 Solution

Every density operator can be written asTherefore $⟨\varphi |\rho |\varphi ⟩={\sum }_j{p}_j|⟨{\psi }_j|\varphi ⟩{|}^2\ge 0$, so $\rho$ is a positive semidefinite operator, and $\mathrm{Tr}\rho ={\sum }_j{p}_j=1$. A pure quantum state has a rank-one density operator $|\psi ⟩⟨\psi |$, equivalently ${\rho }^2=\rho$ and $\mathrm{Tr}({\rho }^2)=1$. A mixed quantum state is not pure and has $\mathrm{Tr}({\rho }^2)<1$ in finite dimension.

The Von Neumann equation and the Pauli matrix commutator identity giveso the Bloch vector obeysWriting $\omega =|\mathbf{\omega }|$ and $\hat{\mathbf{\omega }}=\mathbf{\omega }/\omega$, the assumption $\mathbf{a}\cdot \mathbf{\omega }=0$ givesand henceUnitary evolution rotates the Bloch vector and preserves its length. Since $|\mathbf{a}|<1$, the state remains mixed, whereas $|{\uparrow }_x⟩$ has the unit Bloch vector $(1,0,0)$. The system is therefore never definitely in $|{\uparrow }_x⟩$.