# Separability of density operators on tensor product spaces

Consider a composite system $$\mathcal{H}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}$$ where $$\mathcal{H}_{A}$$ and $$\mathcal{H}_{B}$$ are Hilbert spaces of constituent components (say two qubits).

Let $$\rho_{AB}$$ be a density operator on $$\mathcal{H}$$, i.e., $$\rho_{AB} = \sum_{i}p_{i}|\psi_{i}\rangle\langle\psi_{i}|$$ for $$|\psi_{i}\rangle\in\mathcal{H}$$.

Consider the special case of a pure state, $$\rho_{AB} = |\psi\rangle\langle\psi|$$. The following theorem holds: $$\rho_{AB}\ \text{separable} \Leftrightarrow |\psi\rangle \ \text{separable}$$

By separable, I mean $$\rho_{AB} = \sum_{i}p_{i}^\prime\rho_{A,i}\otimes\rho_{B,i}$$ where $$\rho_{A,i}$$ and $$\rho_{B,i}$$ are density operators on $$\mathcal{H}_{A}$$ and $$\mathcal{H}_{B}$$.

Questions:

1. Is there a similar result for the more general $$\rho_{AB} = \sum_{i}p_{i}|\psi_{i}\rangle\langle\psi_{i}|$$? Given $$\rho_{AB}$$ is separable, can we say anything about the separability of the $$|\psi_{i}\rangle$$s?

2. If $$\rho_{AB}$$ is not separable, does that mean that the systems are entangled?

3. How do the reduced density matrices $$\operatorname{tr}_{A}(\rho_{AB})$$ and $$\operatorname{tr}_{B}(\rho_{AB})$$ figure into all of these? (if they do so at all)

• The wiki page has the same definition for separability of mixed states as the one I have mentioned. It is not obvious to me why a mixed state not being separable means entanglement, that's why I wanted to know what separability meant in terms of the state vectors. – transistor Sep 21 '15 at 11:12

1. $\rho_{AB}$ is called separable if it can be written as $$\rho_{AB}=\sum p_i \rho_{A,i}\otimes \rho_{B,i}\ .$$ You can now further decompose $\rho_{A,i}=\sum_{x} q_{x} \vert\alpha_x\rangle\langle \alpha_x\vert$ and $\rho_{B,i}=\sum_{y} r_{y} \vert\beta_y\rangle\langle \beta_y\vert$; then, $$\rho_{AB}=\sum_{i,x,y} p_i q_{x} r_y \vert\alpha_x\rangle\langle \alpha_x\vert\otimes \vert\beta_y\rangle\langle \beta_y\vert\ ,$$ i.e., $\rho_{AB}$ is of the form $\rho_{AB} =\sum_j w_j\vert\psi_j\rangle\langle\psi_j\vert$ with $\vert\psi_j\rangle$ separable.
Note that in general, separable states also can equally decomposed into a mixture of non-separable states, e.g. $$\tfrac12(\vert0\rangle\langle0\vert+\vert1\rangle\langle1\vert) = \tfrac12(\vert\psi_+\rangle\langle\psi_+\vert+\vert\psi_-\rangle\langle\psi_-\vert)\ ,$$ where $\vert\psi_\pm\rangle = \tfrac{1}{\sqrt{2}}(\vert00\rangle \pm \vert11\rangle)$.
3. The value of the reduced density matrices $\rho_A$ and $\rho_B$ can be used to rule out entanglement, but not separability:
For any given reduced density matrices $\rho_{A}$ and $\rho_{B}$, there is always a separable state (namely $\rho_A\otimes\rho_B$) which has those reduced density matrices.
On the other hand, there are clearly cases where the reduced density matrices are incompatible with an entangled state, such as e.g. $\rho_A=\rho_B=\vert0\rangle\langle0\vert$. Continuity suggests that there should be a finite subset of $\rho_A$ and $\rho_B$ which are incompatible with an entangled state (but I don't know whether this has been studied).
Finally, if we are only given the reduced density matrix of one party, e.g. $\rho_A$, there are always both entangled and separable states compatible with it, except if $\rho_A$ itself is pure.
• Hi Norbert, your last statement confused me a bit. Surely there are no bipartite entangled states compatible with the reduced density matrix $\rho_A = \lvert 0\rangle \langle 0 \rvert$? Could you please give an example of an entangled state compatible with $\rho_A$? – Mark Mitchison Sep 21 '15 at 14:27
• @MarkMitchison Fair point. I guess I wanted to say that for a generic $\rho_A$ (more specifically, a non-pure one), there is always both a pure and an entangled state --- in contrast to the case where both $\rho_A$ and $\rho_B$ are specified, in which case there seems to be a set of finite measure which only admits separable states. (BTW, it would be interesting to know if this has been studied previously.) – Norbert Schuch Sep 21 '15 at 14:32