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Question about quantum observables

  1. Oct 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Given this three operators in the same orthonormal base, [H]=[{0,1,0},{1,0,0},{0,0,1}], [A]=[{1,0,0},{0,1,0},{0,0,1}] and =[{1,1,1},{1,1,1},{1,1,1}], tell which of these observables form a complete set of compatible observables:
    [H], [H,A], [H,B] or [H,A,B]

    2. Relevant equations

    None

    3. The attempt at a solution

    For a complete set of observables has to exist an orthonormal base formed by common eigenvectors to all the operators; here , the eigenvectors of the operators are [H]==>{{-1, 1, 0}, {0, 0, 1}, {1, 1, 0}}
    [A]==>{{0, 0, 1}, {0, 1, 0}, {1, 0, 0}} and ==>{{1, 1, 1}, {-1, 0, 1}, {-1, 1, 0}}
    Since H and A share {0, 0, 1} and A and B share {-1, 1, 0}, I think that the anwser is [H,A,B], can someone please tell me if my gess is right?

    Thanks for reading.
     
  2. jcsd
  3. Oct 16, 2015 #2

    Geofleur

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    You can check your answer by computing the commutators ## [H,A] = HA-AH ##, ## [H,B] ##, and ##[A,B]##. They should all vanish.
     
  4. Oct 17, 2015 #3
    Well, A and B don't commute, I have checked it.
     
  5. Oct 17, 2015 #4

    Geofleur

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    Noting which eigenvectors are shared is a good place to start, like when you noted that ##A## and ##H## share ##[0,0,1]##. However, the eigenspaces of these operators are three dimensional, and so a common eigenbasis must contain three vectors.

    To investigate whether we can find two other shared eigenbasis vectors, we need to look at the eigenvalues too. For suppose an eigenvector, ## |f_1\rangle ##, of ## A ## can be written as a linear combination of two eigenvectors, ## |e_1\rangle ##, and ## |e_2\rangle ##, of ##H##, both with the same eigenvalue ## \lambda ##. We would have ## | f_1 \rangle = c_1 |e_1 \rangle + c_2 |e_2 \rangle ##, and thus

    ##H |f_1 \rangle = c_1 H|e_1 \rangle + c_2 H|e_2 \rangle = c_1 \lambda |e_1 \rangle + c_2 \lambda |e_2 \rangle = \lambda(c_1 |e_1\rangle + c_2 |e_2\rangle) = \lambda |f_1 \rangle##.

    So ## |f_1 \rangle## would be an eigenvector of both ##A## and ##H##. To summarize, in a given basis, if two operators either share the same eigenvectors or if the eigenvectors of one can be written in terms of those for the other with the same eigenvalues, then the two observables are compatible.

    Hopefully these remarks give you an idea of how to solve your problem independently of the commutator approach.
     
  6. Oct 17, 2015 #5
    I will study that; thank you very much for your help.
     
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