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Calculating eigenstates of an operator

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider a two-dimensional space spanned by two orthonormal state vectors [itex] \mid \alpha \rangle [/itex] and [itex] \mid \beta \rangle [/itex]. An operator is expressed in terms of these vectors as
    [tex] A = \mid \alpha \rangle \langle \alpha \mid + \lambda \mid \beta \rangle \langle \alpha \mid + \lambda^* \mid \alpha \rangle \langle \beta \mid + \mu \mid \beta \rangle \langle \beta \mid [/tex]

    Determine the eigenstates of A for the case where (i) [itex] \lambda = 1, \mu = \pm 1 [/itex], (ii) [itex] \lambda = i, \mu = \pm 1 [/itex]. Do this problem also by expressing A as a 2 X 2 matrix with eigenstates as the column vectors.

    2. Relevant equations
    Just linear algebra rules.

    3. The attempt at a solution
    I started with [itex] \lambda = 1, \mu = 1 [/itex]. Then A is:
    [tex] A = \mid \alpha \rangle \langle \alpha \mid + \mid \beta \rangle \langle \alpha \mid + \mid \alpha \rangle \langle \beta \mid + \mid \beta \rangle \langle \beta \mid [/tex]

    [tex] A \mid \alpha \rangle = \mid \alpha \rangle \langle \alpha \mid \alpha \rangle + \mid \beta \rangle \langle \alpha \mid \alpha \rangle + \mid \alpha \rangle \langle \beta \mid \alpha \rangle + \mid \beta \rangle \langle \beta \mid \alpha \rangle = \mid \alpha \rangle + \mid \beta \rangle [/tex]

    [tex] A \mid \beta \rangle = \mid \alpha \rangle \langle \alpha \mid \beta \rangle + \mid \beta \rangle \langle \alpha \mid \beta \rangle + \mid \alpha \rangle \langle \beta \mid \beta \rangle + \mid \beta \rangle \langle \beta \mid \beta \rangle = \mid \alpha \rangle + \mid \beta \rangle [/tex]

    The eigenstate is [itex] \mid a_n \rangle [/itex] with eigenvalue [itex] a_n [/itex]. Then the following holds:
    [tex] A \mid a_n \rangle = a_n \mid a_n \rangle [/tex]

    The eigenstate [itex] \mid a_n \rangle [/itex] can be expressed in the basis vectors [itex] \mid \alpha \rangle [/itex] and [itex] \mid \beta \rangle [/itex]:

    [tex] \mid a_n \rangle = c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle [/tex]

    Then the earlier equation becomes:
    [tex] A \mid a_n \rangle = A( c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle ) = a_n (c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle [/tex]

    But this is also:
    [tex] A( c_1 \mid \alpha \rangle + c_2 \mid \beta \rangle ) = c_1 A \mid \alpha \rangle + c_2 A \mid \beta \rangle = c_1 (\mid \alpha \rangle + \mid \beta \rangle) + c_2 (\mid \alpha \rangle + \mid \beta \rangle) \\ = (c_1 + c_2) \mid \alpha \rangle + (c_1 + c_2) \mid \beta \rangle [/tex]

    This gives the equations :
    [tex] a_n c_1 = c_1 + c_2 [/tex]
    [tex] a_n c_2 = c_1 + c_2 [/tex]

    The only solution is if [itex] c_1 = c_2 = 0 [/itex]. Obviously I am doing something wrong but I can't see it.
     
  2. jcsd
  3. Feb 3, 2015 #2

    TSny

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    Homework Helper
    Gold Member

    Are you sure that's the only solution?
     
  4. Feb 3, 2015 #3
    I see, but I still get one non trivial solution out of it.
    If one is 0, then both are 0. So they are either both nonzero or both 0. If they are nonzero, I can write:
    [tex] a_n = \frac{c_1 + c_2}{c_1} = \frac{c_1 + c_2}{c_2} [/tex]
    If [itex] c_1 + c_2 =0 [/itex] then [itex] a_n =0 [/itex] or the c's are 0. If [itex] c_1 + c_2 [/itex] is nonzero, then I can divide by it and gives me [itex] c_1 = c_2 [/itex] with [itex] a_n = 2 [/itex] Unnormalized I can take as eigenstate [itex] \mid \alpha \rangle + \mid \beta \rangle [/itex].
    Now I can't find the other eigenstate. [itex] \mid \alpha \rangle - \mid \beta \rangle [/itex] is orthogonal to this eigenstate but letting A operate on it gives me 0. Or does this just mean that it is an eigenstate with eigenvalue 0?
     
  5. Feb 3, 2015 #4

    TSny

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    Homework Helper
    Gold Member

    That's it.

    Another approach is to subtract the two equations:
    ##a_nc_1 = c_1+c_2##
    ##a_nc_2 = c_1+c_2##
     
  6. Feb 3, 2015 #5
    Thanks again
     
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