It is a well-known fact that, in more than 2-dimensional spaces, the set of states in the Bloch representation contains non-pure states on its boundary.

An easy way to see this is to observe that if $\rho=\frac{I+\sum_i r_i \sigma_i}{d}$ for a set of traceless Hermitian operators such that $\operatorname{tr}(\sigma_i\sigma_j)=\delta_{ij} d$, with $d$ state dimension, then $\rho$ being pure amounts to $r=\sqrt{d-1}$, whereas $\rho\ge0$ amounts to $r_{\hat{\boldsymbol n}}\le \frac{1}{|\lambda_{\rm min}(\hat{\boldsymbol n}\cdot \boldsymbol\sigma)|}$ for all unit directions $\hat{\boldsymbol n}$. For example, for $d=3$, one can take an operator basis containing $$Z_{12}\equiv \sqrt{\frac23}\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix},$$ and then $\frac{I+\sqrt{3/2} Z_{12}}{3}$ is on the boundary, but also clearly not pure, as $$\frac{I+\sqrt{3/2} Z_{12}}{3} = \frac13\begin{pmatrix}2&0&0\\0&0&0\\0&0&1\end{pmatrix}.$$

My intuition is that these are "flat" parts of the boundary. Indeed, there are directions along which we can move that leave us (for small enough displacements) on the boundary. For example, $$ \frac{I+\sqrt{3/2} Z_{12}}{3} \pm \epsilon \begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} $$ remains an extremal state for all $\epsilon\in\mathbb{R}$ in a small enough interval around $\epsilon=0$.

Is there some general method to find the flat regions in the boundary of the set of $d$-dimensional states? For example, a way to characterise which states sit in flat parts of the boundary, and the number of directions spanning each such flat region.

  • $\begingroup$ The boundary of state space is just given by the density matrices which do not have full rank. It seems easy to characterize their flat pieces (note that "flat" in some sense requires that you specify in how many dimensions): You have to keep the kernel fixed and can change the rest of the spectrum. Any fixed kernel thus defines a dim(ker)-dimensional hypersurface. -- Or am I missing sth., and you are looking for sth. else? $\endgroup$ Commented Feb 20 at 12:29
  • $\begingroup$ @NorbertSchuch yea, that was pretty much my intuition as well, but I've never seen it formalised. Showing that singular states make up those boundaries is the easy part. But then I wonder: how do the boundaries of those flat pieces look like? I'm guessing the boundary of a (d-1)-dim flat region is a set of (d-2)-dim ones corresponding to singular states whose ker contains the one corresponding to the original surface, but what's a formal way to show this? $\endgroup$
    – glS
    Commented Feb 20 at 15:36
  • $\begingroup$ also probably the main reason for the question is the fact that you can always characterise convex regions via the support function. For flat pieces of a surface, this amounts to characterising them via the vector normal to the hyperplane they're contained into. So then, I wondered which directions (meaning here linear combinations of elements of the Herm operator basis) are orthogonal to such flat surfaces. Less specifically, I'm curious about possible formal approaches to answer these kinds of questions more methodically $\endgroup$
    – glS
    Commented Feb 20 at 15:39
  • $\begingroup$ My feeling is that once you write down the equations stating what "flat piece" means, the answer should be pretty straightforward. In the end, you want to linearly interpolate between a set of points, which are all rank-deficient, so all you can change while keeping the rank is the non-zero part of the operator -- rotating the kernel is only a quadratic effect and thus linearly interpolating "into" the kernel will give negative eigenvalues. $\endgroup$ Commented Feb 20 at 16:49

1 Answer 1


Let $\mathbb{C}^d$ be the underlying Hilbert space. We can think of states $\rho\in\operatorname{Herm}(\mathbb{C}^d)$ as elements of the $d^2$-dimensional Hilbert space of Hermitian $d\times d$ matrices. Or more specifically, as elements of the $(d^2-1)$-dimensional affine space of unit-trace Hermitian matrices. The set of states is the convex compact subset of positive semidefinite (PSD) matrices in such affine space. Let us denote this set as $\mathcal D\subset\operatorname{Herm}(\mathbb{C}^d)$. Our goal is to study the (relative) boundary $\partial\mathcal D$ of this set; that is, the boundary of $\mathcal D$ as a subset of its convex hull, which in this case is the set of unit-trace Hermitian matrices.

