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.