I have a beam of electrons prepared in the state $|\Psi\rangle = cos(\theta/2)|+\rangle_z + sin(\theta/2)|-\rangle_z$ passing through a Stern-Gerlach apparatus which is free to rotate along the y-axis, and set with an angle $\alpha$ so as to measure the spin projection along the axis $cos(\alpha)u_z + sin(\alpha)u_x$
I'm supposed to find out wether or not I can determine $\theta$ in three different situations:
- Sending all electrons through the SG apparatus with $\alpha = 0$
- Sending half of the electrons with $\alpha = 0$ and the other half with $\alpha = \pi$
- Sending half of the electrons with $\alpha = 0$ and the other half with $\alpha = \pi/2$
So, how can I do that ?
EDIT:
Using the matrix for the spin projection in an arbitrary direction $$S_u = \left(\begin{array}{ccc} cos(\alpha) & sin(\alpha)e^{-i\phi} \\ sin(\alpha)e^{i\phi} & -cos(\alpha) \\ \end{array}\right)$$
(in the basis $\left\{|+\rangle_z, |-\rangle_z\right\}$, with $\phi = 0$)
I find that
$$S_u |\psi\rangle = \left(\begin{array}{ccc} cos(\alpha - \theta/2) \\ sin(\alpha - \theta/2) \\ \end{array}\right) = cos(\alpha -\theta/2)|+\rangle_z + sin(\alpha -\theta/2)|-\rangle_z$$
So $P_r(\hbar/2) = cos^2(\alpha - \theta/2)$ and $P_r(-\hbar/2) = sin^2(\alpha - \theta/2)$
$\theta = 2\left( \alpha - arcos(\sqrt{P_r(\hbar/2)})\right) = 2\left( \alpha - arcsin(\sqrt{P_r(-\hbar/2)})\right)$
But I don't think it's correct because, if it were the case, I could determine $\theta$ regardless of $\alpha$'s value...