# Lie groups and angular momentum

1. Jan 12, 2010

### sineontheline

As i understand it, the commutation rules for the quantum angular momentum operator in x, y, and z (e.g. Lz = x dy - ydx and all cyclic permutations) are the same as the lie algebras for O3 and SU2. I'm not entirely clear on what the implications of this are. So I can think of Lz as generating physical rotations of the wavefunction? Why is that important? I suspect, if we're working in a spherically symmetric potential, that this has something to do with eigenvalues and ladder functions. But the connections are very murky to me still. Can anyone explain? Or point me to some notes that get to the punchlines really fast? I don't care about proofs so much.

This is all super confusing cause I'm teaching myself all this. I have a *basic* understanding of group theory (level: Artin) and of QM (level: high Griffiths/low Shankar).

2. Jan 12, 2010

### Fredrik

Staff Emeritus
Unfortunately this stuff isn't easy. This post might help you get started.

Edit: I agree with Meopemuk's recommendations in the post below this one. I learned most of this stuff from Weinberg (chapter 2).

Last edited: Jan 12, 2010
3. Jan 12, 2010

### meopemuk

In my opinion, Ballentine's textbook has the best explanation of the role of symmetry groups, generators, commutators, etc in quantum mechanics. Unfortunately, this textbook covers only the Galilei group and non-relativistic QM.

For the relativistic case, I would recommend reading first few chapters in S. Weinberg "The quantum theory of fields" vol. 1 and consult the cited journal articles for more details and explanations.

Eugene.

4. Jan 12, 2010

### sineontheline

"The idea that space is rotationally invariant is incorporated into QM as the assumption that there's a group homomorphism from SO(3) into the group of symmetries."

This was the idea I was looking for.

You go on to say that the representation of SO(3) isn't quite there though.

So would it be right to say:

"The idea that space is rotationally invariant is incorporated into QM as the assumption that there's a lie algebra homomorphism from lie group representation of SO(3) into the lie group representation of symmetries."

5. Jan 12, 2010

### Fredrik

Staff Emeritus
That sounds a bit awkward, and I don't know exactly what you mean by the "lie group representation of" (some group G). Does it mean "representation of the Lie group G"?

If you want to focus on the Lie algebra instead of the Lie group, the best way to do it is probably to use the C*-algebra approach to QM. It starts with the assumption that operationally defined "observables" can be represented mathematically by a C*-algebra with an identity element. This is an alternative to the more common Hilbert space approach, which starts with the assumption that operationally defined "states" can be represented mathematically by the unit rays of a complex separable Hilbert space.

I don't know much about that approach, so chances are pretty good that I won't be able to answer questions you might have about it. This book looks like a good place to learn about it. I hope I will have had time to get through it before 2010 is over.

6. Jan 12, 2010

### sineontheline

Ha, okay.

Basically what I'm thinking:

The angular momentum operator (Lx, Ly, Lz) behave like the generators of SO(3) and SU(2) in that the lie algebras of SO(3) and SU(2) act like Lx, Ly, and Lz. And so because they're all homomorphic, rotation invariance of space expresses itself in QM through Lz, Ly, and Lz.

Is that right?

I probably jumbled up the technical terms in the previous post.

7. Jan 12, 2010

### meopemuk

Yes, you are basically right. The logic is as follows:

Consider a physical system (e.g., an atom). Possible (pure) states of this system are realized as unit vectors in a Hilbert space H. Different observables that can be measured on this system are realized as Hermitian operators in H. The correspondence observable -> Hermitian operator depends on who is the observer. Different (inertial) observers assign different operators to the same observable. For example, two observers rotated wrt each other disagree about the direction of the x-axis. (An alternative, but equivalent, point of view is that all observers use the same operators to describe observables, but they use different state vectors)

We know that different inertial observers are connected by (inertial) transformations that form a group (Galilei group in the non-relativistic physics; Poincare group in the relativistic physics). Infinitesimal transformations of this group form the Lie algebra (Galilei or Poincare). To each (inertial) transformation of observers there should correspond a transformation of Hermitian operators of observables in the Hilbert space (or, equivalently, a transformation of state vectors). In other words, there should exist a representation of the symmetry group (Galilei or Poincare) in the Hilbert space of the system. Then, a couple of theorems (most notably by Wigner and Bargmann) tell us that this representation must be unitary. (To be more precise, this should be a representation of the "universal covering" group). As a consequence, infinitesimal inertial transformations must be represented by Hermitian operators (also called generators). Then we obtain a Hermitian representation of the (Galilei or Poincare) Lie algebra in the Hilbert space of the system. Operators Lx, Ly, Lz are Hermitian representatives of infinitesimal rotations. Commutators of these operators are the same (times a factor) as Lie brackets in the Lie algebra of rotations. This is how angular momentum operators and their properties (e.g., commutators) can be derived from the rotational invariance and from the non-commutativity of rotations in the 3D space.

Eugene.

8. Jan 12, 2010

### sineontheline

Why does unitarity of the representation of Lorentz/Gallile Group in Hilbert space imply that infinitesimal Lorentz/Gallile Transformations in space are generated by the Hilbert space's hermitian operators?

I mean I know I've seen that exp(i*a*Lz) {'a' is parameter) generates the rotations. But why does this construction follow from unitarity? I actually looked this up in Weinberg in the last hour. And he's got U's and I saw Wigner's name in chapter two -- so I know I was looking at the answer. But, alas, Weinberg was too complicated.

9. Jan 12, 2010

### Fredrik

Staff Emeritus
What he meant is that if U=exp(iaX) is unitary for all real numbers a, then X must be hermitian:

$$I=U^\dagger U$$

$$U^{-1}=U^\dagger$$

$$e^{-iaX}=e^{-iaX^\dagger}$$

$$X=X^\dagger$$

The last step follows from a series expansion of both sides of the previous expression. Just match both sides term by term. Alternatively, instead of expanding in a series, just take the derivative of both sides with respect to a, and then set a=0.

10. Jan 12, 2010

### sineontheline

Eugine and Fredick rock my socks.

thx

btw: I'm reading ballentine (ch 3) now...and understanding it!

also
Eugine: that discussion *really* helped

11. Jan 13, 2010

### DrDu

The reason to consider representations of SU(2) and not only representations of SO(3) is that in QM you do not only consider true representations of a group but also so called "ray representations" where the group multiplication may be acompanied by a further phase factor. The reason for this is that the wavefunctions are only defined up to a phase factor anyhow. The ray representations of SO(3) turn out to be just the ordinary representations of SU(2). In case of the Galilei group it is also of utmost importance to use a ray representation in QM. The group, in which the ray representations are true representations is the extended Galilei group which has an additional generator, namely mass. This should be explained in Ballentine.
The proof that unitary operators have Hermitian generators is known as Stones theorem.
Another book on group theory which is very helpful is from Morton Hammermesh, "Group theory and its application to physical problems" from Dover publications.