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Eigenvector of a spin operator

  1. Dec 24, 2012 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=54215&stc=1&d=1356365308.jpg


    2. Relevant equations



    3. The attempt at a solution
    I don't know what's wrong with my work. I can't obtain the eigenvector provided in the model answer.

    My work
    attachment.php?attachmentid=54214&stc=1&d=1356365308.jpg

    Model Answer

    attachment.php?attachmentid=54216&stc=1&d=1356365308.jpg
     

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  2. jcsd
  3. Dec 24, 2012 #2

    vela

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    Your work is fine. Remember that an eigenvector is only unique up to a multiplicative constant. The eigenvector you found can be written
    $$\left\lvert \frac{\hbar}{\sqrt{2}} \right\rangle = \begin{bmatrix} \frac{1-i}{2} \\ \frac{1}{\sqrt{2}} \end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix}\frac{1-i}{\sqrt{2}} \\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix}e^{-i\pi/4} \\ 1 \end{bmatrix}.$$ If you multiply that by ##e^{i\pi/8}## and ignore the normalization factor of ##1/\sqrt{2}##, you'll get the answer in the solution.
     
  4. Dec 24, 2012 #3
    Oh! Thank you very much.
    But why do we bother to have such a complicated eigenvector?
    The one with (1-i)/2 and 1/(sqrt2) is much easier to find. Why do we need to change it to the form of exp?
     
  5. Dec 24, 2012 #4

    vela

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    You don't need to, but you have to admit there's certain symmetry there. And physicists like symmetry. :wink:
     
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