# Spin in arbritary direction

1. Dec 13, 2008

### KFC

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 $$\theta=\pi/2, \phi=\pi/2$$ ???

2. Dec 13, 2008

### Fredrik

Staff Emeritus
How are the angles defined?

3. Dec 13, 2008

### KFC

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.