The Bloch ball is exactly the qubit density-state space

Put $t=\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2$ and $\mathrm{\Sigma }=\mathbf{r}\cdot \mathit{\sigma }$. By [F3], ${\mathrm{\Sigma }}^{\ast }=\mathrm{\Sigma }$ and ${\mathrm{\Sigma }}^2={\sum }_j{r}_j^2{\sigma }_j^2+{\sum }_{j<k}{r}_j{r}_k({\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j)=t^{2}I$. Using the displayed matrices in [F4], $\mathrm{\Sigma }=\left(\begin{array}{cc}{r}_z& r_x-i{r}_y\\ r_x+i{r}_y& -r_z\end{array}\right)$, so $\mathrm{tr}(\mathrm{\Sigma })=0$ and $\det (\mathrm{\Sigma })=-t^2$. Thus its characteristic polynomial is ${\lambda }^2-t^2$. If $t>0$, its two eigenvalues are $t$ and $-t$; if $t=0$, $\mathrm{\Sigma }=0$. Therefore $\rho (\mathbf{r})=\frac{1}{2}(I+\mathrm{\Sigma })$ has eigenvalues $(1+t)\mathrm{/}2$ and $(1-t)\mathrm{/}2$, both $1\mathrm{/}2$ at the zero vector.

Since [F2] gives trace one, [F5]–[F7] show that $\rho (\mathbf{r})$ is a density operator exactly when both eigenvalues are nonnegative. The eigenvalue $(1+t)\mathrm{/}2$ is always positive; $(1-t)\mathrm{/}2\ge 0$ exactly when $t\le 1$. At $t=1$ the zero eigenvalue is allowed by non-negativity; at $t=0$ both eigenvalues are $1\mathrm{/}2$. Hence the condition is precisely $\mathrm{\parallel }\mathbf{r}{\mathrm{\parallel }}_2\le 1$.