There are a certain number of key commutator identities that are useful for
working out the properties of the quantum theory of angular momentum.
And they can be derived using the methods that we introduced in Part 3.
And I strongly encourage you to work these all out for yourself, but I'm going
to go through and outline the steps. Work out a few examples, and then give
you some pointers on how to proceed. Now, we'll keep using the standard
definition of the angular momentum operator and then the Einstein summation
convention to designate it. And in fact, everything that we were
doing in this this part of the lecture derives from this one uncertainty
principal commutator pj xk, this is h bar over i, delta jk.
In fact, you can get all the results by going back to this fundamental form.
Nothing else is needed. However, it's very helpful to derive
some, to build up on this and derive some other fundamental identities, which will
then save you a lot of algebra. So let's try a simple application of
applying this identity to a slightly more complicated expression.
That's an in video quiz. So I hope that was clear.
in other words, by simple application of this, you can derive this and it just
involves an, an explicit calculation of the commutator and then substitution of,
of this, of this relationship whenever there is a product of x and p.
Now, we're going to get the first of the really significant results.
And again, everything precedes from the definition, and this identity, though
one, you know, here, we'll find a case where the use of its this subsidiary
relationship is also important. So our task is calculate the commutator
of the one component of the angular momentum with another component, the
position operator, okay? So to be explicit, this commutator, just
remember, is just computed La xb is just La xb minus xb La.
That's the definition. So now you see, we start using the
specific representation of L in terms of x and p.
So we have here that Lu is epsilon uvw, xp, pw.
xv, pw, [INAUDIBLE] is La is a epsilon acd, cx, pd.
And you see, we use c and d here, because b has already been reserved.
One could use any choice of letters, it's, it's good to, to keep a set that
are handy in the mind. By the way, I, I avoid using i in these
expressions, because that's a conventional description for the square
root of minus one, and one doesn't want to have any confusion about that.
Anyway, okay, so, epsilon acd, xcpd computed with with xb.
Well, then we just apply this and we get epsilon acd, h bar of i, xc delta bd.
And now if you want to go through this yourself, but here's what happens.
La xb is i h bar here, let's start here. La xb is i h bar epsilon abc, xc.
Now, here's a, possibly a helpful mnemonic for you.
You want to check this. You can rewrite this expression in the
following symbolic form, 1 half L plus x minus x cross L is i h bar x.
I consider this to be a really quantum expression, because of course, for
ordinary vectors and classical mechanics, L cross x is necessary, necessarily
perpendicular to x. So this expression says that for the
angular momentum operator, when you apply the cross product to a vector, you get
the vector back itself in a certain sense, with an imaginary prefactor.
And you can think of an other way of looking at this some of you may have seen
this before. And that is for example, if we take, L3
applied to x1, 1, it gives it gives x2. And this is a, is a fact, the action of
L3 rotates, you might say x1 and x2. And then, and in this sense this is way
of displaying the fact that the angle momentum operator is a, is a generative
rotation. So, it's my recommendation now it's my
recommendation as your friend, it's, it's not mandatory that, this is a good time
to pause. We're, the, we're concluding this part.
I'll give you some time to do a few exercises.
which basically, the, the, more and I suggest to do them in this order.
So, this one is just a repetition of what we did for La xb, but now, applying it to
the momentum operator rather than the position operator.
And you should find that you get exactly the same, exactly the same identity if
you exchange x and p in here. Furthermore you get the same identity if
you, you, if you operate on L, so this gives, you know, even nicer expression
with the cross prod of the angular momentum factor with itself.
It's just the angular momentum operator again times an imaginary number.
And finally this important identity which shows that all, any given, so L1, i
squared equal L2, i squared equal L3, L squared equal zero.
Each component of the angular momentum commutes with the sum of the squares.
And this, this square angular momentum operator is something that enters into
the equations of motion. That's actually going to be the subject
of the next part. But meantime, I suggest that you just
pause and work these things about, so that you have some experience with them.
We'll be using them frequently in the next part.