Can someone tell me what is the physical meaning or implication of a system's density matrix having complex eigenvalues?
Do we require the diagonal elements of a density matrix to be real? If so, why and then what about the off diagonal elements?
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This hardly makes sense: if you go to a basis where the density matrix is diagonal, its eigenvalues will appear as the diagonal entries. Since the diagonal entries are populations and thus must be real and non-negative, this pretty much excludes complex eigenvalues.
There is no restriction on the off-diagonal pieces other than $\rho_{ij}=\rho_{ji}^*$ to preserve hermiticity of $\rho$: the coherences can be complex.
(Note that eigenvalues of a hermitian matrix are necessarily real so if you get imaginary parts it could be small roundoff errors, not uncommon if one is not careful and works with very large density matrices.)
answered Apr 22, 2020 at 19:53
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$\begingroup$ 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? $\endgroup$ Commented Apr 22, 2020 at 20:33
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$\begingroup$ Of course not. The hermiticity condition remains even for composite systems. There is no way your $\rho$ is hermitian if you have complex eigenvalues. $\endgroup$ Commented Apr 22, 2020 at 20:40
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$\begingroup$ I have checked that Tr$(\rho) = 1$, can that somehow help? $\endgroup$ Commented Apr 23, 2020 at 7:51
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$\begingroup$ not really helpful. The trace is invariant under conjugation so all it says it that your initial $\rho$ has trace=1. Check hermiticity. $\endgroup$ Commented Apr 23, 2020 at 8:40
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1$\begingroup$ 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 $\det \rho \geq 0$. $\endgroup$ Commented May 14, 2020 at 14:50