= Solution
Let $a=\mathbb E_\psi(A)$, which is real. The dispersion, or <quantum uncertainty>, is the nonnegative <standard deviation>
$$
\boxed{\Delta_\psi(A)=\sqrt{\mathbb E_\psi(A^2)-\mathbb E_\psi(A)^2}=\|(A-aI)\psi\|.}
$$
Indeed, <self-adjointness> and normalization give
$$
\|(A-aI)\psi\|^2=\langle\psi,(A-aI)^2\psi\rangle=\mathbb E_\psi(A^2)-2a\mathbb E_\psi(A)+a^2=\mathbb E_\psi(A^2)-a^2.
$$
A squared <norm> is nonnegative, so the <square root> is defined. It vanishes precisely when $(A-aI)\psi=0$, namely when $\psi$ is an <eigenvector> of $A$. Conversely, an <eigenvector> with <eigenvalue> $\lambda$ has $a=\lambda$ and hence zero dispersion. Thus \b[zero dispersion is equivalent to being an eigenstate].
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