1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Eigenvectors of rotation matrix

  1. Jul 13, 2011 #1
    1. The problem statement, all variables and given/known data
    This question is from Principles of Quantum Mechanics by R. Shankar.

    Given the operator (matrix) [itex] \Omega [/itex] with eigenvalues [itex] e^{i\theta}[/itex] and [itex] e^{-i\theta} [/itex], I am told to find the corresponding eigenvectors.


    2. Relevant equations

    [itex]
    \Omega = \left[ \begin{array}{cc}
    \cos{\theta} & \sin{\theta} \\
    -\sin{\theta} & \cos{\theta} \\
    \end{array} \right]
    [/itex]
    [itex]
    \Omega \left[ \begin{array}{c}
    x_1 \\
    x_2 \\
    \end{array} \right] = \left[ \begin{array}{c}
    x_1 \cos{\theta} + x_2 \sin{\theta} \\
    -x_1 \sin{\theta} + x_2 \cos{\theta} \\
    \end{array}
    \right]
    [/itex]
    [itex]
    e^{i\theta} \left[ \begin{array}{c}
    x_1 \\
    x_2 \\
    \end{array} \right] = \left[ \begin{array}{c}
    x_1 \cos{\theta} + x_1 i\sin{\theta} \\
    x_2 \cos{\theta} + x_2 i\sin{\theta} \\
    \end{array}\right]
    [/itex]

    3. The attempt at a solution

    I let the matrix operate on the generic vector [itex] (x_1, x_2)^T [/itex] and demand that the resulting vector is equal to [itex] (e^{i\theta}x_1, e^{i\theta}x_2)^T [/itex]. From this i get the condition that [itex] x_2 = ix_2 [/itex]and [itex] x_1 = -ix_2 [/itex], which implies that [itex] x_1 = x_2 = 0 [/itex]. Am i missing something crucial?
     
    Last edited: Jul 13, 2011
  2. jcsd
  3. Jul 13, 2011 #2

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Typo? I think you mean [itex] x_2 = ix_1 [/itex] here. In which case these two equations are really the same equation.

    No, it does not imply that. Let x1=1 (and worry about normalization later), then what is x2?
     
  4. Jul 13, 2011 #3
    Thanks for the quick reply. That actually wasn't a typo. I managed to switch the indices the same way in three separate calculations, getting the same erroneous result each time. Are these the correct eigenvectors?

    [itex]
    \frac{1}{\sqrt{2}} \left[ \begin{array}{c}
    1 \\
    i \\
    \end{array} \right] ,
    \frac{1}{\sqrt{2}} \left[ \begin{array}{c}
    i \\
    1 \\
    \end{array} \right]
    [/itex]
     
  5. Jul 13, 2011 #4

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yes, that's right.

    That's weird. From your equations in Post #1, you get
    [tex]x_1 \cos{\theta} + x_2\sin{\theta} = x_1 \cos{\theta} + x_1 i\sin{\theta}[/tex]
    which by inspection implies [itex]x_2 = i x_1[/itex], not [itex]x_2 = i x_2[/itex]
     
  6. Jul 13, 2011 #5
    It is indeed quite amazing that I managed to screw the same trivial manipulation up three times in a row, always in the same manner. Thank you for the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook