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Eigenvalue and eigenvectors, bra-ket

  1. Feb 22, 2016 #1
    Question

    Consider the matrix $$
    \left[
    \matrix
    {
    0&0&-1+i \\
    0&3&0 \\
    -1-i&0&0
    }
    \right]
    $$

    (a) Find the eigenvalues and normalized eigenvectors of A. Denote the eigenvectors of A by |a1>, |a2>, |a3>. Any degenerate eigenvalues?

    (b) Show that the eigenvectors |a1>, |a2>, |a3> form an orthonormal and complete basis ;
    |a1><a1|+|a2><a2|+|a3><a3|= I, where I is the 3x3 unit matrix,
    and that <aj|ak> is the Kronecker delta function

    (c) Find the matrix corresponding to the operator obtained from the ket-bra product of the
    first eigenvector P=|a1><a1|. Is P a projection operator?

    My attempt

    I have done part (a). I got the eigenvalues as 3,√2,√2 with corresponding eigenvectors

    (0 1 0) , ( (1-i)/√2 0 1 ) , ( -(1-i)/√2 0 1 )

    Even after normalizing the vectors, I still can't work out part (b). I just don't get the 3x3 unit matrix.
    Any help would be greatly appreciated
     
  2. jcsd
  3. Feb 22, 2016 #2
    The latter two eigenvectors aren't normalized!

    Also, the eigenvalue corresponding to ##\left(\begin{smallmatrix}\frac{1-i}{\sqrt{2}} & 0 & 1\end{smallmatrix}\right)##should have the opposite sign.
     
  4. Feb 23, 2016 #3
    The eigenvalue should have a -, must have missed it.
    I already normalized the vectors, giving

    1/√(1-i)*((1−i√2) 0 1))

    And it still doesn't seem to work out for me
     
  5. Feb 23, 2016 #4
    I skimmed over that, my bad. Even so, it still isn't normalized--the magnitude is ##2##, not ##1-i##.
    Use the equation ##\|x\|=\sqrt{\langle x\;|\ x\rangle}## to recover the norm on a Hilbert space. It should always be real-valued and nonnegative.
     
  6. Feb 23, 2016 #5
    I managed to figure out where I have been going wrong thanks to you. I have been using Euclidean inner products instead of Hermitian inner products. Thanks for the help
     
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