What is the necessary condition for a quantum system to be in thermal equilibrium?
The quantum systems I have in mind are a bunch of cold atoms, of photons in a cavity.

  • Does a system need to obey a thermal equilibrium distribution (e.g. Bose-Einstein, Fermi-Dirac) in order for it to be in thermal equilibrium? Or the other way round? Is it a necessary or sufficient condition?

  • Is a steady-state the same as a thermal state?

  • $\begingroup$ A steady state is not the same as a thermal equilibrium state, e.g. take two heat baths at different temperature and connect them to a system ("at opposite ends") the system will be in a steady state after some time, but it can't be in equilibrium (it doesn't have a well defined temperature). $\endgroup$ Sep 21, 2018 at 17:00

1 Answer 1


A quantum system in thermal equilibrium is in a mixed state that corresponds to the occupation probabilities of one of the statistical ensembles in equilibrium. For instance, in the canonical ensemble description, the system is in equilibrium if its density matrix is $\mathrm{e}^{-\beta H}$.

  • $\begingroup$ thanks. are there other enemsble distributions apart from Bose and Fermi? Can a system be such. a distribution and not be jn thermal equilibrium? $\endgroup$
    – SuperCiocia
    Oct 6, 2018 at 15:19

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