Does any pure spin state have a direction, that is it is an eigenstate of spin operator (in this direction)? If yes, how to prove it? In other words, suppose we have a spin state $$|S\rangle = \alpha|S_z+\rangle + \beta|S_z-\rangle.$$ Is it true that $$\hat{S_n}|S\rangle = s|S\rangle$$ for some direction $n$? For spin 1/2 it is geometrically clear from the Bloch sphere picture.
PS: I expected a general answer, but if general answer is a problem, I'll accept even answer for spin 1/2 case.
For spin-1/2: yes. You can verify that $$ \cos\textstyle\frac\theta2 \vert + \rangle + e^{i\varphi}\sin\textstyle\frac\theta2\vert -\rangle\, , $$ which parametrizes the most general spin-1/2 state, is an eigenstate of $$ \hat n\cdot\vec S=n_xS_x+n_yS_y+n_zS_z \tag{1} $$ with $$ n_x=\sin\theta\cos\varphi\, ,\quad n_y=\sin\theta\sin\varphi\, ,\quad n_z=\cos\theta\ . \tag{2} $$ Note that Eqs.(1) and (2) actually define the most general linear combination of spin operators.
For larger $S$, no in general. Basically you only have $3$ quantities to play with: $n_x,n_y$ and $n_z$, and they depend on only two parameters: $\theta$ and $\varphi$ because of the constraint $n_x^2+n_y^2+n_z^2=1$. Thus you cannot expect that you will be able to write an arbitrary vector in a large dimension space if you are looking for an eigenstate of a linear combination of at most $3$ operators depending on only two parameters.
A more formal proof uses Wigner $D$-functions and amounts to showing you cannot reach every final state $\vert sm\rangle$ from $\vert ss\rangle$ using rotations alone.
