Hello everyone, welcome back to exploring
quantum physics.
I'm Charles Clark, and we'll now start
using some of the tools that we developed
in the previous section to get the
basic commutation relations for the
angular momentum operators.
Now, once again, I have to emphasize, and
I've
been there myself, the only way to develop
facility with
these techniques is to actually practice
them.
So in the interest of time, I'm going to
outline a number of the
key things that have to be done to develop
a working set of angular momentum tools.
but really, if you want to understand it,
it's not
going to do you good just to look at that.
You will have to do some of the
computations.
And that's, for some people, those sorts
of calculations
turn out to be enjoyable.
Others can't stand them.
So you'll just, you'll, you'll, you'll get
a sense of what it;s like and some of you,
I think will not only benefit from this
but come to enjoy it and value
it, at least I hope so.
Okay, so, once again here's the definition
of the angular momentum operator.
And, now I've, break it down.
So, here's, here's what we started with at
the beginning of the first part.
Here's the Einstein summation condition.
Here's the Einstein summation condition
with the Levi-Civita symbol.
Now by the way i here is the square root
of minus 1.
And I try to avoid try to avoid using i as
a running index, when we have the square
root of minus
1 present.
So just keep your i's out for that.
But, there's no, no significance as to
whether the symbol's
a u, v, w, a, b, c, d, e, f.
Anything that appears here is a running,
is a running index, and the,
the particular choice of symbol that's
used to representative is just something
arbitrary.
Now in fact,
everything that's going to be done in this
part is based on the,
the commutation relationship between
position momentum.
but as you can see when you have, when you
have products, then there's a,
there's a, a bit of bookkeeping that's
required
to make sure that everything turns out
right.
So just to just to start with something
very
basic, let's try a simple application of
this commutation relationship in the,
in-video quiz.
Okay.
So I hope you're able to
to dev-, to, to derive this expression
from the uncertainty
principle, and we'll see, we'll see many
examples like this later on.
Okay, so here's, here's our basic
definition again, just to,
just to keep everything clear before you
on the screen.
here is the fundamental commutation
relation
that, on which everything else is based.
Oh yes, now here is this this identity
that you just derived.
So now here's a very important expression.
What does the angular moment, how, what's
the commutation relation
of the angular momentum with, let's say,
the position operator?
Well, alright that's here's the computated
La,xb.
There's just a very simple expression,
what that means,
and now we, we we now
decompose the angular momentum operator
into its
basic parts, its, you know, r cross p, and
then,
then we, we
apply this
from this expression up here.
We apply contraction, and we get this
this very significant identity, which you
will see is generic.
In other words, if we have any
vector-valued operator
in here, like we've got the same identity
for, for the momentum, we get this common
here.
La with b is ih bar epsilon abc xc.
I, I rather like express, I rather liking
ex,
writing the, expressing that in this way
because you.
It emphasizes the, the difference between
classical mechanics and quantum mechanics.
Do you see that
in class of the mechanics, this expression
would vanish identically because, because
a cross b
is equal to minus b cross a, for classical
operators?
So, here we have like, these two real
observable operators.
The
the sum of these two cross products is an
imaginary number, times the vector, so
this really is, this
is about as quantum as it gets, and indeed
note that in the limit, h bar goes to 0.
This produces the right answer, in that
you get back the same, the classical
cross-product.
So, there are, there are number of
identities that
you'll see that look like this in quantum
mechanics.
So, once again, where are we?
Okay, we had, we had we have here, we
have the basic relations, the one that you
derived.
So we're about to conclude here strongly
recommend, that you, you verify the
following things.
So for example, here's the the analog, the
identity we proved on the previous slide.
With x, we get the same type of thing for
p.
Now here's, and here's the sort of the
most important one.
The commutator of the different components
of angle
[UNKNOWN]
with themselves, again, it has, it is
really of the same
form, because L is a vector.
So this, this
allows us to write another, amusingly
quantum expression.
The cross-product of something with itself
is an imaginary number times that thing.
So, this is, is truly a quantum identity.
And then finally, so, I'd say
these two are more or less the same thing.
But this is, this is of central
importance.
That any com, so any two components of the
angular momentum operator in general do
not com, they,
in general they don't commute but any
component of
the square of the angular momentum
operator, so L squared.
I should have, should have written this as
like Lk, Lk,
just so as not to confuse you.
Or, La,
La.
so let me, let me just write this.
La, Lb, Lb equals 0
is the identity that you are being asked
to
verify in other words, that any component
of the angular momentum
operator commutes with the, the square
magnitude of the angular momentum
operator.
Okay.
Do take some time to work on those.
I think you'll find it rewarding.