Does there exist a state for which $\Delta\sigma_x^2=\Delta\sigma_y^2=0$? If not, how does one prove it? I just realized that the uncertainty principle says that 
$$\Delta\sigma_x^2 \Delta\sigma_y^2 \ge \left(\overline{\hat\sigma_z}\right)^2,$$
where $\Delta\sigma_x^2=\overline{(\hat\sigma_x-\overline{\hat\sigma_x})^2}$ and $\overline{\hat\sigma_x}$ means $\langle\psi|\hat\sigma_x|\psi\rangle$, so there could be a state $\psi$ that gives $0\cdot 0 \ge 0$ which doesn't violate the uncertainty principle.
 A: Yes, this is possible - but only for states with zero total angular momentum.
To see why, the first step is seeing that if $\Delta\sigma_x^2=\Delta\sigma_y^2=0$ on state $\psi$, then $\psi$ is an eigenstate of both $\hat\sigma_x$ and $\hat\sigma_y$:
$$
0=\Delta\sigma_x^2=⟨\psi|(\hat\sigma_x-\overline{\hat\sigma_x})^2|\psi⟩=\left\|(\hat\sigma_x-\overline{\hat\sigma_x})|\psi⟩\right\|^2\quad\text{implies}\quad (\hat\sigma_x-\overline{\hat\sigma_x})|\psi⟩=0.
$$
Thus, you have equations of the form $\hat \sigma_x|\psi⟩=X|\psi⟩$ and  $\hat \sigma_y|\psi⟩=Y|\psi⟩$, and these in turn imply that
$$
\hat \sigma_z|\psi⟩=(\pm)i(\hat \sigma_x\hat \sigma_y-\hat \sigma_y\hat \sigma_x)|\psi⟩=\pm i (XY-YX)|\psi⟩=0
$$
so $\psi$ is also an eigenstate of $\hat \sigma_z$, with eigenvalue zero.
Furthermore, you can now apply the same argument on the other two permutations, to obtain a zero eigenvalue on the other two directions:
$$\hat \sigma_x|\psi⟩=\hat \sigma_y|\psi⟩=\hat \sigma_z|\psi⟩=0.$$
Squaring these and adding them up, you get
$$\hat \sigma^2|\psi⟩=\left(\hat \sigma_x^2+\hat \sigma_y^2+\hat \sigma_z^2\right)|\psi⟩=0;$$
that is, the total angular momentum is zero. In quantum-number language, this is the $l=0$, $m=0$ state.
