# Why do L² and L_z form a CSCO?

1. Dec 17, 2011

### nonequilibrium

CSCO: complete set of commuting observables

Why it is a set of operators, I understand. And why they're commuting, I understand. Given these facts my QM book declares it's a CSCO. But how do we know that they're complete? (even assuming the only relevant operators are L_x, L_y, L_z and functions of these)

2. Dec 17, 2011

### dextercioby

The angular momentum algebra is isomorphic to su(2) which has only one Casimir element. L^2. The irreducible representations in a linear (topological) space are labeled by one parameter, the possible eigenvalues of L^2. A CSCO is then given by any element of the algebra + L^2, since su(2) has a trivial center (that's why one chooses only one element from su(2)). For convenience, from the algebra one chooses L_z.

3. Dec 17, 2011

### nonequilibrium

0_0

I sure hope I'm not supposed to understand that terminology after an undergraduate QM course :( If so, I'm suing my university.

4. Dec 18, 2011

### jrlaguna

I think a "more physical" answer is possible. Suppose that you have a beam of particles, totally random. Now you measure $L^2$ and put the particles in different boxes according to the results, with labels. Now, in each box, you measure $L_z$. This allows you to make smaller boxes within each box. Each small box will have two labels, one for $L^2$ and the other one for $L_z$. This labelling is coherent: you can repeat any measurement of $L^2$ and $L_z$, and the answers will be the same as the labels.

Now, you try to repeat the trick with $L_x$. You measure $L_x$ and make smaller boxes. But, to your surprise, when you measure $L_z$ again... the results are not coherent any more. That's why you can't add $L_x$ to your list of set of commuting observables.

The mathematical reason is very deep: rotations along the X and Z axes do not commute! But you must understand what I said before embarking in that adventure... ;)

5. Dec 18, 2011

### nonequilibrium

Oh I understand that terminology (I'm well acquainted with Dirac notation and commutators), just not the su(2) terminology. The words dextercioby uses I do remember from my Abstract Algebra class, but I've never used them in the context of QM.

Anyway what you explained, jrlaguna, is that {L_z, L²} is a commutating set of operators, which I already understood. What I don't understand is why it is complete. You argue that you can't add L_x and L_y (for reasons that I well understand), but I don't understand how that helps my question. First of all, saying I can't add L_x and L_y tells me nothing about completeness as they are only two of the infinite possibilities that I can try to add (all kinds of functions of L_x, L_y and L_z), and second of all, even if I found an operator I could add, this would tell me nothing about completeness. After all, {L_z, L²} is a CSCO, but I can still add an operator L^4 for example, as it commutes with L² and L_z. This fact doesn't refute the completeness of {L_z, L²}.

But thank you for trying to help :)

6. Dec 18, 2011

### jrlaguna

The question is good: you're never sure that your set is "physically" complete. It is only mathematically complete, according to a certain mathematical model, and to prove that you need some Lie algebra computations. But, physically, your mathematical model (your Hilbert space) may be incomplete, and some further research will get you new commuting observables, i.e.: new labeled boxes.

If you're interested in how to determine the CSCO mathematically, follow dextercioby's suggestion, that was the line of his/her answer. And you'll really have to go into SU(2)... :)

7. Dec 20, 2011

### Morgoth

I thought that Hilbert Space vectors were always forming a complete set.

Do you have any example that completeness breaks?

Last edited: Dec 20, 2011
8. Dec 20, 2011

### kith

Is this generalizable? So if I have a system with Hamiltonian H, is there always a straightforward way to find a CSCO?

The idea of completeness of observables is that you need to measure at least these observables to completely specify your state. Of course, all functions of these observables could be added. But if you have measured your complete set and now measure these new observables, the result is already determined and doesn't yield new information about your state.

The topic here are complete sets of commuting observables. This is something different from complete set of vectors.

9. Dec 20, 2011

### Morgoth

I mean every observable quantity has its hermitian operator, from which you get the eigenvectors and eigenvalues (the values you measure). The eigenvectors are part of the Hilbert Space, and they form a complete set.
That's how i meant it. If I still don't get your point, forget it I am wrong.

10. Dec 20, 2011

### kith

That's right. But this isn't the meaning of completeness which is discussed here.

In general, if you measure a single observable, your state is not completely determined by the obtained eigenvalue because of degeneracy. Informally speaking, measuring a complete set means to break all degeneracy. See wikipedia for example.

11. Dec 20, 2011

### Morgoth

oh well i think I got what you mean...
So:
Generally commuting operators -operators of physical quantities- are being described by the same eigenfunctions which means that Lz and L^2 action on the eigenvector of one of them will give you their eigenvalues, that's why we notate them like |a,b,c,....> depending on the freedom you have.
This notation means that your state is totally determined by the quantities a,b,c,...

Generally one way to write the total |a b c ...> is by using the tensor product of |a>*|b>*|c> (*:tensor product)... Each of them are a complete set for your quantities A,B,C,...
if your A,B,C,... commute then A is going to act on |a> alone and give you eigenvalue a, B is going to act only on |b> and give you eigenvalue b, and so on.
So your system being described by A,B,C,D,... operators has the complete eigenvector |a,b,c,d,...> in the same way each of the operators above determines a complete set alone.

I hope I made my point clear.

Last edited: Dec 20, 2011
12. Dec 20, 2011

### kith

I don't think this is going anywhere, because it probably is just about semantics, but one more comment. ;-)

I don't agree that the two ideas of completeness are the same (if that's what you wanted to say). It is only the same way if there is no degeneracy present. The identity operator for example also determines a complete set of eigenvectors (any complete set of vectors in fact). Yet he is no good for a CSCO because he doesn't break any degeneracy.

13. Dec 20, 2011

### dextercioby

Symmetry algebras and more generally dynamical algebras play an essential role in quantum theory. A CSCO is built with the Casimir elements of the symmetry algebra (Galilei algebra, Poincare algebra, conformal algebra). A CSCO provides the possible states. The Hamiltonian provides the dynamics, how the states are related one to the other.