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.