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Eigenvector of Pauli Matrix (z-component of Pauli matrix)

  1. Oct 30, 2015 #1
    I have had no problem while finding the eigen vectors for the x and y components of pauli matrix. However, while solving for the z- component, I got stuck. The eigen values are 1 and -1. While solving for the eigen vector corresponding to the eigen value 1 using [tex](\sigma _z-\lambda I)X=0[/tex],
    I got [tex]\left( \begin{matrix} 0 & 0 \\ 0 & -2 \end{matrix} \right)\left( \begin{matrix} x \\ y\end{matrix} \right)=0[/tex]
    Now, how can I find the eigen vector for eigen value 1 with this relation since it will give me only [tex]-2y=0[/tex]
     
  2. jcsd
  3. Oct 30, 2015 #2

    HallsofIvy

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    Yes, that says that y= 0. Since you have NO equation restricting x, x can be any thing. An eigenvector corresponding to eigenvalue 1 is (x, 0)= x(1, 0).

    (Perhaps you are missing the fact that the set of all eigenvectors corresponding to a single eigenvalue is a subspace of the vector space. The set of eigenvectors corresponding to eigenvalue 1 is the one dimensional subspace of all multiples of (1, 0).)
     
  4. Oct 30, 2015 #3

    fzero

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    That equation is solved by ##y=0##; ##x## can be anything, but you can determine the normalized eigenvector if you want a concrete answer.
     
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