 Solution

For any normalized vector $|u⟩$, the Born rule assigns the probability $⟨u|\rho |u⟩$ to the corresponding rank-one orthogonal projection. This must be nonnegative for every $u$, which is exactly positive semidefiniteness. For any complete orthonormal basis $\{|j⟩\}$, the outcome probabilities must sum to one:Thus a Hermitian matrix represents a physical density matrix only ifThese conditions are also sufficient in finite dimension. The spectral theorem writes $\rho ={\sum }_j{p}_j|u_j⟩⟨u_j|$ with $p_j\ge 0$ and ${\sum }_j{p}_j=1$, describing a quantum state ensemble of pure states with valid mixing probabilities. A Hermitian matrix alone could have negative eigenvalues, hence negative measurement probabilities.