= Extreme points of the density-operator state space
{title2=$\rho\text{ extreme}\iff\operatorname{rank}\rho=1$}
The <density operators> on a finite-dimensional <Hilbert space> form a convex set. Its <extreme points> are exactly the rank-one projectors, hence exactly the <pure states>. To prove this, suppose $|\psi\rangle\langle\psi|=\sum_i a_i\rho_i$ with $a_i>0$ and $\rho_i$ <positive semidefinite>. For $w\perp\psi$, $0=\sum_i a_i\langle w|\rho_i|w\rangle$ is a sum of nonnegative numbers, so every term vanishes. Diagonalizing $\rho_i$ shows that $\langle w|\rho_i|w\rangle=0$ implies $\rho_iw=0$. Thus each $\rho_i$ is supported on the line spanned by $\psi$, and its unit <trace> forces $\rho_i=|\psi\rangle\langle\psi|$.
Conversely, a <density operator> of <rank> at least two has a <spectral decomposition> with at least two positive <eigenvalues>. If $\lambda$ is one of them, $0<\lambda<1$ and $\rho=\lambda|e\rangle\langle e|+(1-\lambda)\sigma$, where $\sigma$ is the normalized remainder. These are distinct <density operators>, proving that $\rho$ is not an <extreme point>.
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