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 $\mathcal{H}_{B}$.


  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)

  • $\begingroup$ 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. $\endgroup$
    – transistor
    Commented Sep 21, 2015 at 11:12

1 Answer 1

  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 be decomposed into a mixture of non-separable states, e.g. $$ \tfrac12(\vert00\rangle\langle00\vert+\vert11\rangle\langle11\vert) = \tfrac12(\vert\psi_+\rangle\langle\psi_+\vert+\vert\psi_-\rangle\langle\psi_-\vert)\ , $$ where $\vert\psi_\pm\rangle = \tfrac{1}{\sqrt{2}}(\vert00\rangle \pm \vert11\rangle)$.

  2. This is the definition of entangled: A bipartite mixed state is called entangled exactly if it is not separable.

  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.

  • $\begingroup$ 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 $\rho_A$? $\endgroup$ Commented Sep 21, 2015 at 14:27
  • $\begingroup$ @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.) $\endgroup$ Commented Sep 21, 2015 at 14:32

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