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If we define a stationary state as a state in which the probability distributions for every observable are constant in time, is it fair to say that a mixed state can also be a stationary state? One usually sees the particular case of a pure state (in which case we need a Hamiltonian eigenstate). However, I think that if we have a (incoherent) mixture of such eigenstates $\rho = \sum_n p_n |E_n\rangle \langle E_n|$ then the trace Born rule should also show that all probability distributions associated with this (mixed) state are time independent too. I'm not sure about this though.

Thus my questions are (1) is my definition of a stationary state correct and (2) can a mixed state therefore be a stationary state?

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  • $\begingroup$ Stationary states are solutions of Schrodinger equation. $\endgroup$
    – kludg
    Commented Mar 10, 2023 at 14:07
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    $\begingroup$ Have you inspected your non Neumann equation? What have you concluded? $\endgroup$
    – Cosmas Zachos
    Commented Mar 10, 2023 at 14:30
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    $\begingroup$ I know of it. I suppose it supports my assertion (that if every projector in the density operator is built up of Hamiltonian eigenkets then we have a density operator that remains unchanged in time). I think I see now but am hoping you or someone else can confirm that's correct. @CosmasZachos $\endgroup$
    – EE18
    Commented Mar 10, 2023 at 14:49
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    $\begingroup$ Trust your calculation. $\endgroup$
    – Cosmas Zachos
    Commented Mar 10, 2023 at 14:49
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    $\begingroup$ System in thermodynamic equilibrium is a good example. $\endgroup$
    – Roger V.
    Commented Mar 10, 2023 at 17:02

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