Bloch coordinates of |0> and H|0>

By [F1] and [F9], $\mathrm{\mid }0⟩=(1,0{)}^{\mathsf{T}}$ is a unit vector; [F2]–[F3] give the rank-one projector convention, so direct coordinates give ${\rho }_0=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$. Multiplication by [F4] gives $\mathrm{tr}({\rho }_{0}X)=0$, $\mathrm{tr}({\rho }_{0}Y)=0$, and $\mathrm{tr}({\rho }_{0}Z)=1$, so [F6] gives $\mathbf{r}({\rho }_0)=(0,0,1)$. The matrix satisfies ${\rho }_0^2={\rho }_0$ and $\mathrm{tr}({\rho }_0)=1$; by [F2] it is the projector of the pure ray $[\mathrm{\mid }0⟩]$, and [F10] makes it a density operator. Thus also $\mathrm{tr}({\rho }_0^2)=1$.

By [F1] and [F5], $H\mathrm{\mid }0⟩=(\mathrm{\mid }0⟩+\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}=\mathrm{\mid }+⟩$. From [F9], $\mathrm{\parallel }+{\mathrm{\parallel }}^2=\frac{1}{2}(⟨0,0{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}+⟨0,1{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}+⟨1,0{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}+⟨1,1{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}})=1$, so [F2]–[F3] give ${\rho }_{+}=\mathrm{\mid }+⟩⟨+\mathrm{\mid }=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$. Multiplying by [F4] gives $\mathrm{tr}({\rho }_{+}X)=1$, $\mathrm{tr}({\rho }_{+}Y)=0$, and $\mathrm{tr}({\rho }_{+}Z)=0$, so [F6] gives $\mathbf{r}({\rho }_{+})=(1,0,0)$. Direct multiplication gives ${\rho }_{+}^2={\rho }_{+}$ and $\mathrm{tr}({\rho }_{+})=1$; by [F2], ${\rho }_{+}$ is a pure-ray projector, so [F10] makes it a density operator and $\mathrm{tr}({\rho }_{+}^2)=1$.

The computed vectors have norms $\mathrm{\parallel }(0,0,1){\mathrm{\parallel }}_2=\sqrt{0^2+0^2+1^2}=1$ and $\mathrm{\parallel }(1,0,0){\mathrm{\parallel }}_2=\sqrt{1^2+0^2+0^2}=1$. Steps 1.1 and 1.2 prove both matrices are density operators. By [F7] these are the unique Bloch vectors, and [F8] confirms both states are pure and lie on the Bloch sphere.