(up to a phase) in the Hilbert space of the system. If we are given a CSCO, we can choose a basis for the space of states made of common eigenvectors of the corresponding operators. We can uniquely identify each eigenvector (up to a phase) by the set of eigenvalues it corresponds to. Discussion Let us have an operator of an observable , which has all non-degenerate eigenvalues . As a result, there is one unique eigenstate corresponding to each eigenvalue, allowing us to label these by their respective eigenvalues. For example, the eigenstate of corresponding to the eigenvalue can be labelled as . Such an observable is itself a self-sufficient CSCO. However, if some of the eigenvalues of are degenerate (such as having degenerate energy levels), then the above result no longer holds. In such a case, we need to distinguish between the eigenfunctions corresponding to the same eigenvalue. To do this, a second observable is introduced (let us call that ), which is compatible with . The compatibility theorem tells us that a common basis of eigenfunctions of and can be found. Now if each pair of the eigenvalues uniquely specifies a state vector of this basis, we claim to have formed a CSCO: the set . The degeneracy in is completely removed. It may so happen, nonetheless, that the degeneracy is not completely lifted. That is, there exists at least one pair which does not uniquely identify one eigenvector. In this case, we repeat the above process by adding another observable , which is compatible with both and . If the basis of common eigenfunctions of , and is unique, that is, uniquely specified by the set of eigenvalues , then we have formed a CSCO: . If not, we add one more compatible observable and continue the process till a CSCO is obtained. The same vector space may have distinct complete sets of commuting operators. Suppose we are given a finite CSCO . Then we can expand any general state in the Hilbert space as where are the eigenkets of the operators , and form a basis space. That is, , etc If we measure in the state then the probability that we simultaneously measure is given by . For a complete set of commuting operators, we can find a unitary transformation which will simultaneously diagonalize all of them. Examples The hydrogen atom without electron or