# What are the relations between mixed/pure and separable/entangled states?

I read this question which clarified the concepts of mixed/pure/separable/entangled states, but I can't see if one of "mixed/pure" implies one of "separable/entangled" and/or vice versa.

Let $$\mathcal{H}=\mathcal{H}_A\otimes\mathcal{H}_B$$.

• if $$\rho$$ is separable, does it mean it's pure?
• if $$\rho$$ is entangled, does it mean it's mixed?
• if $$\rho$$ is separable, does it mean $$\rho_A$$ and $$\rho_B$$ are pure?
• if $$\rho$$ is entangled, does it mean $$\rho_A$$ and $$\rho_B$$ are mixed?

and vice versa, for each point.
I think point (3) is true (and so whould be point (4)) but I can't see why.

• Too many questions at once! Also, what are $\rho_A$ and $\rho_B$? Jun 6, 2021 at 12:04
• @NorbertSchuch they are all questions on the same subject, only meant to explain in what sense I mean "relations". Jun 6, 2021 at 12:06
• It would make more sense to ask about the definitions of these concepts first ... and then, for those parts where it is unclear why/how they are connected given their definition, ask about those. Jun 6, 2021 at 12:11
• @NorbertSchuch as I wrote, I already read an illuminating question on the different definitions: asking another similar question wouldn't have done any good. Jun 6, 2021 at 12:14
• It would just be good to know where you are stuck in connecting the concepts, given you apparently know the definitions. Otherwise it feels like a "do the work for me"-question. Jun 6, 2021 at 12:21

1. Separability and purity refer to different classes of properties of a state. A generic bipartite state $$\rho$$ being separable means that is can be written as a convex combination of product states, that is, it has a decomposition of the form $$\rho = \sum_i p_i \rho_i^A \otimes \rho_i^B,$$ for some $$p_i\ge0$$ with $$\sum_i p_i=1$$ and collections of states $$\rho_i^A,\rho_i^B$$ on the two systems. This general definition works for arbitrary (i.e. generally non-pure) states. If a state $$\rho$$ is pure, it means that it is a rank-1 projector, and can be written as $$\rho_{\Psi}=|\Psi\rangle\!\langle \Psi|$$ for some ket state $$|\Psi\rangle$$. A pure, separable state is always a product state, meaning $$|\Psi\rangle = |\psi\rangle\otimes|\phi\rangle, \qquad \rho_\Psi = \rho_\psi\otimes\rho_\phi.$$ For example:

• The two-qubit state $$\rho=\frac{1}{2}(|00\rangle\!\langle00|+|+,+\rangle\!\langle+,+|)$$ is separable and non-pure (and not a product state).
• The two qubit maximally mixed state $$\rho=(I\otimes I)/4\equiv \sum_{i=1}^4 |ii\rangle\!\langle ii|$$ is separable and not pure, and it is also a product state.
• A state such as $$\rho=|11\rangle\!\langle11|=|1\rangle\!\langle1|\otimes |1\rangle\!\langle1|$$ is pure and separable (and thus a product state).
2. As above, no, entanglement/separability and purity/mixedness measure different types of properties. An entangled state can be both pure and mixed. For example, a two-qubit Bell state is entangled and pure. A Werner state is (for $$p<1/3$$, or some other range depending on the way it is defined) entangled, but not pure. More explicitly, you can use as an example $$(1-\epsilon)|\Psi^+\rangle\!\langle\Psi^+|+\epsilon \frac{I\otimes I}{4},$$ where $$|\Psi^+\rangle$$ is a(ny) Bell state, and $$\epsilon>0$$ is small enough. The idea is that you are mixing a maximally entangled state with a maximally mixed one. If the mixed component is small enough, the state remains entangled.

One thing that can be said though is that entanglement in pure states is relatively easy to detect and classify, whereas things get way more tricky for general entangled states.

3. No. For example, the maximally mixed state shown above is separable and its partial states are also maximally mixed one-qubit states, which are not pure.

4. If the state is pure and entangled, then the partial states are necessarily mixed, and more precisely have rank equal to the Schmidt rank of the full state. More generally, if the partial states are pure, that means that the original state is not only separable, but also a product state, as it produces no correlations whatsoever. Turning this around, whenever the state is not a product state (and thus in particular whenever the state is entangled), the partial states ought to be mixed.

Here's a brief summary of the examples of pure/mixed separable/entangled states:

pure mixed
product $$|0\rangle\!\langle0|\otimes |0\rangle\!\langle0|$$ $$\frac{1}{4}(I\otimes I)$$
separable always also product $$\frac{1}{2}(|00\rangle\!\langle00|+|11\rangle\!\langle11|)$$
entangled Bell states (some) Werner states
• I don't think I understand what you mean by "product state". In my mind, it means separable, but seeing what you wrote I think I'm wrong. Jun 6, 2021 at 10:11
• @MauroGiliberti it means exactly what I wrote in the second equation. That's the definition. See e.g. the (admittedly not particularly insightful) Wikipedia page. Well, to be clearer, the second equation is a product state in which the components are pure. In general this needs not be the case.
– glS
Jun 6, 2021 at 10:13

The concept of pure and entangled quantum states are decoupled from each other. That is to say, there exists states for all the 4 possible combinations.

Analogous to the pure state scenario, bipartite mixed states $$\rho$$ that can be written as $$\displaystyle \rho=\sum_{i} p_{i}\;\rho_{A}^{i}\otimes \rho_{B}^{i}$$ are defined to be separable and all other mixed states are said to be entangled.

• Ok, so the entanglement of a state doesn't have implications on the purity of the same state. What about the purity of its substates? Can I link the information "$\rho$ is pure/mixed/entangled/separable" to the information "$\rho_A$ is pure/mixed"? Jun 6, 2021 at 7:13
• Yes, if a state is entangled; tracing over one of the sub systems will always give you a mixed state. This is roughly because while tracing over, you have lost some information that was contained in the entanglement between the two subsystems. An intuitive def. of mixed is quantum states for which the maximal information is not available to you. If you trace over a subsystem and still get a pure state (for $\rho_A$), that means the original system($\rho$) was separable. Jun 6, 2021 at 7:36