Let $\mathcal{A}$ be a unital c*-algebra. We know that its state space $\mathcal{S}(\mathcal{A})$ is a non-empty, compact, convex set. By the Krein-Milman theorem we know that $\mathcal{S}(\mathcal{A})$ has extreme points and the convex combinations of them are dense in the state space. We call those extreme points pure states. Now I have found another characterization of pure states:
Let $\varphi\in \mathcal{S}(\mathcal{A})$. Then $\varphi$ is pure, if for any positive linear functional $\psi: \mathcal{A} \to \mathbb{C}$ with $\psi \leq \varphi$ holds that there exists some $\lambda\in (0,1]$ s.t. $\varphi = \lambda \psi$.
I did show that this condition is sufficient:
Assume $\varphi = \lambda \psi_1 + (1-\lambda) \psi_2$ for some states $\psi_1\neq \psi_2 \in \mathcal{S}(\mathcal{A})$ and $\lambda\in(0,1)$. Then $\varphi \geq \lambda \psi_1$ and $\varphi \geq (1-\lambda) \psi_2$, hence by assumption there exist $\mu_1,\mu_2\in (0,1]$ s.t. $\mu_1\varphi = \lambda \psi_1$ and $\mu_2 \varphi = (1-\lambda) \psi_2$. Now since states have norm $1$ we obtain $\mu_1=\lambda, \mu_2 = (1-\lambda)$, therefore $\psi_1=\varphi=\psi_2$ which is a contradiction.
However I was not able to show that the condition is necessary (despite the fact that it totally agrees with my intuition whereas the direction I did show does not).
Kind regards, Sebastian
Here is the other direction (pretty straight forward, just some normalization): Let $\psi\le\phi$ be a positive functional (with $\psi \neq 0$ and $\psi\neq\phi$. Then $\phi-\psi$ is also positive. Now define the two normalized states $$\phi_1 = \frac{1}{||\psi||}\psi$$ $$\phi_2 = \frac{1}{||\phi-\psi||}(\phi-\psi)$$
Then you have
$$ \phi = ||\psi||\phi_1 + ||\phi-\psi||\phi_2$$
where both pre-factors are non-zero by assumption. As $\phi$ is assumed to be pure, it must be $\phi=\phi_1=\phi_2$, i.e. $\phi=\frac{1}{||\psi||}\psi$, which we wanted to show.