words, the eugenfunctions of 
the angular minimum operator do not 
depend upon the, the distance from the, 
the coordinate origin. 
They're just functions of angles and so 
again we have, we have this Lm 
designation and so here are portraits of 
some of the spherical harmonics in the 
usual coordinate system so there's one 
for L equals zero. 
So for L equals zero we have M equals 
zero only because we have this minus L, 
minus L, plus one. 
Up to L, provides the only choice of the 
M values. 
And now for L equal one there are three 
such, terms, and these can be also be 
thought of in, in alternative 
representation. 
Is to think of these, so we, think of 
these as angular momentum oriented along 
the several axis'. 
So we have an X, a Y, and Z axis, so 
what's actually portrayed here, the M 
equals zero state is, is a line that 
looks like a cosine of theta. 
About the zed axis, and then, what I'm 
going to describe now as an alternative 
representation to the, the usual 
spherical harmonics. 
You can also, equally well have the, the 
same distribution along the Y axis, or 
indeed any other axis in space. 
And then a third, a third function which 
provides a complete basis for L equal one 
is to have that type of function along 
the X axis. 
and as, as the, as the value of L gets 
larger. 
There's a, you know, a more, a richer set 
of eigenfunctions, that are given by the 
various, values of M that are possible. 
So the angular momentum and the 
Runge-Lenz lens vector, or the poly 
symmetrized version of the run a lens 
vector, can be combined to form 
generators of rotation group of four 
dimensions. 
And as it turns out the same sort of 
ladder operator approach is used to get 
the, define the angular momentum. 
I can vouch the angular momentum. 
Or is also used to find the spectrum of 
the harmonic oscillator. 
Gives the following spectrum for the 
bound state of the hydrogen atom. 
So it's we, we recover exactly the Bohr 
formula for the energy levels, but with a 
big difference. 
so again it's this universal constant. 
The [UNKNOWN] times Hc times the ratio of 
the reduced mass to the mass of the 
electron, divided by the square of an 
integer. 
But this integer has well, is usually 
thought of have, of having two separate 
contributions. 
one is the so called radial quantum 
number. 
N, and this, this describes increasing 
orbit size, roughly speaking. 
and then this is the angular momentum. 
the angular momentum, quantum number that 
we just discussed previously. 
And you see it enters in a very, in a 
very equivalent way to the radial quantum 
number. 
So in some sense, the angular on the 
radio motions of the electron in a 
hydrogen, in a partition of the energy in 
a, in more or less equivalent way. 
And a major difference so the, the 
spectrum is the same as predicted by 
Bohr, but a major difference is the total 
numbers of states that are available. 
So there are when you be, because so also 
you see for each one of these values of L 
we have an M equal minus L, up to L. 
It turns out that for given energy, we 
have N plus L plus one squared states, 
all at the same energy. 
It's a very high degree of degeneracy. 
So for, N equals zero, we have one. 
For N equals, N equals one, [INAUDIBLE] 
rewrite differently, N plus L equals 
zero. 
We have one state, which is N equals 
zero. 
So for N plus L equals one, we have four 
states and so on. 
So increases roughly is the 
square,exactly is the square. 
And this means that, there, there's a, 
for large values of this integer there 
are quite a large number of degenerate 
item states. 
What I'd like to remind you of this 
amazing data that we, we saw earlier in 
aprevious lecture. 
the radio recombination lines in 
molecular clouds around the supernova, 
around the casiopa a. 
And this showed, this integer that we 
refereed to, it showed transitions of 
value with that integer being up around 
1000. 
So 1000 squared is equal to a million. 
So that means that these, these lines 
that you see, in the, in the absorption 
spectrum are actually, they are combine, 
they're sums of transitions involving 
millions of states, all of the same 
energy. 
And you see them in this one sharp line. 
So this is a very, to me a very, poignant 
demonstration of the high degree of 
degeneracy of all these states of the 
hydrogen atom in quantum mechanics. 
since there are about a million of these 
states that are presumably formed, 
they're being formed by recombination of, 
a very slow, electron in the interstellar 
medium and an ionized carbon atom. 
It's almost certain that these states are 
populated basically, completely randomly 
so the probability of any state being 
popular is one in a million. 
So if there was a significant difference 
in, difference in energy between them, 
you wouldn't see a sharp line, but you'd 
see something blurred. 
So again, the the, the power of quantum 
mechanics to describe atomic transitions 
over such a wide range of frequencies I 
think is really impressive. 
And, it, it validates Bohr's basic idea 
that there's a, you know, sharp 
quantization of atomic motion associated 
with the, with the, the quantization of 
angular minimum . 
Okay, so we'll conclude for now. 
next week will be a set of lectures on 
spin, then we'll return and, and wrap up 
after that.