Here's a few things that can be said.

  1. The boundary of state space is made entirely (and exclusively) of singular states. This is easy to see because if $\rho$ is singular then it has (at least) a zero eigenvalue, and moving in the direction of the corresponding eigenvector gives a non-PSD matrix, i.e. moves you out of state space. Thus any singular $\rho$ is on the boundary. On the other hand, if $\rho$ is not singular, then $\det(\rho)>0$, and for any Hermitian $H$, we have $\det(\rho+\epsilon H)>0$ for small enough $\epsilon$. Thus non-singular states $\rho$ are in the interior.

  2. The extreme points of $\mathcal D$ are pure states. This is clear because any state that is not pure is a nontrivial convex combination of pure states.

  3. Each face of the boundary is characterised by some subspace $K< \mathbb{C}^d$. Namely, the set of states $\rho$ whose support is orthogonal to $K$, $\ker\rho\ge K$, belong to the same face of $\partial\mathcal D$. Let's denote this face with $\partial\mathcal D_K$. Note that "face" is understood here in the sense of convex geometry, as defined e.g. in Rockafellar, meaning in particular that a "face" doesn't need to be "flat" in the usual intuitive sense; it's enough that it has some flat "parts". The general definition of face is here a convex subset that fully contains any line segment whose relative interior it intersects.

    Note that for any pair of singular states $\rho,\sigma$ such that $\ker(\rho),\ker(\sigma)\ge K$, the line $$\{p \rho+(1-p)\sigma : \,\,p\in[0,1]\}$$ joins $\rho$ and $\sigma$, and is fully contained in $\partial\mathcal D_K$, hence $\partial\mathcal D_K$ is convex. To show that $\partial\mathcal D_K$ is a face, we also need to prove that if a line intersects it then it must be completely within it. But if there's $p\in(0,1)$ and states $\rho,\sigma$ such that $\ker(p\rho+(1-p)\sigma)\ge K$, then we must also have $\ker\rho,\ker\sigma\ge K$, hence the line must be entirely in $\partial\mathcal D_K$, hence $\partial\mathcal D_K$ is indeed a face.

    For example, if $d=3$, $\mathbb{C}^3=\operatorname{span}(\{\mathbf e_1,\mathbf e_2,\mathbf e_3\})$, and $K=\mathbb{C} \mathbf e_1$, then any pair of states whose ker contains $K$ belong to the same face. For example, $\rho\equiv \frac12(\mathbb{P}_1+\mathbb{P}_2)$ and $\sigma\equiv \mathbb{P}_1$ are the (relative) extreme points the line $[0,1]\ni p\mapsto p\rho+(1-p)\sigma$. It is interesting to note that this "face" is, topologically, a ball — it's essentially the "filled" Bloch sphere. But it still counts as a face because the entire ball sits on the (relative) boundary of state space, and is "flat" in a higher dimension corresponding to moving along the $\mathbb{P}_3$ direction.

  4. Not all faces "look" the same. Specifically, $\dim(\partial\mathcal D_K)=(d-\dim(K))^2-1$. That's because $d-\ker(K)$ is the dimension of the support of states in $\partial\mathcal D_K$. And some faces are contained in other faces.

    For example, in $d=3$ the face corresponding to $K=\mathbb{C} \mathbf e_1$ is the set of states orthogonal to $\mathbb{P}_3$, that is, the set of states of the form $\begin{pmatrix}0&0\\ 0&\rho\end{pmatrix}$ for any qubit state $\rho$. The pure state $\mathbb{P}_3$ also belongs to this face (in fact, it must be in the boundary of this face, it being an extreme point), but it also belongs to the (trivial) face corresponding to $K'=\operatorname{span}(\{\mathbf e_1,\mathbf e_2\})$.


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