What separable $\rho$ only admit separable pure decompositions with more than $\mathrm{rank}(\rho)$ terms?

Question

As shown e.g. in Watrous' book (Proposition 6.6, page 314), a separable state $\rho$ can always be written as a convex combination of at most $\mathrm{rank}(\rho)^2$ pure, separable states.

More precisely, using the notation in the book, any separable state $\xi\in \mathbb{C}^d\otimes\mathbb{C}^{d'}$ can be decomposed as $$\xi = \sum_{a=1}^m p(a) \, x_a x_a^*\otimes y_a y_a^*,\tag1$$ for some probability distribution $p$, sets of pure states $\{x_a\}_a\subset\mathbb{C}^d$ and $\{y_a\}_a\subset\mathbb{C}^{d'}$, and $$\operatorname{rank}(\xi) \le m\le \mathrm{rank}(\xi)^2.$$ The lower bound is trivial, while the upper bound is shown observing that $\xi$ is an element of the real affine space of hermitian operators $H\in\mathrm{Herm}(\mathcal X\otimes\mathcal Y)$ such that $\mathrm{im}(H)\subseteq\mathrm{im}(\xi)$ and $\mathrm{Tr}(H)=1$. This space has dimension $\mathrm{rank}(\xi)^2-1$, and thus from Carathéodory we get the conclusion.

Consider for example the case of a totally mixed state: $$\xi\equiv \frac{1}{dd'}I = \frac{I}{d}\otimes\frac{I}{d'}.$$ In this case $\mathrm{rank}(\xi)=dd'=m$, with the standard choice of separable pure decomposition. In the case $m=1$, it is trivial to see that we must also always have $m=\mathrm{rank}(\rho)$.

What are examples in which this is not the case? More precisely, what are examples of $\xi$ for which there is no separable pure decomposition with $m\le\mathrm{rank}(\xi)$?

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A related question on physics.SE is What is the minimum number of separable pure states needed to decompose arbitrary separable states?.

Answer 1

Symmetric Werner states in any dimension $n\geq 2$ provide examples.

Let's take $n=2$ as an example for simplicity. Define $\rho\in\mathrm{D}(\mathbb{C}^2\otimes\mathbb{C}^2)$ as $$ \rho = \frac{1}{6}\, \begin{pmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{pmatrix}, $$ which is proportional to the projection onto the symmetric subspace of $\mathbb{C}^2\otimes\mathbb{C}^2$. The projection onto the symmetric subspace is always separable, but here you can see it easily by applying the PPT test. The rank of $\rho$ is 3.

It is possible to write $\rho$ as $$ \rho = \frac{1}{4}\sum_{k = 1}^4 u_k u_k^{\ast} \otimes u_k u_k^{\ast} $$ by taking $u_1,\ldots,u_4$ to be the four tetrahedral states, or any other four states that form a SIC (symmetric information-complete measurement) in $\mathbb{C}^2$. It is, however, not possible to express $\rho$ as $$ \rho = \sum_{k = 1}^3 p_k x_k x_k^{\ast} \otimes y_k y_k^{\ast} $$ for any choice of unit vectors $x_1,x_2,x_3,y_1,y_2,y_3\in\mathbb{C}^2$ and probabilities $p_1, p_2, p_3$. To see why, let us assume toward contradiction that such an expression does exist.

Observe first that because the image of $\rho$ is the symmetric subspace, the vectors $x_k$ and $y_k$ must be scalar multiples of one another for each $k$, so there is no loss of generality in assuming $y_k = x_k$. Next we will use the fact that if $\Pi$ is any rank $r$ projection operator and $z_1,\ldots,z_r$ are vectors satisfying $$ \Pi = z_1 z_1^{\ast} + \cdots + z_r z_r^{\ast}, $$ then it must be that $z_1,\ldots,z_r$ are orthogonal unit vectors. Using the fact that $3\rho$ is a projection operator, we conclude that $p_1 = p_2 = p_3 = 1/3$ and $x_1\otimes x_1$, $x_2\otimes x_2$, $x_3\otimes x_3$ are orthogonal. This implies that $x_1$, $x_2$, $x_3$ are orthogonal. This, however, contradicts the fact that these vectors are drawn from a space of dimension 2, so we have a contradiction and we're done.

More generally, the symmetric Werner state $\rho\in\mathrm{D}(\mathbb{C}^n\otimes\mathbb{C}^n)$ is always separable and has rank $\binom{n+1}{2}$ but cannot be written as a convex combination of fewer than $n^2$ rank one separable states (and that is only possible when there exists a SIC in dimension $n$). This fact is proved in a paper by Andrew Scott [arXiv:quant-ph/0604049].