When speaking about the grand-canonical ensemble of a statistical system, one usually works with a case, when there are several conserved quantities - total number of particles $N$, angular momentum $J$ or some other kind of charge. And the grand-canonical partition function has the form: $$ \mathcal{Z} = \text{Tr } e^{-\beta(H - \mu N - \omega J - \ldots)} $$ This expression can be derived from the point of view of MaxEnt method : by maximization of entropy, constrained by the conservation of energy, number of particles, and maybe some additional integrals of motion.

My question is : what would it look like for the case of integrable system, with an infinite number of conserved quantities?

  1. In general case the convergence of series in the exponent seems to be rather a delicate issue, some conditions have to be imposed for this expression to make sense

  2. Another point, which worries me, is that the infinite number of constraints, restricting the infinite number of degrees of freedom may turn the part of phase space, defined by the equations : $$ I_n = I_{0n} \qquad\forall n $$ where $I_n$ are the integrals of motion, to something finite, that doesn't allow for thermodynamic description.

I apologise for being slighlty messy, I would be grateful for links or any information about thermodynamic description of integrable systems.

  • $\begingroup$ This is a very interesting question. Could you give an example of such a system? $\endgroup$
    – Godzilla
    Jul 19, 2020 at 0:19
  • 2
    $\begingroup$ The keyword you want to look for is "generalised Gibbs ensemble" (GGE). See also my answer at physics.stackexchange.com/a/350819 for a few links. $\endgroup$ Jul 19, 2020 at 5:55
  • 1
    $\begingroup$ @Godzilla123 I have borne in mind ,for instance, spin chains, such as XXZ model, or free quantum field theories - free scalars, fermions, Maxwell theory without matter. $\endgroup$ Jul 19, 2020 at 7:17
  • $\begingroup$ @JulesLamers, thanks! That is what I've been looking for $\endgroup$ Jul 19, 2020 at 7:18


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.