Timeline for Why are accessible states taken as eigenstates in statistical physics? Is the resolution via decoherence?
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9 events
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| Dec 14 at 6:41 | comment | added | GiorgioP-DoomsdayClockIsAt-90 | @Ghorbalchov That's not what I wrote. The logical sequence is: 1) the system is at equilibrium => 2) averages through the density matrix should be time independent => 3) the density matrix should not depend on time => 4) it commutes with the Hamiltonian => 5) there is a common basis of eigenvectors, and this makes the Hamiltonian eigenstates a convenient basis for describing the system. This does not contradict the fact that a state of the system, due to the interaction with the external world, has its own time evolution. | |
| Dec 14 at 0:21 | comment | added | Ghorbalchov | "I never wrote that we use eigenstates of 𝐻 because they make $\rho$ time independent": isn't that the whole point of your answer? | |
| Mar 27 at 22:47 | comment | added | GiorgioP-DoomsdayClockIsAt-90 | @Ghorbalchov I never wrote that we use eigenstates of 𝐻 because they make 𝜌 time independent. If you work with a system at equilibrium able to exchange energy with a thermostat at temperature $T$, the density matrix is the operator $\hat \rho = e^{-\beta \hat H}$. The time derivative of such an operator is zero. | |
| Mar 27 at 17:00 | comment | added | Ghorbalchov | Sorry I don't understand, your argument is that we use eigenstates of $H$ because they make $\rho$ time independent, yet $\rho$ is not time independent if there is interaction with an environment. Maybe the argument is that the interaction with the environment has to be weak enough for this to be negligible? | |
| Mar 27 at 16:52 | comment | added | GiorgioP-DoomsdayClockIsAt-90 | @Ghorbalchov Doesn't matter. The density operator already embodies a time average. | |
| Mar 27 at 16:20 | comment | added | Ghorbalchov | But the eigenstates of $H_{\mathrm{system}}$ will not be stationary states of the full Hamiltonian in this case right? | |
| Mar 27 at 16:03 | comment | added | GiorgioP-DoomsdayClockIsAt-90 | @Ghorbalchov Yes, at the equilibrium, we still work with eigenstates of $H_{system}$, encoding the effect of the interaction with the environment into some environment parameter as the temperature, pressure, or chemical potential. | |
| Mar 27 at 11:08 | comment | added | Ghorbalchov | How does this argument work when the system is interacting with an environment? Then we will have $[H_{\mathrm{system}}+H_{\mathrm{env}}+H_{\mathrm{int}},\rho]=0$ in equilibrium, yet we still work with eigenstates of just $H_{\mathrm{system}}$? | |
| Apr 13, 2023 at 5:15 | history | answered | GiorgioP-DoomsdayClockIsAt-90 | CC BY-SA 4.0 |