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 $m\le\mathrm{rank}(\xi)$?

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


1 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].

  • 2
    $\begingroup$ Ah, I should have known - the symmetric or the antisymmetric projector are always an example! $\endgroup$ Commented Jul 24, 2020 at 15:52
  • 1
    $\begingroup$ 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. $\endgroup$
    – Danylo Y
    Commented Jul 15, 2021 at 8:12
  • 1
    $\begingroup$ Very nice observation! That's a much easier way to argue it. $\endgroup$ Commented Jul 15, 2021 at 13:00
  • $\begingroup$ is arxiv.org/pdf/quant-ph/0604049 the correct reference? I don't see any mention of separable decompositions nor Werner states in there $\endgroup$
    – glS
    Commented yesterday
  • $\begingroup$ Yes, it's just stated using different terminology, as Theorem 4. $\endgroup$ Commented 16 hours ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.