# Clebsch-Gordan Coefficients

1. Jun 22, 2010

### djeikyb

Trying to self-teach myself more quantum mechanics. Reading Zare's "Angular Momentum- Understanding Spatial Aspects in Chemistry and Physics". I don't really understand how to calculate eigenvalues of J using CG coefficients (I mean, I know the eigenvalues of J, just trying to calculate them a different way). Any suggestions of other textbooks that go into more detail?

2. Jun 22, 2010

### djeikyb

Bump.

Any literature suggestions at all?

3. Jun 22, 2010

### djeikyb

The lack of help makes djeikyb very sad... :(

4. Jun 24, 2010

Why do you think you can get the eigenvalues from the Clebsch Gordan coefficients? AFAIK you look them up in a table, and there is is assumed that you know the new eigenvalues.

There is a sketch of the decomposition of the spin group into a Clebsch Gordan series, via the characters of the irreducible representations of SU(2) in "Quantenmechanik: Ein Grundkurs Ã¼ber nichtrelativistische Quantentheorie" by Norbert Straumann. Together with other stuff in that direction.

The hole idea is that J has to carry a representation of the complex rotations. It is given by the tensor product of the individual rotations of the particles. These tensor products decompose into different irreducible representations. Each representation goes along with one eigenvalue. If you know which representations are present, then you know which eigenvalues, are contained in a certain state.

Calculating the eigenvalues should be possible by calculating the eigenstates of a certain representation and then applying J. I don't see how that is helpful though. You just shift the problem from J do some other Operator D^j.

5. Jun 24, 2010

### djeikyb

Yeah.
Turns out my problem was a misunderstanding on my part.

Got it all figured out now.
Thanks.

6. Jun 24, 2010

### djeikyb

What I wasn't understanding was the coupled representation of two angular momentum vectors using a unitary transformation with Clebsch-Gordan coefficients and the uncoupled representation.

Got it now though, and now I can see that my question didn't make sense.