= Solution
An ensemble with normalized quantum states $|\psi_a\rangle$ and probabilities $p_a\geq0$, $\sum_ap_a=1$, has <density matrix>
$$
\boxed{\rho=\sum_a p_a|\psi_a\rangle\langle\psi_a|}.
$$
Each rank-one projector is <Hermitian>, so $\rho^\dagger=\rho$. Taking the <trace> in any <orthonormal basis> gives
$$
\operatorname{Tr}\rho=\sum_a p_a\langle\psi_a|\psi_a\rangle=1.
$$
It is also positive: $\langle v|\rho|v\rangle=\sum_ap_a|\langle\psi_a|v\rangle|^2\geq0$. Its <eigenvalues> $\lambda_a$ are therefore nonnegative and sum to one. Hence
$$
\operatorname{Tr}\rho^2=\sum_a\lambda_a^2\leq\left(\sum_a\lambda_a\right)^2=1,
$$
with equality exactly when one <eigenvalue> is one and the others are zero. Thus the <purity of a density operator> distinguishes
$$
\boxed{\text{pure state: }\operatorname{Tr}\rho^2=1,\qquad
\text{mixed state: }\operatorname{Tr}\rho^2<1}.
$$
Equivalently, a <pure state> has $\rho^2=\rho$ and rank one. In a $d$-dimensional <Hilbert space>, purity is at least $1/d$, attained by the <maximally mixed state> $I/d$.
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