Extreme points of states of matrices

167 Views Asked by At

Consider $M_n(\mathbb{C})$ and the set $S(M_n(\mathbb{C}))$ of states on $M_n(\mathbb{C})$ (linear positive functionals on $M_n(\mathbb{C})$ that preserve the identity).

I'm trying to show that the extreme points $E$ of this state space is equal to the set $$\{\omega_\xi: M_n(\mathbb{C}) \to \mathbb{C}: A \mapsto \langle \xi, A \xi \rangle\mid \Vert \xi \Vert = 1\}.$$

I managed to show that every element $\omega_\xi$ is an extreme point, but now I'm trying to show that any extreme point is of this form. I tried to show the contrapositive: if an element is not of that form, it is not extreme, by looking for a good non-trivial convex combination, but this didn't work out.

Possibly relevant: I know we can write every state $\omega$ as $\omega_P$ where $\omega_P (A) = \operatorname{Tr(PA)}$ and where $P$ is a positive matrix with trace $1$.

Thanks in advance!

2

There are 2 best solutions below

2
On BEST ANSWER

The states $\omega_P$ are precisely your $\omega_\xi$, related by $P=\xi\xi^*$.

As for the proof that extreme states are of the form $\omega_P$, see here.

3
On

Hint: Diagonalize with a unitary matrix.