Entanglement of bi- and tripartite pure and mixed states

since I'm not sure on how to find out whether a system is entangled or not I thought about examples that could clarify the whole thing.

first example: system is in the state $\rho=1/2 (| 000 \rangle \langle 000| + |001 \rangle \langle001|)$
The first thing to do is to find out whether it is a pure or mixed state. Since $\rho^2=\rho$ it is a pure state.

Now, I can write the state as $\rho=|00\rangle \langle00| \bigotimes 1/2 \left(|0\rangle\langle0| + |1\rangle\langle1|\right)=|0\rangle\langle0|_A\bigotimes |0\rangle\langle0|_B\bigotimes 1/2 \left(|0\rangle\langle0|+|1\rangle\langle 1|\right)_C$.

If I take the partial trace now over A and B, then

$\rho_c=Tr_{AB}[\rho_{ABC}]=1/2 (|0\rangle\langle0|+|1\rangle\langle|1)$)

is in a mixed state. This means that the bipartite System AB and C is entangled. But what happens if I only trace over System C?

Then $\rho_{AB}=|00\rangle\langle00|$. But that's a pure state. How is that possible?

If I'm not mistaken, AC is not entangled with B and BC is not entangled with A. Is this correct?

My second example: starting from the density matrix $\rho= 1/2(|\phi^+\rangle\langle \phi^+|+|\phi^-\rangle\langle\phi^-|)=1/2(|00\rangle\langle00|+|11\rangle\langle11|)$.

Now, the criterion "$\text{A and B are entangled} \Rightarrow \rho_A \text{ and } \rho_B$ are mixed" cannot be applied, since $\rho_{AB}$ is not pure.

So, I just guessed and found that the whole state can be written as

$\rho=1/2(|0\rangle\langle0|+|1\rangle\langle1|) \bigotimes (|0\rangle\langle0|+|1\rangle\langle1|)+ 1/2(|0\rangle\langle0| -|1\rangle\langle1|) \bigotimes (0|\rangle\langle0|- |1\rangle\langle1|)$.

This means, that this state is separable as well. Correct?

I can't really think of an example for a mixed entangled state. Does anyone know an example and explain me how I can see that this is a mixed entangled state?

• The first state is a mixed state. I think you missed a factor of $\frac{1}{2}$ – BoundaryGraviton Jul 22 '16 at 18:13
• aaw, you're right. This explains why $\rho_{c}$ is a pure state. But how can I find out then whether it is entangled or not? – anonymous Jul 22 '16 at 18:18
• There are a few entanglement measures for mixed states. I'm not very updated on this. In the cases you have written, it's clear that they are both separable states. You have written the separable form yourselves. NPT I think is a necessary criteria for bi partitite entanglement for N qubit states. But I'm not sure if it's sufficient – BoundaryGraviton Jul 22 '16 at 18:27
• Also, in your comment here, you say "that explains why $\rho_c$ is a pure state". $\rho_c$ is a mixed state. What it explains is why $\rho_c$ is mixed and $\rho_{AB}$ can still be pure – BoundaryGraviton Jul 23 '16 at 6:21
• @SubramanyaHegde NPT is sufficient for bipartite entanglement, but not necessary. (I.e., all NPT states are entangled, but there are non-NPT (=PPT) entangled states). – Norbert Schuch Jul 23 '16 at 21:45