# How is Degeneracy Resolved by Building a CSCO?

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1. Jan 30, 2015

### wgrenard

When finding a complete set of commuting observables, the goal, as I understand it, is to specify enough observables in the set such that, when given an eigenvalue of a state from each observable's matrix, these eigenvalues uniquely identify a corresponding eigenvector.

It seems that every source that I read gives an example similar to the following to outline this concept. Suppose we have a physical observable, $A$, which has a set of $n$ eigenkets $\left|\psi_n\right\rangle$ which correspond to the n-fold degenerate eigenvalue, $a$.

To "resolve" this degeneracy, we can find an observable, $B$, that commutes with $A$ and search for a simultaneous eigenstate of $A$ and $B$. To do this, we can make use of the fact that any linear combination of the degenerate eigenstates of $A$ will also be an eigenstate of $A$ with eigenvalue $a$. We can do some matrix algebra to find a suitable linear combination of the eigenvectors of $A$ which is a simultaneous eigenvector of $B$. This process can be repeated as needed if there turns out to be degeneracies in $B$.

From this point forward, however, I am confused. I cannot find a definitive answer of how exactly finding these simultaneous eigenstates removes degeneracy from the observables. Yes, we have found a single linear combination of the degenerate eigenvectors of $A$, for instance, that is also an eigenstate of $B$. However, this does not change the fact that the eigenvalues of $A$ are degenerate.

My best guess at the answer to this question is that we can use the simultaneous eigenvectors of the CSCO as a new basis, and then rewrite each of the matrices that had degeneracies in this new basis. If this is correct, however, I am unsure exactly how this guarantees that all of the degeneracies will be resolved in the new basis.

My question is: How does adding more commuting observables to the set remove degeneracies?

2. Jan 30, 2015

### Chopin

Hi, and welcome to PhysicsForums!

I guess I'm not sure what you mean when you say "remove degeneracies". The original subspace of states is still going to have the same singular eigenvalue under $\hat{A}$ that it did before, so the operator is still degenerate. The point is just that you've now come up with a convenient basis for spanning that space, by adding extra resolving power that the original operator did not possess. Like you said, this means you can rewrite the matrices in a concrete form. That's really all there is to it. Is there something more that you were looking to see happen as a result of this process?

Last edited: Jan 30, 2015
3. Jan 30, 2015

### wgrenard

Chopin, thank you for your reply. In answer to your question, no there is nothing specifically that I was looking for to happen. My confusion lay in the fact that most of the sources I read use ambiguous language like 'resolve' the degeneracy, or 'remove' the degeneracy, and I was confused as to how finding this new basis served to do this.

It makes much more sense when you say that the original operator still contains degeneracy, because that is what I figured must be true. The only question I still have, then, is this: Would I be correct to say that the usefulness of finding this new basis of common eigenvectors is so that we can diagonalize our operators in this basis, so that the possible measurement values show up on the diagonal entries?

4. Jan 30, 2015

### Chopin

Pretty much. Think of it this way: it's clearly useful to be able to diagonalize an operator in general, because it gives you a useful basis to decompose your states into, where each basis state has a simple transformation rule under the operator. If the operator contains no degeneracy, then you have a one-to-one correspondence between basis states and eigenvalues, so it's easy to name every basis state in the space simply according to its eigenvalue under the operator. If the operator is degenerate, though, then that simple one-to-one relationship is no longer present, so you have to work a little harder to come up with a good naming scheme for your states. You still have the same goal--trying to come up with a good way to name each state which has some simple correspondence to your observables--you just have to use multiple operators to do it, and your state names turn into more complicated compound things like $|p,\uparrow\rangle$ instead of just a simple $|p\rangle$.

5. Jan 30, 2015

### cgk

I'd just like to second what was said (all correct): Choosing as new vector space basis to represent the operators does not change the operators themselves (thus, in particular, not their spectrum: A degenerate operator stays degenerate in whatever basis it is represented). It just offers a, for some applications, more convenient way of writing the operators and states.

If you are happy with non-diagonal operators and expressing states as possibly complex linear combinations of some arbitrary vector space, there is no particular reason to employ a maximal commuting set of operators to define a basis for doing operations in.