As an example I have the density matrix:

$\rho = \frac{1}{3}(| \phi^+ \rangle\langle\phi^+| + | 00 \rangle\langle 00|+| 11 \rangle\langle11| )$

And the two-qubit state is:

$\frac{1}{3}(| \phi^- \rangle\langle\phi^-| + | \psi^+ \rangle\langle \psi^+|+| \psi^- \rangle\langle \psi^-| )$

The trace of $\rho$*state is greater than zero. Does that suffice to show that it is separable?

  • 5
    $\begingroup$ do you have a reason to believe it should(n't)? Where did you get this from? $\endgroup$
    – glS
    Commented Apr 18, 2021 at 17:14

1 Answer 1


No, take Bell state $\sigma = \frac{1}{2} (|00 \rangle+| 11 \rangle)( \langle 00| +\langle11|)$ and $\rho = \frac{1}{4}I$.

Also, if $\rho \ge 0$ then it's always $\text{Tr}(\rho \sigma) \ge 0$.

Though there is a notion of entanglement witness.

We can deduce that state $\sigma$ is separable if (and only if) for every Hermitian operator $\rho$, such that $\text{Tr}(\rho \cdot \sigma_1 \otimes \sigma_2) \ge 0$ for any states $\sigma_1,\sigma_2$, we have that $\text{Tr}(\rho \cdot \sigma) \ge 0$.


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