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 $xiinmathcal Xotimesmathcal Y$ can be decomposed as
$$xi = sum_{ainSigma} p(a) , x_a x_a^*otimes y_a y_a^*,tag1$$
for some probability distribution $p$, sets of pure states ${x_a: ainSigma}subsetmathcal X$ and ${y_a: ainSigma}subsetmathcal Y$, and alphabet $Sigma$ with $lvertSigmarvertle mathrm{rank}(xi)^2$.
This is shown by observing that $xi$ is an element of the real affine space of hermitian operators $Hinmathrm{Herm}(mathcal Xotimesmathcal Y)$ such that $mathrm{im}(H)subseteqmathrm{im}(xi)$ and $mathrm{Tr}(H)=1$. This space has dimension $mathrm{rank}(xi)^2-1$, and thus from Carathéodory we get the conclusion.
Consider the case of the totally mixed state in a space $mathcal Xotimesmathcal Y$ with $mathrm{dim}(mathcal X)=d, mathrm{dim}(mathcal Y)=d'$. For this state, $xiequiv frac{1}{dd'}I = frac{I}{d}otimesfrac{I}{d'}$, we have $mathrm{rank}(xi)=lvertSigmarvert=dd'$ for the standard choice of decomposition.
Generating random convex combinations of product states I also always find $lvertSigmarvert=mathrm{rank}(xi)$ (albeit, clearly, the numerics doesn't check for the existence of an alternative decomposition with less than ${rm rank}(xi)$ components). In the case $lvertSigmarvert=1$, it is trivial to see that we must also always have $lvertSigmarvert=mathrm{rank}(rho)$.
What are examples in which this is not the case?
More precisely, what are examples of states for which there is no alphabet $Sigma$ with $lvertSigmarvertlemathrm{rank}(xi)$, such that $xi=sum_{ainSigma}p(a)x_a x_a^*otimes y_a y_a^*$?

The following is the Mathematica snippet I used to generate random convex combinations of product states:
RandomUnitary[m_] := Orthogonalize[
  Map[#[[1]] + I #[[2]]&, #, {2}]& @ RandomReal[
    NormalDistribution[0, 1], {m, m, 2}
  ]
];
randomPureDM[dim_] := First@RandomUnitary@dim // KroneckerProduct[#, Conjugate@#] &;
With[{numComponents = 4, bigDim = 10},
  With[{
      mats = Table[KroneckerProduct[randomPureDM@bigDim, randomPureDM@bigDim], numComponents],
      probs = RandomReal[{0, 1}, numComponents] // #/Total@# &
    },
    Total[probs*mats] // Eigenvalues // Chop
  ]
]

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

John Watrous · Answer

Symmetric Werner states in any dimension $ngeq 2$ provide examples.
Let's take $n=2$ as an example for simplicity. Define $rhoinmathrm{D}(mathbb{C}^2otimesmathbb{C}^2)$ 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 $mathbb{C}^2otimesmathbb{C}^2$. 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 $x_1,x_2,x_3,y_1,y_2,y_3inmathbb{C}^2$ 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 $3rho$ is a projection operator, we conclude that $p_1 = p_2 = p_3 = 1/3$ and $x_1otimes x_1$, $x_2otimes x_2$, $x_3otimes x_3$ 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 $rhoinmathrm{D}(mathbb{C}^notimesmathbb{C}^n)$ 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].

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

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