Spin in arbritary direction

  • Thread starter KFC
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KFC

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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] ???
 

Fredrik

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How are the angles defined?
 

KFC

488
4
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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