= Solution
A <density operator> is Hermitian, <positive semidefinite> and has <trace> one. Part (a) therefore makes all its coordinates real, and the identity basis vector fixes the final coordinate:
$$
\boxed{r_{N^2}=\operatorname{Tr}(\rho I/\sqrt N)=1/\sqrt N.}
$$
The remaining basis vectors are traceless by orthogonality to $I/\sqrt N$. The <Generalized Bloch representation> is consequently
$$
\rho=\frac IN+\sum_{k=1}^{N^2-1}s_k\sigma_k.
$$
Using orthonormality to expand the squared <Hilbert-Schmidt norm> gives the <purity bound for generalized Bloch vectors>:
$$
\operatorname{Tr}(\rho^2)=\sum_{k=1}^{N^2}r_k^2
=\frac1N+\|s\|^2.
$$
If $p_1,\ldots,p_N$ are its nonnegative <eigenvalues>, then $\sum_jp_j=1$ and
$$
1-\operatorname{Tr}(\rho^2)=\left(\sum_jp_j\right)^2-\sum_jp_j^2
=2\sum_{j<k}p_jp_k\geq0.
$$
It follows that
$$
\boxed{\|s\|\leq\sqrt{1-1/N}.}
$$
Equality requires every pairwise product $p_jp_k$ to vanish, so exactly one <eigenvalue> equals one and all others are zero. Conversely, a rank-one projector has precisely that <spectrum> and attains equality. Thus equality holds exactly for a <pure state>. This normalization is different from the commonly used unit-radius qubit <Bloch vector>.
Back to article page
