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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# How is Degeneracy Resolved by Building a CSCO?

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