Many years before, I had the following problem.

We first give a definition. Given a non-negative definite real-valued definite matrix $n^2\times n^2$ matrix $M$, it is called separable if it can be decomposed in the following way:

$$ M=\sum_{i=1}^{k} \:\: \lambda_i E_i\otimes F_i $$

where $k\le n^2$, $\lambda_i>0$, $\otimes$ denotes the tensor product in the usual sense, and $E_i$, $F_i$ are $n\times n$ non-negative matrices of rank one, having unit length and orthogonal with each other:

$$ Tr(E_i E_j^T) =Tr(F_i F_j^T) = \begin{cases} 1& i=j\cr 0& i\ne j \end{cases}\;. $$

Now the problem is: how to determine whether a non-negative definite matrix is separable or not?

Recently, I have this problem again, but in a more general context. We define the non-negative definite linear operator $\mu \in \mathcal{S}'(R^{2d})$ ($\mathcal{S}'(R^{2d})$ is the space of Schwartz distributions) over the rapidly decreasing functions $\mathcal{S}(R^d)$ as follows:

$$ \langle\mu,\psi\otimes\psi\rangle\ge 0,\quad\forall \psi\in\mathcal{S}(R^{d}), $$ where $\otimes$ denotes the tensor product.

We call a non-negative definite linear operator $\mu \in \mathcal{S}'(R^{2d})$ ($d$ is even), separable, if there exists a sequence of pairs $(\lambda_i,\mu_i,\nu_i)$, with $\lambda_i>0$, and each $\mu_i$ and $\nu_i\in\mathcal{S}'(R^d)$ is non-negative definite, such that

$$ \mu=\sum_i \lambda_i \: \mu_i\otimes\nu_i. $$

Note that the above sum can be integral when the operator has a continuous spectral.

So the similar problem is how to determine whether a non-negative definite linear operator $\mu \in \mathcal{S}'(R^{2d})$ over $\mathcal{S}(R^d)$ is separable or not?

I guess that these two problems are still open. Does anyone have any hints or references?

Thank you very much!


  • 1
    $\begingroup$ Am I missing something, or your first problem is just the spectral theorem for symmetric matrices? $\endgroup$ – Federico Poloni Aug 18 '11 at 10:44
  • $\begingroup$ @Federico, Poloni, you are right. I made a mistake about the statement of the problem. I am reformulating it now...:-) $\endgroup$ – Anand Aug 18 '11 at 10:55
  • $\begingroup$ @Federico Poloni, if you fold the eigenvectors properly to form the matrices $E_i$ above, there is no guarantee that $E_i$ is still a non-negative definite matrix. This is the difficulty of the problem. :-) Thanks for pointing out a mistake in my original formulation. $\endgroup$ – Anand Aug 18 '11 at 11:08

In general the problem is hard and it is widely known in quantum physics as the question if a mixed quantum states is separable or not. The physical formulations is slightly different, as for the decomposition

$$M = \sum_i \lambda_i E_i \otimes F_i$$ there is no restriction on the number of elements in the sum or relation between them (still $\lambda_i\geq0$, while $E_i$ and $F_i$ are nonnegative-definite). Although, as the above decomposition is not unique, one can add more requirements (e.g. we can require that the matrices are range one orthogonal projections).

For $n=2$ there is Peres-Horodecki criterion - the state is separable iff its partial transpose is nonnegative-definite. For $n>2$ the criterion in necessary but not sufficient.

If you require $E_i=F_i$ then an additional necessary condition is the symmetry under relabeling of $M$ in such way that it interchanges the subsystems (i.e. $M[(i_1,i_2),(j_1,j_2)]\mapsto M[(i_2,i_1),(j_2,j_1)]$ and thus $\sum_i \lambda_i E_i \otimes F_i\mapsto \sum_i \lambda_i F_i \otimes E_i$).

Note: the answer is based on an assumption that you meant the dimension $n^2\times n^2$ for $M$ matrix (otherwise the $E_i$ cannot be of the same dimension for every $n$).

  • $\begingroup$ Thanks Piotr Migdal. I found that I made a mistake on the dimensions. It should be $n^2\times n^2$. $\endgroup$ – Anand Oct 17 '11 at 20:06
  • $\begingroup$ I have updated my problem. Thanks a lot :-). The problem is indeed the same as the question of determining the whether a mixed quantum state is separable or not. $\endgroup$ – Anand Oct 17 '11 at 20:21

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