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Finding orthonormal simultaneous eigenkets of two operator matrices

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a thee-dimensional ket space. If a certain set of orthonormal kets - say, |1>, |2>, and |3> - are used as the base kets, the operators A and B are represented by
    A =
    a 0 0
    0 -a 0
    0 0 -a

    B =
    b 0 0
    0 0 -ib
    0 ib 0

    with a and b both real.
    a) Obviously A exhibits a degenerate spectrum. Does B also exhibit a degenerate spectrum?
    b) Show that A and B commute.
    c) Find a new set of orthonormal kets which are simultaneous eigenkets of both A and B. Specify the eigenvalues of A and B for each of the three eigenkets. Does your specification of eigenvalues completely characterize each eigenket?


    2. Relevant equations
    [A,B] = AB - BA = 0 if A & B commute


    3. The attempt at a solution
    a) B has eigenvalues b,b,-b. So yes, B is degenerate.
    b) I have no problem showing that A & B commute.
    c) I know how to find eigenkets (eigenvectors) of a matrix using the matrix eigenvalues, but I do not know how to go about finding eigenkets that are simultaneous eigenkets of both A and B?
    I tried finding eigenvalues of the matrix AB, which come out to ab,ab,-ab (also degenerate) but can only construct 2 eigenkets, and when I construct a third orthonormal eigenket using the cross product |1>[tex]\otimes[/tex]|2> ) it does not give A|3> = a|3> or B|3> = b|3>.

    Any hints or suggestions would be greatly appreciated!

    edit:
    I understand that the eigenkets should satisfy A|a'b'> = a'|a'b'> & B|a'b'> = b'|a'b'>, I just don't know how to go about finding |a'b'>.
     
    Last edited: Sep 27, 2009
  2. jcsd
  3. Sep 28, 2009 #2

    gabbagabbahey

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    If [itex]A[/itex] and [itex]B[/itex] commute, theeigenkets of either will also automatically be eigenkets of the other....I suspect you have already proven this in class, or in your text?
     
  4. Sep 28, 2009 #3
    Thanks for the reply gabbagabbahey,
    Yes, I found that in the textbook. I ended up leaving the eigenkets in a generalized form, then taking a cross product and choosing 3 kets that satisfied the generalized forms for all the eigenkets.
     
  5. Sep 13, 2011 #4
    I know that when two operators A and B commute, any eigenket of A is also an eigenket of B.
    As such, i think(Not certain though) that it suffices in this case to find eigenkets of either A or B and then normalize the vectors. What do you guys think?
     
  6. Sep 14, 2011 #5

    vela

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    That's not quite accurate because of degeneracy. For example, for a free particle, H=p2/2m. The Hamiltonian and the momentum operator p commute, but the state [itex]\vert p \rangle + \vert -p \rangle[/itex] is an eigenstate of H, but it's not an eigenstate of p.
     
  7. Sep 14, 2011 #6
    Oh, that's true.
    In this case, what condition should we place on the statement i did mention above?

    i.e if A and B commute, then any eigenket of A is an eigenket of B.
     
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