# Does $\mathrm{Tr}(\rho\sigma) > 0$ prove that a state $\sigma$ is separable?

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?

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

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 $$\text{Tr}(\rho \sigma) \ge 0$$.
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 $$\text{Tr}(\rho \cdot \sigma) \ge 0$$.