Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 53 / 1 / a / Solution

Take the vectors in the quantum state ensemble to be normalized, with $p_i\ge 0$ and ${\sum }_i{p}_i=1$. The corresponding density matrix isEach rank-one projector is a self-adjoint operator, so ${\rho }^{†}=\rho$. For every $|v⟩$ in the Hilbert space,proving that $\rho$ is positive semidefinite. The trace is $\mathrm{Tr}\rho ={\sum }_i{p}_i⟨{\psi }_i|{\psi }_i⟩=1$.

Conversely, diagonalize a self-adjoint operator $\rho$ in an orthonormal eigenbasis, writing $\rho ={\sum }_j{\lambda }_j|e_j⟩⟨e_j|$. Its positive semidefiniteness implies ${\lambda }_j=⟨e_j|\rho |e_j⟩\ge 0$, and its unit trace gives ${\sum }_j{\lambda }_j=1$. Thus $\{{\lambda }_j,|e_j⟩\}$ is a quantum state ensemble with precisely this density matrix. Terms with zero eigenvalue may be omitted. This argument includes rank-one pure states; an ensemble representation need not describe a genuinely mixed state.