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How do you know when a set of observables form a CSCO?

  1. Dec 11, 2012 #1
    I understand that the operators have to commute, and therefore that the measurement of one has no bearing on the measurement of the other. I know that H, L^2, and L_z form a CSCO for the H atom. Basically, I conceptually understand CSCO. Often though, in my class, we will be working on a problem, finding eigenvalues, and then my teacher says "therefore these form a CSCO."

    How do you know a set is complete? it seems to me that to make claims of completeness one must try to apply every operator you can think of before excluding any from the set...
    My problem is recognizing the completeness in the math language. At which point after finding eigenvalues do you see that these form a CSCO? what are the specific mathematical "pictures" that should trigger "oh, these form a CSCO"? if asked "do these form a CSCO?", how do i check that? (other than checking commutation relations.)

    basically, i think my question can be summed up in: what is the mathematical language of CSCO's, especially in context of the H atom or harmonic oscillator?
     
  2. jcsd
  3. Dec 11, 2012 #2
    This concept is introduced to distinguish eigenstates of the same eigenvalue from each other, i.e., degenerate states. For operator A, if it has degenerate eigenvalues, say a, all the eigenvector for 'a' form a subspace of dimension > 1, so it's not possible to say for sure what state the particle was in once you make an observation of value a, all you know is after the first observation, the particle's state is in this subspace. In order to pin down the exact state, we have to make another observation which corresponds to a commuting operator, say B. Since A and B commute, the observation of B doesn't disturb the fact that the particle is in the above subspace, but hopefully the eigenspace for the two eigenvalues <a, b> will have fewer degeneracy, which means after the 2nd observation, we have fewer choice for the state of the particle. This keeps going until the degenerate space has dimension 1, then you have a complete set of observables
     
    Last edited: Dec 11, 2012
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