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May 14, 2020 at 15:23 comment added ZeroTheHero @MarkMitchison Good point. I know that but somehow overlooked this in crafting my answer as I was focused more on the "complex eigenvalue" part of the question. Thanks for bringing it up.
May 14, 2020 at 14:50 comment added Mark Mitchison Apologies for nitpicking but positivity (as well as hermiticity) does place constraints on the off-diagonal elements as well as the diagonal ones, e.g. in dimension 2 one requires $|\rho_{12}|^2 \leq \rho_{11}\rho_{22}$ so that $\det \rho \geq 0$.
Apr 23, 2020 at 10:05 vote accept self.grassmanian
Apr 23, 2020 at 8:40 comment added ZeroTheHero not really helpful. The trace is invariant under conjugation so all it says it that your initial $\rho$ has trace=1. Check hermiticity.
Apr 23, 2020 at 7:51 comment added self.grassmanian I have checked that Tr$(\rho) = 1$, can that somehow help?
Apr 22, 2020 at 20:40 comment added ZeroTheHero Of course not. The hermiticity condition remains even for composite systems. There is no way your $\rho$ is hermitian if you have complex eigenvalues.
Apr 22, 2020 at 20:33 comment added self.grassmanian Actually, I am working on something an having a $\rho_{4\times4}$ with complex eigenvalues. Is it possible to have such for the case when we have mixed states or entanglement?
Apr 22, 2020 at 20:30 history edited ZeroTheHero CC BY-SA 4.0
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Apr 22, 2020 at 19:58 history edited ZeroTheHero CC BY-SA 4.0
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Apr 22, 2020 at 19:53 history answered ZeroTheHero CC BY-SA 4.0