The extremal points of positive semidefinite operators $\rho$ with trace $=1$ are precisely the rank-one projections

Question

Let $H$ be a complex Hilbert space (finite-dimensional if necessary) and let $S$ denote the set of positive semidefinite operators $\rho$ on $H$ with trace $=1$ so that $S$ is a convex set. Then how do you prove that the extremal points of $S$ are exactly the rank-one projection operators?

Answer 1

Note that in the infinite dimensional case, we are discussing trace-class (and therefore compact) operators.

Every extreme point must be a rank-one projection. In particular, we note that if $\rho$ is not a rank-one projection, then it can be written as a convex combination of rank-one projections (via a spectral decomposition for instance) and therefore fails to be an extreme point.

We argue that every rank-one projection is an extreme point as follows. Note that for positive operators $\rho_1,\rho_2$, $\ker(\rho_1 + \rho_2) \subset \ker \rho_1$. It follows that if $\rho_1 + \rho_2$ is rank $1$, then both $\rho_1,\rho_2$ are rank $1$.

On the other hand, suppose that $\rho_1,\rho_2$ are distinct rank $1$ projections (onto $u_1$ and $u_2$ respectively). It follows that for any $a>b > 0$ with $a+b = 1$, we have $$ a \rho_1 + b \rho_2 = (a-b)\rho_1 + b[\rho_1 + \rho_2]. $$ However, $\rho_1 + \rho_2$ is a rank-2 projection, which means that $a \rho_1 + b \rho_2$ has rank $2$. So, no such convex combination can yield a rank-1 operator. So, the only convex combination that yields a rank-1 projection $\rho$ is of the form $a\rho + b \rho$, which is to say that every rank-1 projection $\rho$ is an extreme point.