- #1
KFC
- 488
- 4
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] ?
[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] ?