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Spin in arbritary direction

  1. Dec 13, 2008 #1


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    If we choose the z direction as the orientation of the spin, then spin-up (|+>) and spin-down (|->) could be written as

    [tex]|+\rangle = \left(\begin{matrix}1 \\ 0\end{matrix}\right)[/tex]
    [tex]|-\rangle = \left(\begin{matrix}0 \\ 1\end{matrix}\right)[/tex]

    From the textbook, if we consider aabritary orientation (n), the new state will be given

    [tex]|+_n\rangle = \left(\begin{matrix}\cos(\theta/2) e^{-i\phi/2} \\ \sin(\theta/2) e^{i\phi/2}\end{matrix}\right)[/tex]

    [tex]|-_n\rangle = \left(\begin{matrix}-\sin(\theta/2) e^{-i\phi/2} \\ \cos(\theta/2) e^{i\phi/2}\end{matrix}\right)[/tex]

    Now, consider the orientation along y direction such that [tex]\theta=\pi/2, \phi=\pi/2[/tex], it will gives the spin-up and spin-down along y direction. However, in some other textbooks, the spin-up/down along y direction is

    [tex]|\pm_y\rangle = \dfrac{\sqrt{2}}{2}\left(\begin{matrix}1 \\ \pm i\end{matrix}\right)[/tex]

    But I cannot get this state if I start from [tex]\theta=\pi/2, \phi=\pi/2[/tex] ???
  2. jcsd
  3. Dec 13, 2008 #2


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    How are the angles defined?
  4. Dec 13, 2008 #3


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    Sorry! I found what's going on now. Since the phase factor doesn't matter in QM, hence, it gives the same state by pulling the phase factor out. Sorry for bothering.
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