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Repeated Eigenvalue Problem

  1. Jun 1, 2010 #1
    1. The problem statement, all variables and given/known data
    There is an Hamiltonian operation which is given by
    (2 1 1)
    (1 2 1) = H ; 3-by-3 matrix
    (1 1 2)
    And let's have an arbtrary eigenvector
    (b) = v ; (3x1) matrix

    Then, from the characteristic equation, the eigenvalues are 1,4. Here eigenvalue 1 is
    repeated one.

    2. Relevant equations
    Now, my question arises. I know that the eigenvectors that corresponds to eigenvalue 1 is two and both are orthgonal to each other. However, I can't find any of them because when
    I substitute eigenvalue into 1, I get a kind of meaningless(?) equation,

    (2 1 1) (a) (a)
    (1 2 1) (b) = (b)
    (1 1 2) (c) (c) .

    And it gives 3-equivalent equation, a+b+c = 0
    I couldn't determine any relation between a,b,c.

    3. The attempt at a solution

    So I quess that do I have to choose a,b,c satisfying the condition a+b+c=0
    but still problem doesn't disappear beacuse there would be a lot of soulutions....

    What did I wrong?????
    Last edited: Jun 1, 2010
  2. jcsd
  3. Jun 1, 2010 #2


    Staff: Mentor

    You didn't take it far enough.

    a + b + c = 0 ==>
    a = -b - c
    b = b
    c = ...... c

    The eigenvectors form a two-dimensional subspace of R3 that is spanned by two vectors. If you look closely at the system above, you can pick out the two vectors. These two vectors are linearly independent but they don't happen to be orthogonal.
  4. Jun 1, 2010 #3
    The operator we handle is Hermitian since Hamlitonian is real physical operator.
    it is no doubt that the eigenvectors of the Hermitian is orthogonal.
    Actually I cannot understand what you are saying. What do you mean by that?
  5. Jun 1, 2010 #4


    User Avatar
    Science Advisor
    Homework Helper

    If it's hermitian the eigenvectors belonging to distinct eigenvalues are orthogonal. Eigenvectors belonging to the same eigenvalue don't have to be. You have to select them to be orthogonal.
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