Spin Uncertainity In the special case that $\langle S_z\rangle =0$. What does this imply? My guess is that it would imply that $\sigma_{S_z}\sigma_{S_y} \geq 0 $ using the general uncertainty formula and [$S_x,S_y$]= $i\hbar S_z$. My gut feelings is that this result does not make sense. In my mind I am envisioning all the spin to be along the $x$ and $y$ axes .  Could someone please help me out?  Thanks
 A: It simply means that, in your state, the probability of getting spin up is the same as the probability of getting spin down (with up/down defined along the $z$ axis).  
It does NOT imply that all your spins are necessarily along $\hat x$ or $\hat y$, although this is one way of getting $\langle S_z\rangle=0$.  You could imagine a state such as 
$$
\vert\psi\rangle = \frac{1}{\sqrt{2}}\vert +\rangle + \frac{e^{i\varphi}}{\sqrt{2}}\vert - \rangle
$$
without additional restriction on $\varphi$.  The resulting state 
is not in general an eigenstate of either $\sigma_x$ or $\sigma_y$, but for which $\langle S_z\rangle=0$ still holds.

Edit: in answer to some further queries:
The most general spin state has the form
$$
\vert\psi\rangle =
\cos \left(\frac{\theta }{2}\right)\vert +\rangle + e^{i \phi } \sin \left(\frac{\theta }{2}\right)\vert -\rangle
$$
with average values
$$
\langle S_z\rangle=\cos\theta \, ,\qquad
\langle S_x\rangle=\sin\theta\cos\phi\, ,\qquad
\langle S_y\rangle=\sin\theta\sin\phi\, .
$$
It is not hard to see that an appropriate choice of angles $\theta,\phi$ can lead to various triples of average values.  In general, if 
$\hat n=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$
then the state $\vert\psi\rangle$ will be an eigenstate of 
$\hat n\cdot \vec S:= n_xS_x+n_yS_y+n_zS_z$, and so not of any single spin operator in general.
