KFC
Dec13-08, 12:04 AM
If we choose the z direction as the orientation of the spin, then spin-up (|+>) and spin-down (|->) could be written as
|+\rangle = \left(\begin{matrix}1 \\ 0\end{matrix}\right)
|-\rangle = \left(\begin{matrix}0 \\ 1\end{matrix}\right)
From the textbook, if we consider aabritary orientation (n), the new state will be given
|+_n\rangle = \left(\begin{matrix}\cos(\theta/2) e^{-i\phi/2} \\ \sin(\theta/2) e^{i\phi/2}\end{matrix}\right)
|-_n\rangle = \left(\begin{matrix}-\sin(\theta/2) e^{-i\phi/2} \\ \cos(\theta/2) e^{i\phi/2}\end{matrix}\right)
Now, consider the orientation along y direction such that \theta=\pi/2, \phi=\pi/2, 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
|\pm_y\rangle = \dfrac{\sqrt{2}}{2}\left(\begin{matrix}1 \\ \pm i\end{matrix}\right)
But I cannot get this state if I start from \theta=\pi/2, \phi=\pi/2 ???
|+\rangle = \left(\begin{matrix}1 \\ 0\end{matrix}\right)
|-\rangle = \left(\begin{matrix}0 \\ 1\end{matrix}\right)
From the textbook, if we consider aabritary orientation (n), the new state will be given
|+_n\rangle = \left(\begin{matrix}\cos(\theta/2) e^{-i\phi/2} \\ \sin(\theta/2) e^{i\phi/2}\end{matrix}\right)
|-_n\rangle = \left(\begin{matrix}-\sin(\theta/2) e^{-i\phi/2} \\ \cos(\theta/2) e^{i\phi/2}\end{matrix}\right)
Now, consider the orientation along y direction such that \theta=\pi/2, \phi=\pi/2, 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
|\pm_y\rangle = \dfrac{\sqrt{2}}{2}\left(\begin{matrix}1 \\ \pm i\end{matrix}\right)
But I cannot get this state if I start from \theta=\pi/2, \phi=\pi/2 ???