Is it possible to find all the non-separable vectors in a tensor product of $n$ Hilbert spaces? How many are there? Do they live in a specific subspace?

To be more precise, if we have two complex Hilbert spaces $H_1$ and $H_2$ of the same dimension $d$ and we form their tensor product $H=H_1\otimes H_2$, $H$ is itself a vector space, of dimension $2d$ (is that correct?), however in $H$ some vectors can be written as the tensor product of a vector in $H_1$ and a vector in $H_2$ while others cannot.

To illustrate what is meant by "separable/non-separable", let's take $H_1$ and $H_2$ of dimension 2. If the pair ($u_1, u_2$) form a basis of $H_1$ and the pair ($v_1,v_2$) a basis of $H_2$, the general form of the (tensor) product of a vector in $H_1$ and a vector in $H_2$ is, with $(\alpha_1, \alpha_2, \beta_1, \beta_2) \in \mathbb{C}^4$: $$(\alpha_1 u_1 + \alpha_2 u_2)\otimes(\beta_1 v_1 + \beta_2 v_2) = \alpha_1 \beta_1 u_1\otimes v_1 + \alpha_1 \beta_2 u_1\otimes v_2 + \alpha_2\beta_1 u_2\otimes v_1 + \alpha_2\beta_2 u_2\otimes v_2,$$ Now if we consider the following vector in $H$: $$w=u_1\otimes v_1+u_2\otimes v_2,$$ we see that we cannot write it as a (tensor) product of a vector in $H_1$ and a vector in $H_2$, as it leads to contradiction when trying to find values for $\alpha_1, \beta_1, \alpha_2, \beta_2$. We call $w$ "non-separable".

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    $\begingroup$ The term you're looking for is "pure tensor" (a tensor which can be written as a tensor product). Neither the pure tensors nor the non-pure tensors form a subspace, but the pure tensors form a Zariski closed subvariety. You can identify the pure tensors in $H_1 \otimes H_2$ with the linear maps $H_1^{\ast} \to H_2$ of rank at most $1$, so the non-pure tensors correspond to linear maps of rank at least $2$. $\endgroup$ – Qiaochu Yuan Sep 18 '17 at 18:04
  • $\begingroup$ @QiaochuYuan - thanks! That's exactly what I was looking for. I was asking physicists about how this is called in maths, but they couldn't say :-) $\endgroup$ – Frank Sep 18 '17 at 19:03

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