= Solution
Every <density operator> can be written as
$$
\rho=\sum_jp_j|\psi_j\rangle\langle\psi_j|,
\qquad p_j\geq0,
\qquad\sum_jp_j=1.
$$
Therefore $\langle\phi|\rho|\phi\rangle=\sum_jp_j|\langle\psi_j|\phi\rangle|^2\geq0$, so $\rho$ is a <positive semidefinite operator>, and $\operatorname{Tr}\rho=\sum_jp_j=1$. A <pure quantum state> has a rank-one density operator $|\psi\rangle\langle\psi|$, equivalently $\rho^2=\rho$ and $\operatorname{Tr}(\rho^2)=1$. A <mixed quantum state> is not pure and has $\operatorname{Tr}(\rho^2)<1$ in finite dimension.
The <Von Neumann equation> and the <Pauli matrix commutator identity> give
$$
\dot\rho=-\frac{i}{\hbar}[H,\rho],
\qquad
[\boldsymbol\omega\mathbin\cdot\boldsymbol\sigma,
\mathbf a\mathbin\cdot\boldsymbol\sigma]
=2i(\boldsymbol\omega\times\mathbf a)\mathbin\cdot\boldsymbol\sigma,
$$
so the <Bloch vector> obeys
$$
\dot{\mathbf a}=2\boldsymbol\omega\times\mathbf a.
$$
Writing $\omega=|\boldsymbol\omega|$ and $\widehat{\boldsymbol\omega}=\boldsymbol\omega/\omega$, the assumption $\mathbf a\mathbin\cdot\boldsymbol\omega=0$ gives
$$
\mathbf a(t)=\mathbf a\cos(2\omega t)
+(\widehat{\boldsymbol\omega}\times\mathbf a)\sin(2\omega t),
$$
and hence
$$
\boxed{\rho(t)=\frac12\left[1_{\mathcal H}+\left(\mathbf a\cos(2\omega t)+(\widehat{\boldsymbol\omega}\times\mathbf a)\sin(2\omega t)\right)\mathbin\cdot\boldsymbol\sigma\right]}.
$$
Unitary evolution rotates the Bloch vector and preserves its length. Since $|\mathbf a|<1$, the state remains mixed, whereas $|\!\uparrow_x\rangle$ has the unit Bloch vector $(1,0,0)$. The system is therefore never definitely in $|\!\uparrow_x\rangle$.
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