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Consider the following 4-qubit entangled state $$\left|\psi\right>=\left|0000\right>+\left|1110\right>+\left|1101\right>+\left|1011\right> $$

By tracing out qubits A and B (where I am using $\left|ABCD\right>$), we are left with the reduced density matrix $$ \rho^{CD} = \left|00\right>\left<00\right|+\left|11\right>\left<11\right|+\left|01\right>\left<01\right|+\left|01\right>\left<10\right|+\left|10\right>\left<01\right|+\left|10\right>\left<10\right|, $$ or, in a later more useful way, $$\rho^{CD} =\big(\left|0\right>\left<0\right|\otimes\left|0\right>\left<0\right|\big)+\big(\left|1\right>\left<1\right|\otimes\left|1\right>\left<1\right|\big)+\big(\left|0\right>\left<0\right|\otimes\left|1\right>\left<1\right|\big)+\big(\left|0\right>\left<1\right|\otimes\left|1\right>\left<0\right|\big)+\big(\left|1\right>\left<0\right|\otimes\left|0\right>\left<1\right|\big)+\big(\left|1\right>\left<1\right|\otimes\left|0\right>\left<0\right|\big)$$

I was hoping that this would lead to an evidently separable solution, but that does not seem to be the case.

What I would like to know is if there is any generic way of checking if such a density matrix is separable or not, rather than simply looking at the expression. This because the process becomes near impossible for a density matrix of more than two qubits.

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    $\begingroup$ But is this not a sum of 2-qubit density matrices? If yes (as I think it is) then it's a simple application of the Peres-Horodecki criterion en.wikipedia.org/wiki/Peres%E2%80%93Horodecki_criterion $\endgroup$ Mar 17, 2017 at 19:58
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    $\begingroup$ For two qubits (or one qubit and one qutrit), you can use the PPT criterion. But generally, this problem is computationally hard. There should be an answer somewhere here; if you can't find it, I can try to write on later. Of course, there are always criteria which are necessary or sufficient. $\endgroup$ Mar 17, 2017 at 20:01
  • $\begingroup$ Thank you! This is exactly what I was looking for. Since I am new to the topic, I had never heard of the Peres-Horodecki (PPT) criterion. It turns out that the state is indeed separable. $\endgroup$
    – Rmaa
    Mar 17, 2017 at 21:41
  • $\begingroup$ If I may add a follow up question... Were this a three qubit density matrix, the criterion would still be a necessary condition, but not sufficient, correct? $\endgroup$
    – Rmaa
    Mar 17, 2017 at 21:42

1 Answer 1

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For 2×2 (which is your case) and 2×3 dimensional bipartite systems the PPT condition can be used to check for separability. For higher dimensions separability testing has been proven to be a computationally hard problem (NP-hard). In general there exists an infinite hierarchy of separability tests based on symmetric extensions of the state. The PPT condition is just the first level of this hierarchy. An entangled state of any dimension is guaranteed to fail at some level of this hierarchy. For details and derivations refer this paper.

From a purely geometrical perspective, separability testing boils down to characterizing the convex body of separable states. The above paper gives a way to approximate this convex body using techniques from semi-definite programming.

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  • $\begingroup$ This paper is going to be really helpful, not only for the 2 qubit case, but also for the 3 and 4 qubits case, which I am also working with. Thank you! $\endgroup$
    – Rmaa
    Mar 19, 2017 at 19:49

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