A. Neumaier
Science Advisor
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No. Your observable ##H/N## is the mass-based energy density, which is quite independent from the temperature.Demystifier said:Yes, but q-expectations of what? Of all observables. In particular, there is an observable (essentially Hamiltonian times inverse number of particles) the q-expectation of which is the temperature.
The fundamental observable quantities are the q-expectations of quantities. The latter are defined as the densely defined operators. The temperature is not such a quantity. It is instead a parameter in the grand canonical definition of the density operator.