The state space of a two-level finite-spin system is the Bloch ball

[(i) parametrization] A general Hermitian $2\times 2$ matrix with trace one has the form $A=\left(\begin{array}{cc}a& b-ic\\ b+ic& 1-a\end{array}\right)$ with $a,b,c\in \mathbb{R}$; expanding the three Pauli matrices shows $A=\frac{1}{2}I+b{\sigma }_x+c{\sigma }_y+(a-\frac{1}{2}){\sigma }_z=\frac{1}{2}(I+r\cdot \sigma ),r=(2b,2c,2a-1),$ so every Hermitian trace-one matrix is of the displayed form with a unique real vector $r$, and conversely every $r\in {\mathbb{R}}^3$ gives a Hermitian matrix of trace one.

[(i) positivity exactly for $\mathrm{\mid }r\mathrm{\mid }\le 1$] If $r=0$, then $\rho =\frac{1}{2}I$ and $⟨\rho x,x⟩=\frac{1}{2}\mathrm{\parallel }x{\mathrm{\parallel }}^2\ge 0$ for every $x$, while $\mathrm{\mid }r\mathrm{\mid }=0\le 1$. Now suppose $r\mathrm{\ne }0$ and set $n=r\mathrm{/}\mathrm{\mid }r\mathrm{\mid }$. By [L1], $(n\cdot \sigma {)}^2={\sum }_{i,j}{n}_i{n}_j{\sigma }_i{\sigma }_j=\frac{1}{2}{\sum }_{i,j}{n}_i{n}_j({\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i)=\mathrm{\mid }n{\mathrm{\mid }}^{2}I=I$. Put $P:=\frac{1}{2}(I+n\cdot \sigma )$; then $P^2=\frac{1}{4}(I+2n\cdot \sigma +(n\cdot \sigma {)}^2)=\frac{1}{4}(2I+2n\cdot \sigma )=P$ and $P^{\ast }=P$, so $P$ and $I-P$ are orthogonal projections, and $I+r\cdot \sigma =(1+\mathrm{\mid }r\mathrm{\mid })P+(1-\mathrm{\mid }r\mathrm{\mid })(I-P)$. For every $x$ the quadratic form of $I+r\cdot \sigma$ is $(1+\mathrm{\mid }r\mathrm{\mid })\mathrm{\parallel }Px{\mathrm{\parallel }}^2+(1-\mathrm{\mid }r\mathrm{\mid })\mathrm{\parallel }(I-P)x{\mathrm{\parallel }}^2$ by [L5], which is nonnegative if $\mathrm{\mid }r\mathrm{\mid }\le 1$. If $\mathrm{\mid }r\mathrm{\mid }>1$, then $\mathrm{tr}P=\frac{1}{2}(\mathrm{tr}I+\mathrm{tr}(n\cdot \sigma ))=1$, whereas $\mathrm{tr}I=2$, so $P\mathrm{\ne }I$. If $\ker P=\{0\}$, the two columns of the matrix of $P$ would be linearly independent and hence form a basis of ${\mathbb{C}}^2$; thus $P$ would be invertible. The identity $P^2=P$ would then imply $P=I$, a contradiction. Therefore there is a nonzero $x\in \ker P$, and for this vector $⟨\rho x,x⟩=\frac{1}{2}(1-\mathrm{\mid }r\mathrm{\mid })\mathrm{\parallel }x{\mathrm{\parallel }}^2<0$. Hence $\rho =\frac{1}{2}(I+r\cdot \sigma )$ is positive exactly when $\mathrm{\mid }r\mathrm{\mid }\le 1$; together with step 1.1 and the finite-rank trace-class fact in [L4], this proves (i).

[(ii) purity is $\mathrm{\mid }r\mathrm{\mid }=1$] For $\rho =\frac{1}{2}(I+r\cdot \sigma )$ one computes, using $(r\cdot \sigma {)}^2=\mathrm{\mid }r{\mathrm{\mid }}^{2}I$ from [L1], ${\rho }^2=\frac{1}{4}(I+2r\cdot \sigma +(r\cdot \sigma {)}^2)=\frac{1}{4}((1+\mathrm{\mid }r{\mathrm{\mid }}^2)I+2r\cdot \sigma ),$ so $\mathrm{tr}({\rho }^2)=\frac{1}{4}(1+\mathrm{\mid }r{\mathrm{\mid }}^2)\mathrm{tr}I=\frac{1}{2}(1+\mathrm{\mid }r{\mathrm{\mid }}^2)$ because $\mathrm{tr}(r\cdot \sigma )=0$ by [L1]. By [L2] the state is pure exactly when this number equals $1$, that is exactly when $\mathrm{\mid }r{\mathrm{\mid }}^2=1$, which is the first part of (ii).

[(ii) every spin-up unit vector gives its rank-one state] Assume $\mathrm{\mid }r\mathrm{\mid }=1$ and put $A:=r\cdot \sigma$, so $A^2=I$, $A^{\ast }=A$, and $\rho =\frac{1}{2}(I+A)$ is a self-adjoint idempotent, hence an orthogonal projection by [L5]. Since $\mathrm{tr}\rho =1$, $\rho \mathrm{\ne }0$; choose $y$ with $\rho y\mathrm{\ne }0$ and set ${\psi }_0:=\rho y\mathrm{/}\mathrm{\parallel }\rho y\mathrm{\parallel }$. Then ${\psi }_0$ is a unit vector and $\rho {\psi }_0={\psi }_0$, so $A{\psi }_0={\psi }_0$. This proves that the spin-up ray in direction $r$ exists. Now let $\psi$ be any unit eigenvector of $A$ with eigenvalue $+1$, as in (ii). There is a nonzero vector $\chi \in \ker (A+I)$: if $A+I$ were injective, its two columns in ${\mathbb{C}}^2$ would be linearly independent and hence a basis, making $A+I$ invertible; but $(A-I)(A+I)=A^2-I=0$ would then imply $A=I$, contradicting $\mathrm{tr}A=0\mathrm{\ne }2=\mathrm{tr}I$ by [L1]. Normalize $\chi$ to have norm one. The vectors $\psi$ and $\chi$ belong to distinct eigenspaces of the self-adjoint operator $A$, so they are orthogonal by [L4]; as two nonzero orthogonal vectors in ${\mathbb{C}}^2$, they form an orthonormal basis. On this basis, $\rho$ and $T_{\psi ,\psi }$ both act as the identity on $\psi$ and as zero on $\chi$ by [L3]. Since a linear map is determined by its values on a basis by [L4], $\rho =T_{\psi ,\psi }$. Thus every boundary point is the pure state of every unit vector in its spin-up ray, and with step 2.2 every state with $\mathrm{\mid }r\mathrm{\mid }<1$ is mixed.

[(iii) the two displayed values] For $r=0$ the operator $\rho =\frac{1}{2}I$ is a state by (i) with purity $\frac{1}{2}(1+0)=\frac{1}{2}$ by step 2.2, and it is mixed because $\frac{1}{2}\mathrm{\ne }1$ by [L2]; for $r=e_3$ one has $\mathrm{\mid }r\mathrm{\mid }=1$ and $\frac{1}{2}(I+{\sigma }_z)=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)=T_{e_1,e_1}$ by the definition of $T_{u,u}$, so it is a pure state by [L3]. This verifies (iii) and completes the computation.