Does spin $0$ violate the uncertainty principle? It seems that if we have some spin eigenstate $|0\rangle$ for which the square of the total spin operator is $\hat{S}^2|0\rangle = 0$, then we simultaneously know the spin in all three directions, since $\hat{S}^2 = \hat{S}_x^2 + \hat{S}_y^2 + \hat{S}_z^2$, and the only way to get zero on the right is for each of the three quantities on the left to give zero when applied to $|0\rangle$. That looks like a violation of the uncertainty principle. Is it?
(If there's some caveat about elementary vs. composite particles or systems, please consider an elementary particle, like the Higgs boson, that has spin 0.)
 A: The uncertainty principle tells us that 
$$ \Delta A \Delta B \geq \frac{1}{2}|\langle [A,B]\rangle|$$
where $\Delta A = \sqrt{\langle A^2\rangle-\langle A \rangle^2}$ and similarly for $B$. It means that the uncertainty is related to the commutation relations between the operators via the expectation value of the commutator. For spin-$0$ the expectation value of every spin operator is $0$, so for any pair of spin operators we will get on the right hand side of the inequality another spin operator, whose expectation value will be zero, and therefore the uncertainty principle will say $$\Delta S_i \Delta S_j \geq \frac{1}{2}|\langle S_j \rangle| =0$$
and we can have zero uncertainty, without violating the uncertainty principle.
In a sense, since the spin-$0$ space is spanned by a single vector, we cannot have any uncertainty in it. We always know exactly what the state is -- it is the only one available.
A: Indeed the component spin operators do in general not commute. However, for S=0  they do commute. There is no uncertainty in this particular case.
