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.


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.

  • $\begingroup$ I'm not sure this applies, since we seem to be working with spin $1/2$ here. $\endgroup$ – Javier Oct 1 '14 at 11:36
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    $\begingroup$ That may or may not be the case; this answer applies to the most general case with the uncertainty relation stated in the question. Obviously for spin 1/2 the relation is impossible. If the OP wants to refine their question to specify it, I'll further refine my answer. $\endgroup$ – Emilio Pisanty Oct 1 '14 at 11:57

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