Why is the boundary of the set of states in the generalised Bloch representation comprised of singular matrices?

Question

Consider an arbitrary qudit state $\rho$ over $d$ modes. Any such state can be represented as a point in $\mathbb R^{d^2-1}$ via the standard Bloch representation:

$$\rho=\frac{1}{d}\left(\mathbb I +\sum_k c_k\sigma_k\right)$$ with $\sigma_k$ traceless Hermitian matrices satisfying $\text{Tr}(\sigma_j\sigma_k)=d\delta_{jk}$.

In the case of a qubit, $d=2$, we can take the $\sigma_k$ matrices to be the standard Pauli matrices, and the set of states is thus mapped into a sphere in $\mathbb R^3$, which is the standard Bloch sphere representation of qubits. For $d>2$, this picture becomes significantly more complicated, and the states are not in general mapped into a (hyper-)sphere.

In particular, the special case $d=3$ is analysed in this paper (arXiv). At the beginning of page 6, the authors make the following statement:

It is clear that the boundary of $\Omega_3$ comprises density matrices which are singular.

Here, $\Omega_3$ represents the set of qutrit states in their standard Bloch representation (thus $\Omega_3\subset\mathbb R^8$).

This statement does not, however, seem so obvious to me. I understand that if $\rho$ is not singular, then it can be written as convex combinations of other states, and it would therefore seem that it could not lie on the boundary of the set of states. However, is it not in principle possible that at least a section of $\Omega_3$ is "flat" (as in, is a subsection of a hyperplane), so that convex combinations of states on this boundary still lie on the boundary?

Intuitively this is most likely not the case, but is there an easy way to prove it?

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Related questions:

1. Geometry of qutrit and Gell-Mann matrices
2. What is the dimension of the space of all pure qudit states of dimension $D$?

Answer 1

Your intuition is correct, since rank-2 density matrices will be convex combinations of rank-1 density matrices, but they will be singular and hence will still be on the boundary.

We can prove that non-singular $\rho$ are in the interior of $\Omega_3$ with a fairly direct proof. Let $\epsilon>0$ and let $\rho'$ be some matrix (the Bloch representation of some vector of $\mathbb{R}^8$) such that $\Vert\rho-\rho'\Vert<\epsilon$. Because $\rho'$ is a Bloch representation of something, it has trace 1, so if we want to show that it's a density matrix, we just need to show that it's positive. To do this we show that $\langle v,\rho'v\rangle\geq 0$ for all vectors $v$.

Note that since $\rho$ is non-singular, it has a minimum eigenvalue $\lambda>0$. Thus, $\langle v,\rho v\rangle\geq\lambda\Vert v\Vert^2$ for all vectors $v$. We also have that $\vert \langle v,(\rho-\rho')v\rangle\vert\leq \Vert \rho-\rho'\Vert\Vert v\Vert^2$. Combining these facts: $$\begin{aligned}\langle v,\rho'v\rangle=&\langle v,(\rho'-\rho) v\rangle+\langle v,\rho v\rangle \\ \geq & -\Vert \rho-\rho'\Vert\Vert v\Vert^2+\lambda \Vert v\Vert^2\\ >&-\epsilon\Vert v\Vert^2+\lambda\Vert v\Vert^2 \end{aligned}$$ If we choose $\epsilon<\lambda$, this will be positive for all such $\rho'$. Thus, the ball of radius $\epsilon$ around $\rho$ is contained in $\Omega_3$, so $\rho$ is an interior point.

Answer 2

Non-singular density matrices $\rho$ have no zero eigenvalues, $\lambda=\lambda_{\mathrm{min}}(\rho)>0$, with $\lambda_{\min}$ the smallest eigenvalue. Then, the ball $$ \{\rho+M|-\lambda I \le M \le \lambda I\} $$ will also be in the set of density operators, and thus, $\rho$ is an interior point.