= Solution
Write $c=\cos(\theta/2)$ and $s=\sin(\theta/2)$. Then
$$
|\uparrow_\theta\downarrow_\theta\rangle
-|\downarrow_\theta\uparrow_\theta\rangle
=(c^2+s^2)
\bigl(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\bigr).
$$
Thus $|\psi\rangle$ is unchanged. It is the <spin-one-half singlet state>, so joint rotations leave it invariant. Equivalently, direct use of the <Pauli matrix>[Pauli matrices] shows
$$
(S_i^{(1)}+S_i^{(2)})|\psi\rangle=0
\qquad(i=x,y,z).
$$
It is therefore an <eigenstate> of every Cartesian component of the combined spin, with eigenvalue zero.
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