# Can vectorization lead to mixed states?

Given an operator $$L = \sum_{ij}L_{ij}\vert i\rangle\langle j\vert$$, in some basis, the definition of vectorization is $$vec(L) = \sum_{ij}L_{ij}\vert i\rangle\vert j\rangle$$. The operation is invertible and heavily used in quantum information.

This definition always leads to a pure bipartite state but my notes, Wikipedia, etc. have no indication about whether one can also map operators (of some sort) to a mixed bipartite state. Can anyone shed some light on if this makes sense and there exist extensions of the definition of vectorization?

• Your map maps linear maps to pure states. You can also map linear superoperators (i.e. operators acting on operators) to matrices. In particular, CP maps get mapped to physical states (i.e. density operators). This is the Choi-Jamiolkowski-isomorphism. – Norbert Schuch Dec 2 '18 at 22:52
• So is it correct to say that the Choi-Jamiolkowski isomorphism is the generalization of vectorization? – user1936752 Dec 2 '18 at 23:24
• If you wish, you can say so. – Norbert Schuch Dec 2 '18 at 23:43

You have given $$L$$ as a general linear operator between Hilbert spaces $$H_1$$ to $$H_2$$ (the one indexed by $$j$$ and $$i$$ respectively). That is all possible linear operators, the entirety of $$Mor(H_1,H_2)$$. Mor here is short for the word morphism, a word for whatever the notion of whatever sorts of functions/operators/etc that are allowed. In the present context of vector spaces, this just means what are the linear operators $$H_1 \to H_2$$.
So we have all of the linear operators and for each of them $$\mathbf{vec}$$ produces a pure state in $$H_2 \otimes H_1$$. There is nothing missing. There are no other linear operators. There is no way to get states $$\rho$$ on $$H_2 \otimes H_1$$ that aren't pure here. Only mixed.
You already noticed this is invertible. That is we have $$Mor(H_1,H_2) \simeq H_2 \otimes H_1$$. The LHS is where $$L$$ is an element and the RHS is where $$\mathbf{vec}(L)$$ is an element.
There are extensions of this concept. For example suppose we have 3 Hilbert spaces. $$H_1$$, $$H_2$$ and $$H_3$$. Giving a linear map from $$H_1$$ to the space of linear maps from $$H_2$$ to $$H_3$$, which would be denoted as $$Mor(H_1, Mor(H_2,H_3))$$ is the same as giving a linear operator from $$H_1 \otimes H_2 \to H_3$$. See how this implies the above. Take $$H_3=\mathbb{C}$$. Then a way of thinking about a bra for $$H_2$$ is as a linear map $$H_2 \to H_3=\mathbb{C}$$. I'll leave you to ponder this fact some more to fully flush out how it implies the above.