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

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 $$\xi\in\mathcal X\otimes\mathcal Y$$ can be decomposed as $$\xi = \sum_{a\in\Sigma} p(a) \, x_a x_a^*\otimes y_a y_a^*,\tag1$$ for some probability distribution $$p$$, sets of pure states $$\{x_a: a\in\Sigma\}\subset\mathcal X$$ and $$\{y_a: a\in\Sigma\}\subset\mathcal Y$$, and alphabet $$\Sigma$$ with $$\lvert\Sigma\rvert\le \mathrm{rank}(\xi)^2$$. This is shown by 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 the case of the totally mixed state in a space $$\mathcal X\otimes\mathcal Y$$ with $$\mathrm{dim}(\mathcal X)=d, \mathrm{dim}(\mathcal Y)=d'$$. For this state, $$\xi\equiv \frac{1}{dd'}I = \frac{I}{d}\otimes\frac{I}{d'}$$, we have $$\mathrm{rank}(\xi)=\lvert\Sigma\rvert=dd'$$ for the standard choice of decomposition. Generating random convex combinations of product states I also always find $$\lvert\Sigma\rvert=\mathrm{rank}(\xi)$$ (albeit, clearly, the numerics doesn't check for the existence of an alternative decomposition with less than $${\rm rank}(\xi)$$ components). In the case $$\lvert\Sigma\rvert=1$$, it is trivial to see that we must also always have $$\lvert\Sigma\rvert=\mathrm{rank}(\rho)$$.

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 $$\lvert\Sigma\rvert\le\mathrm{rank}(\xi)$$, such that $$\xi=\sum_{a\in\Sigma}p(a)x_a x_a^*\otimes y_a y_a^*$$?

The following is the Mathematica snippet I used to generate random convex combinations of product states:

RandomUnitary[m_] := Orthogonalize[
Map[#[] + I #[]&, #, {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?.

• In the title, "less" means "less or equal"? Jul 23 '20 at 16:52
• @NorbertSchuch indeed. Should be phrased better now
– glS
Jul 23 '20 at 16:58
• Definitely, much better without the negation. Interesting question, by the way. Jul 23 '20 at 17:01

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].

• Ah, I should have known - the symmetric or the antisymmetric projector are always an example! Jul 24 '20 at 15:52
• One can also use Example 6.10 of the mentioned chapter of your book and notice that the partial transpose of the symmetric state is an isotropic state with full rank $n^2$. So, it can't have less than $n^2$ elements in the separable pure decomposition. Thus the same is true for the symmetric state. Jul 15 at 8:12
• Very nice observation! That's a much easier way to argue it. Jul 15 at 13:00