Lowest energy state possess a symmetry but Gibbs state does not?

Question

I believe the Bose condensate to be a system where above the critical temperature the only lowest energy eigenstate breaks the U(1) symmetry but the Gibbs state does not.

On the contrary; Can there exist a system where the lowest energy eigenstates all obey a symmetry of the Hamiltonian but the Gibbs state does not.

_{^{This question is related to the following - which I asked a few hours before asking this one: Finite temperature spontanous symmetry breaking and Goldstone bosons}}

A ground state is the same thing as a lowest energy eigenstate. Also, the lowest energy eigenstate does not depend on temperature. Please edit your question to clarify what you're asking. — tparker, Mar 29 '18 at 15:20
@tparker But 'ground states' I was referring to Gibbs states $\rho=e^{-\beta H}/Z$ — Quantum spaghettification, Mar 29 '18 at 15:26
I see. At zero temperature that's usually called the "ground state manifold". — tparker, Mar 29 '18 at 16:07

score 1 · Answer 1 · answered Mar 29 '18 at 16:35

A Gibbs state $e^{-\beta H} / Z$ always has the same symmetry as the Hamiltionian. This follows simply from the fact that the Boltzmann weight depends only on the energy, which is preserved by symmetry operations. But below the critical temperature, the physical state is not a Gibbs state, but rather a nonergodic subset of the Gibbs state. For this reason symmetry breaking is sometimes called "ergodicity breaking".

What can happen is that the ground state manifold has a symmetry which is not a symmetry of the Hamiltonian at all. This is called an "emergent" or "accidental symmetry".

Lowest energy state possess a symmetry but Gibbs state does not?

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