Briefly explain why the density operator $\rho$ obeys $\rho \geqslant 0$ and $\operatorname{Tr}(\rho)=1$. What is meant by a pure state and a mixed state?

A two-state system evolves under the Hamiltonian $H=\hbar \boldsymbol{\omega} \cdot \boldsymbol{\sigma}$, where $\boldsymbol{\omega}$ is a constant vector and $\sigma$ are the Pauli matrices. At time $t$ the system is described by a density operator

$$\rho(t)=\frac{1}{2}\left(1_{\mathcal{H}}+\mathbf{a}(t) \cdot \boldsymbol{\sigma}\right)$$

where $1_{\mathcal{H}}$ is the identity operator. Initially, the vector $\mathbf{a}(0)=\mathbf{a}$ obeys $|\mathbf{a}|<1$ and $\mathbf{a} \cdot \boldsymbol{\omega}=0$. Find $\rho(t)$ in terms of a and $\boldsymbol{\omega}$. At what time, if any, is the system definitely in the state $\left|\uparrow_{x}\right\rangle$ that obeys $\sigma_{x}\left|\uparrow_{x}\right\rangle=+\left|\uparrow_{x}\right\rangle ?$