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Following the notation in Watrous' book, page 353, an unextendable product set in a bipartite space $\mathcal X\otimes \mathcal Y$ is a set of orthonormal product vectors of the form $$\mathcal A\equiv \{u_1\otimes v_1,...,u_m\otimes v_m\},$$ with $u_i,v_i$ unit vectors, such that

  1. $m< \dim(\mathcal X\otimes \mathcal Y)$ (that is, $\mathcal A$ spans a strict subspace of $\mathcal X\otimes \mathcal Y$);
  2. If $x\otimes y\perp \mathcal A$, then $x\otimes y=0$.

An explicit example of such a set with $m=5$ is given for $\mathbb{C}^3\otimes \mathbb{C}^3$ (Example 6.40). Is there any smaller example of such sets? In particular, does an unextendable product set exist for $\mathbb{C}^2\otimes \mathbb{C}^2$?

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Upon some reflection I realised that, no, there can't be any such set in $\mathbb{C}^2\otimes\mathbb{C}^2$ or $\mathbb{C}^2\otimes\mathbb{C}^3$. That's because, as discussed shortly thereafter in the book: the projection onto the subspace orthogonal to an unextendable product set must be both PPT and entangled.

But in $2\times2$ and $2\times3$, it is well known that PPT is the same as separability. Thus in such dimensions the subspace orthogonal to an unextendable product set must be trivial, as otherwise one could build an entangled PPT state in such dimensions.

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  • $\begingroup$ I was about to write that before I scrolled down ... ;) $\endgroup$ Commented Mar 11 at 18:20
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    $\begingroup$ (More specifically, the point is that unextentible product bases have arisen exactly as a way to construct PPT bound entangled states. So their existence implies PPT bound entangled states.) $\endgroup$ Commented Mar 11 at 19:09

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