What is the spin of an electron along the x-axis? I know that an electron or any other particle for that matter, has a measured spin which is either up or down. This spin is along the z-axis. But what if we do not measure it along the z-axis and do it along the x-axis instead. What will the result be?
Will it get deflected along the x-axis in a Stern Gerlach experiment?
Lastly, will it always get deflected along the axis we measure its spin?
 A: Up or down, obviously. However, the representation of the x-up state or the x-down state using z-up and z-down states depends on how you represent the spin operator. For the most favoured choice
$S_x=\frac{1}{2}\left( \begin{array}{cc}
0 & 1\\
1 & 0 \end{array} \right)$
gives up/down state as (not normalised)
$\left|x_{up}\right>=\left( \begin{array}{c}
1\\
1\end{array} \right), \left|x_{down}\right>=\left( \begin{array}{c}
1\\
-1\end{array} \right)$
A: It's important that the spin need not be up or down, but may be in some superposition of the two, $|\psi\rangle = a |\uparrow\rangle+b|\downarrow\rangle$. If you choose to measure it in the $z$-direction, the squared magnitudes of $a$ and $b$ will give relative probabilities of getting each answer.
But given any direction, you can always rewrite the state as a superposition of up and down spin in that direction instead (technical bit: the up and down are the eigenstates of $\vec{n}\cdot\vec{\sigma}$ where $\vec{n}$ is the unit direction vector and $\vec{\sigma}$ the Pauli matrices). There will now be different probabilities associated to the possible outcomes of the measurement. For example, if the original state had definite spin up in the $z$-direction, $|\psi\rangle = |\uparrow\rangle$, the probability of getting 'up' when measuring the spin in a direction at an angle $\theta$ to the $z$-axis will be $\cos^2(\theta/2)$.
