I know that given two qubits, $A$ and $B$, in a state, $|\psi\rangle_{AB}$, you can find out if $A$ is entangled with $B$ by partial tracing and seeing if $tr_{A}\left(|\psi\rangle\langle\psi|\right)$ is a mixed state.

However, what I want to know is if you're given a state of many qubits, including $A$ and $B$; e.g., $|\phi\rangle_{ABCD\cdots}$, how can you determine if $A$ is entangled specifically with $B$.

It cannot be a simple matter of tracing out $A$ to see if it leaves behind a mixed state as that does not tell you with whom $A$ is entangled. For example:

$$ |\phi\rangle_{ACBD} = |\Phi^{+}\rangle_{AC}\ \otimes\ |\Phi^{+}\rangle_{BD}. $$

If we trace over $A$ then we get a mixed state. If we trace over $B$ then we get a mixed state. However, this does not tell us with whom $A$ and $B$ are entangled. Even if we trace over ever qubit, we just get four mixed states, it still provides no information as to which qubits are entangled with which. The only way I can see to determine this is to trace over every combination of qubits. Then when you trace over $BD$ you'll realise that you get a pure state.

Is there any more straightforward way to find out the entanglement relation between any arbitrary qubits in an ensemble in a general state?

  • $\begingroup$ Not sure I understand how you determine if A and B are entangled even if there are no other qubits, as in first part of your discussion. The necessary and sufficient condition is on the partial transpose, as per en.m.wikipedia.org/wiki/Peres%E2%80%93Horodecki_criterion $\endgroup$ Sep 2, 2017 at 0:15
  • $\begingroup$ @ZeroTheHero Well if a qubit is in a mixed state then it must be entangled with something $\endgroup$
    – gautampk
    Sep 2, 2017 at 2:41
  • $\begingroup$ I don't think this is correct. You can check the example in the wiki page en.m.wikipedia.org/wiki/Peres%E2%80%93Horodecki_criterion for an example of a density matrix that is a combination of a maximally entangled and a maximally mixed state: the result is not always entangled. $\endgroup$ Sep 2, 2017 at 6:42
  • $\begingroup$ @gautampk In such a setting with multiple parties you have to specify what you mean by "A and B are entangled". There is more than one way to define this property. $\endgroup$ Sep 2, 2017 at 9:28

1 Answer 1


If I understand your question correctly you have a multipartite pure state $|\psi\rangle_{ABCD..}$, but you are only concerned with the properties of the bipartite subsystem $AB$. In that case it is sufficient to consider the reduced density matrix $\rho_{AB}=Tr_{CD..}(\psi_{ABCD..})$. You then want to know if there is a simple way to test if this bipartite state $\rho_{AB}$ is entangled. As ZeroTheHero has suggested, in the case where systems $A$ and $B$ are qubits (or if one system is a qutrit and the other a qubit), then the Peres-Horodecki criterion is both necessary and sufficient for testing if $\rho_{AB}$ is entangled. For higher dimensions the problem becomes harder.

As Norbert Schuch has suggested there are multiple different kinds of entanglement among multipartied systems when you consider more than two parties. However it is not clear to me that you are concerned with any other situation than the two qubit case.

So there is a more straightforward way to find out the entanglement relation between any two arbitrary qubits in an ensemble in a general state. But not between any n arbitrary qubits, because there is no definition of the entanglement between n qubits.

Hope that answers your question! :)

  • 1
    $\begingroup$ I was referring more that there are different concepts of entanglement (with being inseparable and being distillable being the most extremal onces). Now for qubits, all of those are indeed equivalent since and non-separable state can be distilled, but this is a bit further from the fact that PPT detects all inseparable (="entangled") states. $\endgroup$ Sep 3, 2017 at 16:33
  • $\begingroup$ Thanks for clearing up what you meant Norbert. I have been under the impression that the gold standard definition of entanglement is inseparable, and that bound entanglement is a further sub categorization of the entangled states. $\endgroup$ Sep 3, 2017 at 16:54
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    $\begingroup$ I'm not sure how much "gold" the standard is - it really depends on the setup. What is clear is that separable states have no entanglement of any kind. But it is not clear that inseparable states have interesting entanglement in any scenario -- the "upper bound" here would be the distillable entanglement. $\endgroup$ Sep 3, 2017 at 17:56
  • $\begingroup$ Just to confirm, the criterion still applies if the two qubits in question are a part of a larger entangled state right? E.g., if A, B, and C are in a GHZ or a W state, you can trace out C and apply to criterion to AB and it will return the correct result. $\endgroup$
    – gautampk
    Sep 3, 2017 at 19:20
  • $\begingroup$ Yes, if you are only concerned with the entanglement between systems A and B then it doesn't matter what relationship they have to system C. The Peres-Horodecki criterion only cares about the reduced density matrix $\rho_{AB}$. $\endgroup$ Sep 3, 2017 at 20:17

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