The extreme points of the state space are exactly the pure states

[(pure $\Rightarrow$ extreme) the quadratic forms are pinned to one] Assume $\rho =T_{u,u}$ with $u\in S(H)$ and $\rho =(1-t)\sigma +t\tau$ with $0<t<1$. Then $⟨\rho u,u⟩=⟨T_{u,u}u,u⟩=⟨u,u{⟩}^2=\mathrm{\parallel }u{\mathrm{\parallel }}^2=1$ by [F5] and the definition of $T_{u,u}$, so $1=(1-t)⟨\sigma u,u⟩+t⟨\tau u,u⟩$. Every state $\eta \in D(H)$ satisfies $⟨\eta u,u⟩\ge 0$ by positivity and $⟨\eta u,u⟩\le \mathrm{\mid }⟨\eta u,u⟩\mathrm{\mid }\le \mathrm{\parallel }\eta u\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }=\mathrm{\parallel }\eta u\mathrm{\parallel }\le \mathrm{\parallel }\eta \mathrm{\parallel }\le \mathrm{\parallel }\eta {\mathrm{\parallel }}_1=1$ by [F4]; a real convex combination with weights $1-t,t>0$ of two numbers $\le 1$ equals $1$ only if both numbers equal $1$. Hence $⟨\sigma u,u⟩=⟨\tau u,u⟩=1$.

[(pure $\Rightarrow$ extreme) the first component fixes $u$] For the state $\sigma$ we have the chain $1=⟨\sigma u,u⟩\le \mathrm{\mid }⟨\sigma u,u⟩\mathrm{\mid }\le \mathrm{\parallel }\sigma u\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }=\mathrm{\parallel }\sigma u\mathrm{\parallel }\le \mathrm{\parallel }\sigma \mathrm{\parallel }\le 1$ by [F4]; all inequalities are therefore equalities, so equality holds in the Cauchy–Schwarz inequality for the pair $(\sigma u,u)$, and the equality case of [F4] gives $\sigma u=\zeta u$ for a scalar $\zeta$ because $u\mathrm{\ne }0$. Taking the pairing with $u$ gives $1=⟨\sigma u,u⟩=\zeta ⟨u,u⟩=\zeta$, so $\sigma u=u$. The same argument gives $\tau u=u$.

[(pure $\Rightarrow$ extreme) a state fixing $u$ equals $T_{u,u}$] Let $\eta \in D(H)$ with $\eta u=u$. By [F3] its positive eigenvalues ${\lambda }_n$, counted with multiplicity, satisfy ${\sum }_n{\lambda }_n=1$, and $u$ is an eigenvector with eigenvalue $1$. If $m$ is the multiplicity of eigenvalue $1$, then $1={\sum }_n{\lambda }_n=m+{\sum }_{{\lambda }_n\mathrm{\ne }1}{\lambda }_n$ with a nonnegative second sum, so $m=1$ and there is no other positive eigenvalue. Thus $E_1(\eta )$ is one-dimensional and contains $u$, so $E_1(\eta )=\mathbb{C}u$. By the distinct-eigenvalue decomposition in [F3], the closed span $M$ of the positive eigenspaces is now $E_1(\eta )=\mathbb{C}u$ and $H=M\oplus \ker \eta$; hence $\eta$ is the orthogonal projection $P_1$ onto $\mathbb{C}u$. By [F7] and [F5] this projection is $T_{u,u}$. Applying this to $\sigma$ and to $\tau$ gives $\sigma =\tau =T_{u,u}=\rho$; hence every representation of $\rho$ as a nontrivial convex combination is trivial, and $\rho$ is an extreme point by [F6].

[(mixed $\Rightarrow$ not extreme) an explicit splitting] Assume now that $\rho$ is not pure. By [F1] the state $\rho$ has at least two positive eigenvalues; list them with multiplicity as ${\lambda }_1\ge {\lambda }_2\ge \cdots >0$ with orthonormal eigenvectors $u_1,u_2$, so ${\lambda }_2>0$. Put $\epsilon :=\frac{{\lambda }_2}{2}>0,{\sigma }_{\pm }:=\rho \pm \epsilon (T_{u_1,u_1}-T_{u_2,u_2})\mathrm{.}$ The expansion of [F3] for $\rho$ turns the displayed definitions into countable mixtures of the pure states $T_{u_n,u_n}$ with weights ${\lambda }_1\pm \epsilon$, ${\lambda }_2\mp \epsilon$, ${\lambda }_3,{\lambda }_4,\dots$; these weights are nonnegative because ${\lambda }_2-\epsilon =\epsilon >0$ and ${\lambda }_1\pm \epsilon \ge 0$, and they sum to ${\sum }_n{\lambda }_n\pm \epsilon \mp \epsilon =1$; hence ${\sigma }_{\pm }\in D(H)$ by [F2]. Moreover $\rho =\frac{1}{2}{\sigma }_{+}+\frac{1}{2}{\sigma }_{-}$, and ${\sigma }_{+}-{\sigma }_{-}=2\epsilon (T_{u_1,u_1}-T_{u_2,u_2})$, which is nonzero because $u_1,u_2$ are orthogonal unit vectors and hence not phase multiples of one another, so that $T_{u_1,u_1}\mathrm{\ne }{T}_{u_2,u_2}$ by [F5]. Thus the representation $\rho =\frac{1}{2}{\sigma }_{+}+\frac{1}{2}{\sigma }_{-}$ with ${\sigma }_{+}\mathrm{\ne }{\sigma }_{-}$ is a nontrivial convex representation with $0<\frac{1}{2}<1$, and $\rho$ is not an extreme point by [F6].

[Conclusion] Steps 1.1, 2.1 and 3.1 show that every pure state is extreme, and step 4.1 shows that every non-pure state is not extreme. Hence the extreme points of $D(H)$ are exactly the pure states.